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10300	https://www.kreo.net/news-2d-takeoff/what-is-a-scale-drawing	What is a Scale Drawing? — Kreo Customers Solutions for Construction Takeofffor Cost PlanningAI Agentic Workflowfor Quantity Surveyorsfor Cost Estimatorsfor Contractorsfor Measuring blueprintsBIM Takeoff Trades FlooringConcreteElectricalFramingDrywallPaintingSteelMasonry Resources EventsHelp CenterPromo PresentationAcademy Blog Pricing Talk to a HumanBook a Demo Get StartedLogin Sign Up Book a Call AI-Driven Takeoff: Your Blueprint to Accurate Estimates! Let technology elevate your project planning. Try AI Takeoff Now! Blogs What is a Scale Drawing? What is a Scale Drawing? October 12, 2023 Karolina Construction In an age where precision and detail dictate the success of projects across myriad industries, scale drawings stand as paramount tools. These drawings, which reduce or enlarge real-world objects to a size that's manageable on paper or screens, are the backbone of fields ranging from architecture to engineering. Their ubiquitous presence testifies to their importance, making every constructed entity around us a testimony to their critical role. What is a Scale Drawing? A scale drawing is a representation of an object or place that retains its proportions but is adjusted in size according to a specific ratio. It's akin to shrinking or enlarging an object without distorting its shape. The significance of such drawings is manifold. By providing a proportional perspective, they offer clarity, precision, and a means to visualize large-scale projects on a standard medium. Whether you're crafting the blueprint of a building or designing a piece of machinery, scale drawings ensure that every detail is captured, proportioned, and represented correctly. Components Every scale drawing comprises certain elements that make it comprehensible and usable. Central to this is the scale or ratio, which indicates the relationship between the drawing and the actual object. For instance, a scale of 1:100 means that every unit on the drawing equates to 100 units in reality. This scale ensures that, although the drawing's size is adjusted, its proportions remain true to the original, allowing for accurate extrapolation of dimensions and measurements. How to Read Scale Reading a scale drawing involves understanding its proportions and measurements. This comprehension ensures that you can translate what's on paper to the real world and vice versa. To aid this process, various scale drawing tools are employed. Traditional tools include rulers, compasses, and protractors. In today's digital age, we also have scale drawing apps and software, which offer added precision and convenience. Examples The application of scale drawings examples spans across multiple fields, some of which include: Architecture: Blueprint of buildings, structures, and homes. Engineering: Designs of machinery, vehicles, and infrastructure. Cartography: Maps that represent vast terrains on compact sheets. Interior Design: Layouts of rooms, furniture arrangement, and decor. And many more. The Most Common Engineering Drawing Scales In engineering, scale drawings are indispensable. Common scales here include 1:1 (life-size), 1:2 (half size), 1:10, 1:100, and so forth. These specific scales are preferred because they offer a standardized reference, ensuring consistency across projects and ease of interpretation. AI for Scale Drawing Kreo Software is a cloud-based construction takeoff and estimating software that uses artificial intelligence (AI) to help users save time and win more business. It is designed for quantity surveyors, estimators, contractors, and architects. One of Kreo's key features is its ability to scale drawings using AI. This feature can automatically suggest a scale for a particular drawing based on its peculiarities once the drawing is uploaded. All users need to do is confirm the suggestion by pressing the "Set" button. This feature is particularly useful for users who need to create scale drawings on a regular basis, such as architects, engineers, and draftsmen. It can also be helpful for users who are new to scale drawing, as it can help them to avoid common mistakes. To use the AI-powered scale suggestion feature, users simply need to upload their drawing to Kreo. Kreo will then analyze the drawing and suggest a scale based on its peculiarities. Users can then review the suggestion and confirm it by pressing the "Set" button. Overall, Kreo's AI-powered scale suggestion feature is a valuable tool for anyone who needs to create scale drawings. It can save users a significant amount of time and effort, improve accuracy, and reduce errors. Conclusion Scale drawings, with their unmatched precision and adaptability, underpin many professions. Their influence, evident in every constructed space and object around us, is a testament to their importance. As we advance technologically, the tools and methods may evolve, but the essence of scale drawing remains unchanged — to represent with accuracy and clarity. It's an invitation for every professional to delve deeper into this realm and harness its potential. ‍ Table of contents What is a Scale Drawing? Components How to Read Scale Examples The Most Common Engineering Drawing Scales AI for Scale Drawing Conclusion 24 Shares 20 Recent blogs AI Andrew September 17, 2025 ###### AI-Powered Office Space Planning: A Definitive Guide to Automating Drawing Analysis for Strategic Advantage AI Andrew September 17, 2025 ###### The Blueprint for Global Real Estate: A Deep Dive into BOMA's International Measurement Standards AI Andrew September 17, 2025 ###### Beyond the Blueprint: What is the UFA – Usable Floor Area? Try Kreo Free Get Started Talk to a Human No installation required Works on both Windows & Mac Timely customer support Subscribe to Twitter ≈ 1.3K Subscribe to Youtube ≈ 1K Follow us on Instagram ≈ 300 Subscribe to Linkedin ≈ 3200 Don't leave yet! See how Kreo can boost your speed and save time! Discover how much faster Kreo's AI can boost your workflow! What is your primary professional role? Quantity Surveyor Cost Estimator Contractor Architect Other What types of projects do you work on most frequently? Residential Commercial Industrial Infrastructure Other What tools do you currently use for takeoff and estimating? Excel Specialized Software Manual Calculations Other How often do you perform takeoff and estimates for projects? Daily Weekly Monthly Less frequently What is your main challenge in takeoff and estimating? Errors and Inaccuracies Takes Too Long to Complete Tasks Lack of Automation High Cost of Existing Solutions How much time do you usually spend on a single estimate? Less than 1 Working Day 1-2 Days 2-5 Days 1-2 Weeks More than 2 weeks Depends on Project Complexity What is the most important factor when choosing estimating software? Accuracy Speed Intuitive Interface Cost Support and Integrations Enter your email Enter an email to get the results See How Kreo Could Transform Your Workflow! Insights are calculated based on a 40-hour workweek. Work Speed Boost Time Saved Improving accuracy Ready to see the results? Try Kreo and transform your workflow! Get Started Oops! Something went wrong while submitting the form. Ready to see how Kreo Software can benefit you? Just 15 minutes to see if our solution is right for you Email Address Thank you! Your submission has been received! Oops! Something went wrong while submitting the form. Privacy policyTerms & Conditions Copyright © 2025 Kreo Software COMPANY Home Contact Us About Us Blog Customers Partners Solutions for Constuction Takeoff for Cost Planning for Quantity Surveyors for Cost Estimators for Contractors for Measuring Blueprints for BIM Takeoff trades Flooring Concrete Electrical Framing Drywall Painting Steel Masonry learn Events Help Center Book a Demo Glossary lATEST BLOGS AI-Powered Office Space Planning: A Definitive Guide to Automating Drawing Analysis for Strategic Advantage September 17, 2025 The Blueprint for Global Real Estate: A Deep Dive into BOMA's International Measurement Standards September 11, 2025 Beyond the Blueprint: What is the UFA – Usable Floor Area? September 11, 2025 Read all blogs
10301	https://en.wikipedia.org/wiki/Bernstein%27s_theorem_on_monotone_functions	Jump to content Search Contents (Top) 1 Bernstein functions 2 See also 3 References Bernstein's theorem on monotone functions Esperanto Français 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 Mathematical theorem In real analysis, a branch of mathematics, Bernstein's theorem states that every real-valued function on the half-line [0, ∞) that is totally monotone is a mixture of exponential functions. In one important special case the mixture is a weighted average, or expected value. Total monotonicity (sometimes also complete monotonicity) of a function f means that f is continuous on [0, ∞), infinitely differentiable on (0, ∞), and satisfies for all nonnegative integers n and for all t > 0. Another convention puts the opposite inequality in the above definition. The "weighted average" statement can be characterized thus: there is a non-negative finite Borel measure on [0, ∞) with cumulative distribution function g such that the integral being a Riemann–Stieltjes integral. In more abstract language, the theorem characterises Laplace transforms of positive Borel measures on [0, ∞). In this form it is known as the Bernstein–Widder theorem, or Hausdorff–Bernstein–Widder theorem. Felix Hausdorff had earlier characterised completely monotone sequences. These are the sequences occurring in the Hausdorff moment problem. Bernstein functions [edit] Nonnegative functions whose derivative is completely monotone are called Bernstein functions. Every Bernstein function has the Lévy–Khintchine representation: where and is a measure on the positive real half-line such that See also [edit] Absolutely and completely monotonic functions and sequences References [edit] S. N. Bernstein (1928). "Sur les fonctions absolument monotones". Acta Mathematica. 52: 1–66. doi:10.1007/BF02592679. D. Widder (1941). The Laplace Transform. Princeton University Press. Rene Schilling, Renming Song and Zoran Vondraček (2010). Bernstein functions. De Gruyter. Retrieved from " Categories: Theorems in real analysis Theorems in measure theory Hidden categories: Articles with short description Short description matches Wikidata Bernstein's theorem on monotone functions Add topic
10302	https://gist.github.com/Miserlou/11500b2345d3fe850c92	1000 Largest US Cities By Population · GitHub Skip to content Search Gists Search Gists All gistsBack to GitHubSign inSign up Sign inSign up You signed in with another tab or window. Reload to refresh your session.You signed out in another tab or window. Reload to refresh your session.You switched accounts on another tab or window. Reload to refresh your session.Dismiss alert {{ message }} Instantly share code, notes, and snippets. Miserlou/gist:11500b2345d3fe850c92 Created April 27, 2015 21:20 Show Gist options Download ZIP Star 59(59)You must be signed in to star a gist Fork 25(25)You must be signed in to fork a gist Embed Embed Embed this gist in your website. Share Copy sharable link for this gist. Clone via HTTPS Clone using the web URL. Learn more about clone URLs Clone this repository at <script src=" Save Miserlou/11500b2345d3fe850c92 to your computer and use it in GitHub Desktop. CodeRevisions 1Stars 57Forks 25 Embed Embed Embed this gist in your website. Share Copy sharable link for this gist. Clone via HTTPS Clone using the web URL. Learn more about clone URLs Clone this repository at <script src=" Save Miserlou/11500b2345d3fe850c92 to your computer and use it in GitHub Desktop. Download ZIP 1000 Largest US Cities By Population Raw gistfile1.txt This file contains hidden or bidirectional Unicode text that may be interpreted or compiled differently than what appears below. To review, open the file in an editor that reveals hidden Unicode characters. Learn more about bidirectional Unicode characters Show hidden characters Largest 1000 Cities in America 2013 popuation data - Biggest US Cities By Population rank,city,state,population,2000-2013 growth 1,New York,New York,8405837,4.8% 2,Los Angeles,California,3884307,4.8% 3,Chicago,Illinois,2718782,-6.1% 4,Houston,Texas,2195914,11.0% 5,Philadelphia,Pennsylvania,1553165,2.6% 6,Phoenix,Arizona,1513367,14.0% 7,San Antonio,Texas,1409019,21.0% 8,San Diego,California,1355896,10.5% 9,Dallas,Texas,1257676,5.6% 10,San Jose,California,998537,10.5% 11,Austin,Texas,885400,31.7% 12,Indianapolis,Indiana,843393,7.8% 13,Jacksonville,Florida,842583,14.3% 14,San Francisco,California,837442,7.7% 15,Columbus,Ohio,822553,14.8% 16,Charlotte,North Carolina,792862,39.1% 17,Fort Worth,Texas,792727,45.1% 18,Detroit,Michigan,688701,-27.1% 19,El Paso,Texas,674433,19.4% 20,Memphis,Tennessee,653450,-5.3% 21,Seattle,Washington,652405,15.6% 22,Denver,Colorado,649495,16.7% 23,Washington,District of Columbia,646449,13.0% 24,Boston,Massachusetts,645966,9.4% 25,Nashville-Davidson,Tennessee,634464,16.2% 26,Baltimore,Maryland,622104,-4.0% 27,Oklahoma City,Oklahoma,610613,20.2% 28,Louisville/Jefferson County,Kentucky,609893,10.0% 29,Portland,Oregon,609456,15.0% 30,Las Vegas,Nevada,603488,24.5% 31,Milwaukee,Wisconsin,599164,0.3% 32,Albuquerque,New Mexico,556495,23.5% 33,Tucson,Arizona,526116,7.5% 34,Fresno,California,509924,18.3% 35,Sacramento,California,479686,17.2% 36,Long Beach,California,469428,1.5% 37,Kansas City,Missouri,467007,5.5% 38,Mesa,Arizona,457587,13.5% 39,Virginia Beach,Virginia,448479,5.1% 40,Atlanta,Georgia,447841,6.2% 41,Colorado Springs,Colorado,439886,21.4% 42,Omaha,Nebraska,434353,5.9% 43,Raleigh,North Carolina,431746,48.7% 44,Miami,Florida,417650,14.9% 45,Oakland,California,406253,1.3% 46,Minneapolis,Minnesota,400070,4.5% 47,Tulsa,Oklahoma,398121,1.3% 48,Cleveland,Ohio,390113,-18.1% 49,Wichita,Kansas,386552,9.7% 50,Arlington,Texas,379577,13.3% 51,New Orleans,Louisiana,378715,-21.6% 52,Bakersfield,California,363630,48.4% 53,Tampa,Florida,352957,16.0% 54,Honolulu,Hawaii,347884,-6.2% 55,Aurora,Colorado,345803,24.4% 56,Anaheim,California,345012,4.7% 57,Santa Ana,California,334227,-1.2% 58,St. Louis,Missouri,318416,-8.2% 59,Riverside,California,316619,22.5% 60,Corpus Christi,Texas,316381,14.1% 61,Lexington-Fayette,Kentucky,308428,18.0% 62,Pittsburgh,Pennsylvania,305841,-8.3% 63,Anchorage,Alaska,300950,15.4% 64,Stockton,California,298118,21.8% 65,Cincinnati,Ohio,297517,-10.1% 66,St. Paul,Minnesota,294873,2.8% 67,Toledo,Ohio,282313,-10.0% 68,Greensboro,North Carolina,279639,22.3% 69,Newark,New Jersey,278427,2.1% 70,Plano,Texas,274409,22.4% 71,Henderson,Nevada,270811,51.0% 72,Lincoln,Nebraska,268738,18.0% 73,Buffalo,New York,258959,-11.3% 74,Jersey City,New Jersey,257342,7.2% 75,Chula Vista,California,256780,46.2% 76,Fort Wayne,Indiana,256496,1.0% 77,Orlando,Florida,255483,31.2% 78,St. Petersburg,Florida,249688,0.3% 79,Chandler,Arizona,249146,38.7% 80,Laredo,Texas,248142,38.2% 81,Norfolk,Virginia,246139,5.0% 82,Durham,North Carolina,245475,29.9% 83,Madison,Wisconsin,243344,15.8% 84,Lubbock,Texas,239538,19.6% 85,Irvine,California,236716,61.3% 86,Winston-Salem,North Carolina,236441,16.9% 87,Glendale,Arizona,234632,5.7% 88,Garland,Texas,234566,8.5% 89,Hialeah,Florida,233394,3.2% 90,Reno,Nevada,233294,26.8% 91,Chesapeake,Virginia,230571,15.1% 92,Gilbert,Arizona,229972,96.0% 93,Baton Rouge,Louisiana,229426,0.4% 94,Irving,Texas,228653,19.1% 95,Scottsdale,Arizona,226918,11.0% 96,North Las Vegas,Nevada,226877,92.2% 97,Fremont,California,224922,10.0% 98,Boise City,Idaho,214237,9.5% 99,Richmond,Virginia,214114,8.2% 100,San Bernardino,California,213708,13.0% 101,Birmingham,Alabama,212113,-12.3% 102,Spokane,Washington,210721,7.0% 103,Rochester,New York,210358,-4.1% 104,Des Moines,Iowa,207510,3.9% 105,Modesto,California,204933,7.7% 106,Fayetteville,North Carolina,204408,2.4% 107,Tacoma,Washington,203446,4.9% 108,Oxnard,California,203007,18.2% 109,Fontana,California,203003,38.3% 110,Columbus,Georgia,202824,8.7% 111,Montgomery,Alabama,201332,-0.1% 112,Moreno Valley,California,201175,40.4% 113,Shreveport,Louisiana,200327,-0.1% 114,Aurora,Illinois,199963,38.4% 115,Yonkers,New York,199766,1.8% 116,Akron,Ohio,198100,-8.6% 117,Huntington Beach,California,197575,3.9% 118,Little Rock,Arkansas,197357,7.6% 119,Augusta-Richmond County,Georgia,197350,1.1% 120,Amarillo,Texas,196429,12.8% 121,Glendale,California,196021,0.3% 122,Mobile,Alabama,194899,-1.9% 123,Grand Rapids,Michigan,192294,-2.8% 124,Salt Lake City,Utah,191180,5.1% 125,Tallahassee,Florida,186411,21.8% 126,Huntsville,Alabama,186254,16.3% 127,Grand Prairie,Texas,183372,43.1% 128,Knoxville,Tennessee,183270,3.9% 129,Worcester,Massachusetts,182544,5.8% 130,Newport News,Virginia,182020,0.9% 131,Brownsville,Texas,181860,26.8% 132,Overland Park,Kansas,181260,19.4% 133,Santa Clarita,California,179590,15.3% 134,Providence,Rhode Island,177994,2.3% 135,Garden Grove,California,175140,5.8% 136,Chattanooga,Tennessee,173366,10.5% 137,Oceanside,California,172794,6.6% 138,Jackson,Mississippi,172638,-6.8% 139,Fort Lauderdale,Florida,172389,0.7% 140,Santa Rosa,California,171990,15.2% 141,Rancho Cucamonga,California,171386,32.7% 142,Port St. Lucie,Florida,171016,91.7% 143,Tempe,Arizona,168228,5.8% 144,Ontario,California,167500,5.5% 145,Vancouver,Washington,167405,14.2% 146,Cape Coral,Florida,165831,60.4% 147,Sioux Falls,South Dakota,164676,31.1% 148,Springfield,Missouri,164122,7.8% 149,Peoria,Arizona,162592,46.5% 150,Pembroke Pines,Florida,162329,17.4% 151,Elk Grove,California,161007,97.1% 152,Salem,Oregon,160614,16.4% 153,Lancaster,California,159523,33.8% 154,Corona,California,159503,23.6% 155,Eugene,Oregon,159190,14.4% 156,Palmdale,California,157161,33.7% 157,Salinas,California,155662,8.4% 158,Springfield,Massachusetts,153703,1.1% 159,Pasadena,Texas,152735,7.5% 160,Fort Collins,Colorado,152061,26.6% 161,Hayward,California,151574,7.5% 162,Pomona,California,151348,2.1% 163,Cary,North Carolina,151088,55.1% 164,Rockford,Illinois,150251,-1.0% 165,Alexandria,Virginia,148892,15.0% 166,Escondido,California,148738,10.7% 167,McKinney,Texas,148559,165.3% 168,Kansas City,Kansas,148483,1.1% 169,Joliet,Illinois,147806,36.5% 170,Sunnyvale,California,147559,11.9% 171,Torrance,California,147478,6.6% 172,Bridgeport,Connecticut,147216,5.4% 173,Lakewood,Colorado,147214,1.9% 174,Hollywood,Florida,146526,4.8% 175,Paterson,New Jersey,145948,-2.2% 176,Naperville,Illinois,144864,12.0% 177,Syracuse,New York,144669,-0.9% 178,Mesquite,Texas,143484,14.7% 179,Dayton,Ohio,143355,-13.5% 180,Savannah,Georgia,142772,7.5% 181,Clarksville,Tennessee,142357,36.9% 182,Orange,California,139969,7.7% 183,Pasadena,California,139731,3.8% 184,Fullerton,California,138981,9.8% 185,Killeen,Texas,137147,52.1% 186,Frisco,Texas,136791,287.7% 187,Hampton,Virginia,136699,-6.6% 188,McAllen,Texas,136639,27.6% 189,Warren,Michigan,134873,-2.3% 190,Bellevue,Washington,133992,19.1% 191,West Valley City,Utah,133579,22.2% 192,Columbia,South Carolina,133358,11.7% 193,Olathe,Kansas,131885,40.4% 194,Sterling Heights,Michigan,131224,5.2% 195,New Haven,Connecticut,130660,5.5% 196,Miramar,Florida,130288,74.7% 197,Waco,Texas,129030,12.5% 198,Thousand Oaks,California,128731,9.5% 199,Cedar Rapids,Iowa,128429,5.4% 200,Charleston,South Carolina,127999,29.2% 201,Visalia,California,127763,33.6% 202,Topeka,Kansas,127679,3.4% 203,Elizabeth,New Jersey,127558,5.5% 204,Gainesville,Florida,127488,12.8% 205,Thornton,Colorado,127359,52.9% 206,Roseville,California,127035,56.2% 207,Carrollton,Texas,126700,14.9% 208,Coral Springs,Florida,126604,5.7% 209,Stamford,Connecticut,126456,7.6% 210,Simi Valley,California,126181,12.6% 211,Concord,California,125880,2.9% 212,Hartford,Connecticut,125017,0.6% 213,Kent,Washington,124435,54.3% 214,Lafayette,Louisiana,124276,11.0% 215,Midland,Texas,123933,30.4% 216,Surprise,Arizona,123546,281.9% 217,Denton,Texas,123099,47.1% 218,Victorville,California,121096,87.6% 219,Evansville,Indiana,120310,-0.8% 220,Santa Clara,California,120245,17.4% 221,Abilene,Texas,120099,3.6% 222,Athens-Clarke County,Georgia,119980,19.0% 223,Vallejo,California,118837,1.2% 224,Allentown,Pennsylvania,118577,11.2% 225,Norman,Oklahoma,118197,22.0% 226,Beaumont,Texas,117796,3.7% 227,Independence,Missouri,117240,3.2% 228,Murfreesboro,Tennessee,117044,65.1% 229,Ann Arbor,Michigan,117025,2.0% 230,Springfield,Illinois,117006,4.2% 231,Berkeley,California,116768,13.3% 232,Peoria,Illinois,116513,3.0% 233,Provo,Utah,116288,10.0% 234,El Monte,California,115708,-0.4% 235,Columbia,Missouri,115276,34.0% 236,Lansing,Michigan,113972,-4.4% 237,Fargo,North Dakota,113658,24.9% 238,Downey,California,113242,5.3% 239,Costa Mesa,California,112174,2.4% 240,Wilmington,North Carolina,112067,24.8% 241,Arvada,Colorado,111707,9.2% 242,Inglewood,California,111542,-1.0% 243,Miami Gardens,Florida,111378,10.5% 244,Carlsbad,California,110972,39.7% 245,Westminster,Colorado,110945,9.4% 246,Rochester,Minnesota,110742,23.9% 247,Odessa,Texas,110720,22.3% 248,Manchester,New Hampshire,110378,2.9% 249,Elgin,Illinois,110145,16.0% 250,West Jordan,Utah,110077,38.4% 251,Round Rock,Texas,109821,81.0% 252,Clearwater,Florida,109703,0.1% 253,Waterbury,Connecticut,109676,2.2% 254,Gresham,Oregon,109397,20.7% 255,Fairfield,California,109320,12.8% 256,Billings,Montana,109059,18.6% 257,Lowell,Massachusetts,108861,3.4% 258,San Buenaventura (Ventura),California,108817,7.4% 259,Pueblo,Colorado,108249,5.9% 260,High Point,North Carolina,107741,24.3% 261,West Covina,California,107740,2.3% 262,Richmond,California,107571,7.9% 263,Murrieta,California,107479,107.4% 264,Cambridge,Massachusetts,107289,5.5% 265,Antioch,California,107100,16.9% 266,Temecula,California,106780,55.4% 267,Norwalk,California,106589,1.9% 268,Centennial,Colorado,106114,3.5% 269,Everett,Washington,105370,9.4% 270,Palm Bay,Florida,104898,31.7% 271,Wichita Falls,Texas,104898,0.7% 272,Green Bay,Wisconsin,104779,1.9% 273,Daly City,California,104739,1.0% 274,Burbank,California,104709,4.2% 275,Richardson,Texas,104475,13.2% 276,Pompano Beach,Florida,104410,4.0% 277,North Charleston,South Carolina,104054,27.4% 278,Broken Arrow,Oklahoma,103500,28.2% 279,Boulder,Colorado,103166,9.0% 280,West Palm Beach,Florida,102436,23.5% 281,Santa Maria,California,102216,30.9% 282,El Cajon,California,102211,7.4% 283,Davenport,Iowa,102157,3.9% 284,Rialto,California,101910,9.8% 285,Las Cruces,New Mexico,101324,37.6% 286,San Mateo,California,101128,9.0% 287,Lewisville,Texas,101074,28.9% 288,South Bend,Indiana,100886,-6.8% 289,Lakeland,Florida,100710,18.3% 290,Erie,Pennsylvania,100671,-2.8% 291,Tyler,Texas,100223,18.6% 292,Pearland,Texas,100065,117.2% 293,College Station,Texas,100050,45.2% 294,Kenosha,Wisconsin,99889,9.5% 295,Sandy Springs,Georgia,99770,17.4% 296,Clovis,California,99769,42.6% 297,Flint,Michigan,99763,-20.0% 298,Roanoke,Virginia,98465,3.8% 299,Albany,New York,98424,4.1% 300,Jurupa Valley,California,98030, 301,Compton,California,97877,4.5% 302,San Angelo,Texas,97492,10.2% 303,Hillsboro,Oregon,97368,36.4% 304,Lawton,Oklahoma,97151,4.9% 305,Renton,Washington,97003,88.4% 306,Vista,California,96929,7.7% 307,Davie,Florida,96830,17.7% 308,Greeley,Colorado,96539,23.1% 309,Mission Viejo,California,96346,2.9% 310,Portsmouth,Virginia,96205,-4.2% 311,Dearborn,Michigan,95884,-2.0% 312,South Gate,California,95677,-0.8% 313,Tuscaloosa,Alabama,95334,21.1% 314,Livonia,Michigan,95208,-5.4% 315,New Bedford,Massachusetts,95078,1.2% 316,Vacaville,California,94275,5.4% 317,Brockton,Massachusetts,94089,-0.3% 318,Roswell,Georgia,94034,15.2% 319,Beaverton,Oregon,93542,17.0% 320,Quincy,Massachusetts,93494,5.8% 321,Sparks,Nevada,93282,39.4% 322,Yakima,Washington,93257,11.7% 323,Lee's Summit,Missouri,93184,31.2% 324,Federal Way,Washington,92734,8.8% 325,Carson,California,92599,2.9% 326,Santa Monica,California,92472,9.6% 327,Hesperia,California,92147,46.1% 328,Allen,Texas,92020,104.0% 329,Rio Rancho,New Mexico,91956,74.4% 330,Yuma,Arizona,91923,16.2% 331,Westminster,California,91739,3.9% 332,Orem,Utah,91648,8.5% 333,Lynn,Massachusetts,91589,2.6% 334,Redding,California,91119,11.9% 335,Spokane Valley,Washington,91113,12.6% 336,Miami Beach,Florida,91026,3.3% 337,League City,Texas,90983,98.3% 338,Lawrence,Kansas,90811,12.7% 339,Santa Barbara,California,90412,0.9% 340,Plantation,Florida,90268,8.6% 341,Sandy,Utah,90231,1.3% 342,Sunrise,Florida,90116,4.6% 343,Macon,Georgia,89981,-7.3% 344,Longmont,Colorado,89919,24.4% 345,Boca Raton,Florida,89407,7.5% 346,San Marcos,California,89387,60.0% 347,Greenville,North Carolina,89130,41.9% 348,Waukegan,Illinois,88826,0.5% 349,Fall River,Massachusetts,88697,-3.7% 350,Chico,California,88077,14.2% 351,Newton,Massachusetts,87971,4.9% 352,San Leandro,California,87965,10.3% 353,Reading,Pennsylvania,87893,8.0% 354,Norwalk,Connecticut,87776,5.6% 355,Fort Smith,Arkansas,87650,8.6% 356,Newport Beach,California,87273,10.4% 357,Asheville,North Carolina,87236,19.6% 358,Nashua,New Hampshire,87137,0.4% 359,Edmond,Oklahoma,87004,26.9% 360,Whittier,California,86635,3.3% 361,Nampa,Idaho,86518,57.9% 362,Bloomington,Minnesota,86319,1.3% 363,Deltona,Florida,86290,23.1% 364,Hawthorne,California,86199,2.3% 365,Duluth,Minnesota,86128,-0.1% 366,Carmel,Indiana,85927,60.4% 367,Suffolk,Virginia,85728,33.5% 368,Clifton,New Jersey,85390,7.9% 369,Citrus Heights,California,85285,-0.1% 370,Livermore,California,85156,15.1% 371,Tracy,California,84691,45.9% 372,Alhambra,California,84577,-0.7% 373,Kirkland,Washington,84430,87.5% 374,Trenton,New Jersey,84349,-1.2% 375,Ogden,Utah,84249,8.6% 376,Hoover,Alabama,84126,32.7% 377,Cicero,Illinois,84103,-1.6% 378,Fishers,Indiana,83891,114.8% 379,Sugar Land,Texas,83860,29.1% 380,Danbury,Connecticut,83684,11.4% 381,Meridian,Idaho,83596,127.6% 382,Indio,California,83539,66.0% 383,Concord,North Carolina,83506,47.4% 384,Menifee,California,83447,95.0% 385,Champaign,Illinois,83424,18.3% 386,Buena Park,California,82882,6.1% 387,Troy,Michigan,82821,2.2% 388,O'Fallon,Missouri,82809,62.6% 389,Johns Creek,Georgia,82788,36.5% 390,Bellingham,Washington,82631,21.8% 391,Westland,Michigan,82578,-4.7% 392,Bloomington,Indiana,82575,16.1% 393,Sioux City,Iowa,82459,-2.9% 394,Warwick,Rhode Island,81971,-4.6% 395,Hemet,California,81750,37.6% 396,Longview,Texas,81443,11.6% 397,Farmington Hills,Michigan,81295,-0.9% 398,Bend,Oregon,81236,54.3% 399,Lakewood,California,81121,2.1% 400,Merced,California,81102,25.4% 401,Mission,Texas,81050,74.5% 402,Chino,California,80988,15.6% 403,Redwood City,California,80872,7.1% 404,Edinburg,Texas,80836,65.1% 405,Cranston,Rhode Island,80566,1.4% 406,Parma,Ohio,80429,-5.9% 407,New Rochelle,New York,79446,9.9% 408,Lake Forest,California,79312,4.2% 409,Napa,California,79068,8.4% 410,Hammond,Indiana,78967,-4.6% 411,Fayetteville,Arkansas,78960,32.9% 412,Bloomington,Illinois,78902,20.1% 413,Avondale,Arizona,78822,111.5% 414,Somerville,Massachusetts,78804,1.6% 415,Palm Coast,Florida,78740,137.2% 416,Bryan,Texas,78709,19.3% 417,Gary,Indiana,78450,-23.4% 418,Largo,Florida,78409,5.1% 419,Brooklyn Park,Minnesota,78373,16.0% 420,Tustin,California,78327,15.6% 421,Racine,Wisconsin,78199,-4.4% 422,Deerfield Beach,Florida,78041,4.8% 423,Lynchburg,Virginia,78014,19.5% 424,Mountain View,California,77846,10.1% 425,Medford,Oregon,77677,17.1% 426,Lawrence,Massachusetts,77657,7.5% 427,Bellflower,California,77593,6.3% 428,Melbourne,Florida,77508,5.9% 429,St. Joseph,Missouri,77147,4.1% 430,Camden,New Jersey,76903,-3.6% 431,St. George,Utah,76817,53.1% 432,Kennewick,Washington,76762,29.1% 433,Baldwin Park,California,76635,0.8% 434,Chino Hills,California,76572,13.6% 435,Alameda,California,76419,5.4% 436,Albany,Georgia,76185,-0.6% 437,Arlington Heights,Illinois,75994,-0.6% 438,Scranton,Pennsylvania,75806,0.0% 439,Evanston,Illinois,75570,1.9% 440,Kalamazoo,Michigan,75548,-1.9% 441,Baytown,Texas,75418,13.1% 442,Upland,California,75413,9.5% 443,Springdale,Arkansas,75229,57.1% 444,Bethlehem,Pennsylvania,75018,5.2% 445,Schaumburg,Illinois,74907,-0.5% 446,Mount Pleasant,South Carolina,74885,53.2% 447,Auburn,Washington,74860,34.9% 448,Decatur,Illinois,74710,-8.7% 449,San Ramon,California,74513,65.8% 450,Pleasanton,California,74110,15.2% 451,Wyoming,Michigan,74100,6.5% 452,Lake Charles,Louisiana,74024,3.0% 453,Plymouth,Minnesota,73987,12.0% 454,Bolingbrook,Illinois,73936,29.7% 455,Pharr,Texas,73790,55.7% 456,Appleton,Wisconsin,73596,4.5% 457,Gastonia,North Carolina,73209,8.2% 458,Folsom,California,73098,38.6% 459,Southfield,Michigan,73006,-6.7% 460,Rochester Hills,Michigan,72952,5.7% 461,New Britain,Connecticut,72939,1.9% 462,Goodyear,Arizona,72864,271.0% 463,Canton,Ohio,72535,-10.3% 464,Warner Robins,Georgia,72531,45.7% 465,Union City,California,72528,7.4% 466,Perris,California,72326,98.7% 467,Manteca,California,71948,42.7% 468,Iowa City,Iowa,71591,13.8% 469,Jonesboro,Arkansas,71551,28.3% 470,Wilmington,Delaware,71525,-1.6% 471,Lynwood,California,71371,2.0% 472,Loveland,Colorado,71334,37.4% 473,Pawtucket,Rhode Island,71172,-2.5% 474,Boynton Beach,Florida,71097,17.3% 475,Waukesha,Wisconsin,71016,8.0% 476,Gulfport,Mississippi,71012,-0.6% 477,Apple Valley,California,70924,29.9% 478,Passaic,New Jersey,70868,4.3% 479,Rapid City,South Dakota,70812,17.9% 480,Layton,Utah,70790,20.2% 481,Lafayette,Indiana,70373,14.5% 482,Turlock,California,70365,23.5% 483,Muncie,Indiana,70316,-0.7% 484,Temple,Texas,70190,27.1% 485,Missouri City,Texas,70185,31.1% 486,Redlands,California,69999,9.4% 487,Santa Fe,New Mexico,69976,10.5% 488,Lauderhill,Florida,69813,4.2% 489,Milpitas,California,69783,11.0% 490,Palatine,Illinois,69350,4.5% 491,Missoula,Montana,69122,19.7% 492,Rock Hill,South Carolina,69103,36.0% 493,Jacksonville,North Carolina,69079,5.0% 494,Franklin,Tennessee,68886,48.5% 495,Flagstaff,Arizona,68667,29.3% 496,Flower Mound,Texas,68609,32.5% 497,Weston,Florida,68388,34.5% 498,Waterloo,Iowa,68366,-0.5% 499,Union City,New Jersey,68247,1.7% 500,Mount Vernon,New York,68224,-0.2% 501,Fort Myers,Florida,68190,31.2% 502,Dothan,Alabama,68001,16.6% 503,Rancho Cordova,California,67911,26.4% 504,Redondo Beach,California,67815,6.7% 505,Jackson,Tennessee,67685,12.9% 506,Pasco,Washington,67599,98.5% 507,St. Charles,Missouri,67569,11.3% 508,Eau Claire,Wisconsin,67545,8.7% 509,North Richland Hills,Texas,67317,20.2% 510,Bismarck,North Dakota,67034,20.1% 511,Yorba Linda,California,67032,13.4% 512,Kenner,Louisiana,66975,-4.8% 513,Walnut Creek,California,66900,3.5% 514,Frederick,Maryland,66893,25.9% 515,Oshkosh,Wisconsin,66778,5.3% 516,Pittsburg,California,66695,16.6% 517,Palo Alto,California,66642,13.7% 518,Bossier City,Louisiana,66333,17.4% 519,Portland,Maine,66318,3.2% 520,St. Cloud,Minnesota,66297,10.9% 521,Davis,California,66205,11.9% 522,South San Francisco,California,66174,9.1% 523,Camarillo,California,66086,14.9% 524,North Little Rock,Arkansas,66075,9.0% 525,Schenectady,New York,65902,6.7% 526,Gaithersburg,Maryland,65690,24.2% 527,Harlingen,Texas,65665,11.6% 528,Woodbury,Minnesota,65656,39.8% 529,Eagan,Minnesota,65453,2.6% 530,Yuba City,California,65416,27.9% 531,Maple Grove,Minnesota,65415,27.3% 532,Youngstown,Ohio,65184,-20.2% 533,Skokie,Illinois,65176,2.8% 534,Kissimmee,Florida,65173,32.6% 535,Johnson City,Tennessee,65123,16.2% 536,Victoria,Texas,65098,7.5% 537,San Clemente,California,65040,28.6% 538,Bayonne,New Jersey,65028,5.1% 539,Laguna Niguel,California,64652,2.8% 540,East Orange,New Jersey,64544,-7.4% 541,Shawnee,Kansas,64323,32.2% 542,Homestead,Florida,64079,100.7% 544,Rockville,Maryland,64072,34.0% 543,Delray Beach,Florida,64072,6.1% 545,Janesville,Wisconsin,63820,5.6% 546,Conway,Arkansas,63816,46.1% 547,Pico Rivera,California,63771,0.4% 548,Lorain,Ohio,63710,-7.2% 549,Montebello,California,63495,2.0% 550,Lodi,California,63338,10.1% 551,New Braunfels,Texas,63279,64.0% 552,Marysville,Washington,63269,115.7% 553,Tamarac,Florida,63155,12.9% 554,Madera,California,63105,44.4% 555,Conroe,Texas,63032,61.9% 556,Santa Cruz,California,62864,12.5% 557,Eden Prairie,Minnesota,62603,13.3% 558,Cheyenne,Wyoming,62448,16.9% 559,Daytona Beach,Florida,62316,-2.3% 560,Alpharetta,Georgia,62298,33.6% 561,Hamilton,Ohio,62258,2.7% 562,Waltham,Massachusetts,62227,5.0% 563,Coon Rapids,Minnesota,62103,0.6% 564,Haverhill,Massachusetts,62088,5.0% 565,Council Bluffs,Iowa,61969,6.2% 566,Taylor,Michigan,61817,-6.3% 567,Utica,New York,61808,2.2% 568,Ames,Iowa,61792,21.3% 569,La Habra,California,61653,3.6% 570,Encinitas,California,61588,5.8% 571,Bowling Green,Kentucky,61488,24.1% 572,Burnsville,Minnesota,61434,1.9% 573,Greenville,South Carolina,61397,8.2% 574,West Des Moines,Iowa,61255,29.8% 575,Cedar Park,Texas,61238,134.3% 576,Tulare,California,61170,33.3% 577,Monterey Park,California,61085,1.5% 578,Vineland,New Jersey,61050,9.3% 579,Terre Haute,Indiana,61025,2.5% 580,North Miami,Florida,61007,2.0% 581,Mansfield,Texas,60872,114.2% 582,West Allis,Wisconsin,60697,-0.6% 583,Bristol,Connecticut,60568,0.4% 584,Taylorsville,Utah,60519,2.9% 585,Malden,Massachusetts,60509,7.4% 586,Meriden,Connecticut,60456,3.7% 587,Blaine,Minnesota,60407,32.8% 588,Wellington,Florida,60202,55.0% 589,Cupertino,California,60189,14.3% 590,Springfield,Oregon,60177,12.4% 591,Rogers,Arkansas,60112,50.6% 592,St. Clair Shores,Michigan,60070,-4.6% 593,Gardena,California,59957,3.4% 594,Pontiac,Michigan,59887,-11.4% 595,National City,California,59834,10.1% 596,Grand Junction,Colorado,59778,30.9% 597,Rocklin,California,59738,60.3% 598,Chapel Hill,North Carolina,59635,24.1% 599,Casper,Wyoming,59628,19.9% 600,Broomfield,Colorado,59471,50.3% 601,Petaluma,California,59440,8.4% 602,South Jordan,Utah,59366,100.1% 603,Springfield,Ohio,59357,-9.8% 604,Great Falls,Montana,59351,3.9% 605,Lancaster,Pennsylvania,59325,4.5% 606,North Port,Florida,59212,154.6% 607,Lakewood,Washington,59097,1.1% 608,Marietta,Georgia,59089,-3.8% 609,San Rafael,California,58994,5.0% 610,Royal Oak,Michigan,58946,-1.7% 611,Des Plaines,Illinois,58918,3.2% 612,Huntington Park,California,58879,-4.1% 613,La Mesa,California,58642,6.9% 614,Orland Park,Illinois,58590,13.9% 615,Auburn,Alabama,58582,26.4% 616,Lakeville,Minnesota,58562,34.3% 617,Owensboro,Kentucky,58416,7.7% 618,Moore,Oklahoma,58414,41.5% 619,Jupiter,Florida,58298,46.2% 620,Idaho Falls,Idaho,58292,14.0% 621,Dubuque,Iowa,58253,0.9% 622,Bartlett,Tennessee,58226,31.7% 623,Rowlett,Texas,58043,28.6% 624,Novi,Michigan,57960,22.0% 625,White Plains,New York,57866,8.5% 626,Arcadia,California,57639,8.3% 627,Redmond,Washington,57530,26.0% 628,Lake Elsinore,California,57525,96.5% 629,Ocala,Florida,57468,20.8% 630,Tinley Park,Illinois,57282,16.3% 631,Port Orange,Florida,57203,22.8% 632,Medford,Massachusetts,57170,2.7% 633,Oak Lawn,Illinois,57073,3.3% 634,Rocky Mount,North Carolina,56954,-3.1% 635,Kokomo,Indiana,56895,21.3% 636,Coconut Creek,Florida,56792,28.4% 637,Bowie,Maryland,56759,8.6% 638,Berwyn,Illinois,56758,5.1% 639,Midwest City,Oklahoma,56756,4.5% 640,Fountain Valley,California,56707,3.0% 641,Buckeye,Arizona,56683,480.9% 642,Dearborn Heights,Michigan,56620,-3.0% 643,Woodland,California,56590,13.8% 644,Noblesville,Indiana,56540,88.1% 645,Valdosta,Georgia,56481,22.3% 646,Diamond Bar,California,56449,0.1% 647,Manhattan,Kansas,56143,22.8% 648,Santee,California,56105,5.7% 649,Taunton,Massachusetts,56069,0.0% 650,Sanford,Florida,56002,42.8% 651,Kettering,Ohio,55870,-3.1% 652,New Brunswick,New Jersey,55831,15.5% 653,Decatur,Alabama,55816,3.1% 654,Chicopee,Massachusetts,55717,1.7% 655,Anderson,Indiana,55670,-6.6% 656,Margate,Florida,55456,2.7% 657,Weymouth Town,Massachusetts,55419, 658,Hempstead,New York,55361,4.0% 659,Corvallis,Oregon,55298,11.8% 660,Eastvale,California,55191, 661,Porterville,California,55174,20.1% 662,West Haven,Connecticut,55046,5.1% 663,Brentwood,California,55000,122.3% 664,Paramount,California,54980,-0.7% 665,Grand Forks,North Dakota,54932,11.5% 666,Georgetown,Texas,54898,91.9% 667,St. Peters,Missouri,54842,6.5% 668,Shoreline,Washington,54790,2.9% 669,Mount Prospect,Illinois,54771,-2.5% 670,Hanford,California,54686,30.3% 671,Normal,Illinois,54664,19.7% 672,Rosemead,California,54561,1.7% 673,Lehi,Utah,54382,176.3% 674,Pocatello,Idaho,54350,5.4% 675,Highland,California,54291,21.0% 676,Novato,California,54194,13.3% 677,Port Arthur,Texas,54135,-6.0% 678,Carson City,Nevada,54080,2.9% 679,San Marcos,Texas,54076,48.5% 680,Hendersonville,Tennessee,54068,31.7% 681,Elyria,Ohio,53956,-3.7% 682,Revere,Massachusetts,53756,13.4% 683,Pflugerville,Texas,53752,123.4% 684,Greenwood,Indiana,53665,46.0% 685,Bellevue,Nebraska,53663,20.5% 686,Wheaton,Illinois,53648,-3.4% 687,Smyrna,Georgia,53438,20.0% 688,Sarasota,Florida,53326,1.4% 689,Blue Springs,Missouri,53294,9.9% 690,Colton,California,53243,10.8% 691,Euless,Texas,53224,15.1% 692,Castle Rock,Colorado,53063,153.5% 693,Cathedral City,California,52977,23.2% 694,Kingsport,Tennessee,52962,16.7% 695,Lake Havasu City,Arizona,52844,24.6% 696,Pensacola,Florida,52703,-6.0% 697,Hoboken,New Jersey,52575,35.8% 698,Yucaipa,California,52536,26.8% 699,Watsonville,California,52477,12.7% 700,Richland,Washington,52413,34.6% 701,Delano,California,52403,31.8% 702,Hoffman Estates,Illinois,52398,5.4% 703,Florissant,Missouri,52363,-2.8% 704,Placentia,California,52206,11.8% 705,West New York,New Jersey,52122,13.3% 706,Dublin,California,52105,70.0% 707,Oak Park,Illinois,52066,-0.8% 708,Peabody,Massachusetts,52044,7.5% 709,Perth Amboy,New Jersey,51982,9.7% 710,Battle Creek,Michigan,51848,-2.8% 711,Bradenton,Florida,51763,3.4% 712,Gilroy,California,51701,23.9% 713,Milford,Connecticut,51644,1.8% 714,Albany,Oregon,51583,25.5% 715,Ankeny,Iowa,51567,86.9% 716,La Crosse,Wisconsin,51522,-0.8% 717,Burlington,North Carolina,51510,12.1% 718,DeSoto,Texas,51483,36.0% 719,Harrisonburg,Virginia,51395,27.1% 720,Minnetonka,Minnesota,51368,0.4% 721,Elkhart,Indiana,51265,-2.5% 722,Lakewood,Ohio,51143,-9.4% 723,Glendora,California,51074,3.1% 724,Southaven,Mississippi,50997,72.8% 725,Charleston,West Virginia,50821,-4.7% 726,Joplin,Missouri,50789,11.2% 727,Enid,Oklahoma,50725,8.1% 728,Palm Beach Gardens,Florida,50699,39.6% 729,Brookhaven,Georgia,50603, 730,Plainfield,New Jersey,50588,5.7% 731,Grand Island,Nebraska,50550,16.0% 732,Palm Desert,California,50508,13.2% 733,Huntersville,North Carolina,50458,92.9% 734,Tigard,Oregon,50444,17.8% 735,Lenexa,Kansas,50344,24.6% 736,Saginaw,Michigan,50303,-18.2% 737,Kentwood,Michigan,50233,10.5% 738,Doral,Florida,50213,137.6% 739,Apple Valley,Minnesota,50201,9.2% 740,Grapevine,Texas,50195,17.6% 741,Aliso Viejo,California,50175,25.4% 742,Sammamish,Washington,50169,44.1% 743,Casa Grande,Arizona,50111,86.0% 744,Pinellas Park,Florida,49998,5.9% 745,Troy,New York,49974,1.5% 746,West Sacramento,California,49891,55.6% 747,Burien,Washington,49858,56.7% 748,Commerce City,Colorado,49799,135.4% 749,Monroe,Louisiana,49761,-6.1% 750,Cerritos,California,49707,-3.6% 751,Downers Grove,Illinois,49670,0.0% 752,Coral Gables,Florida,49631,16.1% 753,Wilson,North Carolina,49628,10.1% 754,Niagara Falls,New York,49468,-10.8% 755,Poway,California,49417,2.4% 756,Edina,Minnesota,49376,4.1% 757,Cuyahoga Falls,Ohio,49267,-0.2% 758,Rancho Santa Margarita,California,49228,4.6% 759,Harrisburg,Pennsylvania,49188,0.6% 760,Huntington,West Virginia,49177,-5.0% 761,La Mirada,California,49133,4.6% 762,Cypress,California,49087,5.3% 763,Caldwell,Idaho,48957,77.1% 764,Logan,Utah,48913,14.5% 765,Galveston,Texas,48733,-15.2% 766,Sheboygan,Wisconsin,48725,-3.9% 767,Middletown,Ohio,48630,-5.7% 768,Murray,Utah,48612,6.6% 769,Roswell,New Mexico,48611,7.5% 770,Parker,Colorado,48608,96.4% 771,Bedford,Texas,48592,2.9% 772,East Lansing,Michigan,48554,4.2% 773,Methuen,Massachusetts,48514,10.3% 774,Covina,California,48508,3.3% 775,Alexandria,Louisiana,48426,4.1% 776,Olympia,Washington,48338,12.1% 777,Euclid,Ohio,48139,-8.4% 778,Mishawaka,Indiana,47989,2.0% 779,Salina,Kansas,47846,4.5% 780,Azusa,California,47842,6.7% 781,Newark,Ohio,47777,3.1% 782,Chesterfield,Missouri,47749,1.9% 783,Leesburg,Virginia,47673,66.0% 784,Dunwoody,Georgia,47591, 785,Hattiesburg,Mississippi,47556,3.1% 786,Roseville,Michigan,47555,-1.0% 787,Bonita Springs,Florida,47547,43.8% 788,Portage,Michigan,47523,5.7% 789,St. Louis Park,Minnesota,47411,7.3% 790,Collierville,Tennessee,47333,43.4% 791,Middletown,Connecticut,47333,3.6% 792,Stillwater,Oklahoma,47186,20.1% 793,East Providence,Rhode Island,47149,-3.3% 794,Lawrence,Indiana,47135,20.5% 795,Wauwatosa,Wisconsin,47134,0.0% 796,Mentor,Ohio,46979,-6.6% 797,Ceres,California,46714,34.0% 798,Cedar Hill,Texas,46663,42.4% 799,Mansfield,Ohio,46454,-10.1% 800,Binghamton,New York,46444,-1.7% 801,Coeur d'Alene,Idaho,46402,32.8% 802,San Luis Obispo,California,46377,4.4% 803,Minot,North Dakota,46321,26.6% 804,Palm Springs,California,46281,7.7% 805,Pine Bluff,Arkansas,46094,-16.2% 806,Texas City,Texas,46081,10.3% 807,Summerville,South Carolina,46074,62.9% 808,Twin Falls,Idaho,45981,31.5% 809,Jeffersonville,Indiana,45929,53.3% 810,San Jacinto,California,45851,91.8% 811,Madison,Alabama,45799,53.7% 812,Altoona,Pennsylvania,45796,-7.3% 813,Columbus,Indiana,45775,16.4% 814,Beavercreek,Ohio,45712,19.0% 815,Apopka,Florida,45587,63.9% 816,Elmhurst,Illinois,45556,5.7% 817,Maricopa,Arizona,45508,2503.4% 818,Farmington,New Mexico,45426,18.1% 819,Glenview,Illinois,45417,5.2% 820,Cleveland Heights,Ohio,45394,-10.3% 821,Draper,Utah,45285,77.4% 822,Lincoln,California,45237,285.2% 823,Sierra Vista,Arizona,45129,19.3% 824,Lacey,Washington,44919,41.7% 825,Biloxi,Mississippi,44820,-11.5% 826,Strongsville,Ohio,44730,1.9% 827,Barnstable Town,Massachusetts,44641,-7.1% 828,Wylie,Texas,44575,185.2% 829,Sayreville,New Jersey,44412,9.6% 830,Kannapolis,North Carolina,44359,18.6% 831,Charlottesville,Virginia,44349,10.5% 832,Littleton,Colorado,44275,9.4% 833,Titusville,Florida,44206,7.8% 834,Hackensack,New Jersey,44113,2.9% 835,Newark,California,44096,3.3% 836,Pittsfield,Massachusetts,44057,-3.6% 837,York,Pennsylvania,43935,6.4% 838,Lombard,Illinois,43907,2.9% 839,Attleboro,Massachusetts,43886,4.6% 840,DeKalb,Illinois,43849,11.8% 841,Blacksburg,Virginia,43609,9.4% 842,Dublin,Ohio,43607,37.6% 843,Haltom City,Texas,43580,11.4% 844,Lompoc,California,43509,5.5% 845,El Centro,California,43363,13.7% 846,Danville,California,43341,3.7% 847,Jefferson City,Missouri,43330,6.7% 848,Cutler Bay,Florida,43328,42.9% 849,Oakland Park,Florida,43286,2.7% 850,North Miami Beach,Florida,43250,3.6% 851,Freeport,New York,43167,-1.4% 852,Moline,Illinois,43116,-1.9% 853,Coachella,California,43092,88.4% 854,Fort Pierce,Florida,43074,6.9% 855,Smyrna,Tennessee,43060,54.9% 856,Bountiful,Utah,43023,3.9% 857,Fond du Lac,Wisconsin,42970,1.7% 858,Everett,Massachusetts,42935,12.1% 859,Danville,Virginia,42907,-11.0% 860,Keller,Texas,42907,53.3% 861,Belleville,Illinois,42895,1.2% 862,Bell Gardens,California,42889,-2.7% 863,Cleveland,Tennessee,42774,14.1% 864,North Lauderdale,Florida,42757,10.8% 865,Fairfield,Ohio,42635,1.2% 866,Salem,Massachusetts,42544,5.1% 867,Rancho Palos Verdes,California,42448,2.9% 868,San Bruno,California,42443,5.6% 869,Concord,New Hampshire,42419,4.1% 870,Burlington,Vermont,42284,6.1% 871,Apex,North Carolina,42214,98.8% 872,Midland,Michigan,42181,0.9% 873,Altamonte Springs,Florida,42150,2.0% 874,Hutchinson,Kansas,41889,0.1% 875,Buffalo Grove,Illinois,41778,-3.4% 876,Urbandale,Iowa,41776,41.5% 877,State College,Pennsylvania,41757,8.7% 878,Urbana,Illinois,41752,10.3% 879,Plainfield,Illinois,41734,203.6% 880,Manassas,Virginia,41705,19.5% 881,Bartlett,Illinois,41679,13.1% 882,Kearny,New Jersey,41664,2.8% 883,Oro Valley,Arizona,41627,27.0% 884,Findlay,Ohio,41512,5.8% 885,Rohnert Park,California,41398,0.0% 887,Westfield,Massachusetts,41301,3.0% 886,Linden,New Jersey,41301,4.7% 888,Sumter,South Carolina,41190,1.3% 889,Wilkes-Barre,Pennsylvania,41108,-4.3% 890,Woonsocket,Rhode Island,41026,-5.2% 891,Leominster,Massachusetts,41002,-1.1% 892,Shelton,Connecticut,40999,7.3% 893,Brea,California,40963,15.2% 894,Covington,Kentucky,40956,-4.7% 895,Rockwall,Texas,40922,117.2% 896,Meridian,Mississippi,40921,-0.9% 897,Riverton,Utah,40921,61.6% 898,St. Cloud,Florida,40918,86.2% 899,Quincy,Illinois,40915,0.5% 900,Morgan Hill,California,40836,19.5% 901,Warren,Ohio,40768,-15.2% 902,Edmonds,Washington,40727,2.9% 903,Burleson,Texas,40714,85.3% 904,Beverly,Massachusetts,40664,2.0% 905,Mankato,Minnesota,40641,24.7% 906,Hagerstown,Maryland,40612,10.4% 907,Prescott,Arizona,40590,18.1% 908,Campbell,California,40584,4.2% 909,Cedar Falls,Iowa,40566,12.0% 910,Beaumont,California,40481,254.5% 911,La Puente,California,40435,-1.6% 912,Crystal Lake,Illinois,40388,5.3% 913,Fitchburg,Massachusetts,40383,3.5% 914,Carol Stream,Illinois,40379,-0.2% 915,Hickory,North Carolina,40361,7.0% 916,Streamwood,Illinois,40351,10.1% 917,Norwich,Connecticut,40347,11.6% 918,Coppell,Texas,40342,10.3% 919,San Gabriel,California,40275,0.9% 920,Holyoke,Massachusetts,40249,0.9% 921,Bentonville,Arkansas,40167,97.7% 922,Florence,Alabama,40059,10.2% 923,Peachtree Corners,Georgia,40059, 924,Brentwood,Tennessee,40021,51.9% 925,Bozeman,Montana,39860,41.9% 926,New Berlin,Wisconsin,39834,3.6% 927,Goose Creek,South Carolina,39823,26.1% 928,Huntsville,Texas,39795,13.2% 929,Prescott Valley,Arizona,39791,62.9% 930,Maplewood,Minnesota,39765,12.3% 931,Romeoville,Illinois,39650,79.5% 932,Duncanville,Texas,39605,9.7% 933,Atlantic City,New Jersey,39551,-2.2% 934,Clovis,New Mexico,39508,21.3% 935,The Colony,Texas,39458,45.7% 936,Culver City,California,39428,1.3% 937,Marlborough,Massachusetts,39414,7.6% 938,Hilton Head Island,South Carolina,39412,16.0% 939,Moorhead,Minnesota,39398,21.3% 940,Calexico,California,39389,44.0% 941,Bullhead City,Arizona,39383,15.9% 942,Germantown,Tennessee,39375,4.1% 943,La Quinta,California,39331,59.9% 944,Lancaster,Ohio,39325,10.7% 945,Wausau,Wisconsin,39309,1.7% 946,Sherman,Texas,39296,11.6% 947,Ocoee,Florida,39172,57.9% 948,Shakopee,Minnesota,39167,85.7% 949,Woburn,Massachusetts,39083,4.4% 950,Bremerton,Washington,39056,4.9% 951,Rock Island,Illinois,38877,-1.9% 952,Muskogee,Oklahoma,38863,-0.7% 953,Cape Girardeau,Missouri,38816,9.4% 954,Annapolis,Maryland,38722,7.6% 955,Greenacres,Florida,38696,35.5% 956,Ormond Beach,Florida,38661,5.8% 957,Hallandale Beach,Florida,38632,12.4% 958,Stanton,California,38623,2.8% 959,Puyallup,Washington,38609,11.8% 960,Pacifica,California,38606,0.5% 961,Hanover Park,Illinois,38510,0.6% 962,Hurst,Texas,38448,5.8% 963,Lima,Ohio,38355,-8.1% 964,Marana,Arizona,38290,166.2% 965,Carpentersville,Illinois,38241,22.8% 966,Oakley,California,38194,47.7% 967,Huber Heights,Ohio,38142,-0.2% 968,Lancaster,Texas,38071,46.4% 969,Montclair,California,38027,12.1% 970,Wheeling,Illinois,38015,4.8% 971,Brookfield,Wisconsin,37999,-1.9% 972,Park Ridge,Illinois,37839,0.1% 973,Florence,South Carolina,37792,19.8% 974,Roy,Utah,37733,13.3% 975,Winter Garden,Florida,37711,142.5% 976,Chelsea,Massachusetts,37670,7.3% 977,Valley Stream,New York,37659,3.6% 978,Spartanburg,South Carolina,37647,-6.2% 979,Lake Oswego,Oregon,37610,5.3% 980,Friendswood,Texas,37587,28.6% 981,Westerville,Ohio,37530,5.7% 982,Northglenn,Colorado,37499,15.5% 983,Phenix City,Alabama,37498,31.9% 984,Grove City,Ohio,37490,35.6% 985,Texarkana,Texas,37442,7.4% 986,Addison,Illinois,37385,2.6% 987,Dover,Delaware,37366,16.0% 988,Lincoln Park,Michigan,37313,-6.7% 989,Calumet City,Illinois,37240,-4.5% 990,Muskegon,Michigan,37213,-7.1% 991,Aventura,Florida,37199,47.2% 992,Martinez,California,37165,3.4% 993,Greenfield,Wisconsin,37159,4.8% 994,Apache Junction,Arizona,37130,15.7% 995,Monrovia,California,37101,0.2% 996,Weslaco,Texas,37093,28.8% 997,Keizer,Oregon,37064,14.4% 998,Spanish Fork,Utah,36956,78.1% 999,Beloit,Wisconsin,36888,2.9% 1000,Panama City,Florida,36877,0.1% Source: !/usr/bin/env python from mechanize import Browser from BeautifulSoup import BeautifulSoup mech = Browser() url = "cities.html" # originally from: page = mech.open(url) html = page.read() soup = BeautifulSoup(html) table = soup.find("table", 'table-condensed') for row in table.findAll('tr')[0:]: col = row.findAll('td') rank = col.string city = col.find('a').text.split('\n') state = col.find('a').text population = str(col.text.replace(',', '')) growth = col.text.replace(',', '') row = (rank, city, state, population, growth) print ','.join(row) Copy link adifficultgameaboutclimbingunblocked commented Aug 23, 2023 Thanks for share this helpful information. Besides research about US Cities, I love to play uno online with my friends. Sorry, something went wrong. Uh oh! There was an error while loading. Please reload this page. Copy link alexwrinner commented Sep 18, 2023 Exellent file 👍👍🐱‍🏍 Sorry, something went wrong. Uh oh! There was an error while loading. Please reload this page. Copy link jrfent commented Oct 8, 2023 Thank you for sharing. My intention is to use it to find cities big enough to support my hood cleaning companies like New Orleans Hood Cleaning. I find that the top 50 cities are too competitive. So I want to concentrate on the next 50 on the list. Sorry, something went wrong. Uh oh! There was an error while loading. Please reload this page. Copy link miky1460 commented Oct 10, 2023 Thank you for sharing the information Sorry, something went wrong. Uh oh! There was an error while loading. Please reload this page. Copy link elegancedesign commented Nov 29, 2023 Did you know that New York previously was owned by The Netherlands? Sorry, something went wrong. Uh oh! There was an error while loading. Please reload this page. Copy link johnnymross commented Feb 8, 2024 Thanks, I have an activewear shop in the USA. Sorry, something went wrong. Uh oh! There was an error while loading. Please reload this page. Copy link TTUTTU commented Feb 16, 2024• edited Loading Uh oh! There was an error while loading. Please reload this page. The transition from one administration TU娛樂城冠利娛樂城 TUVN Sorry, something went wrong. Uh oh! There was an error while loading. Please reload this page. Copy link EstellaDavis commented Mar 1, 2024 I have Clothing stores in US Cities. Sorry, something went wrong. Uh oh! There was an error while loading. Please reload this page. Copy link mggger commented Mar 10, 2024 Good file 👍 Sorry, something went wrong. Uh oh! There was an error while loading. Please reload this page. Copy link CrazyAboutBetting commented Apr 16, 2024 Thank you for these infos. Read this article if you want to know more about the Betano in Argentina and this if you want to know more about the Meilleures Sites de Paris Sportifs Suisse Thank you Crazy About Betting Sorry, something went wrong. Uh oh! There was an error while loading. Please reload this page. Copy link KingGame0 commented Apr 27, 2024 Kinggame -- Gold99PGasia Sorry, something went wrong. Uh oh! There was an error while loading. Please reload this page. Copy link KingGame0 commented May 14, 2024 KingGame -- Gold99PGasia Sorry, something went wrong. Uh oh! There was an error while loading. Please reload this page. Copy link turd1204 commented Jun 21, 2024 2024歐洲杯美國天天樂 at99娛樂城戰神賽特訊號 2024奧運 Sorry, something went wrong. Uh oh! There was an error while loading. Please reload this page. Copy link ademlasmer commented Oct 24, 2024 Thanks you, got office 2021 pro plus and windows 11 pro Sorry, something went wrong. Uh oh! There was an error while loading. Please reload this page. Copy link kimjimin198765-coder commented Aug 15, 2025 Wow, analyzing code like this can get intense After long coding sessions, it’s important to relax your body too! Services like Hony Massage ( offer on-site massages so you can unwind comfortably at home or in the office Coding break = self-care! Sorry, something went wrong. Uh oh! There was an error while loading. Please reload this page. Copy link williamzdebra-sudo commented Aug 15, 2025 Thanks for sharing this useful information! Along with exploring topics like US cities, I’m always interested in how businesses can protect their digital presence. That’s where Cyberscout comes in — offering smart solutions to safeguard data, manage risk, and strengthen trust with customers. In today’s connected world, security is as important as strategy. Sorry, something went wrong. Uh oh! There was an error while loading. Please reload this page. Copy link nournour2690 commented Aug 31, 2025 I decided to purchase my Windows 11 Pro key from after seeing their perfect review score. The process couldn’t have been smoother — I received the key almost instantly after completing my order. Activation was quick and hassle-free; I simply entered the key, followed the steps, and within minutes, my Windows 11 Pro was fully activated with no issues at all. It was a huge relief not having to worry about activation problems or expiry concerns. Sorry, something went wrong. Uh oh! There was an error while loading. Please reload this page. Copy link chaesoobin9987-droid commented Sep 7, 2025 New York, LA, Chicago… with such big populations, life must be hectic. A massage( at home is the perfect way to relax! Sorry, something went wrong. Uh oh! There was an error while loading. Please reload this page. Sign up for freeto join this conversation on GitHub. Already have an account? Sign in to comment Footer © 2025 GitHub,Inc. 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10303	https://www.inchcalculator.com/convert/from-kilonewton/	Inch Calculator Skip to Content Popular Searches calculate body fat what is today's date? calculate yards of concrete how long until 3:30? calculate a pay raise 30 minute timer calculate board and batten wall layout how to do long division calculate the best TV size how many days until Christmas? Kilonewton Conversion Calculator Enter a value in kilonewtons below to convert to another unit of force. Have a Question or Feedback? Results: Learn how we calculate this below Kilonewton Conversion Calculators You can also convert force using one of our kilonewton converters below. SI Units kilonewtons to nanonewtons converter kilonewtons to micronewtons converter kilonewtons to millinewtons converter kilonewtons to newtons converter kilonewtons to meganewtons converter Gravitational Units kilonewtons to gram-force converter kilonewtons to kilogram-force converter kilonewtons to pound-force converter kilonewtons to ounce-force converter CGS/FPS Units kilonewtons to dynes converter kilonewtons to poundals converter What Is a Kilonewton? By Joe Sexton Joe is the creator of Inch Calculator and has over 20 years of experience in engineering and construction. He holds several degrees and certifications. Full bio Share on Facebook Share on X Share on Pinterest Cite This Page Sexton, J. (n.d.). Kilonewton Conversion Calculator. Inch Calculator. Retrieved September 22, 2025, from A kilonewton is a unit used to measure force. The kilonewton is a multiple of the newton, which is the SI derived unit for force. In the metric system, "kilo" is the prefix for thousands, or 103. Kilonewtons can be abbreviated as kN; for example, 1 kilonewton can be written as 1 kN. How Much Is a Kilonewton? One kilonewton is equal to 1,000 newtons, which are equal to the force needed to move one kilogram of mass at a rate of one meter per second squared. How to Convert Kilonewtons To convert kilonewtons to another unit of force, you need to multiply the value by a conversion factor. A conversion factor is a numerical value used to change the units of a measurement without changing the value. You can find the conversion factors for kilonewtons in the conversion table below. Then, multiply the force measurement by the conversion factor to find the equivalent value in the desired unit of measurement. kilonewtons × conversion factor = result You can also use a calculator, such as one of the converters below, for the conversion. Kilonewton Conversion Table Common kilonewton values and equivalent imperial and metric force measurements \| kilonewtons \| nanonewtons \| micronewtons \| millinewtons \| newtons \| meganewtons \| gram-force \| kilogram-force \| pound-force \| ounce-force \| dynes \| poundals \| \| 1 kN \| 1,000,000,000,000 nN \| 1,000,000,000 μN \| 1,000,000 mN \| 1,000 N \| 0.001 MN \| 101,972 gf \| 101.97162 kgf \| 224.80894 lbf \| 3,597 ozf \| 100,000,000 dyn \| 7,233 pdl \| \| 2 kN \| 2,000,000,000,000 nN \| 2,000,000,000 μN \| 2,000,000 mN \| 2,000 N \| 0.002 MN \| 203,943 gf \| 203.94324 kgf \| 449.61788 lbf \| 7,194 ozf \| 200,000,000 dyn \| 14,466 pdl \| \| 3 kN \| 3,000,000,000,000 nN \| 3,000,000,000 μN \| 3,000,000 mN \| 3,000 N \| 0.003 MN \| 305,915 gf \| 305.91486 kgf \| 674.426819 lbf \| 10,791 ozf \| 300,000,000 dyn \| 21,699 pdl \| \| 4 kN \| 4,000,000,000,000 nN \| 4,000,000,000 μN \| 4,000,000 mN \| 4,000 N \| 0.004 MN \| 407,886 gf \| 407.88648 kgf \| 899.235759 lbf \| 14,388 ozf \| 400,000,000 dyn \| 28,932 pdl \| \| 5 kN \| 5,000,000,000,000 nN \| 5,000,000,000 μN \| 5,000,000 mN \| 5,000 N \| 0.005 MN \| 509,858 gf \| 509.8581 kgf \| 1,124 lbf \| 17,985 ozf \| 500,000,000 dyn \| 36,165 pdl \| \| 6 kN \| 6,000,000,000,000 nN \| 6,000,000,000 μN \| 6,000,000 mN \| 6,000 N \| 0.006 MN \| 611,830 gf \| 611.82972 kgf \| 1,349 lbf \| 21,582 ozf \| 600,000,000 dyn \| 43,398 pdl \| \| 7 kN \| 7,000,000,000,000 nN \| 7,000,000,000 μN \| 7,000,000 mN \| 7,000 N \| 0.007 MN \| 713,801 gf \| 713.80134 kgf \| 1,574 lbf \| 25,179 ozf \| 700,000,000 dyn \| 50,631 pdl \| \| 8 kN \| 8,000,000,000,000 nN \| 8,000,000,000 μN \| 8,000,000 mN \| 8,000 N \| 0.008 MN \| 815,773 gf \| 815.77296 kgf \| 1,798 lbf \| 28,776 ozf \| 800,000,000 dyn \| 57,864 pdl \| \| 9 kN \| 9,000,000,000,000 nN \| 9,000,000,000 μN \| 9,000,000 mN \| 9,000 N \| 0.009 MN \| 917,745 gf \| 917.74458 kgf \| 2,023 lbf \| 32,372 ozf \| 900,000,000 dyn \| 65,097 pdl \| \| 10 kN \| 10,000,000,000,000 nN \| 10,000,000,000 μN \| 10,000,000 mN \| 10,000 N \| 0.01 MN \| 1,019,716 gf \| 1,020 kgf \| 2,248 lbf \| 35,969 ozf \| 1,000,000,000 dyn \| 72,330 pdl \| \| 11 kN \| 11,000,000,000,000 nN \| 11,000,000,000 μN \| 11,000,000 mN \| 11,000 N \| 0.011 MN \| 1,121,688 gf \| 1,122 kgf \| 2,473 lbf \| 39,566 ozf \| 1,100,000,000 dyn \| 79,563 pdl \| \| 12 kN \| 12,000,000,000,000 nN \| 12,000,000,000 μN \| 12,000,000 mN \| 12,000 N \| 0.012 MN \| 1,223,659 gf \| 1,224 kgf \| 2,698 lbf \| 43,163 ozf \| 1,200,000,000 dyn \| 86,796 pdl \| \| 13 kN \| 13,000,000,000,000 nN \| 13,000,000,000 μN \| 13,000,000 mN \| 13,000 N \| 0.013 MN \| 1,325,631 gf \| 1,326 kgf \| 2,923 lbf \| 46,760 ozf \| 1,300,000,000 dyn \| 94,029 pdl \| \| 14 kN \| 14,000,000,000,000 nN \| 14,000,000,000 μN \| 14,000,000 mN \| 14,000 N \| 0.014 MN \| 1,427,603 gf \| 1,428 kgf \| 3,147 lbf \| 50,357 ozf \| 1,400,000,000 dyn \| 101,262 pdl \| \| 15 kN \| 15,000,000,000,000 nN \| 15,000,000,000 μN \| 15,000,000 mN \| 15,000 N \| 0.015 MN \| 1,529,574 gf \| 1,530 kgf \| 3,372 lbf \| 53,954 ozf \| 1,500,000,000 dyn \| 108,495 pdl \| \| 16 kN \| 16,000,000,000,000 nN \| 16,000,000,000 μN \| 16,000,000 mN \| 16,000 N \| 0.016 MN \| 1,631,546 gf \| 1,632 kgf \| 3,597 lbf \| 57,551 ozf \| 1,600,000,000 dyn \| 115,728 pdl \| \| 17 kN \| 17,000,000,000,000 nN \| 17,000,000,000 μN \| 17,000,000 mN \| 17,000 N \| 0.017 MN \| 1,733,518 gf \| 1,734 kgf \| 3,822 lbf \| 61,148 ozf \| 1,700,000,000 dyn \| 122,961 pdl \| \| 18 kN \| 18,000,000,000,000 nN \| 18,000,000,000 μN \| 18,000,000 mN \| 18,000 N \| 0.018 MN \| 1,835,489 gf \| 1,835 kgf \| 4,047 lbf \| 64,745 ozf \| 1,800,000,000 dyn \| 130,194 pdl \| \| 19 kN \| 19,000,000,000,000 nN \| 19,000,000,000 μN \| 19,000,000 mN \| 19,000 N \| 0.019 MN \| 1,937,461 gf \| 1,937 kgf \| 4,271 lbf \| 68,342 ozf \| 1,900,000,000 dyn \| 137,427 pdl \| \| 20 kN \| 20,000,000,000,000 nN \| 20,000,000,000 μN \| 20,000,000 mN \| 20,000 N \| 0.02 MN \| 2,039,432 gf \| 2,039 kgf \| 4,496 lbf \| 71,939 ozf \| 2,000,000,000 dyn \| 144,660 pdl \| You might also find our other electrical calculators useful. References National Institute of Standards & Technology, Unit Conversion, Other Force Units SI Units nanonewton conversions micronewton conversions millinewton conversions newton conversions meganewton conversions Gravitational Units gram-force conversions kilogram-force conversions pound-force conversions ounce-force conversions CGS/FPS Units dyne conversions poundal conversions
10304	https://artofproblemsolving.com/wiki/index.php/Cauchy-Schwarz_Inequality?srsltid=AfmBOorHI5FXiWC6zw9PLwqyG8HhiT4ALdec4AxM-46JHxBcYb6Iodcd	Art of Problem Solving Cauchy-Schwarz Inequality - 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 Cauchy-Schwarz Inequality Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Cauchy-Schwarz Inequality In algebra, the Cauchy-Schwarz Inequality, also known as the Cauchy–Bunyakovsky–Schwarz Inequality or informally as Cauchy-Schwarz, is an inequality with many ubiquitous formulations in abstract algebra, calculus, and contest mathematics. In high-school competitions, its applications are limited to elementary and linear algebra. Its elementary algebraic formulation is often referred to as Cauchy's Inequality and states that for any list of reals and , with equality if and only if there exists a constant such that for all , or if one list consists of only zeroes. Along with the AM-GM Inequality, Cauchy-Schwarz forms the foundation for inequality problems in intermediate and olympiad competitions. It is particularly crucial in proof-based contests. Its vector formulation states that for any vectors and in , where is the dot product of and and is the norm of , with equality if and only if there exists a scalar such that , or if one of the vectors is zero. This formulation comes in handy in linear algebra problems at intermediate and olympiad problems. The full Cauchy-Schwarz Inequality is written in terms of abstract vector spaces. Under this formulation, the elementary algebraic, linear algebraic, and calculus formulations are different cases of the general inequality. Contents [hide] 1 Proofs 2 Lemmas 2.1 Complex Form 2.2 A Useful Inequality 3 Real Vector Spaces 3.1 Proof 1 3.2 Proof 2 3.3 Proof 3 4 Complex Vector Spaces 4.1 Proof 5 Problems 5.1 Introductory 5.2 Intermediate 5.3 Olympiad 6 Other Resources 6.1 Books Proofs Here is a list of proofs of Cauchy-Schwarz. Consider the vectors and . If is the angle formed by and , then the left-hand side of the inequality is equal to the square of the dot product of and , or .The right hand side of the inequality is equal to . The inequality then follows from , with equality when one of is a multiple of the other, as desired. Lemmas Complex Form The inequality sometimes appears in the following form. Let and be complex numbers. Then This appears to be more powerful, but it follows from A Useful Inequality Also known as Sedrakyan's Inequality, Bergström's Inequality, Engel's Form or Titu's Lemma the following inequality is a direct result of Cauchy-Schwarz inequality: For any real numbers and where the following is true: Real Vector Spaces Let be a vector space, and let be an inner product. Then for any , with equality if and only if there exist constants not both zero such that . The following proofs assume the inner product to be real-valued and commutative, and so only apply to vector spaces over the real numbers. Proof 1 Consider the polynomial of This must always be greater than or equal to zero, so it must have a non-positive discriminant, i.e., must be less than or equal to , with equality when or when there exists some scalar such that , as desired. Proof 2 We consider Since this is always greater than or equal to zero, we have Now, if either or is equal to , then . Otherwise, we may normalize so that , and we have with equality when and may be scaled to each other, as desired. Proof 3 Consider for some scalar . Then: (by the Trivial Inequality) . Now, let . Then, we have: . Complex Vector Spaces For any two vectors in the complex vector space , the following holds: with equality holding only when are linearly dependent. Proof The following proof, a geometric argument that uses only the algebraic properties of the inner product, was discovered by Tarung Bhimnathwala in 2021. Define the unit vectors , as and . Put . In other words, is the complex argument of and lies on the unit circle. If any of the denominators are zero, the entire result follows trivially. Let and . Importantly, we have Since and , this calculation shows that and form an orthogonal basis of the linear subspace spanned by and . Thus we can think of and as lying on the unit sphere in this subspace, which is isomorphic to . Another thing to note is that The previous two calculations established that and are orthogonal, and that the sum of their squared norms is . Now we have Equality holds when either or , or equivalently when and . Lastly, multiplying each side by , we have Problems Introductory Consider the function , where is a positive integer. Show that . (Source) (APMO 1991 #3) Let , , , , , , , be positive real numbers such that . Show that Intermediate Let be a triangle such that where and denote its semiperimeter and inradius, respectively. Prove that triangle is similar to a triangle whose side lengths are all positive integers with no common divisor and determine those integers. (Source) Olympiad is a point inside a given triangle . are the feet of the perpendiculars from to the lines , respectively. Find all for which is least. (Source) Other Resources Wikipedia entry Books The Cauchy-Schwarz Master Class: An Introduction to the Art of Mathematical Inequalities by J. Michael Steele. Problem Solving Strategies by Arthur Engel contains significant material on inequalities. Retrieved from " Categories: Algebra Inequalities 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.
10305	https://visitmath.eu/?sdm_process_download=1&download_id=2383	1 Thal es’ theorem Topic Geometry Learning Objec0ves Using Thales' theorem in different ways Age s 12-16 years old (to be adapted by each country) Dura0on es0mate 45 min s Ac0vi0es Discover Thales and applicaDons of his theorem . Associated visits Agen, Montauban Pr evious knowledge require ments • Know the basic concepts of geometry, such as points, lines, segments, rays, angles, etc. • Know the properDes of triangles . • Understand the concepts of raDos and proporDons, as well as the ability to solve simple proporDonal equaDons. Step by step : th e classroom learning sequence Step 1: Topic introduc0on A li%le backstory Thales of Miletus was born around 620 BC in Miletus, Greece. He is considered the first pre -SocraAc philosopher, the first of the seven sages of anAquity. He was a mathemaAcian, physicist, astronomer, engineer, and meteorologist. He is the founder of the Ionian School of Natural Philosophy in Miletus. Aristotle and other ancient philosophers considered Thales to be the first Greek philosopher . Thales refused to go along with previous interpretaAons of natural phenomena, which unAl then had only been based on myths, legends, and religious beliefs. Therefore, he managed to approach and explain natural phenomena through scienAfic logic . He Thales of Miletus is therefore considered the first to have opened the way to scienAfic research. 2 Reminders We will use the following notaAon convenAons: AB designaAng a length between a point A and a point B (AB) designaAng the line of unlimited dimension passing through point A and point B [AB] designaAng the segment, parAAon of the line (AB), whose ends are point A on the one hand and point B on the other . Thales’ theorem: a defini9on Thal es of Miletus was widely known for his theorems and rules among the geometry field. One of them is presented below. If we consider three straight and parallel lines , l1, l2, and l3, intersecDng two other lines, namely R1 and R2, then they produce proporDonal segments. Either if l1 / / l2 / / l3 and cuTng segments through R1 and R2, then: 𝐴𝐵 𝐴 ′𝐵 ′ = 𝐵𝐶 𝐵 ′𝐶 ′ = 𝐴𝐶 𝐴 ′𝐶 ′3 Here are two figures for which Thales' theorem c ould be used: At least three length measurements must be known in this type of figure for us to use Thales’ theorem. Applying Thales' theorem , therefore, consists of wriDng equal raDos of lengths in this type of figure .4 Step 2: classroom ac0vi0es Thales’ theorem : e xample of applica9on Thales' theorem states that, in this type of configuraDon, the sides ’ lengths of one triangle are proporDonal to the associated sides of an other triangle. And indeed we have: 𝐴𝑀 𝐴𝐵 = 𝐴𝑁 𝐴𝐶 = 𝑀𝑁 𝐵𝐶 Remar ks Thales’ theorem cannot be applied if the figure does not include parallel lines . Here : ( MN ) / / ( BC ). Upon wriDng the equality of the three quoDents, we put: • one side of the first triangle as the numerator, • the associated side of the second triangle as the denominator .5 Then, 𝐷𝐸 𝐷𝐹 = 𝐷𝐻 𝐷𝐺 = 𝐸𝐻 𝐹𝐺 Ex ample : In the layout above, suppose we know the lengths of the following segments: DE =8cm DF =12cm DH =4cm We want to find out the length of DG . Using Thales’ theorem, we can write: 𝐷𝐸 𝐷𝐹 = 𝐷𝐻 𝐷𝐺 = 𝐸𝐻 𝐹𝐺 Let’s subsDtute now with known values : 8 12 = 4 𝐷𝐺 By solving this proporDonal equaDon, we can calculate the value of DG. We successfully used Thales' theorem to solve a geometric problem. 6 DG = !∗#$ % = 6 cm A soluDon is obtained. How to verify the parallelism of two lines In order to check whether two lines are parallel or not, we use the converse of Thales' theorem. Ex ample: In order to demonstrate that ( BC ) and ( DE ) are parallel, we calculate the two quoDents &' &( and ') ( separately . &' &( = + +,$ = - + ') ( = - = - + &' &( = ') ( So , according to the converse of Thales' theorem, the lines (DE) and (BC) are parallel. 7 St ep 3: homework ideas and idea development This theorem can be used to measure many things in your establishment or the pupil environment! Here is a first exercise that can help develop this idea. Exercise : We will use a method called the woodcuder’s cross, based on Thales' theorem, to esDmate the height of a tree. This method can also be used to esDmate any inaccessible height. This method requires a T formed by two sDcks of the same length. We consider a tree of height AB at a distance BC from the observer . Take two sDcks of the same size (for example , 20cm) and straight (ab=cd) Place the first one horizontally (parallel to the ground) . Place the second sDck perpendicular to the first one . Then , stand facing the tree at a distance , approximately close to its height. 8 Then, move forward or backward and slide the verDcal sDck to align: • the foot of the tree, the bodom of the verDcal sDck and your eye on the same line (cB) • the top of the tree, the top of the verDcal sDck and your eye on the same line (cA) When the two ends of the tree correspond to the ends of the verDcal sDck , measure the distance separaDng you from the tree BC. ð The height of the tree AB is then equal to the distance BC Why so? The lines (ab) and (AB) are parallel due to the recommended posiDoning, so we can use Thales' theorem. In the ABc triangle , Thales ’ rule can be wriden as this : 𝑐𝑎 𝑐𝐴 = 𝑎𝑏 𝐴𝐵 And we can write Thales ‘s theorem as well for the cHA triangle : 𝑐𝑑 𝑐𝐻 = 𝑐𝑎 𝑐𝐴 Hence : 𝑐𝑑 𝑐𝐻 = 𝑎𝑏 𝐴𝐵 𝑅𝑒𝑚𝑒𝑚𝑏𝑒𝑟 , 𝑎𝑠 𝑡 ℎ𝑒 𝑡𝑤𝑜 𝑠𝑡𝑖𝑐𝑘𝑠 𝑎𝑟𝑒 𝑡 ℎ𝑒 𝑠𝑎𝑚𝑒 𝑙𝑒𝑛𝑔 ℎ𝑡 , ab = dc . And s ince [cH] is perpendicular to [AB], and the observer is standing verDcally, then : 9 cH = BC. That means : 𝑎𝑏 𝐴𝐵 = 𝑎𝑏 𝐵𝐶 ð AB = BC Hence, using Thales' theorem and the so -called "lumber's cross" method, the tree's height is equal to the distance of said tree from the observer. Now, thanks to Thal es’ theorem , have fun measuring the height of your school, your city's bridge, a statue... These videos can complete the sequence: hdps:/ /www.youtube.com/watch?v=EOBMCvDMo4M hdps:/ /www.youtube.com/watch?v=u_bpbsFZqAA 10 This project has been funded with support from the European Commission. This publica9on reflects the views only of the author, and the Commission cannot be held responsible for any use which may be made of the informa9on contained therein. Project code: 1-FR01 -KA220 -SCH -00027771 Learn more about Visit Math at: hIps:/ /visitmath.eu This work is licensed under the Crea9ve Commons AIribu9on -NonCommercial -ShareAlike 4.0 Interna9onal License ( hIp:/ /crea9vecommons.org/licenses/by -nc - sa/4.0/ ).
10306	https://www.quora.com/What-is-the-formula-for-evaluating-the-nth-term-of-the-sequence-0-1-3-7-15	What is the formula for evaluating the nth term of the sequence�0, 1, 3, 7, 15? - Quora Something went wrong. Wait a moment and try again. Try again Skip to content Skip to search Sign In Mathematics The Nth Term Patterns Integer Sequences Formulae Arithmetic Series and Sequences Number Pattern Sequence in Mathematics 5 What is the formula for evaluating the nth term of the sequence0, 1, 3, 7, 15? All related (39) Sort Recommended Enrico Gregorio Associate professor in Algebra · Author has 18.4K answers and 16M answer views ·Jul 13 1!=1 1!=1 has 0 0 proper divisors 2!=2 2!=2 has 1 1 proper divisors 3!=6 3!=6 has 3 3 proper divisors 4!=24 4!=24 has 7 7 proper divisors 5!=120 5!=120 has 15 15 proper divisors 6!=720 6!=720 has 29 29 proper divisors 7!=5040 7!=5040 has 59 59 proper divisors If we denote by D(m)D(m) the number of divisors of m,m, then your sequence is “obviously” D(n!)−1 D(n!)−1 If you want an “explicit formula”, ∏p≤n p prime(1+∑k≥0⌊n p k⌋)−1∏p≤n p prime(1+∑k≥0⌊n p k⌋)−1 There’s NO way to tell what a sequence is when a few terms are given. Upvote · 99 10 Sponsored by Best Gadget Advice Here Are The 33 Coolest Gifts For This Year. We've put together a list of incredible gifts that are selling out fast. Get these before they're gone! Learn More 999 160 Related questions More answers below What is the nth term formula for the sequence 0, 3, 7, 15? What is the nth term for the sequence 1, 3, 7, 15…? What is the next term of this sequence 1, 3, 7, 15, 31, 63, _? What is the nth term of 1,1/3,1/5,1/7? What is the formula to find the nth term of the sequence 3, 15, 35, and 63? Alexander Budianto An Indonesian who fell in love with mathematics · Author has 2.2K answers and 3.4M answer views ·3y 0,1,3,7,15,…0,1,3,7,15,… a 1=0 a 1=0 a 2=1=a 1+2 0 a 2=1=a 1+2 0 a 3=3=a 2+2 1 a 3=3=a 2+2 1 a 4=7=a 3+2 2 a 4=7=a 3+2 2 a 5=15=a 4+2 3 a 5=15=a 4+2 3 ⋮⋮ a n=a n−1+2 n−2 a n=a n−1+2 n−2 As you can see, we end up with a recurrence relation. If you’re content with this, you can stop reading this answer. If you want a closed-form solution to the relation — a non-recursive function that’s equivalent to it — keep reading. a n−a n−1=2 n−2 a n−a n−1=2 n−2 From here, we’ll separate it into two parts: the homogeneous solution and the particular solution. We’ll get the actual solution by adding them together. This is the homogeneous solution: a h n−a h n−1=0 a n h−a n−1 h=0 Let a h n=r n Let a n h=r n r n−r n−1=0 r n−r n−1=0 r 1−r 0=0 r 1−r 0=0 r−1 r−1 Continue Reading 0,1,3,7,15,…0,1,3,7,15,… a 1=0 a 1=0 a 2=1=a 1+2 0 a 2=1=a 1+2 0 a 3=3=a 2+2 1 a 3=3=a 2+2 1 a 4=7=a 3+2 2 a 4=7=a 3+2 2 a 5=15=a 4+2 3 a 5=15=a 4+2 3 ⋮⋮ a n=a n−1+2 n−2 a n=a n−1+2 n−2 As you can see, we end up with a recurrence relation. If you’re content with this, you can stop reading this answer. If you want a closed-form solution to the relation — a non-recursive function that’s equivalent to it — keep reading. a n−a n−1=2 n−2 a n−a n−1=2 n−2 From here, we’ll separate it into two parts: the homogeneous solution and the particular solution. We’ll get the actual solution by adding them together. This is the homogeneous solution: a h n−a h n−1=0 a n h−a n−1 h=0 Let a h n=r n Let a n h=r n r n−r n−1=0 r n−r n−1=0 r 1−r 0=0 r 1−r 0=0 r−1=0 r−1=0 r=1 r=1 a h n=1 n=1 a n h=1 n=1 This is the particular solution: a p n−a p n−1=2 n−2 a n p−a n−1 p=2 n−2 Let a p n=A 2 n Let a n p=A 2 n A 2 n−A 2 n−1=2 n−2 A 2 n−A 2 n−1=2 n−2 A 2 2−A 2 1=2 0 A 2 2−A 2 1=2 0 4 A−2 A=1 4 A−2 A=1 2 A=1 2 A=1 A=1 2=2−1 A=1 2=2−1 a p n=2−1 2 n=2 n−1 a n p=2−1 2 n=2 n−1 Now, for the actual solution: a n=c a h n+a p n a n=c a n h+a n p a n=c+2 n−1 a n=c+2 n−1 a 1=0⟹0=c+1⟹c=−1 a 1=0⟹0=c+1⟹c=−1 a n=2 n−1−1 a n=2 n−1−1 Actually, now that I look at it again, you could literally just observe that the numbers are 1 unit below powers of 2, meaning that you can easily get the formula without the mess I’ve made here, but whatever. Upvote · 9 2 Chris Spencer Lives in England · Author has 17.7K answers and 16.6M answer views ·4y Originally Answered: What is the nth term formula for the sequence 0, 3, 7, 15? · Are you sure that is correct - there appears to be no sequence because of the initial 0. That means your sequence must be by simple addition (as there is not other way to go from 0 to 3) but the run 3, 4, 8, …. has no pattern 0, 1, 3, 7, 15 makes more sense. - because then you are simply adding the powers of 2 (1, 2, 4, 8,) Upvote · 9 3 9 2 Lawson Lutchman Former Jack of All Trades (2014–2018) · Author has 536 answers and 283.7K answer views ·3y assuming the n term is t then n = 2^(t-1) x = 2^(t-1) -1 + 2^(t-1) in this case the 5th term n = 2^4 → 16 x = 15 + 16 x = 31 6th term n = 2^5 x = 31 + 32 → 63 9th term n= 2^8 = 256 x= 255 + 256 → 511 0,1,3,7,15,31,63…511 Upvote · Sponsored by MRPeasy Powerful yet simple MRP software for growing manufacturers. MRPeasy makes it easier for growing manufacturing businesses to keep on expanding successfully. Free Trial 99 65 Related questions More answers below What is the rule for the nth term of the sequence 1, 3, 7, and 15? How do I find the nth term for the sequence 1, 3, 6, 10, and 15? What is the nth term rule of 3, 5, 7, 9, 11, 13, and 15? What is the 7th term of the sequence -5, -3, 1, 7, and 15? What is the formula for the nth term of the series 2, 5, 3, 10, 4, 15? Sanjiv Soni Former Pvt Tutor · Author has 9.7K answers and 1.9M answer views ·Feb 13 Originally Answered: What is the formula of the nth term in this number 0,3,8,15? · N th term=n^2–1 First term=(1)^2–1=1–1=0 Second term=2^2–1=4–1=3 Third term =3^2–1=9–1=8 Fourth term=4^2–1=16–1=15 Fifth term=5^2–1=25–1=24 Upvote · 9 1 Sanjiv Soni Former Pvt Tutor · Author has 9.7K answers and 1.9M answer views ·Updated Jul 1 The formula for evaluating the n th term of the sequence is Tn=2^n-1.where n is a natural number (n=1,2,3,..) 1=2^1–1 3=2^2–1 7=2^3–1 15=2^4–1 Upvote · 9 2 9 3 Promoted by Bata India Dhruti Shah Visualiser \| Graphic Designer (2018–present) ·Sep 12 What are the best professional affordable and comfortable shoes for women? I usually look at three things when I’m buying work shoes: comfort, cushioning and arch support; how sturdy the sole is; and whether I can actually afford to get more than one pair if I want them in different colours. Ballerinas by Bata though, are what I wear the most. I didn’t know about them until recently, when a coworker recommended them to me, also spotted my favorite creator Siddhi Karwa styling them across Europe and I have been absolutely loving them. They’re professional enough for work wear but don’t feel heavy and keep me comfortable throughout the day, even when I’m commuting to the Continue Reading I usually look at three things when I’m buying work shoes: comfort, cushioning and arch support; how sturdy the sole is; and whether I can actually afford to get more than one pair if I want them in different colours. Ballerinas by Bata though, are what I wear the most. I didn’t know about them until recently, when a coworker recommended them to me, also spotted my favorite creator Siddhi Karwa styling them across Europe and I have been absolutely loving them. They’re professional enough for work wear but don’t feel heavy and keep me comfortable throughout the day, even when I’m commuting to the office. I got mine for around ₹999 from Bata, which felt like a steal compared to some other brands I looked at. They’ve held up really well, and I can easily pair them with trousers, skirts for my work outfits. If you’re on a budget but still want something that is comfortable and follows fashion trends, Ballerinas by Bata are the perfect choice. I picked up mine from a Bata store near me, you can grab yours too. Upvote · 1.1K 1.1K 99 86 99 13 Arnaldo Mandel Mildly successful at the theorem production business · Author has 249 answers and 165.3K answer views ·4y Originally Answered: What is the nth term formula for the sequence 0, 3, 7, 15? · Of course, there are infinitely many. My favorite is, using the Iverson bracket: a n=3[n=1]+7[n=2]+15[n=3]a n=3[n=1]+7[n=2]+15[n=3] Hard to be simpler than this. Upvote · 9 2 9 1 Micheal Nwaorah B.ENG from Federal University of Technology, Owerri (FUTO) (Graduated 2006) · Author has 73 answers and 45.6K answer views ·4y Related What is the rule for the nth term of the sequence 1, 3, 7, and 15? nth = 1 + 2^1 + 2^2 + … + 2^n . where 1 as in the given series 1 , 3 , 7 and 15 is the zero term and constant in all terms. Therefore, for nth = T7 (7 term) ,then T7 = 1 + 2^1 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 nth = nth = 1 + 2^1 + 2^2 + … + 2^n ( Take T1 = 3 and T0 = 1 constant in all terms)Notice that the last power of two in each term, equals the term itself. Continue Reading nth = 1 + 2^1 + 2^2 + … + 2^n . where 1 as in the given series 1 , 3 , 7 and 15 is the zero term and constant in all terms. Therefore, for nth = T7 (7 term) ,then T7 = 1 + 2^1 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 nth = nth = 1 + 2^1 + 2^2 + … + 2^n ( Take T1 = 3 and T0 = 1 constant in all terms)Notice that the last power of two in each term, equals the term itself. Upvote · 9 1 Sponsored by VAIZ.com Tool Like Asana — Try For Free Our Better Alternative! VAIZ — Better Asana Alternative That Boosts Productivity with Built-in Doc Editor & Task Management. Sign Up 99 14 John R. Clymer PhD in Electrical Engineering&Mathematics, Arizona State University (Graduated 1971) · Author has 1.2K answers and 490.6K answer views ·Feb 12 Originally Answered: What is the formula of the nth term in this number 0,3,8,15? · A+B+C =. 0. 3. 8. 15 3A+B =. 3. 5. 7 2A. =. 2. 2 A = 1 B = 3 - 3 = 0 C = 0 - 1. =. -1 Qn = n^2 - 1 5 24 Upvote · Roger Asdale Author has 2.3K answers and 327.8K answer views ·Feb 12 Originally Answered: What is the formula of the nth term in this number 0,3,8,15? · By inspection, n: 1, 2, 3, 4, … R: 0, 3, 8, 15, … Formula for position n in this SEQUENCE (or SERIES)… R = n² - 1 Upvote · Ellis Cave 40+ years as an Electrical Engineer · Author has 7.9K answers and 4.3M answer views ·6y Related What is the nth term for the sequence 1, 3, 7, 15…? Using the J programming language: Build the function x f n that will implement the series, starting with x for n terms: f =.h/.&.\|.@,(1$~ ]) Test it: 1 f 5 1 3 7 15 31 63 So the answer is the next term is 31. <<<>>> 1 f 15 1 3 7 15 31 63 127 255 511 1023 2047 4095 8191 16383 32767 65535 Upvote · 9 7 Hisham Tumiron Stupid as stupid does · Author has 103 answers and 45.2K answer views ·5y Related How do I derive the formula for the nth term of the sequence 0, 2, 5, 7, 10,12,15? Call me lazy but I don't want to think too long. There must be 2 answers, one for n that is even and another that is odd. n=1 and n=2 is exclusive. n (even) is… 5, 10, 15,… n (odd) is…. 7, 12,….. So you can see the patterns there. So, for Tn (odd), T3 = 5, T5 = 10, T7 = 15 or T3 = 1x5, T5 = 2x5, T7 = 3x5 3 as in T3 correlates to 1, 5 to 2, and 7 to 3. So, n as in Tn correlates to (n/2) - 0.5 So, Tn = ((n/2) - 0.5) x 5 = 5((n/2) - 0.5) For Tn (even), T4 = 7, T6 = 12 or T4 = (5x1)+2 T6 = (5x2)+2, and you may assume T8 = (5x3)+2 4 correlates to 1, 6 to 2, and 8 to 3. So, n correlates to (n/2)-1. So Tn = 5((n/2) - Continue Reading Call me lazy but I don't want to think too long. There must be 2 answers, one for n that is even and another that is odd. n=1 and n=2 is exclusive. n (even) is… 5, 10, 15,… n (odd) is…. 7, 12,….. So you can see the patterns there. So, for Tn (odd), T3 = 5, T5 = 10, T7 = 15 or T3 = 1x5, T5 = 2x5, T7 = 3x5 3 as in T3 correlates to 1, 5 to 2, and 7 to 3. So, n as in Tn correlates to (n/2) - 0.5 So, Tn = ((n/2) - 0.5) x 5 = 5((n/2) - 0.5) For Tn (even), T4 = 7, T6 = 12 or T4 = (5x1)+2 T6 = (5x2)+2, and you may assume T8 = (5x3)+2 4 correlates to 1, 6 to 2, and 8 to 3. So, n correlates to (n/2)-1. So Tn = 5((n/2) - 1) + 2 In short, n is odd, Tn = 5((n/2) - 0.5) n is even, Tn = 5((n/2) - 1) + 2 Maybe I think too long…. Upvote · 9 2 9 2 Martyn Hathaway BSc in Mathematics, University of Southampton (Graduated 1986) · Upvoted by Yusuf Dadkhah , master degree Mathematics (2024) · Author has 4.7K answers and 6.7M answer views ·6y Related What is the nth term for the sequence 1, 3, 7, 15…? What is the nth term for the sequence 1, 3, 7, 15…? The way to solve this is to find a function that generates the given sequence and use it to find the nth term. The problem here is that there is an infinite number of possible generating functions and therefore an infinite number of ‘correct’ answers. For example, I could use a quartic formula in the form: f(x)=a x 4+b x 3+c x 2+d x+e f(x)=a x 4+b x 3+c x 2+d x+e Thus f(1)=a+b+c+d+e=1 f(1)=a+b+c+d+e=1 f(2)=16 a+8 b+4 c+2 d+e=3 f(2)=16 a+8 b+4 c+2 d+e=3 f(3)=81 a+27 b+9 c+3 d+e f(3)=81 a+27 b+9 c+3 d+e f(4)=256 a+64 b+16 c+4 d+e f(4)=256 a+64 b+16 c+4 d+e Thus f(n)=a n 4+b n 3+c n 2+d n+e f(n)=a n 4+b n 3+c n 2+d n+e For any chosen value of of a a, there will b Continue Reading What is the nth term for the sequence 1, 3, 7, 15…? The way to solve this is to find a function that generates the given sequence and use it to find the nth term. The problem here is that there is an infinite number of possible generating functions and therefore an infinite number of ‘correct’ answers. For example, I could use a quartic formula in the form: f(x)=a x 4+b x 3+c x 2+d x+e f(x)=a x 4+b x 3+c x 2+d x+e Thus f(1)=a+b+c+d+e=1 f(1)=a+b+c+d+e=1 f(2)=16 a+8 b+4 c+2 d+e=3 f(2)=16 a+8 b+4 c+2 d+e=3 f(3)=81 a+27 b+9 c+3 d+e f(3)=81 a+27 b+9 c+3 d+e f(4)=256 a+64 b+16 c+4 d+e f(4)=256 a+64 b+16 c+4 d+e Thus f(n)=a n 4+b n 3+c n 2+d n+e f(n)=a n 4+b n 3+c n 2+d n+e For any chosen value of of a a, there will be a unique set of values for the other coefficients. As there are an infinite number of possible values for a a, we thus have an infinite set of answers. Generally, the idea is to find a (relatively) simple generating function by looking for simple patterns. Of course, while you might find a function quite quickly, someone else could find a different function as quickly; both functions would be ‘right’. We have: f(1)=1=2−1=2 1−1 f(1)=1=2−1=2 1−1 f(2)=3=4−1=2 2−1 f(2)=3=4−1=2 2−1 f(3)=7=8−1=2 3−1 f(3)=7=8−1=2 3−1 f(4)=15=16−1=2 4−1 f(4)=15=16−1=2 4−1 Thus, using this pattern, f(n)=2 n−1 f(n)=2 n−1 Upvote · 99 20 Related questions What is the nth term formula for the sequence 0, 3, 7, 15? What is the nth term for the sequence 1, 3, 7, 15…? What is the next term of this sequence 1, 3, 7, 15, 31, 63, _? What is the nth term of 1,1/3,1/5,1/7? What is the formula to find the nth term of the sequence 3, 15, 35, and 63? What is the rule for the nth term of the sequence 1, 3, 7, and 15? How do I find the nth term for the sequence 1, 3, 6, 10, and 15? What is the nth term rule of 3, 5, 7, 9, 11, 13, and 15? What is the 7th term of the sequence -5, -3, 1, 7, and 15? What is the formula for the nth term of the series 2, 5, 3, 10, 4, 15? What is the formula for the nth term of the sequence 3, 15, 35, and 32? What is a formula for the nth term of the given sequence? 15,21,27? What is the possible formula for the nth term of a sequence whose four terms are 7, 11, 15, and 19? In (-2, 4, -8, 16 , -32, 64, -128), what is the formula that describes the nth term of the given sequence? How can we find the nth term formula of the sequence "2, 10, 30, 68"? Related questions What is the nth term formula for the sequence 0, 3, 7, 15? What is the nth term for the sequence 1, 3, 7, 15…? What is the next term of this sequence 1, 3, 7, 15, 31, 63, _? What is the nth term of 1,1/3,1/5,1/7? What is the formula to find the nth term of the sequence 3, 15, 35, and 63? What is the rule for the nth term of the sequence 1, 3, 7, and 15? 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. 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10307	https://www.turito.com/learn/math/arithmetic-sequence-recursive-explicit-formula	Need Help? Get in touch with us Earth and space Maths English Physics Chemistry more.. Biology Science Earth and space Arithmetic Sequence: Recursive & Explicit Formula Grade 10 Aug 21, 2023 Connect Sequences and Function a. Is the Ordered List 26, 39, 52, 65, 78 an Arithmetic Sequence? A sequence is an ordered list of numbers that often forms a pattern. Each number is a term of the sequence. In an arithmetic sequence, the difference between any two consecutive terms is a constant called the common difference. Find the differences between pairs of consecutive terms. There is a common difference of 13. So, this is an arithmetic sequence. b. How Are the Sequences Related to Functions? You can think of a sequence as a function where the domain is restricted to the natural numbers and the The range is the term of the sequence. For the sequence 26,39,52,65,78 Let n = the term number in the sequence. Let A(n) = the value of the nth term of the sequence. The 1st term is 26. A (1) =26 A (2) =39 The 2nd term is 39. c. How Do You Represent Sequences Using Subscript Notation? The subscript notation is commonly used to describe sequences. a₂=39. The 2nd term is 39. You can use either function or subscript notation to represent sequences. Try It! Is the domain of the function in Part B of Example 1 continuous-discrete? Explain. Solution: The domain is discrete because it consists of whole numbers or integers. Recursive Formula Recursive, in mathematics, means to repeat a process over and over again, using the output of each step as the next input. A recursive formula relates each term of a sequence to the previous term. It is composed of an initial value and a rule for generating the sequence. The recursive formula for an arithmetic sequence is: A recursive formula describes the pattern of a sequence and can be used to find the next term in a sequence. Apply the Recursive Formula Example 2 What is a recursive formula for the height above the ground of the nth step of the pyramid shown? Use the recursive formula. The formula an an = an-1 + 26 gives the height above the ground of the nth step with a₁ = 26. 2. Use the Recursive Formula to Find the Height above the Ground of the 3rd Step. Find the height above the ground of the 2nd step. a1 =26 a2 = a1 + 26 a2 = 26 + 26 =52 Use a1 , to find a2 . Find the height above the ground of the 3rd step. a2 = 52 a3 =a2 + 26 Use a2 to find a3 a3 = 52 +26 = 78 The 3rd step is 78 cm above the ground. Try It! Write a recursive formula to represent the total height of the nth stair above the ground if the height of each stair is 18 cm. Solution: Use the recursive formula of an arithmetic sequence given by an = an-1 + where d is the common difference. Each step of the stair is 18 cm so d = 18. nth term: an = an-1 + 18 Explicit Formula Definition An explicit formula expresses the nth term of a sequence in terms of n. The explicit formula for an arithmetic sequence is: Apply the Explicit Formula Example 3 a. The cost of renting a bicycle is given in the table. How can you represent the rental cost using an explicit formula? b. What is the cost of renting the bicycle for 10 days? c. How is the explicit formula of an arithmetic sequence related to a linear function Solution: To find the rental cost for n days, write an explicit formula for the nth term of the sequence. Use the explicit formula. a n = a1 + (n-1) d a n = 26 + (n-1).12 ………………Substitute 26 for a₁ and 12 for d. = 26+ 12n – 12 ………………. Distributive Property = 14 + 12n ……………………… Simplify. The explicit formula an= 14 + 12n gives the rental cost for n days. b.Solution: Use the explicit formula to find the 10th term in the sequence. Use the explicit formula to find the 10th term in the sequence an = 14 + 12n a10 = 14 + 12(10) ……………………Substitute 10 for n. a10 = 134 The 10th term in a sequence is 134. It costs $134 to rent the bicycle for 10 days. c.Solution: The formula a, = 14 + 12n shows that the cost, an, is a function of the number of days, n, the bicycle is rented. You can write this as a linear function, f(x)= 12x + 14, or as an equation in slope-intercept form, y = 12x + 14. The common difference, 12, corresponds to the slope of the graph COMMON ERROR You might think that the initial value of the sequence and the y-intercept of the graph of the equivalent linear function are the same, but they will not be the same if the initial value of the sequence is not zero. Try It! 3. The cost to rent a bike is $28 for the first day plus $2 for each day after that. Write an explicit formula for the rental cost for n days. What is the cost of renting the bike for 8 days? Solution: Use the explicit formula of an arithmetic sequence given by where a₁ is the first term and d is the common difference. an = a1 + (n-1) d The cost on the first day is $28 so a1 = 28 and increases by $2 for each day after that so d = 2 an = a1 +(n-1) d =28 + (n-1)2 =28+2n-2 =an = 2n+28 a8=2 x 8 +28 = 16 +28 a8=$42 The cost of renting the bike for 8 days=$42 Write an Explicit Formula from a Recursive Formula Example 4 The recursive formula for the height above the ground of the nth step of the stairs shown is an = an-1 + 4 with a1 = 7 What explicit formula finds the height above the ground of the nth step? Solution: Use the recursive formula to find information about the sequence. a1 = 7 an =an-1 + 4 common difference Write the explicit formula. an = a1 + (n-1) d an = 7 + (n-1)4 ………………. substitute 7 for a1 and 4 for d The explicit formula an = 7+ (n-1)4 can be used to find the height above the ground of the nth step. an = 7 + (n-1) 4 = 7+ 4n-4 an=4n+3 Try It! Write an explicit formula for each arithmetic sequence. a. an = a n-1 – 3; a1 = 10 b.an = a n-1 + 2.4; a1 = -1 Solution: The explicit formula of an arithmetic sequence is given where a₁ is the first term and d is a common difference. an = a1 + (n-1) d From the given, an = 10 and d = -3 so, the explicit formula is: an =10+ (n − 1) (-3) =10-3n+3 = -3n+13 b. From the given, a1= −1 and d = 2.4 so, the explicit formula is: an= −1 + (n − 1) (2.4) = −1+2.4n – 2. an=2.4n – 3.4 Write a Recursive Formula from an Explicit Formula Example 5 The explicit formula for an arithmetic sequence is an= 1 + 1/2 n. What is the recursive formula for the sequence? Solution: Step 1: Identify the common difference. an= 1 + 1/2 n. d =1/2 Step 2: Find the first term of the sequence. an = 1 + ½ n a1= 1 + ½ (1) ………………………..Substitute 1 for n an = 3/2 …………………..Simplify. Step 3: Write the recursive formula. an = an-1+ d an = an-1+1/2 ……………………substitute ½ for d The recursive formula for the sequence is : First term a1= 3/2 ; nth term an = an-1+ 1/2 Try It! Write a recursive formula for each explicit formula. a. an = 8 +3n b. an = 12 – 5n a. Solution: Use the recursive formula of an arithmetic sequence given by: an = a1 + (n-1) d where d is a common difference. Given the explicit formula, we can easily identify the common difference, which is the coefficient of the variable, n. Here, we have d = 3 Next, we identify the first term by substituting n = 1 to the recursive formula: a₁ = 8 + 3(1) a₁ = 11 When writing a recursive formula for an arithmetic sequence, including the value of the first term, a₁: first term: a₁ =11; nth term: an= an-1 + 3 b.Solution: Use the recursive formula of an arithmetic sequence given by: an = a1 + (n-1) d where d is a common difference. Given the explicit formula, we can easily identify the common difference, which is the coefficient of the variable, n. Here, we have d = -5 Next, we identify the first term by substituting n = 1 to the recursive formula: a₁ = 12 – 5(1) a₁ = 7 When writing a recursive formula for an arithmetic sequence, including the value of the first term, a1: first term: a₁ = 7; nth term: an= an-1 – 5 Make Sense and Persevere The lowest and leftmost note on a piano keyboard is an A. The next lowest A is seven white keys to the right. This pattern continues. Write an explicit formula for an arithmetic sequence to represent the position of each A key on the piano, counting from the left. If a piano has 52 white keys, in what position is the key that plays the highest A? Solution: Use the explicit formula of an arithmetic sequence given by: an = a₁ + (n − 1)d – where a₁ is the first term and d is a common difference. The leftmost A corresponds to a₁ = 1 and since the A’s are 7 keys apart, the d = 7: an = 1 + (n − 1)(7) = 1+7n–7 an = 7n – 6 Writing out the A keys that are less than 52, we have: Hence, the highest A is at the 50th key. a1= 1 a2=7(2) – 6=8 a3= 7(3) – 6 = 15 a4= 7(4) – 6 = 22 a5= 7(5) – 6 = 29 a6= 7(6) – 6 = 36 a7= 7(7) – 6 = 43 a8= 7(8) – 6 = 50 Hence, the highest A is at the 50th key. Check your knowledge Make Sense and Persevere: After the first raffle drawing, 497 tickets remain. After the second raffle drawing, 494 tickets remain. Assuming that the pattern continues, write an explicit formula for an arithmetic sequence to represent the number of raffle tickets that remain after each drawing. How many tickets remain in the bag after the seventh raffle drawing? In a video game, you must score 5,500 points to complete level 1. To move through each additional level, you must score an additional 3,250 points. What number would you use as a₁ when writing an arithmetic sequence to represent this situation? What would n represent? Write an explicit formula to represent this situation. Write a recursive formula to represent this situation. Tell whether each sequence is an arithmetic sequence or not. 4, 7, 10, 14,… Write a recursive formula for each sequence. 81, 85, 89, 93, 97,…………… Check Your Knowledge-Answers Solution 1: Use the explicit formula of an arithmetic sequence given by: an= a1+ (n − 1) d where a₁ is the first term and d is a common difference. Based on the sequence, the common difference is d = 494 – 497 = -3. Given that 497 tickets remain after the first raffle drawing, then a₁ = 497. Hence, the explicit formula is: an= 497 + (n − 1) (−3) = 497-3n+3 an = 500 – 3n Substitute n=7 representing the seventh raffle drawing an= 500 – 3(7) = 500 – 21 an = 479 So, 479 tickets remain after the seventh raffle drawing. Solution 2: We would use a₁ = 5500, the required points to complete level 1 because succeeding levels require 3250 more points than the previous level, which is the common difference. n would represent the level (level 1, 2, 3, etc.) in this case. Use the explicit formula of an arithmetic sequence given by: an= a1+ (n − 1) d where a₁ is the first term, and d is the common difference. So, the explicit formula is: an= 5500+ (n – 1).3250 = 5500 + 3250n – 3250 an= 3250n + 2250 The recursive formula of an arithmetic sequence given by: an = an-1 + d where d is the common difference. Writing together with the first term, the recursive formula is: first term: a₁ = 5500; nth term: an= an-1 + 3250 Solution 3: Given series; 4,7,10,14 ,………….. Check the difference: 7-4=3 10-7=3 14-10=4 So, the difference between any two consecutive terms is not constant. So, the series is not an arithmetic sequence Solution 4: Given series: 81,85,89,93,97. First find the common difference: 85 – 81 = 4 89 – 85= 4…….. So, the common difference is d = 4 ecursive formula for an arithmetic sequence is an = an-1+ d. whereis nth term, d is the common difference. an = an-1 + 4 Exercise Tell whether each sequence is an arithmetic sequence or not. 4,10,16, 22,………… -2,2, -2,2,-2,…………. 1,1,1,2,2,2,3,3,3…………. Write a recursive formula for each explicit formula and find the first term of the sequence. an = 12 +4n an = 102-n Write an explicit formula for each recursive formula. an = an-1 + 17 and a1 = 7 an = an-1 + 5 and a1 = 4 Write a recursive formula for each explicit formula and find the first term of the sequence. an = 12 + 4n an = 102 -n Concept Summary Arithmetic Sequences An arithmetic sequence is a sequence of numbers that follows a pattern. The difference between two consecutive terms is a constant called the common difference. Recursive Formula Used to describe a sequence and find the next few terms an = an-1+ d an=nth term of the sequence an-1=previous term of the sequence d=common difference Explicit Formula Used to find a specific term in the sequence an = a1 + (n-1) d an =nth term of the sequence a1=first term of the sequence d=common difference Numbers 1, 7, 13, 19, 25,…………….. Use the recursive formula to describe the sequence and find the next two terms. d=7-1=6 an = an-1 + 6 a6 = a5 + 6 = 25+6 a6 = 31 a7 = a6 + 6 = 31+6 a7 = 37 The next two terms are 31 and 37 Use the explicit formula to find the 15th term in the sequence. an = 1+ (n-1) 6 a15 =1 + (14)6 a15 = 85 Concept Map: Related topics Addition and Multiplication Using Counters & Bar-Diagrams Introduction: We can find the solution to the word problem by solving it. Here, in this topic, we can use 3 methods to find the solution. 1. Add using counters 2. Use factors to get the product 3. Write equations to find the unknown. Addition Equation: 8+8+8 =? Multiplication equation: 3×8=? Example 1: Andrew has […] Read More >> Dilation: Definitions, Characteristics, and Similarities Understanding Dilation A dilation is a transformation that produces an image that is of the same shape and different sizes. Dilation that creates a larger image is called enlargement. Describing Dilation Dilation of Scale Factor 2 The following figure undergoes a dilation with a scale factor of 2 giving an image A’ (2, 4), B’ […] Read More >> How to Write and Interpret Numerical Expressions? Write numerical expressions What is the Meaning of Numerical Expression? A numerical expression is a combination of numbers and integers using basic operations such as addition, subtraction, multiplication, or division. The word PEMDAS stands for: P → Parentheses E → Exponents M → Multiplication D → Division A → Addition S → Subtraction Some examples […] Read More >> System of Linear Inequalities and Equations Introduction: Systems of Linear Inequalities: A system of linear inequalities is a set of two or more linear inequalities in the same variables. The following example illustrates this, y < x + 2…………..Inequality 1 y ≥ 2x − 1…………Inequality 2 Solution of a System of Linear Inequalities: A solution of a system of linear inequalities […] Read More >> Other topics #### How to Find the Area of Rectangle? Mar 3, 2022 #### How to Solve Right Triangles? Nov 26, 2022 #### Ways to Simplify Algebraic Expressions Nov 26, 2022 callback button
10308	http://uamte.math.byu.edu/~bakker/Math346/Lectures/M346Lec33.pdf	Math 346 Lecture #33 12.3 The Resolvent The Jordan Canonical Form, or spectral decomposition, of a linear operator on a ﬁnite dimension vector space has important applications in many areas such as diﬀerential equations and dynamical systems. Computing the Jordan Canonical form depends on ﬁnding bases for the generalized eigenspaces, and as mentioned before, this is a poorly conditioned numerical algorithm. A more powerful approach to ﬁnding a spectral decomposition of a linear operator on a ﬁnite dimensional vector space uses the tools of complex analysis. This theoretical approach is basis-free, meaning we do not have to ﬁnd bases of the generalized eigenspaces to get a spectral decomposition. 12.3.1 Properties of the Resolvent Deﬁnition 12.3.1. The resolvent set of A ∈Mn(C), denoted by ρ(A), is the set of points z ∈C for which zI −A is invertible. Note that the complement σ(A) = C \ ρ(A) is the spectrum of A. The resolvent of A is the map R(A, ·) : ρ(A) →Mn(C) deﬁned by R(A, z) = (zI −A)−1. We will sometimes make use of the notation RA(z) for R(A, z) to make clear the depen-dence of the resolvent on A. When there is no ambiguity, we denote R(A, z) simply as R(z). Example (in lieu of 12.3.2). The resolvent RA for A = 1 1 4 1 ∈M2(C) is RA(z) = z −1 −1 −4 z −1 −1 = 1 (z −1)2 −4 z −1 1 4 z −1 = 1 z2 −2z −3 z −1 1 4 z −1 = 1 (z −3)(z + 1) z −1 1 4 z −1 The resolvent has simple poles precisely on σ(A) = {3, −1}. The domain of the resolvent RA is the resolvent set ρ(A) = C \ σ(A). Remark 12.3.3. Each entry of the resolvent is a rational function by Cramer’s Rule: RA(z) = (zI −A)−1 = 1 det(zI −A)adj(zI −A) where det(zI −A) is the characteristic polynomial of A, and adj(zI −A) is the adjugate matrix, i.e., the transpose of the matrix of signed minors of zI −A, which satisﬁes (zI −A)adj(zI −A) = det(zI −A)I (see Deﬁnition 2.9.19 and Theorem 2.9.22). The rational function nature of the entries of the resolvent shows that the resolvent has poles, some possibly not simple, precisely on σ(A). Example (in lieu of 12.3.4). (i) Find the resolvent for A =   6 1 0 0 6 7 0 0 4  . Because A is upper triangular, the characteristic polynomial of A is det(zI −A) = (z −6)2(z −4). The adjugate of A is adj(zI −A) =   (z −6)(z −4) z −4 7 0 (z −6)(z −4) 7(z −6) 0 0 (z −6)2  . We verify this by computing (zI −A)adj(zI −A) = det(zI −A)I. ✓ The resolvent of A is R(z) =   (z −6)−1 (z −6)−2 7(z −6)−2(z −4)−1 0 (z −6)−1 7(z −6)−1(z −4)−1 0 0 (z −4)−1  . (ii) Find the resolvent of A =     2 1 0 0 0 2 1 0 0 0 2 3 0 0 0 5    . Because A is upper triangular, the characteristic polynomial of A is det(zI −A) = (z −2)3(z −5). The required adjugate of A is adj(zI −A) =     (z −2)2(z −5) (z −2)(z −5) z −5 3 0 (z −2)2(z −5) (z −2)(z −5) 3(z −2) 0 0 (z −2)2(z −5) 3(z −2)2 0 0 0 (z −2)3    . The resolvent has several basic properties as expressed next. Lemma 12.3.5. Let A, A1, A2 ∈Mn(C). (i) If z1, z2 ∈ρ(A), then R(z2) −R(z1) = (z1 −z2)R(z2)R(z1). [This is known as Hilbert’s Identity.] (ii) If z ∈ρ(A1) ∩ρ(A2), then R(A2, z) −R(A1, z) = R(A1, z)(A2 −A1)R(A1, z). (iii) If z ∈ρ(A), then R(z)A = AR(z). (iv) If z1, z2 ∈ρ(A), then R(z1)R(z2) = R(z2)R(z1). Proof. (i) Multiplying the identity (z1 −z2)I = (z1I −A) −(z2I −A) on the left by R(z2) and on the right by R(z1) gives R(z2)(z1 −z2)R(z1) = R(z2)(z1I −A)R(z1) −R(z2)(z2I −A)R(z1). Since R(z1) = (z1I −A)−1 and R(z2) = (z2I −A)−1 the equation simpliﬁes to R(z1)(z1 −z2)R(z2) = R(z2) −R(z1). The term z1 −z2 is a complex scalar and so commutes with the matrix R(z2) to give (z1 −z2)R(z2)R(z1) = R(z2) −R(z1). (ii) Multiply the identity A2 −A1 = (zI −A1) −(zI −A2) on the left by R(A1, z) and on the right by R(A2, z) to obtain R(A1, z)(A2 −A1)R(A2, z) = R(A1, z)(zI −A1)R(A2, z) −R(A1, z)(zI −A2)R(A2, z). Since R(A1, z) = (zI −A1)−1 and R(A2, z) = (zI −A2)−1 the equation simpliﬁes to R(A1, z)(A2 −A1)R(A2, z) = R(A2, z) −R(A1, z). (iii) Since R(z)(zI −A) = I and (zI −A)R(z) = I, we have R(z)(zI −A) = (zI −A)R(z). Multiplying this out gives zR(z) −R(z)A = zR(z) −AR(z). Cancelling the common term zR(z) gives R(z)A = AR(z). (iv) When z1 = z2 it follows trivially that R(z1)R(z2) = R(z2)R(z1). When z1 ̸= z2, then Hilbert’s identity gives R(z2)R(z1) = R(z2) −R(z1) z1 −z2 = R(z1) −R(z2) z2 −z1 = R(z1)R(z2). This gives the desired result. □ 12.3.2 Local Properties We show that the resolvent RA is a matrix-valued holomorphic function on ρ(A) by ﬁnding power series expansions of RA at all points z ∈ρ(A). The original lower bounds on the radii of convergence for these power series are enlarged by means of a limit quantity called the spectral radius of A. Throughout this subsection we assume that ∥·∥is a matrix norm on Mn(C), i.e., a norm on Mn(C) that for all A, B ∈Mn(C) satisﬁes ∥AB∥≤∥A∥∥B∥. Each induced p-norm ∥A∥p = sup ∥Ax∥p ∥x∥p : x ∈Cn \ {0} is a matrix norm, as is the Frobenius norm ∥· ∥F. Theorem 12.3.6. For A ∈Mn(C), the resolvent set ρ(A) is open, and R is holomorphic on ρ(A) where for each z0 ∈ρ(A), the resolvent is given by the power series R(z) = ∞ X k=0 (−1)k(z −z0)kRk+1(z0), whose the radius of convergence is at least as large as ∥R(z0)∥−1. Proof. For B ∈Mn(C), if ∥B∥< 1, then by Proposition 5.7.4, I −B is invertible and there holds the Neumann series ∞ X k=0 Bk = (I −B)−1. For A ∈Mn(C), using Hilbert’s Identity for points z0, z ∈ρ(A), the resolvent R satisﬁes R(z0) = R(z) + (z −z0)R(z0)R(z) = [I + (z −z0)R(z0)]R(z). Setting B = −(z −z0)R(z0) the condition ∥B∥< 1 implies that \|z −z0\| ∥R(z0)∥= ∥B∥< 1. Since z0 ∈ρ(A), the matrix R(z0) is invertible so that ∥R(z0)∥> 0. Thus for z satisfying \|z −z0\| ≤∥R(z0)∥−1, the matrix I −B = I + (z −z0)R(z0) is invertible and [I + (z −z0)R(z0)]−1 = ∞ X k=0 (−1)k(z −z0)kRk(z0). Since R(z0) = R(z) + (z −z0)R(z0)R(z) = [I + (z −z0)R(z0)]R(z) we obtain R(z) = [I −(z −z0)R(z0)]−1R(z0) = ∞ X k=0 (−1)k(z −z0)kRk+1(z0), which power series converges on the open disk D(z0) = \|z −z0\| ≤∥R(z0)∥−1. The union ∪z0∈ρ(A)D(z0) is the same as ρ(A), implying that ρ(A) is open. Therefore, by Remark 11.2.10, the resolvent R is holomorphic on the open ρ(A). □ Remark 12.3.7. Comparing the power series R(z) = ∞ X k=0 (−1)k(z −z0)kRk+1(z0) in Theorem 12.3.6 and the Taylor series R(z) = ∞ X k=0 R(k)(z0) k! (z −z0)k reveals, by the uniqueness of the Taylor series, a relationship between the powers of R and its derivatives, namely that (−1)kRk+1(z0) = R(k)(z0) k! . As this holds for every z0 ∈ρ(A), we may replace z0 by z to get R(k)(z) = k!(−1)kRk+1(z) = k!(−1)k(zI −A)−(k+1). Theorem 12.3.8. For A ∈Mn(C), the Laurent series of R(z) on the open annulus \|z\| > ∥A∥exists and is given by R(z) = ∞ X k=0 Ak zk+1. Proof. The resolvent is R(z) = (zI −A)−1 = (z(I −z−1A))−1 = z−1(I −z−1A)−1. To express (I −z−1A)−1 as a Neumann series requires that ∥z−1A∥< 1. This condition gives the annulus \|z\| > ∥A∥, on which we have R(z) = z−1(I −z−1A)−1 = z−1 ∞ X k=0 Ak zk = ∞ X k=0 Ak zk+1 = I z + A z2 + A2 z3 + · · · . This is the Laurent series of R(z) on the open annulus \|z\| > ∥A∥. □ Remark. One might be tempted to say that the resolvent R(z) has an essential singu-larity at z = 0, but the open annulus \|z\| > ∥A∥is not a punctured disk unless A = 0 in which case the resolvent has a simple pole at 0. But we can use the Laurent series of R(z) on \|z\| > ∥A∥to say something about the behavior of R(z) as \|z\| →∞. For this we speak of R(z) being holomorphic in a neighbourhood of ∞, which means that the function (R ◦g)(w), for the change of variables z = g(w) = 1/w, is holomorphic on an open disk centered at w = 0. Here \|z\| = 1/\|w\| so that \|w\| →0 if and only if \|z\| →∞. Corollary 12.3.9. For any A ∈Mn(C), the resolvent R is holomorphic in a neighbour-hood of ∞, and moreover there holds lim \|z\|→∞∥R(z)∥= 0. Proof. From Theorem 12.3.8, the Laurent series R(z) = ∞ X k=0 Ak zk+1 = ∞ X k=0 Ak 1 z (k+1) converges on \|z\| > ∥A∥. If A = 0, then R(z) = z−1I and under the change of variables z = g(w) = 1/w we obtain (R ◦g)(w) = wI which is entire. Now suppose A ̸= 0. Then ∥A∥> 0. Under the change of variables z = g(w) = 1/w, the annulus \|z\| > ∥A∥becomes 1/\|w\| > ∥A∥which is the disk \|w\| < 1/∥A∥. With the change of variables z = g(w) = 1/w we obtain the power series (R ◦g)(w) = ∞ X k=0 Akwk+1 that converges on the disk \|w\| < 1/∥A∥. Now from the Laurent series for the resolvent and \|z\| > ∥A∥we have that ∥A∥/\|z\| < 1 and that ∥R(z)∥≤ ∞ X k=0 ∥A∥k \|z\|k+1 = 1 \|z\| ∞ X k=0 ∥A∥ \|z\| k = 1 \|z\| 1 −∥A∥ \|z\| −1 = 1 \|z\| −∥A∥. This shows that lim\|z\|→∞∥R(z)∥= 0. □ Remark 12.3.11. We saw in Remark 12.3.3 that the values of z for which (zI −A) is not invertible are precisely the eigenvalues of A, i.e., the roots of the characteristic polynomial det(zI −A). The Fundamental Theorem of Algebra implies that the charac-teristic polynomial has n roots (counting repeated eigenvalues if any). This means that the resolvent set ρ(A) is not all of C, that σ(A) = C \ ρ(A) ̸= ∅. Next is another proof that the spectrum of A is not empty. Corollary 12.3.12. For any A ∈Mn(C), the spectrum σ(A) is not empty. Proof. Suppose to the contrary that σ(A) = ∅. Then ρ(A) = C, and hence that the resolvent R is entire. Corollary 12.3.9 implies that ∥R∥is a bounded entire function. Liouville’s Theorem implies that R is a constant function. Since lim\|z\|→∞∥R(z)∥= 0, the resolvent R is the zero function. But this is a contradiction because the resolvent satisﬁes I = (zI −A)R(z). □ Note. The radius of convergence of the power series of R(z) about z0 ∈ρ(A) is at least as large as ∥R(z0)∥−1 by Theorem 12.3.6. The inner radius of the annulus for the Laurent series of R(z) about 0 is at least as small as ∥A∥. We show that these radii can be improved through a limit quantity known as the spectral radius. Lemma 12.3.13. For any A ∈Mn(C), the limit r(A) = lim k→∞∥Ak∥1/k exists and is bounded above by ∥A∥. Proof. If there exists L ∈N such that ∥AL∥= 0, then AL = 0 and ∥Al∥= 0 for all l ≥L, and hence ∥Al∥= 0 for all l ≥L, implying that r(A) = 0. Suppose that ∥Al∥> 0 for all l ∈N. Using the submultiplicative property of the given norm, we have for integers 1 ≤s < t that ∥As∥= ∥At−sAs∥≤∥At−s∥∥As∥. Since none of the norms are zero, we can apply the logarithm function to the inequality to get log ∥At∥≤log ∥At−s∥+ log ∥As∥. Setting al = log ∥Al∥for each l ∈N gives whenever 1 ≤s < t that at ≤at−s + as. Now ﬁx integers k and m satisfying 1 ≤m < k. Since k > m ≥1, there exists by the Division Algorithm unique integers q and p where q ≥1 and 0 ≤p < m such that k = mq + p. We can express q = ⌊k/m⌋, the greatest integer equal to or less then k/m = q + p/m. Setting t = k and s = p in at ≤at−s + as gives ak ≤ak−p + ap = amq + ap. Setting t = mq and s = m(q −1) in at ≤at−s + as gives amq ≤am + am(q−1). Setting t = m(q −1) and s = m(q −2) in at ≤at−s + as gives am(q−1) ≤am + am(q−2). Continuing this gives amq ≤qam. Thus for 1 ≤m < k we obtain ak = amq+p ≤qam + ap. This implies that ak k ≤q kam + 1 kap. Note that exp ak k = exp log ∥Ak∥ k = exp ∥Ak∥1/k = ∥Ak∥1/k. Leaving m ﬁxed (for now) and letting k > m be arbitrary we have that ⌊k/m⌋= q for all k = mq, mq + 1, . . . , mq + (m −1), i.e., ⌊(mq + p)/m⌋= q for all p = 0, 1, . . . , m −1. Hence q k = q mq + p ≤ q mq = 1 m with equality realized when p = 0. This implies that lim sup k→∞ q k = 1 m. We then have lim sup k→∞ ak k ≤lim sup k→∞ q kam + 1 kap = am m . This holds for each m, so that lim sup k→∞ ak k ≤lim inf m→∞ am m . This implies that the limit of (1/k)ak and hence that the limit of exp((1/k)ak) = ∥Ak∥1/k exists by the continuity of the exponential function. To get the upper bound of ∥A∥on ρ(A), we ﬁx m = 1 to get log r(A) = lim k→∞ log ∥Ak∥ k = lim sup k→∞ ak k ≤a1 1 = log ∥A∥. Exponentiating the inequality gives r(A) ≤∥A∥. □ Remark. We will see in the next section that the quantity r(A) is independent of the matrix norm used in the limit. Deﬁnition. The spectral radius of A ∈Mn(C) is deﬁned to be the quantity r(A). Theorem 12.3.14. For A ∈Mn(C), the power series R(z) = ∞ X k=0 (−1)k(z −z0)kRk+1(z0) converges on \|z −z0\| < [r(R(z0))]−1, and the Laurent series R(z) = ∞ X k=0 Ak zk+1 converges on the annulus \|z\| > r(A). Proof. The power series R(z) = ∞ X k=0 (−1)k(z −z0)kRk+1(z0) = R(z0) ∞ X k=0 (−1)k(z −z0)kRk(z0). converges if there exists N ∈N such that for all k ≥N there holds (z −z0)R(z0) k < 1. To simplify notation set r = r(R(z0)). Let z ∈C satisfy \|z −z0\| < r−1. Then there exists ϵ > 0 such that \|z −z0\| < (r + 2ϵ)−1. For this same ϵ, since ∥R(z0)k∥1/k →r, there exists N ∈N such that ∥R(z0)k∥1/k < r +ϵ, or ∥R(z0)k∥≤(r + ϵ)k, for all k ≥N. Thus \|z −z0\|k < (r + 2ϵ)−k for all k ≥N, so that (z −z0)R(z0) k = \|z −z0\|k∥R(z0)k∥< r + ϵ r + 2ϵ k < 1. Thus the power series converges on \|z −z0\| < r−1. The Laurent series R(z) = ∞ X k=0 Ak zk+1 = 1 z ∞ X k=0 A z k converges if there exists N ∈N such that for all k ≥N there holds A z k < 1. To simplify notation set r = r(A), and let z ∈C satisfy \|z\| > r. There exists ϵ > 0 such that \|z\| > r + 2ϵ. Hence \|z\|k > (r + 2ϵ)k for all k which implies that 1 \|z\|k < 1 (r + 2ϵ)k for all k. For this same ϵ, since ∥Ak∥1/k →r, there exists N ∈N such that for all k ≥N there holds ∥Ak∥1/k < r + ϵ, or ∥Ak∥< (r + ϵ)k. Thus for all k ≥N there holds A z k < ∥Ak∥ \|z\|k < r + ϵ r + 2ϵ k < 1. Therefore the Laurent series converges on \|z\| > r. □ Remark. The radii of convergence given in Theorem 12.3.14 do improve the radii ∥R(z0)∥−1 and ∥A∥given in Theorem 12.3.6 and Theorem 12.3.8. For the ﬁrst radii this is because r(R(z0)) ≤∥R(z0)∥implies ∥R(z0)∥−1 ≤[r(R(z0))]−1, and for the second radii this is because r(A) ≤∥A∥, i.e., a smaller inner radius of the open annulus is possible. Remark 12.3.15. A lower bound on the spectral radius r(A) is given by quantity σM = sup{\|λ\| : λ ∈σ(A)}. Justiﬁcation of this lower bound follows from Theorem 12.3.14 because the open annulus on which the Laurent series converges cannot include any point in σ(A) where the resol-vent has poles. In the next section we will show that r(A) = σM which implies that the spectral radius is independent of the matrix norm used to compute r(A).
10309	https://tutorial.math.lamar.edu/extras/algebratrigreview/triggraphs.aspx	Paul's Online Notes Show/Hide Show all Solutions/Steps/etc. Hide all Solutions/Steps/etc. Sections Trig Function Evaluation Trig Formulas Chapters Algebra Exponentials & Logarithms Classes Algebra Calculus I Calculus II Calculus III Differential Equations Extras Algebra & Trig Review Common Math Errors Complex Number Primer How To Study Math Cheat Sheets & Tables Misc Contact Me MathJax Help and Configuration Extras Download Complete Book Problems Only (No Solutions) Other Items Get URL's for Download Items Print Page in Current Form (Default) Show all Solutions/Steps and Print Page Hide all Solutions/Steps and Print Page Paul's Online Notes Home / Algebra Trig Review / Trigonometry / Graphs of Trig Functions Prev. Section Notes Next Section Show Mobile Notice Show All Notes Hide All Notes Mobile Notice You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best viewed in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (you should be able to scroll/swipe to see them) and some of the menu items will be cut off due to the narrow screen width. Section 2.2 : Graphs of Trig Functions There is not a whole lot to this section. It is here just to remind you of the graphs of the six trig functions as well as a couple of nice properties about trig functions. Before jumping into the problems remember we saw in the Trig Function Evaluation section that trig functions are examples of periodic functions. This means that all we really need to do is graph the function for one periods length of values then repeat the graph. Graph the following function. Show All Solutions Hide All Solutions (y = \cos \left( x \right)) Show Solution There really isn’t a whole lot to this one other than plotting a few points between 0 and (2\pi ), then repeat. Remember cosine has a period of (2\pi ) (see Problem 5 in Trig Function Evaluation). Here’s the graph for ( - 4\pi \le x \le 4\pi ). Notice that graph does repeat itself 4 times in this range of (x)’s as it should. Let’s also note here that we can put all values of (x) into cosine (which won’t be the case for most of the trig functions) and let’s also note that [ - 1 \le \cos \left( x \right) \le 1] It is important to notice that cosine will never be larger than 1 or smaller than -1. This will be useful on occasion in a calculus class. 2. (y = \cos \left( {2x} \right)) Show Solution We need to be a little careful with this graph. (\cos \left( x \right)) has a period of (2\pi ), but we’re not dealing with (\cos \left( x \right)) here. We are dealing with (\cos \left( {2x} \right)). In this case notice that if we plug in (x = \pi ) we will get [\cos \left( {2\left( \pi \right)} \right) = \cos \left( {2\pi } \right) = \cos \left( 0 \right) = 1] In this case the function starts to repeat itself after (\pi ) instead of (2\pi )! So, this function has a period of (\pi ). So, we can expect the graph to repeat itself 8 times in the range ( - 4\pi \le x \le 4\pi ). Here is that graph. Sure enough, there are twice as many cycles in this graph. In general, we can get the period of (\cos \left( {\omega \,x} \right)) using the following. [{\rm{Period}} = \frac{{2\pi }}{\omega }] If (\omega > 1) we can expect a period smaller than (2\pi ) and so the graph will oscillate faster. Likewise, if (\omega < 1) we can expect a period larger than (2\pi ) and so the graph will oscillate slower. Note that the period does not affect how large cosine will get. We still have [ - 1 \le \cos \left( {2x} \right) \le 1] 3. (y = 5\cos \left( {2x} \right)) Show Solution In this case I added a 5 in front of the cosine. All that this will do is increase how big cosine will get. The number in front of the cosine or sine is called the amplitude. Here’s the graph of this function. Note the scale on the (y)-axis for this problem and do not confuse it with the previous graph. The (y)-axis scales are different! In general, [ - R \le R\cos \left( {\omega \,x} \right) \le R] 4. (y = \sin \left( x \right)) Show Solution As with the first problem in this section there really isn’t a lot to do other than graph it. Here is the graph on the range ( - 4\pi \le x \le 4\pi ). From this graph we can see that sine has the same range that cosine does. In general [ - R \le R\sin \left( {\omega \,x} \right) \le R] As with cosine, sine itself will never be larger than 1 and never smaller than -1. 5. (\displaystyle y = \sin \left( {\frac{x}{3}} \right)) Show Solution So, in this case we don’t have just an (x) inside the parenthesis. Just as in the case of cosine we can get the period of (\sin \left( {\omega \,x} \right)) by using [\text{Period}=\frac{2\pi }{\omega }=\frac{2\pi }{{}^{1}/{}_{3}}=6\pi ] In this case the curve will repeat every (6\pi ). So, for this graph I’ll change the range to ( - 6\pi \le x \le 6\pi ) so we can get at least two traces of the curve showing. Here is the graph. 6. (y = \tan \left( x \right)) Show Solution In the case of tangent, we have to be careful when plugging (x)’s in since tangent doesn’t exist wherever cosine is zero (remember that (\tan x = \frac{{\sin x}}{{\cos x}})). Tangent will not exist at [x = \cdots , - \frac{{5\pi }}{2}, - \frac{{3\pi }}{2}, - \frac{\pi }{2},\frac{\pi }{2},\frac{{3\pi }}{2},\frac{{5\pi }}{2}, \ldots ] and the graph will have asymptotes at these points. Here is the graph of tangent on the range ( - \frac{{5\pi }}{2} < x < \frac{{5\pi }}{2}). Finally, a couple of quick properties about (R\tan \left( {\omega \,x} \right)). [\begin{array}{c} - \infty < R\tan \left( {\omega \,x} \right) < \infty \ {\rm{Period}} = \displaystyle \frac{\pi }{\omega }\end{array}] For the period remember that (\tan \left( x \right)) has a period of (\pi ) unlike sine and cosine and that accounts for the absence of the 2 in the numerator that was there for sine and cosine. 7. (y = \sec \left( x \right)) Show Solution As with tangent we will have to avoid (x)’s for which cosine is zero (remember that (\sec x = \frac{1}{{\cos x}})). Secant will not exist at [x = \cdots , - \frac{{5\pi }}{2}, - \frac{{3\pi }}{2}, - \frac{\pi }{2},\frac{\pi }{2},\frac{{3\pi }}{2},\frac{{5\pi }}{2}, \ldots ] and the graph will have asymptotes at these points. Here is the graph of secant on the range ( - \frac{{5\pi }}{2} < x < \frac{{5\pi }}{2}). Notice that the graph is always greater than 1 or less than -1. This should not be terribly surprising. Recall that ( - 1 \le \cos \left( x \right) \le 1). So, 1 divided by something less than 1 will be greater than 1. Also, ({1}/{\pm 1}=\pm 1) and so we get the following ranges out of secant. [R\sec \left( {\omega \,x} \right) \ge R\hspace{0.5in}{\rm{and}}\hspace{0.55in}R\sec \left( {\omega \,x} \right) \le - R] 8. (y = \csc \left( x \right)) Show Solution For this graph we will have to avoid (x)’s where sine is zero (\left( {\csc x = \frac{1}{{\sin x}}} \right)). So, the graph of cosecant will not exist for [x = \cdots , - 2\pi , - \pi ,0,\pi ,2\pi , \cdots ] Here is the graph of cosecant. Cosecant will have the same range as secant. [R\csc \left( {\omega \,x} \right) \ge R\hspace{0.5in}{\rm{and}}\hspace{0.5in}R\csc \left( {\omega \,x} \right) \le - R] 9. (y = \cot \left( x \right)) Show Solution Cotangent must avoid [x = \cdots , - 2\pi , - \pi ,0,\pi ,2\pi , \cdots ] since we will have division by zero at these points. Here is the graph. Cotangent has the following range. [ - \infty < R\cot \left( {\omega \,x} \right) < \infty ] [Contact Me] [Privacy Statement] [Site Help & FAQ] [Terms of Use] © 2003 - 2025 Paul Dawkins Page Last Modified : 11/17/2022
10310	https://mathoverflow.net/questions/73129/interpolating-between-piecewise-linear-functions-with-a-family-of-smooth-functi	real analysis - Interpolating between piecewise linear functions, with a family of smooth functions - MathOverflow Join MathOverflow By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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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 Interpolating between piecewise linear functions, with a family of smooth functions Ask Question Asked 14 years, 1 month ago Modified4 years, 7 months ago Viewed 2k times This question shows research effort; it is useful and clear 15 Save this question. Show activity on this post. Let [a,b)⊂R[a,b)⊂R, and F,G:[a,b)→R F,G:[a,b)→R two decreasing piecewise linear functions so that F(x)≤G(x)F(x)≤G(x) for any x∈[a,b)x∈[a,b). We assume that: there is a number k∈N−{0}k∈N−{0} and a set of k+1 k+1 numbers a=x 0<…<x k=b a=x 0<…<x k=b partitioning [a,b)[a,b) in k k intervals [x j−1,x j)[x j−1,x j) on which the restrictions of F F and G G are linear. the equalizer set E=E q(F,G):={x∈[a,b)\|F(x)=G(x)}E=E q(F,G):={x∈[a,b)\|F(x)=G(x)} is included in the set {x 0,x 1,…,x k}{x 0,x 1,…,x k}. the restriction of F F to the set E E is strictly decreasing for any j j, 0<j<k 0<j<k, there is an open neighborhood (x j−ϵ,x j+ϵ)(x j−ϵ,x j+ϵ), and a linear function L:(x j−ϵ,x j+ϵ)→R L:(x j−ϵ,x j+ϵ)→R, so that G≥L≥F G≥L≥F on (x j−ϵ,x j+ϵ)(x j−ϵ,x j+ϵ). (conditions 2-4 forbid some cases when the problem has no solution) Problem: Find a continuous family of functions f t:[a,b)→R f t:[a,b)→R, t∈[0,1]t∈[0,1] satisfying the conditions: f t f t is smooth and strictly decreasing for any t∈(0,1)t∈(0,1). For any fixed x∈[a,b)−E x∈[a,b)−E, the application ϑ x:[0,1]→[F(x),G(x)]ϑ x:[0,1]→[F(x),G(x)], ϑ x(t)=f t(x)ϑ x(t)=f t(x) is a strictly increasing and bijective smooth function. It is easy to see that from the condition 2 it follows that: if 0≤s<t≤1 0≤s<t≤1, then f(s)<f(t)f(s)<f(t) on [a,b)−E[a,b)−E (and of course f(s)=f(t)=F=G f(s)=f(t)=F=G on E E). f 0=F f 0=F and f 1=G f 1=G. If possible, please also provide some references which can help solving this problem. Update 1: So far I tried to construct the functions from known analytical functions, splines and sigmoids, and to use Schwarz-Christoffel to map the region between F F and G G to a rectangle in the complex plane. While these methods appeared to have some advantages, it seems difficult to show that they really satisfy the required conditions. Anyway, I don't want to reinvent the wheel. real-analysis interpolation smoothness splines Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Improve this question Follow Follow this question to receive notifications edited Feb 20, 2021 at 7:15 gmvh 3,741 6 6 gold badges 33 33 silver badges 54 54 bronze badges asked Aug 18, 2011 at 8:50 Cristi StoicaCristi Stoica 4,322 4 4 gold badges 40 40 silver badges 59 59 bronze badges 4 Interesting problem! I believe some minor condition may be missing, since it seems to me that the problem as posed does not necessarily have a solution. Set for example F:[0,2)→R F:[0,2)→R, F\|[0,1)≡1 F\|[0,1)≡1, F\|[1,2)≡0 F\|[1,2)≡0 and G:[0,2)→R G:[0,2)→R, G\|[0,1)≡2 G\|[0,1)≡2, G\|[1,2)≡1 G\|[1,2)≡1. These two functions satisfy conditions 1-4 but a smooth family you're looking for would have to contain some functions discontinuous at point x=1 x=1 because of condition 2 which asks that the family be strictly increasing. Maybe the inequalities Dejan Govc –Dejan Govc 2011-08-23 16:07:15 +00:00 Commented Aug 23, 2011 at 16:07 How about something like this: fix ϕ ϕ a smooth symmetric nonnegative bump function with total integral one, extend F,G F,G by zero to all of R R, and then form f t(x)=(1−t)2 t∫R F(x+y)ϕ(y 1−t t)d y+t 2 1−t∫R G(x+y)ϕ(y t 1−t)d y.f t(x)=(1−t)2 t∫R F(x+y)ϕ(y 1−t t)d y+t 2 1−t∫R G(x+y)ϕ(y t 1−t)d y. This is clearly decreasing as a function of x x (differentiating under the integral sign gives nonpositive integrands), and it satisfies f 0=F,f 1=G f 0=F,f 1=G, but I haven't examined what properties of ϕ ϕ are required for monotonicity in t t to be true or plausible...David Hansen –David Hansen 2011-08-24 03:55:37 +00:00 Commented Aug 24, 2011 at 3:55 1 You may find something interesting in the so called Nurbs: en.wikipedia.org/wiki/Non-uniform_rational_B-splineKirill Shmakov –Kirill Shmakov 2012-03-01 17:44:23 +00:00 Commented Mar 1, 2012 at 17:44 Since my comment appears to have been cut in half during conversion, I'll add the ending here: "Maybe the inequalities for L L should be strict? Or maybe you should add to E E those points where F(x)F(x) equals the left or right limit of G G there?"Dejan Govc –Dejan Govc 2013-07-09 09:18:26 +00:00 Commented Jul 9, 2013 at 9:18 Add a comment\| 1 Answer 1 Sorted by: Reset to default This answer is useful 6 Save this answer. +250 This answer has been awarded bounties worth 250 reputation by Dejan Govc Show activity on this post. As noted by Dejan Govc, the set E E should also contain those points where the right limit of F(x)F(x) equals the left limit of G(x)G(x), because any continuous function bounded between F F and G G will be forced to the unique limit in these points. Let's first describe a construction for a family of continuous functions, which will later be refined to give a family of smooth functions. In the picture below, the parts where the smooth construction is identical to the continuous construction are drawn with bold lines. The drawn parts are f t(x)f t(x) for t=1 4 t=1 4, t=1 2 t=1 2 and t=3 4 t=3 4. The construction separates an interval into four different regions, either using the diagonal if F F and G G don't meet each other in the interval, or cutting the interval in half an using the diagonals of the subinterval where F F and G G don't meet. The piecewise linear family p t(x)=(1−t)F(x)+t G(x)p t(x)=(1−t)F(x)+t G(x) corresponds to a family of straight lines inside of each interval. This family either consists of parallel lines, or these lines meet in a common point which is not inside the interval. The bold lines in the image are the parts where we have f t(x)=p t(x)f t(x)=p t(x). Let y j:=lim x→x−j F(x)y j:=lim x→x j−F(x) be the relevant limit of F F at x j x j and Y j:=lim x→x+j G(x)Y j:=lim x→x j+G(x) be the relevant limit of G G. We have y j≤Y j y j≤Y j. We use f t(x j)=(1−t)y j+t Y j f t(x j)=(1−t)y j+t Y j. For the continuous construction, these two different parts of the construction are simply connected by straight lines. Let's denote the resulting family by f c o n t t f t c o n t. It's easy to check that this construction satisfies to requirements of the question, except that f c o n t t f t c o n t (and the corresponding ϑ c o n t x)ϑ x c o n t) is only continuous instead of smooth, and that the set E E had to be enlarged as described above. If we want to smooth out the continuous family f c o n t t f t c o n t to get a smooth family f t f t, we need to "specify" how f t f t should behave near the non-smooth points of f c o n t t f t c o n t. It's clear how it should behave near the parts with f t(x)=p t(x)=f c o n t t(x)f t(x)=p t(x)=f t c o n t(x), so let's focus on the regions around x j x j. If y j<Y j y j<Y j, we can use the family of straight lines generated by the straight line given by F F on the left side of x j x j and by the straight line given by G G on the right side of x j x j. If y j=Y j y j=Y j, we can use the linear function L L from assumption 4 to define how f 1/2 f 1/2 should behave near x j x j. Because f t f t should not cross f 1/2 f 1/2, we use L(x)+(t−1 2)x 2 L(x)+(t−1 2)x 2 to define how f t f t should behave near x j x j. At the left side, we see that line L L for the case y j=Y j y j=Y j can lead to problems with how we separated the interval into different regions. The image also suggests how these problems can be fixed. Regarding the missing details of this construction, I had started to explicitly construct the smoothing for the case y j<Y j y j<Y j, but inadvertently ignored the condition that ϑ x(t)ϑ x(t) should be a smooth function of t t instead of just a continuous function. For the case with y j=Y j y j=Y j, I haven't started to think about an explicit smoothing, and I also don't know for sure whether the behavior of f t f t around x j x j prescribed above will always allow such a smoothing (but a more "generic" linear function L L with strict inequalities should fix this, in case it really is a problem). Regarding references, this is one of the reasons why I decided to work out a detailed solution. A long time ago, I wrote a German text that starts by construction smooth and differential curves satisfying various conditions (these curves are used as "test curves" later in the text). That text cites normal introductory and slightly advanced analysis texts as main references. I really think this is the place where one gets taught how to construct curves adapted to various purposes. It is laborious to do so, but I fear no theory will save us from that. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Improve this answer Follow Follow this answer to receive notifications edited Oct 26, 2014 at 22:16 answered Jul 13, 2013 at 12:29 Thomas KlimpelThomas Klimpel 2,514 21 21 silver badges 43 43 bronze badges 1 The explicit smoothing construction I thought about constructs the first derivative. The piecewise linear functions gives me a constant derivative, which is the average value of the function I want to construct. The values at the borders are specified, and with a smooth function I can go from these values to a constant value close to the average value in any specified distance. Using a sufficient symmetric smooth function as building block, the exact values for the constant value only depend on the distance (and not the smooth function). A picture might add some value here, but...Thomas Klimpel –Thomas Klimpel 2013-07-16 07:35:06 +00:00 Commented Jul 16, 2013 at 7:35 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 real-analysis interpolation smoothness splines See similar questions with these tags. 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10312	https://webbook.nist.gov/cgi/cbook.cgi?InChI=1/CHNS.H3N/c2-1-3;/h3H;1H3	Ammonium thiocyanate Jump to content National Institute of Standards and Technology NIST Chemistry WebBook, SRD 69 Home Search Name Formula IUPAC identifier CAS number More options NIST Data SRD Program Science Data Portal Office of Data and Informatics About FAQ Credits More documentation Ammonium thiocyanate Formula: CH 4 N 2 S Molecular weight: 76.121 IUPAC Standard InChI:InChI=1S/CHNS.H3N/c2-1-3;/h3H;1H3 Copy IUPAC Standard InChIKey:SOIFLUNRINLCBN-UHFFFAOYSA-N Copy CAS Registry Number: 1762-95-4 Chemical structure: This structure is also available as a 2d Mol file Permanent link for this species. Use this link for bookmarking this species for future reference. Information on this page: Notes Other data available: Condensed phase thermochemistry data Phase change data IR Spectrum Data at other public NIST sites: X-ray Photoelectron Spectroscopy Database, version 5.0 Options: Switch to calorie-based units Data at NIST subscription sites: NIST / TRC Web Thermo Tables, professional edition (thermophysical and thermochemical data) NIST subscription sites provide data under the NIST Standard Reference Data Program, but require an annual fee to access. The purpose of the fee is to recover costs associated with the development of data collections included in such sites. Your institution may already be a subscriber. Follow the links above to find out more about the data in these sites and their terms of usage. Notes Go To:Top Data from NIST Standard Reference Database 69: NIST Chemistry WebBook The National Institute of Standards and Technology (NIST) uses its best efforts to deliver a high quality copy of the Database and to verify that the data contained therein have been selected on the basis of sound scientific judgment. However, NIST makes no warranties to that effect, and NIST shall not be liable for any damage that may result from errors or omissions in the Database. Customer support for NIST Standard Reference Data products. © 2025 by the U.S. Secretary of Commerce on behalf of the United States of America. All rights reserved. Copyright for NIST Standard Reference Data is governed by the Standard Reference Data Act. Privacy Statement Privacy Policy Security Notice Disclaimer (Note: This site is covered by copyright.) Accessibility Statement FOIA Contact Us
10313	https://wiki.roadsafetyanalysis.org/wiki/index.php?title=Double_counting	f Double counting - Wiki Double counting From Wiki Jump to: navigation, search Double counting is a phenomenon which sometimes occurs in report presentation, when the structure of a report results in a single item qualifiying for inclusion within more than one distinct category. As a consequence, any totals shown in reports which contain double counting may appear not to correspond to the report detail. It is the responsibility of any person using or presenting statistics to minimise the danger of misunderstandings which may arise from this technique. For a detailed academic discussion of double counting, see . Contents [hide] 1 Double counting in MAST 2 Example of double counting 3 Real life example of double counting Double counting in MAST Because all MAST reports are based on pre-calculated data cubes, double counting is impossible in MAST. The main technique used to avoid double counting, while still exposing the data to rigorous analysis, is the use of boolean values such as the top level of the Crash Involved Vehicle type dimensions. Example of double counting In road safety, a common example of double counting can occur when crashes are categorised by the types of vehicle involved in them. For instance, suppose that four crashes had occurred on a particular stretch of road, involving the following vehicles: a car and a motorcyle a single car a bus and a motorcycle a car and a pedal cycle This information could legitimately be reported by asserting that there were: three crashes involving cars two crashes involving motorcycles one crash involving buses one crash involving pedal cycles However if this report was carelessly presented, the unwary reader might jump to an incorrect conclusion: that there were seven crashes on the stretch of road, rather than only four. In addition, if the report showed the correct total of four, the same reader might notice that the individual values did not add up to the total and erroneously conclude that the information shown in the report was inaccurate or unreliable. Real life example of double counting Double counting is sometimes a valid approach to reporting certain sorts of information. for example, table 23c of RRCGB 2008 would report a crash involving a car and a motorcycle in two different places: once in cell H29, and again in cell G38. As a consequence, the totals reported in row 83 do not add up to the sum of the preceding individual values. Retrieved from " Views Page Discussion View source History Personal tools Log in / create account Navigation Main Page Community portal Current events Recent changes Random page Help Search Toolbox What links here Related changes Special pages Printable version Permanent link This page was last modified on 21 January 2011, at 11:26. This page has been accessed 83,025 times. Privacy policy About Wiki Disclaimers
10314	https://stackoverflow.com/questions/61856365/transpose-column-vector-to-row-vector	arrays - Transpose Column vector to row vector - Stack Overflow Join Stack Overflow By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google Sign up with GitHub OR Email Password Sign up Already have an account? Log in Skip to main content Stack Overflow 1. About 2. Products 3. 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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 Transpose Column vector to row vector Ask Question Asked 5 years, 4 months ago Modified5 years, 4 months ago Viewed 324 times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. I supposed to face with a simple task, but I finding some problems to transpose a 1D array/Column vector [0..n, 0..0] to a 1D array/Row vector [0..0, 0..n]. I tried using the Application.WorksheetFunction.Transpose built-in function without success. It only seems to work with nD array/matrix. The context is: - the 1D array/Column vector comes from a Recordset.GetRows method (if Recordset.Recordcount=1 => the array is a 1D array/Column vector) - the 1D array/Row vector (obtained by the transpose function) is used to populate the listbox.list property of a listbox object Is there a smart way to transpose a 1D array (from Column vector to Row vector and viceversa) ? Thanks in advance for any help arrays vba transpose Share Share a link to this question Copy linkCC BY-SA 4.0 Improve this question Follow Follow this question to receive notifications asked May 17, 2020 at 18:01 Jumpy73Jumpy73 115 1 1 silver badge 11 11 bronze badges Add a comment\| 2 Answers 2 Sorted by: Reset to default This answer is useful 0 Save this answer. Show activity on this post. Transpose Zero-Based Arrays The problem with Application.Transpose is that it transposes a 1D any-based one-row array to a 2D one-based one-column array. Now when you try to transpose back you will end up with a 1D one-based one-row array (see TransposeIssue). Toggle Transpose will 'recognize' if the array is vertical or horizontal and will transpose accordingly (see toggleTransposeTest). It will accept only zero-based arrays. The Code ```vbnet Option Explicit '''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''' ' Purpose: Transposes a 1D zero-based (one-row) array ' ' to a 2D zero-based one-column array and vice versa. ' '''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''' Function toggleTranspose0(SourceArray As Variant) As Variant Dim Transpose, i As Long On Error Resume Next i = UBound(SourceArray, 2) If Err.Number <> 0 Then On Error GoTo 0 If LBound(SourceArray) <> 0 Then Exit Function GoSub transposeVertical Else If i <> 0 Then Exit Function GoSub transposeHorizontal End If toggleTranspose0 = Transpose Exit Function transposeVertical: ReDim Transpose(UBound(SourceArray), 0) For i = 0 To UBound(SourceArray) Transpose(i, 0) = SourceArray(i) Next i Return transposeHorizontal: ReDim Transpose(UBound(SourceArray)) For i = 0 To UBound(SourceArray) Transpose(i) = SourceArray(i, 0) Next i Return End Function '''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''' Sub toggleTransposeTest() Dim v, t, i As Long ReDim v(9) ' Populate data to 1D array. For i = 0 To 9 v(i) = i + 1 Next i ' Transpose to 2D zero-based one-column array. t = toggleTranspose0(v) For i = 0 To 9 Debug.Print t(i, 0) Next i ' Transpose back to 1D array. v = toggleTranspose0(t) For i = 0 To 9 Debug.Print v(i) Next i End Sub Sub TransposeIssue() Dim v, t, i As Long ReDim v(9) ' Populate data to 1D zero-based one-row array. For i = 0 To 9 v(i) = i + 1 Debug.Print i, v(i) Next i ' Convert 1D array to a 1D one-based one-row array. t = Application.Transpose(Application.Transpose(v)) For i = 1 To 10 Debug.Print i, t(i) Next ' Transpose to 2D one-based one-column array. t = Application.Transpose(v) For i = 1 To 10 Debug.Print i, t(i, 1) Next ' Transpose to 1D one-based one-row array. v = Application.Transpose(t) For i = 1 To 10 Debug.Print i, v(i) Next End Sub ``` Share Share a link to this answer Copy linkCC BY-SA 4.0 Improve this answer Follow Follow this answer to receive notifications edited May 17, 2020 at 20:06 answered May 17, 2020 at 19:55 VBasic2008VBasic2008 56.7k 5 5 gold badges 21 21 silver badges 38 38 bronze badges 1 Comment Add a comment Jumpy73 Jumpy73Over a year ago Thanks a lot for putting me in the right direction... I don't know if this behaviour is a bug or only a poor design, but manipulating your code I wrote an UDF function which manage all transpose scenarios (I hope :) ) 2020-05-18T14:23:20.383Z+00:00 0 Reply Copy link This answer is useful 0 Save this answer. Show activity on this post. Starting from the code suggested by @VBasic2008, I post the UDF function I wrote to manage all transpose scenarios. Features: any-based arrays management no modification of the base of input array at the exit from the UDF function 2 options for managing 1D (one-row) array/1D (one-column) array ```vbnet Function Transpose(sAr As Variant, Optional Force2DOneRowArray As Boolean = True, Optional Force2DOneClmArray As Boolean = True) As Variant ''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''' ' Purpose: Transposes any-based array, manages in the right way the case of ' ' 2D (one-row) array/1D array to a 2D (one-column) array and vice versa ' ' Arguments: ' ' - sAr Source Array ' ' - Force2DOneRowArray Force function to transpose 2D matrix [n x 0]/1D (one-column) array to ' ' 2D matrix [0 x n]/1D (one-row) array rather than to a simple 1D array ' ' - Force2DOneClmArray Force function to transpose 2D matrix [0 x n]/1D (one-row) array to ' ' 2D matrix [n x 0]/1D (one-column) array rather than to a simple 1D array ' ''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''' Dim tAr As Variant Dim i As Long, j As Long On Error Resume Next i = UBound(sAr, 2) If Err.Number <> 0 Then '1D (one-row) array --> Vertical transpose On Error GoTo 0 ReDim tAr(LBound(sAr) To UBound(sAr), 0) For i = LBound(sAr) To UBound(sAr) tAr(i, 0) = sAr(i) Next i Else '2D array If i <> 0 Then If UBound(sAr) <> 0 Then '2D matrix [n x m] ReDim tAr(LBound(sAr, 2) To UBound(sAr, 2), LBound(sAr) To UBound(sAr)) For i = LBound(sAr, 2) To UBound(sAr, 2) For j = LBound(sAr) To UBound(sAr) tAr(i, j) = sAr(j, i) '2D matrix [n x m] --> 2D matrix [m x n] Next j Next i Else '2D matrix [0 x n]/1D (one-row) array --> Vertical transpose If Force2DOneClmArray Then ReDim tAr(LBound(sAr, 2) To UBound(sAr, 2), 0) For i = LBound(sAr, 2) To UBound(sAr, 2) tAr(i, 0) = sAr(0, i) '2D matrix [n x 0]/1D (one-column) array Next i Else ReDim tAr(LBound(sAr, 2) To UBound(sAr, 2)) For i = LBound(sAr, 2) To UBound(sAr, 2) tAr(i) = sAr(0, i) '1D array Next i End If End If Else '2D matrix [n x 0]/1D (one-column) array --> Horizontal transpose If Force2DOneRowArray Then ReDim tAr(0, LBound(sAr) To UBound(sAr)) For i = LBound(sAr) To UBound(sAr) tAr(0, i) = sAr(i, 0) '2D matrix [0 x n]/1D (one-row) array Next i Else ReDim tAr(LBound(sAr) To UBound(sAr)) For i = LBound(sAr) To UBound(sAr) tAr(i) = sAr(i, 0) '1D array Next i End If End If End If Transpose = tAr End Function ``` Share Share a link to this answer Copy linkCC BY-SA 4.0 Improve this answer Follow Follow this answer to receive notifications answered May 18, 2020 at 14:31 Jumpy73Jumpy73 115 1 1 silver badge 11 11 bronze badges Comments Add a comment Your Answer Thanks for contributing an answer to Stack Overflow! 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10315	https://www.cgm.northwestern.edu/cores/nuseq/services/bioinformatics/rna-seq-analysis.html	Northwestern University Feinberg School of Medicine Center for Genetic Medicine Skip to main content RNA-Seq Analysis RNA-seq is an approach to estimate transcript abundance by sequencing the transcriptome of a cell type or tissue. It is primarily used to identify genes that are either upregulated or downregulated in relation to a control. Deliverables Differential Gene Expression: For each comparison, we provide a spreadsheet containing all the genes from the annotation file. Included in the spreadsheet is the average normalized gene expression value for each gene along with the log fold change, p-value, and FDR-adjusted p-value. Principle Component Analysis (PCA): PCA is an analytical technique that clusters samples based on their gene expression profiles. We run a PCA analysis on all samples in the experiment to evaluate how similar biological replicates are to one another. There is also a PCA analysis performed for each individual comparison. Hierarchical Clustering: Similar to a PCA, hierarchical clustering arranges samples on a dendrogram according to their gene expression profiles. This is another effective way to evaluate how similar samples are to one another. Clustering is also performed for each individual comparison. Heatmap of Differentially Expressed Genes: For each comparison, a heatmap is created containing only differentially expressed genes. Differential expression is determined using an FDR-adjusted p-value less than 0.05. However, if there are fewer than 20 genes based on the FDR-adjusted value, the heatmap will include genes with a regular p-value less than 0.05. Volcano Plot: This is a plot of the log2 fold change versus the –log10(p-value). Genes that are significantly upregulated or downregulated are highlighted red. Optional Analysis Motif Analysis Given a list of differential genes, a motif analysis will extract their known promoters and search for similar sequences. This predictive analysis might identify possible transcription factors that may be responsible for the observed changes in gene expression. Small RNA-Seq / microRNA-Seq In a small RNA-seq (or microRNA-seq) experiment, short (< 200 nt) transcripts are selected during library construction. The workflow is similar to an mRNA-seq experiment with an additional option. Instead of aligning to the reference genome, reads from a small RNA-seq library may be aligned directly to the mature small RNA sequences, such as those found in miRBase. After alignment, a count file is generated which represents the number of reads that have aligned to each small RNA. From there, differential expression can be performed on the small RNA count files. Non-coding RNA-Seq / Total RNA-Seq Small RNAs are non-coding, but because of their short length, they must be processed differently during library construction. However, other non-coding transcripts can be assessed by RNA-seq, particularly long non-coding RNA. Since many non-coding RNA molecules lack a poly-A tail, ribosomal RNA is removed with capture probes rather than with oligo-dT columns. Show Details What quality checks are performed on RNA before library construction? Each RNA sample that is submitted to the core undergoes a Bioanalyzer analysis to determine the integrity of the RNA. A RIN score is generated for each sample, ranging from 0 to 10. A RIN score of 7 or higher indicates that the RNA sample is of sufficient quality to proceed with library construction. How much does an RNA-seq project cost? The billing of most RNA-seq projects can be divided into 3 parts. First is library construction. The charge for library construction is per sample, and depends on how much RNA is in your sample. The second part is sequencing, which is charged per lane or per flowcell as opposed to per sample. Because the charge is per lane or flowcell, the more samples that are sequenced in the project will result in lower per-sample costs. The third part is analysis. This is charged by the hour and the cost will depend on the type and extent of analysis requested. What does each column of the differential expression spreadsheet mean? Ensembl ID: the gene ID provided by Ensembl. Gene Symbol: the symbol that is generally attributed to the particular gene. Entrez ID: the gene number assigned by Entrez, which is part of NCBI. Description: the gene name, which is more descriptive than the gene symbol. Location: the chromosomal location of the gene, from beginning to end including introns. Strand: the strand on which the gene is found, (+ or -). Log2 Fold Change: the log2 fold change (LFC) of the gene in the experimental group relative to the control group. Positive LFC values indicate upregulation relative to control, while negative LFC values indicate downregulation. In naming the spreadsheet, the control group is listed first and the experimental group second, such as “Control_vs_Experimental_Differential_Expression.xlsx”. It’s important to note that DESeq2 computes LFC values differently than other programs. DESeq2 applies a shrinkage estimation that reduces initial LFC calculations depending on the expression of the gene. Therefore, highly expressed genes are reduced slightly, while lowly expressed genes are reduced more. LFC Standard Error: because the LFC values are determined through a shrinkage estimation, the LFC Standard Error exists to express the confidence in the LFC calculation. Wald Statistic: the statistic used to calculate the p-value, much like a t-statistic for t-tests. The greater the difference between the Wald statistic and 0, the lower the p-value. p Value: a p-value is calculated for every gene, reflecting the probability of incorrectly rejecting the null hypothesis that the mean expression of the gene is not different between the two groups. FDR Adj p Value: the p-value after correction for the false discovery rate. This is the p-value most often used to determine significance. Significant: a simple yes/no designation whether the FDR-adjusted p-value is less than 0.05. Status: an indication of whether there were sufficient reads to confidently determine the expression of the gene. This field will be “OK” if there are at least ten reads mapping to the gene, “LOW” otherwise. Base Mean: the mean expression of the gene across both groups. Control Group Mean: the mean expression of the gene in the control group. The label of the actual control group will be displayed in the spreadsheet. Experimental Group Mean: the mean expression of the gene in the experimental group. The label of the actual experimental group will be displayed in the spreadsheet. Individual Expression: the remainder of the columns indicate the gene expression in each sample. What is the difference between the regular p-value and the FDR-adjusted p-value, and which one should I use? During differential gene expression, the mean expression values of each gene are compared between each group in a test similar to a t-test. As a result, a p-value is calculated for each test/gene. A p-value cutoff of 0.05 indicates that there is a 5% of incorrectly rejecting the null hypothesis. For a single test, this is acceptable. However, an RNA-seq analysis involves several thousands tests, one for each gene. Therefore, as more tests are performed, the likelihood that some null hypotheses are incorrectly rejected increases. For example, image a situation in which there are 1,000 differentially expressed genes, all with a p-value of 0.05. On average, we can expect 5%, or 50 genes, are found to be significant when they are actually not. Therefore, each p-value is adjusted using a calculated false discovery rate (FDR), which is analogous to a post-hoc test. The most common cutoff to determine significance is an FDR-adjusted p-value of 0.05 or less. Does NUSeq offer a pathway analysis too? Yes. A pathway analysis will identify common functions or pathways among significant genes. Most pathway analysis tools are free, web-based, and simply require a list of genes. One pathway analysis tool that we use often in NUSeq is MetaScape, which is fast and easy to use. Can NUSeq help me submit my project to GEO (Gene Expression Omnibus)? Yes. If you wish to submit your RNA-seq data to GEO, first download and complete this spreadsheet up to and including the row labeled "extract protocol". This is row 29 on a blank form, but it may change if additional rows are added above it. For guidance, consult the tab labeled "EXAMPLE 2" on the spreadsheet. The core will complete the rest of the form. If the sequencing was performed in NUSeq, then we already possess the raw data, but if it was sequenced elsewhere, the raw FASTQ files must be provided to the core. After completing the spreadsheet, submit it to the core along with the following: The GEO user account. Every GEO submission must be associated with a GEO account, which will be designated as the owner of the data. Individuals within NUSeq should not use their own GEO accounts for this purpose, because we do not own the data. The date on which you would like the data made public. This date should be far out enough to ensure that the manuscript will be published. A good recommendation is between 6 and 9 months. The release date may be changed, so a good suggestion would be to over-estimate, and then release the data to the public earlier if necessary. The genome of my organism is not well characterized and the genes are not well annotated, how will that impact the analysis? Many bioinformatics analyses depend on a well-annotated reference genome. These are usually readily available for model organisms such as human and mouse. If your organism is not well annotated, the results may be difficult to interpret. For example, the final results of an RNA-seq project will use the gene names provided in the gene annotation file. If the gene annotation file uses gene names that are unfamiliar to the user, the results will be difficult to interpret. For non-model organisms, we suggest becoming familiar with the gene models available and indicating to NUSeq exactly which gene model you would prefer us to use. If you're unsure, please contact us. Can I use RNA-seq to identify differential isoforms or alternative splicing? Yes, but there are caveats. NUSeq currently only houses "short read" sequencing instruments. Short read sequencing platforms are suitable for measuring gene expression, but are not ideal for isoform expression. If you require isoform expression, a long read sequencing platform such as Nanopore or PacBio is preferable. However, if only short read platforms are available, we suggest sequencing with PE100 or PE150 with 50 million to 100 million reads per sample. There are bioinformatics workflows available that will accommodate isoform expression, but their reliability is not certain and we recommend validating any interesting results. Does NUSeq perform RNA extraction too? No. NUSeq requires total RNA to be submitted for RNA-seq services (with the exception of single cell RNA-seq). However, RNA extraction is offered by the Pathology Core. Back to top Follow Center for Genetic Medicine on TwitterFlickr
10316	https://math.fandom.com/wiki/Equivalent_fractions	Equivalent fractions \| \| \| In order to understand the information on this page, it can be helpful to already know: Fractions Multiply Fractions Factors and Multiples \| Contents Outline[] Equivalent fractions proportions expressed as one fraction to be expressed with a different fraction of the same value. For example 1 3 {\displaystyle \frac{1}{3}} is equal in value to 2 6 {\displaystyle \frac{2}{6}} as shown below: Arithmetically, this is achieved by multiplying the fraction by a fraction that has the same numerator and denominator (so is equal to 1). In this case, we multiply by 2 2 {\displaystyle \frac{2}{2}} : 1 3 × 2 2 = 1 × 2 3 × 2 = 2 6 {\displaystyle {\frac {1}{3}}\times {\frac {2}{2}}={\frac {1\times 2}{3\times 2}}={\frac {2}{6}}} Graphically, this represents all pieces (the denominator) including the shaded pieces (the numerator) into two pieces each. This gives twice as many pieces in total (six) and twice as many pieces shaded (two). Worked Examples[] Sample question:[] Express 2 3 {\displaystyle \frac{2}{3}} as twelfths (ie. as a fraction over 12) Solution:[] To convert thirds into twelfths, each piece must be divided into four because 3 × 4 = 12 {\displaystyle 3 \times 4 = 12 } . We can achieve this by multiplying the original fraction by 4 4 {\displaystyle \frac{4}{4}} 2 3 × 4 4 = 2 × 4 3 × 4 = 8 12 {\displaystyle {\frac {2}{3}}\times {\frac {4}{4}}={\frac {2\times 4}{3\times 4}}={\frac {8}{12}}} This shows that 2 3 {\displaystyle \frac{2}{3}} expressed as twelfths is 8 12 {\displaystyle \frac{8}{12}} This can be confirmed visually using the diagram below: Practice[] You can practice skills in percent using the following references: Khan Academy[] Make sure you are logged in to Khan Academy when doing these exercises so your practice is recorded. Application[] Equivalent fractions are used whenever objects or quantities can be broken down into different sized parts. Equivalent fractions can help you to work out how many pieces you would need for each different size to get the same total amount. For example: Next Steps[] Comfortable with Equivalent Fractions? Check out the follow-on mathematics you can now do: Fandom logo Explore properties Follow Us Overview Community Advertise Fandom Apps
10317	https://bmccancer.biomedcentral.com/articles/10.1186/s12885-024-13242-1	BMC Cancer Treatment options for unilateral vestibular schwannoma: a network meta-analysis Download PDF Download PDF Systematic Review Open access Published: Treatment options for unilateral vestibular schwannoma: a network meta-analysis Xianhao Huo1, Xu Zhao2, Xiaozhuo Liu2, Yifan Zhang1, Jihui Tian1 & … Mei Li3 BMC Cancer volume 24, Article number: 1490 (2024) Cite this article 1470 Accesses Metrics details Abstract This study aimed to explore the effect of observation, microsurgery, and radiotherapy for patients with vestibular schwannoma (VS). We searched PubMed, Medline, Embase, Web of Science, and Cochrane library from their establishment to July 31, 2024. 34 non-RCTs and 1 RCT that included 6 interventions were analyzed. We found the MS, and different SRS all had better tumor local control rates. Regarding preserved hearing, the order from the highest to the lowest was FSRT 5 fractions, FSRT 3 fractions, SRS, ConFSRT, Observation, and MS. Regarding improvement in the rate of tinnitus, the order from the highest to the lowest was ConFSRT, FSRT 3 fractions, SRS, Observation, MS, and FSRT 5 fractions. In terms of improving the rate of disequilibrium/vertigo, the order from the highest to the lowest was SRS, Observation, FSRT 3 fractions, FSRT 5 fractions, MS, and ConFSRT. In terms of protection of the trigeminal nerve, the order from the highest to lowest was observation, SRS, ConFSRT, FSRT 3 fractions, FSRT 5 fractions, and MS. Lastly, in terms of protection of the facial nerve, the order from the highest to lowest was SRS, ConFSRT, Observation, FSRT 3 fractions, FSRT 5 fractions, and MS. In patients with VS, MS and radiosurgery showed better local tumor control rates; however, compared with MS, different SRS all provided better protection of nerve function and improved the symptoms of vestibular function and tinnitus, among which the best was SRS. Therefore, in these patients, SRS may be a promising alternative treatment. Peer Review reports What’s new? Current treatment options for VS are divided into three categories as follows: microsurgery, radiotherapy, and observation. Unfortunately, all these options have inherent risks, and the vestibular function is inevitably compromised. Moreover, in the literature, there is a relative lack of relevant studies directly comparing these three treatment regimens. In this study, we conducted a network meta-analysis to explore the advantages and disadvantages of surgery, radiotherapy, and observation. It found that in patients with VS, MS and radiosurgery showed better local tumor control rates than observation; however, compared with MS, different SRS all provided better protection of nerve function and improved the symptoms of vestibular function and tinnitus, among which the best was SRS. Therefore, in patients with unilateral VS, SRS may be a promising alternative treatment. Background Vestibular schwannoma (VS), also known as acoustic neuroma, is a benign Schwann cell-derived tumor originating from the vestibulocochlear nerve . It accounts for approximately 8% of intracranial tumors and is the most common tumor in the cerebellopontine angle region, with an annual incidence of 10.4/1000,000 . With the advances in imaging technology and increased access to magnetic resonance imaging, more VS cases are being diagnosed, allowing for the detection of VS at an early stage when the tumor is smaller [3, 4]. Owing to its deep location, which is adjacent to the brain stem and cerebellum, and its close relationship with important structures such as the trigeminal nerve, facial nerve, cochlear nerve, and posterior cranial nerve, when the tumor, including benign ones, persists or grows continuously, it can become invasive and compress the surrounding structures. This results in the development of various clinical symptoms, with more than 60% of patients with VS having progressive hearing loss and tinnitus . Large tumors can also cause hydrocephalus and brain stem compression, which can lead to facial paresthesia, vertigo, headache, and other symptoms, thus seriously affecting the patient’s daily activities and quality of life [6, 7]. Current treatment options for VS are divided into three categories as follows: microsurgery (MS), radiotherapy, and observation [8,9,10,11,12]. Unfortunately, all these options have inherent risks, and the vestibular function is inevitably compromised, as the tumor originates from the vestibular nerve. In particular, in radiotherapy, tumors and their potentially deleterious effects on vestibular function are not eliminated, and the simultaneous exposure of vestibular organs and vestibular nerves to radiation may cause additional vestibular toxicity [13, 14]. In observation treatment, the tumor is left in situ, and vestibular function may deteriorate owing to the natural course of the disease. Therefore, the main goal of VS treatment is not only to improve vestibular symptoms but also to prevent future complications due to tumor progression. Nonetheless, the reported effects of VS treatment on vestibular symptoms are diverse. Some studies have reported no significant difference in the development of vestibular symptoms in patients with VS undergoing surgery, radiotherapy, or observation treatment, whereas others have noted a reduction in the incidence of vestibular symptoms after surgery [15, 16]. In the long term, because of the central compensation of unilateral VS, regardless of the treatment strategy used, vestibular symptoms are expected to be relieved. However, the current treatment options for these symptoms are mainly based on the tumor size, presence of tumor growth, age of the patient, hearing ability, complications, and patient selection. These three treatment options are supported by level II or III quality of evidence only, and there is a lack of high-level evidence to support which treatment option is the most advantageous . Moreover, in the literature, there is a relative lack of relevant studies directly comparing these three treatment regimens. This study aimed to conduct a network meta-analysis to explore the advantages and disadvantages of surgery, radiotherapy, and observation as treatment options for VS patients from the aspects of tumor control rate and vestibular function. Patients and methods Study design The systematic review and network meta-analysis were performed according to the Preferred Reporting Items for Systematic Reviews and Meta-analyses (PRISMA) checklist for network meta-analysis. This network meta-analysis study was registered with PROSPERO under the registration number: CRD42024574320. Search strategy The following retrieval formula was used: (((((((((((((((((((single dose radiosurgery) OR (single dose)) OR (radiosurgical)) OR (radiosurgically)) OR (stereotactic radiation therapy)) OR (stereotactic radiation therapies)) OR (stereotactic radiotherapy)) OR (stereotactic radiotherapies)) OR (stereotactic radiosurgery)) OR (stereotactic radiosurgeries)) OR (gamma knife)) OR (linacs)) OR (particle)) OR (accelerators)) OR (particle accelerators)) OR (linac)) OR (linear accelerator)) OR (CyberKnife)) OR (radiosurgery)) AND ((((((((((((vestibular schwannoma) OR (vestibular schwannomas)) OR (acoustic neuroma)) OR (acoustic neuromas)) OR (acoustic schwannoma)) OR (acoustic schwannomas)) OR (acoustic neurilemoma)) OR (acoustic neurilemomas)) OR (acoustic tumor)) OR (acoustic tumors)) OR (acoustic neurinoma)) OR (acoustic neurinomas)). Medline, PubMed, Web of Science, Embase, and Cochrane Library were searched from their establishment until July 31, 2024. Moreover, we also manually searched the published articles (such as, systematic reviews, meta-analysis), some unpublished trials, which was registried in the World Health Organization clinical registries, and so on. Inclusion and exclusion criteria We made the criteria of inclusion and exclusion, which were based on the PICOS strategy (P: patient/population, I: intervention, C: comparison/control, O: outcome, S: study design). In terms of patients, the aged ≤ 70 years with newly diagnosed unilateral VS. In terms of interventions, studies with treatment options, including involved stereotactic radiosurgery (SRS), fractionated stereotactic radiotherapy (FRST), gamma knife surgery (GKS), microsurgery (MS), and conservative management (CM), were included. In terms of study design, randomized controlled clinical trials or non-randomized controlled clinical trials were included. In terms of outcomes: the outcome indicators were local control rate of tumor, preserved hearing, trigeminal nerve toxicity, facial nerve toxicity, tinnitus, vertigo/disequilibrium. The exclusion criteria were as follows: type 2 neurofibromatosis, trials including patients with other types of tumors (such as meningeoma, craniopharyngioma, metastatic encephaloma, etc.), pregnant women, lactating patients, patients had severe complications and could not tolerate treatment, single-arm trials, single case reports, protocol, and animal experiments. Study endpoints The outcome measures were local control rate of the tumor, preserved hearing, trigeminal nerve toxicity, facial nerve toxicity, tinnitus, vertigo/disequilibrium. Data screening and quality evaluation The literature-retrieval results were screened by two reviewers, independently. And then, they assessed all randomized trials based on six aspects by using the Cochrane quality evaluation method, and evaluated all non-randomized trials by using the Newcastle–Ottawa Scale (NOS). If there had some problems or disagreements during the process, we can resolve by two reviewers discussion or consulted by a third person. At last, we also assessed the evidence quality in our analysis by using Grading of Recommendations Assessment and Development and Evaluation (GRADE) framework. Data extraction The data of all included studies were extracted, including the name of first or corresponding author, the year of publication, nation, study type, age of the patients, intervention indicators, total number in each intervention group, the treatment details, and local control rates and rates of preserved hearing, trigeminal nerve toxicity, facial nerve toxicity, tinnitus, and disequilibrium/vertigo in patients with VS in different treatment groups. If data were missing, we contacted the authors of this study wherever possible. Statistical analysis Before making the network meta-analysis, we performed a heterogeneity test, transitivity and consistency test for all the included studies. The fixed-effects model was adopted when p > 0.1 and I2 < 50%, which means the results were non-heterogeneous. Otherwise, the heterogeneous was adopted the random-effects model. The clinical and methodological variables (such as: sex, age, the percent of male, the tumor size and the tumor volume) were compared between the different interventions for transitivity. We also used a two-tailed statistical test, p < 0.05 was considered statistically significant. In the network meta-analysis, the surface under the cumulative ranking curve was used to rank the outcome measures. We also performed a consistency test by using the node-splitting method. The software of RevMan version 5.3 (Cochrane Collaboration, London, UK) and Stata 16.0 (StataCorp, Texas, USA) were performed to statistical analyses. Results Literature search results Figure 1 presents a flow diagram of the literature search and inclusion of relevant studies. The initial search identified 10,820 studies, of which 8642 duplicates were excluded by screening the title and abstract. Then, 2178 studies were screened by reading the research objective and study type, as a result, 1985 studies were excluded (not relevant, letter to editors or commentary, review articles, animal experiments, and case reports, and so on). Of the 193 studies, 110 were excluded because they were not published in English, lacking the main outcome indicators, not control group, and so on. Finally, after excluding 48 studies due to without outcome indicators, inappropriate results, included patients with other tumors. 35 were analyzed in this network meta-analysis. All were retrospective or prospective studies [6,7,8,9,10,11,12,13,14, 18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42], except for one randomized control trial . Of the included studies, 15 were published in the USA, 5 in Germany, 3 in Norway, 2 in Canada, 2 in Korea, 1 in France, 1 in Netherlands, 1 in Thailand, 1 in Australia, 1 in Belgium, 1 in Japan, 1 in Singapore, and 1 in Sweden. The percent of sex in our analysis also has no statistically significant difference (95% CI: [− 22.52, 11.49], p > 0.05). More details on the included studies, including the study design, publication year, and type of interventions, are shown in Tables 1 and 2. The total sample size was 5069 cases and included 6 interventions (SRS, ConFSRT, FSRT 5 fractions, MS, FSRT 3 fractions, and observation treatment). Quality evaluation The non-randomized trials (n = 34) were assessed using the NOS based on selection, comparability, and outcome, with a total score of > 5, indicating high quality (Table 3). Meanwhile, the randomized controlled trial (RCT) was assessed by the Cochrane risk of bias tool, revealing the use of the correct randomization method, use of the correct blinding method in the conduct of the study and the assessment of the results, complete outcome data, and no selective reporting. Therefore, the included RCT trial was of high quality. Traditional meta-analysis Subgroup analysis was performed on 30 studies reporting local tumor rates in patients with VS after treatment with different interventions (Supplementary Fig. 1a). Owing to significant heterogeneity, a random-effects model was used (I2 > 50%, p < 0.1). Compared with FSRT 5 fractions (RR: 1.01, 95% CI: 0.98–1.04, P = 0.61), FSRT 3 fractions (RR: 0.97, 95% CI: 0.86–1.09, P = 0.60), ConFSRT (RR: 0.99, 95% CI: 0.97–1.02, P = 0.68), and MS (RR: 1.02, 95% CI: 0.98–1.06, P = 0.32), SRS was not significantly associated with higher local tumor rates. There was also no significant difference when MS compared with FSRT 3 (RR: 1.00, 95% CI: 0.86–1.17, P = 1.00) and FSRT 5 fractions (RR: 1.04, 95% CI: 0.90–1.21, P = 0.58), respectivelly and between those undergoing ConFSRT and FSRT 5 fractions (RR: 1.00, 95% CI: 0.90–1.12, P = 0.96). However, compared with the patients receiving observation treatment, those receiving SRS (RR: 1.82, 95% CI: 1.57–2.12, P < 0.00001) and ConFSRT (RR: 1.86, 95% CI: 1.25–2.76, P = 0.002) had better local tumor control rates. Subgroup analysis was performed on 26 studies reporting preserved hearing in patients with VS after treatment with different interventions (Supplementary Fig. 1b). Owing to significant heterogeneity, a random-effects model was used (I2 > 50%, p < 0.1). Compared with FSRT 5 fractions (RR: 1.04, 95% CI: 0.92–1.18, P = 0.49), FSRT 3 fractions (RR: 0.98, 95% CI: 0.53–1.79, P = 0.94), ConFSRT (RR: 0.91, 95% CI: 0.76–1.08, P = 0.28), and observation treatment (RR: 0.90, 95% CI: 0.76–1.07, P = 0.23), SRS was not significantly associated with the preservation of hearin. Compared with the observation treatment and FSRT 5 fractions (RR: 0.89, 95% CI: 0.56–1.42, P = 0.63), ConFSRT (RR: 1.74, 95% CI: 0.87–3.48, P = 0.12) showed no significant difference. However, compared with the observation treatment, MS was significantly associated with improvement in preserved hearing (RR: 0.56, 95% CI: 0.43–0.73, P < 0.0001). Subgroup analysis was performed on 16 studies reporting tinnitus in patients with VS after treatment with different interventions (Supplementary Fig. 1c). Owing to significant heterogeneity, a random-effects model was used (I2 > 50%, p < 0.1). Compared with FSRT 3 fractions (RR: 1.05, 95% CI: 0.37–2.98, P = 0.92), FSRT 5 fractions (RR: 0.87, 95% CI: 0.55–1.37, P = 0.54), ConFSRT (RR: 0.90, 95% CI: 0.75–1.07, P = 0.24), MS (RR: 1.27, 95% CI: 0.91–1.78, P = 0.16), and observation treatment (RR: 1.03, 95% CI: 0.86–1.24, P = 0.74), SRS was not significantly associated with the occurrence of tinnitus. There was also no significant difference in the development of tinnitus between patients treated with MS and observation treatment (RR: 1.06, 95% CI: 0.59–1.89, P = 0.85) and between those undergoing ConFSRT and FSRT 5 fractions (RR: 0.57, 95% CI: 0.17–1.88, P = 0.35). Subgroup analysis was performed on 16 studies reporting disequilibrium/vertigo in patients with VS following different treatment interventions (Supplementary Fig. 1d). Owing to significant heterogeneity, a random-effects model was used (I2 > 50%, p < 0.1). Compared with FSRT 3 fractions (RR: 0.68, 95% CI: 0.33–1.37, P = 0.28), FSRT 5 fractions (RR: 0.67, 95% CI: 0.39–1.16, P = 0.15), MS (RR: 1.18, 95% CI: 0.86–1.62, P = 0.31), ConFSRT (RR: 0.81, 95% CI: 0.56–1.18, P = 0.28), and observation treatment (RR: 1.07, 95% CI: 0.74–1.54, P = 0.71), SRS was not significantly associated with the development of disequilibrium/vertigo. There was no significant difference in the occurrence of disequilibrium/vertigo between patients treated with MS and observation treatment (RR: 1.57, 95% CI: 0.96–2.56, P = 0.07) and between those treated with ConFSRT and FSRT 5 fractions (RR: 1.15, 95% CI: 0.13–10.13, P = 0.90). Subgroup analysis was performed on 15 studies reporting trigeminal nerve toxicity in patients with VS undergoing different treatment interventions (Supplementary Fig. 1e). Owing to unobserved heterogeneity, a fixed-effects model was used (I2 < 20%, p > 0.1). Compared with FSRT 3 fractions (RR: 0.63, 95% CI: 0.16–2.46, P = 0.51), FSRT 5 fractions (RR: 0.86, 95% CI: 0.38–1.94, P = 0.71), ConFSRT (RR: 1.37, 95% CI: 0.86–2.19, P = 0.18), and MS (RR: 0.70, 95% CI: 0.16–3.04, P = 0.63), SRS was not significantly associated with the development of trigeminal nerve toxicity. There was no significant difference in the occurrence of trigeminal nerve toxicity between patients undergoing ConFSRT and FSRT 5 fractions (RR: 0.68, 95% CI: 0.16–3.02, P = 0.62) or observation treatment (RR: 1.04, 95% CI: 0.67–1.62, P = 0.87). Lastly, subgroup analysis was performed on 19 studies reporting facial nerve toxicity in patients with VS following different treatment interventions (Supplementary Fig. 1f). Owing to significant heterogeneity, a random-effects model was used (I2 > 20%, p < 0.1). Compared with FSRT 3 fractions (RR: 0.45, 95% CI: 0.13–1.62, P = 0.22), FSRT 5 fractions (RR: 0.84, 95% CI: 0.38–1.86, P = 0.66), ConFSRT (RR: 1.64, 95% CI: 0.95–2.84, P = 0.07), and observation treatment (RR: 0.18, 95% CI: 0.02–1.50, P = 0.11), SRS was not significantly associated with the development of facial nerve toxicit. However, compared with MS, SRS (RR: 0.12, 95% CI: 0.02–0.72, P = 0.02) could reduce the incidence of facial nerve toxicity. There was no significant difference in the occurrence of facial nerve toxicity between ConFSRT and FSRT 5 fractions (RR: 0.21, 95% CI: 0.01–4.11, P = 0.30) or observation treatment (RR: 0.68, 95% CI: 0.04–10.28, P = 0.78). The assessment of evidence quality in this analysis We also evaluated the quality in this analysis by using the GRADE. We found that the credibility of the evidence was moderate or very low in the aspects of local tumor rates, improved preserved hearing, improved tinnitus, improved disequilibrium/vertigo, reduced tigeminal nerve toxicity, and reduced facial nerve toxicity for the observation study. The reason for these ratings of evidence quality were mainly driven by the design of study and the indirectness comparisons between several interventions (supplementary Table 1a). The credibility of the evidence was moderate in the aspects of local tumor rates, improved preserved hearing, improved tinnitus, and improved disequilibrium/vertigo for the RCT. The reason for these ratings of evidence quality were mainly due to the indirectness comparisons between several interventions and the small sample size (supplementary Table 1b). Network meta-analysis Network diagram of different intervention measures A direct comparison is shown if there is a direct line between the two intervention groups, otherwise, the absence of lines indicates a lack of evidence for direct comparison. The size of the dots represents the sample size, and the thickness of the lines represents the number of study items. The SRS, ConFSRT, and FSRT 5 fractions had the largest sample sizes and the largest number of entries in direct or indirect comparative trials (Fig. 2). Transitivity In this analysis, we also made a transitivity in the aspects of percent of male, age, tumor size, and tumor volume (supplementary Fig. 2a-d). We found that, there were no significant differences in baseline of the percentage of male patients, mean age, and tumor size for most comparisons. And there were slightly differences in baseline tumor volume for most comparisons. Inconsistency test Because both direct and indirect evidence were available, an inconsistency test was required, prior to the integration analysis. We found no inconsistency in the network meta-analysis, and the difference was not statistically significant (p > 0.05), that is, the direct and indirect evidence included could be combined (Supplementary Fig. 3). Sequence diagram The rate of tumor local control from the highest to lowest was MS, FSRT 3 fractions, ConFSRT, FSRT 5 fractions, SRS, and observation treatment, indicating that except for the observation treatment, the rate of tumor local control was good in the other five interventions (Fig. 3a). In terms of preserved hearing, the order from highest to lowest was FSRT 5 fractions, FSRT 3 fractions, SRS, ConFSRT, observation treatment, and MS, indicating that SRS is associated with a higher rate of preserved hearing than observation treatment (Fig. 3b). Regarding the improved rate of tinnitus, the order from the highest to lowest was ConFSRT, FSRT 3 fractions, SRS, Observation, MS, and FSRT 5 fraction (Fig. 3c). This indicates that radiosurgery is associated with a lower rate of tinnitus. Regarding the improved rate of disequilibrium/vertigo, the order from the highest to lowest was SRS, observation treatment, FSRT 3 fractions, FSRT 5 fractions, MS, and ConFSRT. SRS had a lower rate of disequilibrium/vertigo than the observation treatment, and the other four interventions had a higher rate (Fig. 3d). In terms of improving trigeminal nerve toxicity, the order from the highest to lowest was observation treatment, SRS, ConFSRT, FSRT 3 fractions, FSRT 5 fractions, and MS. Compared with the observation treatment, the other four interventions had higher trigeminal nerve toxicity (Fig. 3e). Lastly, in terms of improving facial nerve toxicity, the order from the highest to lowest was SRS, ConFSRT, observation treatment, FSRT 3 fractions, FSRT 5 fractions, and MS (Fig. 3f). Compared with MS, SRS had better facial nerve protection. Discussion VS is a benign but potentially devastating tumor with aggressive growth and is associated with significant complications (including deafness and lesions of the facial nerve) . Thus, it is necessary to seek effective interventions for VS. Historically, surgical resection has been a valued treatment option because complete resection represents the greatest degree of tumor control . However, owing to the anatomical relationship between VS and multiple cranial nerves, surgical resection needs to be very precise and fine. Some studies have shown that the decline in cranial nerve function after surgical treatment can have serious clinical consequences in patients with VS . Hence, researchers have explored other alternative treatment options, such as SRS, surgical excision is different, the main purpose of which is to control tumor growth and, owing to its advantages, and makes the increase in the number of patients treated by this intervention. However, the efficacy of these interventions in terms of tumor control and function preservation is uneven. In this study, we performed a traditional meta-analysis for a preliminary exploration of tumor control and neurological protection of VS patients after different interventions. Considering that VS is a benign tumor, many researchers and patients choose to have regular examinations (i.e., observation treatment) in the early stage of the disease, which is the basic treatment strategy for VS . In this study, traditional meta-analysis was used to explore the benefits of MS, ConFSRT, and SRS in comparison with observation treatment. Although tinnitus and disequilibrium/vertigo were not improved in the MS group, the rate of preserved hearing was better than that in observation treatment. Compared with the observation treatment, there was no significant difference between ConFSRT and SRS in terms of preservation of hearing, tinnitus, and trigeminal and facial nerve toxicity, however, it has a better rate of tumor local control. This is consistent with the findings of Dhayalan D, et al., and Leon J, et al., , who pointed out that ConFSRT and SRS have benefits in the local tumor control rate but have no significant benefits in improving vestibular function (such as vertigo and tinnitus) and neuroprotection. Therefore, the choice of immediate removal of the tumor or regular examination for the early detection of the lesion remains to be discussed. Surgery is the most common treatment for VS. With developments in microsurgical technology, the operating microscope has been used for VS. Thus, surgery in these patients not only prevents death, but also allows the complete resection of tumors while preserving hearing and facial nerve function [48, 49]. Moreover, SRS is among the optional methods for surgical treatment of VS. Therefore, in this study, we used a traditional meta-analysis to explore the efficacy of SRS and MS. We found no significant difference in the aspects of tumor local control, preserved hearing, tinnitus, disequilibrium/vertigo, and trigeminal nerve toxicity, however, but the incidence of facial nerve toxicity was reduced. Similar to the results of a meta-analysis conducted by Jakubeit et al., compared with MS, SRS improved not only facial nerve toxicity, but also trigeminal nerve toxicity. This difference may be attributed to the low sample size of this study. Therefore, the apparent trigeminal nerve toxicity associated with MS and SRS requires further exploration. Subsequently, in this study, compared with FSRT 5 fractions, FSRT 3 fractions, and ConFSRT, SRS showed no significant association between local tumor control rate, preserved hearing, tinnitus, disequilibrium/vertigo, trigeminal nerve toxicity, and facial nerve toxicity. This is consistent with the findings of Söderlund Diaz et al., who found that SRS, FSRT 3 fractions, and FSRT 5 fractions had no significant differences in improving the rate of tumor local control, preserved hearing, trigeminal nerve toxicity, facial nerve toxicity, tinnitus, and disequilibrium/vertigo. In addition, we explored the effects of ConFSRT and FSRT 5 fractions on tumor control and the protection of neural and vestibular function. There were no statistically significant differences between the two interventions in these aspects. Similar to the results of Anderson BM, et al., we confirmed that ConFSRT and FSRT 5 fractions had comparable efficacy in terms of tumor control, neuroprotection, and vestibular function protection, with no statistically significant differences. Slane BG, et al. found that both interventions had similar good local tumor control rates and minimal toxicity, however, they suggested that FSRT 5 fractions are more convenient for patients to treat and may, therefore, be more suitable. Considering that traditional meta-analysis is based on the pairwise direct comparison of effect sizes, data for direct comparison are relatively limited . Nevertheless, there is a growing need for indirect comparisons between different treatment interventions with the same efficacy used in clinical practice [51, 52]. This network meta-analysis includes allowed both direct comparisons and indirect comparisons based on logical reasoning. To the best of our knowledge, this study was the first attempt at a network meta-analysis for a comprehensive analysis of direct and indirect comparative evidence for the three major treatment interventions for VS. We found that the rate of tumor local control was better with surgery and radiotherapy than with observation treatment, with MS, FSRT 3 fractions, and ConFSRT having the best tumor local control rate. In terms of preserved hearing, radiotherapy had a better rate than observation treatment and MS, and the best radiotherapy protocol was FSRT 5 fractions, FSRT 3 fractions, and SRS in this order. In terms of neuroprotection, various radiotherapy protocols have better neuroprotection than MS, and SRS was the best intervention. In terms of improvement of vestibular function (i.e., disequilibrium and vertigo) and tinnitus, radiotherapy intervention also had a good effect. This study had some limitations. First, most of the included studies were non-randomized controlled trials, which reduced the strength of evidence in our analysis to some extent. Therefore, high-quality, large-scale, multicenter RCTs are required to further verify the efficacy and safety of the current treatment options, including observation, microsurgery, and radiotherapy, for VS. Second, the radiotherapy methods included in this study were GammaKnife, CyberKnife, and linear accelerator, which were not divided in detail according to specific radiotherapy methods in our analysis. Thus, whether different radiotherapy methods affect the outcome of patients should be explored. We plan to address these limitations in our follow-up study. Conclusion This network meta-analysis indicated that in patients with VS, MS, FSRT 3 fractions, ConFSRT, FSRT 5 fractions, and SRS showed better local tumor control rates, however, compared with MS, FSRT 3 fractions, FSRT 5 fractions, SRS, and ConFSRT were associated with better protection of nerve function and improved symptoms of vestibular function and tinnitus, among which the best one was SRS. Therefore, in patients with unilateral VS, SRS may be a promising alternative treatment. However, there are only a few high-quality randomized controlled clinical trials on MS, different SRS, and observation, thus, the feasibility of SRS for clinical practice requires further exploration. Data availability Data is provided within the manuscript. Abbreviations VS: : Vestibular schwannoma MS: : Microsurgery SRS: : Stereotactic radiosurgery FRST: : Fractionated stereotactic radiotherapy GKS: : Gamma knife surgery CM: : Conservative management NOS: : Newcastle–Ottawa Scale ConFSRT: : Conventionally-fractionated stereotactic radiotherapy; y NR: : Not report LC: : Local control RCT: : Randomized controlled trial References Gupta VK, Thakker A, Gupta KK. Vestibular Schwannoma: what we know and where we are heading. Head Neck Pathol. 2020;14:1058–66. Article PubMed PubMed Central Google Scholar 2. Carlson ML, Link MJ. Vestibular Schwannomas. N Engl J Med. 2021;384:1335–48. Article PubMed Google Scholar 3. Jakubeit T, Sturtz S, Sow D, Groß W, Mosch C, Patt M, Weingärtner V, Boström J, Goldbrunner R, Markes M. Single-fraction stereotactic radiosurgery versus microsurgical resection for the treatment of vestibular schwannoma: a systematic review and meta-analysis. Syst Reviews. 2022;11:265. Article Google Scholar 4. 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Article CAS PubMed Google Scholar Download references Funding Ningxia Natural Science Foundation (2024AAC03699), Ningxia Key Research and Development Program Project (2022BEG03081, 2023BEG2009), and Research Program of Ningxia Medical University (XJ2023036) were received for this study or publication of this article. Author information Authors and Affiliations Department of Neurosurgery, General Hospital of Ningxia Medical University, Yinchuan, China Xianhao Huo, Yifan Zhang & Jihui Tian 2. Department of Neurosurgery, North China University of Science and Technology Affiliated Hospital, Tangshan, Hebei, China Xu Zhao & Xiaozhuo Liu 3. Department of Neurosurgery, The First Medical Center of Chinese PLA General Hospital, Beijing, China Mei Li Authors Xianhao Huo View author publications Search author on:PubMed Google Scholar 2. Xu Zhao View author publications Search author on:PubMed Google Scholar 3. Xiaozhuo Liu View author publications Search author on:PubMed Google Scholar 4. Yifan Zhang View author publications Search author on:PubMed Google Scholar 5. Jihui Tian View author publications Search author on:PubMed Google Scholar 6. Mei Li View author publications Search author on:PubMed Google Scholar Contributions L.M. and H.X.H. performed the study subject and design, data extraction, statistical analysis, interpretation of data and manuscript drafting T.J.H., L.M., and H.X.H. contributed to the study design, statistical analysis and manuscript revising Z.X. contributed to the study design, interpretation of data L.X.Z. performed statistical analysis Z.Y.F. and H.X.H. extracted the data. L.M. and T.J.H. performed quality assessment H.X.H. and Z.X. was involved in critical revision of manuscript. Corresponding authors Correspondence to Jihui Tian or Mei Li. Ethics declarations Ethics approval and consent to participate Not Applicable. Consent for publication Not applicable. Competing interests 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. Electronic supplementary material Below is the link to the electronic supplementary material. Supplementary Material 1 Supplementary Material 2 Rights and permissions Open Access This article is licensed under a Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License, which permits any non-commercial use, sharing, 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 you modified the licensed material. You do not have permission under this licence to share adapted material derived from this article or parts of it. 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Provided by the Springer Nature SharedIt content-sharing initiative Keywords Vestibular schwannoma Microsurgery Stereotactic radiosurgery Fractionated stereotactic radiotherapy Conventionally fractionated stereotactic radiotherapy BMC Cancer ISSN: 1471-2407
10318	https://latin.stackexchange.com/questions/7346/malo-in-motto-maelstrom	vocabulary - "Malo" in Motto Maelstrom - Latin Language Stack Exchange Join Latin Language By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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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 "Malo" in Motto Maelstrom Ask Question Asked 6 years, 11 months ago Modified6 years, 3 months ago Viewed 418 times This question shows research effort; it is useful and clear 4 Save this question. Show activity on this post. The motto for Concordia University Saint Paul (MN) reads: "In litteris proficere volo, malo diligere Jesum." The CSP website, magazine (Spring 2009), and various internet sources offer these translations—some literal and some interpretive: I wish to be proficient in academics, but even more I wish to know Jesus. It is good to pursue knowledge, better to know Jesus. Will Prosper in Academics and Be Diligent in Christ. And even this interesting adaptation: Lord, give us joy in education especially knowing the love of Christ. I'm having a hard time fitting "malo" into this, because everything I've found or learned relates it to bad or evil. I've tried my Cassell's, and the Latin Dictionary. Is there a comparative or superlative at work (or play) here, that I am not comprehending? Any insight would be much appreciated! Thank you in advance. vocabulary meaning motto translation-explanation Share Share a link to this question Copy linkCC BY-SA 4.0 Improve this question Follow Follow this question to receive notifications edited Jun 25, 2019 at 17:41 Joonas Ilmavirta♦ 119k 23 23 gold badges 212 212 silver badges 638 638 bronze badges asked Oct 21, 2018 at 21:28 VerbiwhoreVerbiwhore 43 4 4 bronze badges 0 Add a comment\| 2 Answers 2 Sorted by: Reset to default This answer is useful 7 Save this answer. Show activity on this post. malo here is the first-person singular active indicative form of malle, which means “I prefer”. It has nothing to do with either malus “bad” (or, for that matter, malum “apple”). I believe that the verb is a contraction of maius “better” and velle “to want”. EDIT: Lewis & Short says that it's actually from magis "more" + velle. Share Share a link to this answer Copy linkCC BY-SA 4.0 Improve this answer Follow Follow this answer to receive notifications edited Jun 25, 2019 at 18:03 answered Oct 21, 2018 at 22:42 NRitHNRitH 204 1 1 silver badge 6 6 bronze badges 2 Thank you, NRitH, for your answer. Please see disclaimer in my comment above. However, I wanted to let you know that I learned something else (beyond my momentary myopia when attempting a search this arvo) interesting from your post, namely the idea of the "contraction of maius 'better' and velle 'to want'" that I had not considered. Cheers!Verbiwhore –Verbiwhore 2018-10-22 01:59:57 +00:00 Commented Oct 22, 2018 at 1:59 I'm confused by the rollback of the edit--any particular reason you want it to stand this way?brianpck –brianpck 2019-06-25 20:10:29 +00:00 Commented Jun 25, 2019 at 20:10 Add a comment\| This answer is useful 8 Save this answer. Show activity on this post. To add to this, there's an old mnemonic rhyme for the word malo: Mālō, I would rather be Mālō, in an apple tree Mālō, than a ship at sea Mălō, in adversity It's a confusingly ambiguous word! In the first line, it's a form of mālle, "to want"; in the second line, it's the ablative of position of mālus, "apple tree"; in the third line, it's the ablative of comparison of mālus, "upright beam, post, mast" (or "ship" by synecdoche); in the fourth line, it's the masculine ablative of mălus, "bad". The fourth of these is the only one that looks any different, since it has a short ă rather than a long ā. Share Share a link to this answer Copy linkCC BY-SA 4.0 Improve this answer Follow Follow this answer to receive notifications answered Oct 22, 2018 at 19:15 Draconis♦Draconis 72.6k 6 6 gold badges 138 138 silver badges 308 308 bronze badges 3 This is wonderful!NRitH –NRitH 2018-10-22 22:34:20 +00:00 Commented Oct 22, 2018 at 22:34 2 What a fascinating word pond! Charmed I am to be / sprouting cognitively / at sea in synecdoche / enlightened incredulity!Verbiwhore –Verbiwhore 2018-10-24 01:31:22 +00:00 Commented Oct 24, 2018 at 1:31 I just saw this, but the fourth line would actually be neuter, no? I think I remember learning this (or something like this?) way back in high school, but I had entirely forgotten it!cmw –cmw♦ 2021-07-22 18:47:51 +00:00 Commented Jul 22, 2021 at 18:47 Add a comment\| Your Answer Reminder: Answers generated by AI tools are not allowed due to Latin Language Stack Exchange's artificial intelligence policy Thanks for contributing an answer to Latin Language Stack Exchange! Please be sure to answer the question. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. Making statements based on opinion; back them up with references or personal experience. To learn more, see our tips on writing great answers. 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10319	https://www.quora.com/How-is-PV-CONSTANT-in-a-isothermal-process	How is PV CONSTANT in a isothermal process? - Quora Something went wrong. Wait a moment and try again. Try again Skip to content Skip to search Sign In Physics Isothermal Change Gas Laws PV Thermodynamics Energetics Physical Processes Ideal Gas Isothermal Curve Isothermal Process 5 How is PV CONSTANT in a isothermal process? All related (36) Sort Recommended Bharat Sales and Service Engineer (2016–present) ·8y Ideal gas Equation states that Product of Pressure and Volume is directly proportional to Temperature. To eliminate the proportionality sign a Constant R is used, known as Boltzmann Constant. PV=nRT n=No.of Moles Consider a substance in state 1: Then , Ideal gas equation for state 1 is : P1V1=nRT1 When conditions change, i.e. Pressure, volume or temperature, the substance reaches state 2. Ideal gas equation for state 2 is : P2V2=nRT2 Change during this process : P2V2-P1V1=nR(T2-T1) If the Process happens to be Isothermal, then there wont be any change in temperature. So T2=T1. then, T2-T1=0 then, P2V2-P Continue Reading Ideal gas Equation states that Product of Pressure and Volume is directly proportional to Temperature. To eliminate the proportionality sign a Constant R is used, known as Boltzmann Constant. PV=nRT n=No.of Moles Consider a substance in state 1: Then , Ideal gas equation for state 1 is : P1V1=nRT1 When conditions change, i.e. Pressure, volume or temperature, the substance reaches state 2. Ideal gas equation for state 2 is : P2V2=nRT2 Change during this process : P2V2-P1V1=nR(T2-T1) If the Process happens to be Isothermal, then there wont be any change in temperature. So T2=T1. then, T2-T1=0 then, P2V2-P1V1=nR( 0 ) = 0 therefore, P2V2=P1V1. So, we can see that PV at both the states is same. Thus, It has been constant all through, in an isothermal process. Hence, PV= Constant, in isothermal process. Upvote · 99 11 Sponsored by Grammarly Stuck on the blinking cursor? Move your great ideas to polished drafts without the guesswork. Try Grammarly today! Download 99 34 Related questions More answers below Why is the PV constant in the adiabatic process? Is the equation PV/T=constant applicable for an isothermal or adaibatic process or both? Why temperature is constant in isothermal process? What is an adiabatic process, an isothermal process, and a cyclic process? In a PV diagram, how do you know the process is isothermal or adiabatic? Darshan Pitroda M.E. in Thermodynamics&Quantum Mechanics, The Maharaja Sayajirao University of Baroda ·8y Thanks for A2A Isothermal process is generally defined as process in which heat is transferred through infinitely small temperature difference between system and surroundings. To transfer heat through infinitely small temperature difference process need to be as slow as possible in order to adjust the system to the temperature changes. The characteristic gas equation for ideal gas is given by PV/T = Constant; Now if T = Constant then equation reduces to, PV = Constant This can be demonstrated on P-V diagram as hyperbolas, If you go from A to B (expansion process) the volume of gas increases and pr Continue Reading Thanks for A2A Isothermal process is generally defined as process in which heat is transferred through infinitely small temperature difference between system and surroundings. To transfer heat through infinitely small temperature difference process need to be as slow as possible in order to adjust the system to the temperature changes. The characteristic gas equation for ideal gas is given by PV/T = Constant; Now if T = Constant then equation reduces to, PV = Constant This can be demonstrated on P-V diagram as hyperbolas, If you go from A to B (expansion process) the volume of gas increases and pressure decreases relatively thereby keeping product PV constant. If you go from B to A ( Compression Process) the volume of gas decreases and pressure increases relatively. At any point on graph you will find PV constant. Upvote · 9 2 9 1 Shubham Kashyap M. Tech from National Institute of Technology, Raipur (Graduated 2019) ·8y PV = constant is termed as isothermal, only and only when gas used in the process is IDEAL GAS. Since we know for ideal gas; PV=mRT. So if PV =constant then T should also be constant, making the process “Isothermal” Continue Reading PV = constant is termed as isothermal, only and only when gas used in the process is IDEAL GAS. Since we know for ideal gas; PV=mRT. So if PV =constant then T should also be constant, making the process “Isothermal” Upvote · 9 1 9 2 Shashank Tiwari B.E (EE) from Jabalpur Engineering College (Graduated 2018) ·Updated 8y PV is constant in isothermal process.. It can be understood in 3 logical ways which have their own maturity…. 1: PV=nRT Isothermal process is a process in which temperature remains constant ,so from above equation it can be seen tht only temperature is variable on the right side of equation,thus if it becomes constant PV becomes constant. 2:It can also be understood that for performing isothermal process,Values of P and V are so chosen that its product remains constant (equal to nRT).. Qualitatively ,in an isothermal process there is change in temperature( practically ) but this process is too Continue Reading PV is constant in isothermal process.. It can be understood in 3 logical ways which have their own maturity…. 1: PV=nRT Isothermal process is a process in which temperature remains constant ,so from above equation it can be seen tht only temperature is variable on the right side of equation,thus if it becomes constant PV becomes constant. 2:It can also be understood that for performing isothermal process,Values of P and V are so chosen that its product remains constant (equal to nRT).. Qualitatively ,in an isothermal process there is change in temperature( practically ) but this process is too slow that any change in temperature results in the flow of heat in and out of the system such that its temperature remains constant..( if temp. Rises by del(T) then this heat is released in the surrounding or vice versa) Upvote · 9 5 Sponsored by RedHat Know what your AI knows, with open source models. Your AI should keep records, not secrets. Learn More 99 36 Related questions More answers below How can one draw the curve of an isothermal process on a PV diagram? In an isothermal process, is the net heat change zero? Why isothermal process is a reversible process? What is isothermal process? How is the temperature practically constant in an isothermal process? Abhishek Mishra Former Assistant Manager at Sesa Sterlite Limited - Vedanta Aluminium & Power (2017–2018) ·8y It's not that when PV= C then it is known as Isothermal process. A process in which the temperature is constant thus making PV=C accorting to the ideal gas equation i.e pv=mrt.This process is known as Isothermal Process. Your response is private Was this worth your time? This helps us sort answers on the page. Absolutely not Definitely yes Upvote · 9 4 9 1 Kim Aaron Has PhD from Caltech in fluid dynamics · Author has 8.4K answers and 27.4M answer views ·5y Related Is the equation PV/T=constant applicable for an isothermal or adaibatic process or both? Is the equation PV/T=constant applicable for an isothermal or adaibatic process or both? This is the equation of state for an ideal gas. It is true for any state of the material. By “state,” I’m not talking about gas, liquid, solid. Those are phases. I’m talking about the conditions in the gas. The combination of pressure, temperature and density is the state of the gas. Given an equation of state, if you know any two of these, the equation tells you the third one. A process is a changing of the state of the gas. When you declare that you are following an isothermal process, you are saying that Continue Reading Is the equation PV/T=constant applicable for an isothermal or adaibatic process or both? This is the equation of state for an ideal gas. It is true for any state of the material. By “state,” I’m not talking about gas, liquid, solid. Those are phases. I’m talking about the conditions in the gas. The combination of pressure, temperature and density is the state of the gas. Given an equation of state, if you know any two of these, the equation tells you the third one. A process is a changing of the state of the gas. When you declare that you are following an isothermal process, you are saying that the state is changing in such a way that the temperature is constant. But for every step along the way, the state still satisfies the equation of state. Similarly, if you say it’s an adiabatic process, you are constraining the process such that as the state changes, it does so in a way that has no heat added or subtracted from the substance. Each step along the way is a state and each state satisfies the equation of state. So, in answer to your question, that equation of state that you wrote is applicable for both processes as well as all other processes. Upvote · 9 5 Promoted by JH Simon JH Simon Author of 'How To Kill A Narcissist' ·Updated Fri Is there a way to prevent a narcissistic triangulation? Stopping narcissistic triangulation requires first understanding how triangulation works, how it is maintained within us, and how we transcend it through a strong spiritual practice. What Is Triangulation? Triangulation is the conscious or unconscious introduction of a third person into the relationship dynamic between two people. Examples of triangulation are: Physical: Your significant other invites somone over or to mutual events, or spends increasingly more time with that person. In other cases, the other person is a core part of your significant other’s life. Verbal: Your significant other tal Continue Reading Stopping narcissistic triangulation requires first understanding how triangulation works, how it is maintained within us, and how we transcend it through a strong spiritual practice. What Is Triangulation? Triangulation is the conscious or unconscious introduction of a third person into the relationship dynamic between two people. Examples of triangulation are: Physical: Your significant other invites somone over or to mutual events, or spends increasingly more time with that person. In other cases, the other person is a core part of your significant other’s life. Verbal: Your significant other talks about the other person favorably or compares you to them unfavourably. Triangulation can either be situational, or it can be manipulative. In the situational case, the presence of another person simply happens to be a part of relating with your significant other. For example, your parnter might still have a relationship with their ex-husband or ex-wife in order to keep things civilised for the children. A third person can also be brought up in conversation. A partner can discuss or even compare you with an ex or a sexual competitor. A mother can compare you unfavourably with a sibling. Triangulation becomes manipulative when it is introduced as a way to impact you negatively or to pressure you into reacting from a place of threat or inferiority. The line between the two types of triangulation can quickly grow blurry in relationships. In either case, whether it is situational or manipulative, triangulation can make you feel the following: That you are less than. Jealous. Threatened. That you are not as important to your significant other as you had hoped. That your significant other is a valuable commodity who you must fight to keep or please. Why Do People Triangulate? As already stated, triangulation can be situational or manipulative. Therefore, there are many reasons people triangulate. Often the triangle simply exists without anyone doing anything. Other times your significant other actively feeds it. Then you have narcissistic triangulation, where a three-corner opposition is actively and manipulatively used against you to ensure your significant other’s sense of grandiosity, control and safety. Some reasons people might triangulate are: It makes them feel wanted and in demand: Having multiple people show interest makes a person feel like a valuable commodity. Telling the people about each other then creates competition and increases insecurity, making the ‘competitors’ feel inferior and unworthy. It communicates “You are not special, let alone the only one.” They want to control you: Triangulation can trigger a person’s jealousy and sense of abandonment. This makes them feel needy and insecure. When someone is in that state, they become desperate to feel close to the person at the centre of the triangle. These feelings of insecurity make you reactive and panicky, and so easier to control. You are always on edge, doing everthing you can to ensure the person you need does not make the other person more important than you, or drop you altogether. A mother or father can also triangulate to get their way and direct their children how they want. In all such cases, it is about control. It helps them win an argument: If so-and-so also thinks the same way as your significant other, then that’s two against one. Often it can be dozens against one, since “everyone” thinks the same way. Fear of commitment and vulnerability: They keep multiple people around to avoid being ‘stuck’ with all their eggs in one basket. Grandiosity: They can never make someone as special or more special than them. Triangulation keeps others feeling insignificant and tips the power distribution in your significant other’s favour. They simply want that person in their life: People do not need to drop others out of their life when they meet you. They have a unique constellation of connections, some of which might not make sense to you, especially when exes become friends. That means you might have company in your intimate space. Point 6 is the key to to this entire puzzle. You may well be in a relationship with a narcissist who simply wants to control you and make you feel small. It might be a protection or coping mechanism. Your significant other might diversify their emotional interests to ensure they don’t get too close to you, and therefore avoid being vulnerable. Or, they might just like having that peson in their life. It can be tempting to simply see all triangulation as manipulative, but that is not the case. In the last example, it has nothing to do with you. Other people simply exist in that person’s life. However, even if that person is purposely triangulating you, such as is the case with narcissism, you still have the choice to decide whether it affects you. How To Stop Triangulation Ultimately, triangulation exists only in your mind. It is created and reinforced by the thoughts, meanings and emotions you apply to the situation. Triangulation, above all, says less about the other person than it does about you. This is a difficult pill to swallow, but it is also a doorway to enlightenment. It is a way to break codependent habits, learn the art of detachment, and above all, to deepen your connection with yourself. Yes, a triangle has three points, but every single point leads directly back to you. Triangulation stops with you. By shifting your paradigms, perspective, beliefs and focus, you can free yourself of the triangle for good. While triangulation can be painful, it is also a chance to move closer toward spiritual growth, freedom and actualisation. Triangulation is also a natural psychological phenomenon which narcissists exploit constantly. This makes your spiritual journey a kind of holy war when the two forces clash. Triangulation can be a great teacher. It can help you understand that: We are born alone, and die alone: Accepting and leaning into this reality can help diminish the power of the triangle by bringing your focus deep into your greatest fear. Nobody is in our keeping: You are your own Self, and others are their own Self. Your Self is sufficient onto itself, and answers only to God. We always have the option of directing the focus back on ourselves: The power to shift your focus is always available to you. By focussing on your significant other and who they bring into your relationship dynamic, you are ceding control to them. Enmeshment stunts growth: We must constantly let go of the other person and relate to them from a healthy distance if we are to grow with them. They will do or say what they must. What is it that you must attend to? Control is an illusion: You cannot make the other person ‘yours’. You cannot make them give up anyone. You can only make sure you DO NOT GIVE UP YOURSELF in the panic, shame, jealousy and anger of triangulation. The triangle is ultimately in your head: It only seeps into your heart and body if you let it. Reject it. Call that person out. Or simply see it for what it is, and divest your emotions away from it and into Self growth instead. We should compare our current Self to our previous Self, and not to others: Every day is an opportunity to connect deeply with our True Self and let it guide us toward growth. Triangulation distracts us from this accessible reality by drawing us into the ego-based duality realm. Less than, more than. Special, not special. Mine, not mine. You have an unhealthy desire to be the ‘special’ one: This comes from childhood, where the first triangle emerged between your mother, father and you. Understand that your emotions usually stem from this time. Yet you are no longer a child. Having the focus of your parent meant life and death when you were a baby. Now you are an adult with numerous resources. You can always develop more resources, both within and without. Your significant other is important, but they are not the difference between life and death. Those existential feelings are not from the present moment, but from the past. You needed to be special to survive. Now, you are inherently special, but you do not need another person’s complete focus and positive regard to access that. It is there either way. And there is a way to find it. Inner Freedom From Triangulation Overcoming triangulation begins within. A spiritual approach to breaking the triangle includes the following: Self-Rememberance: Orient with your world. Notice the details around you. What objects are there in the room you are in? What can you hear, feel and smell? Orienting with the world around you allows you to shift your focus out of your mind, where duality rules. You stop comparing and analysing, and start being. From this place of neutral observing, you will notice your current emotional state. Allow those feelings to arise. Sadness, grief, shame, anger. Regardless of what you feel, allow it to come up. Now ask yourself: Who is noticing these details? Who is feeling these feelings? In this awareness you will remember your Self. There is no more triangle in your mind. There is simply you and your feeling Self, i.e. Your True Self. And by asking the question of who is feeling and noticing, you introduce your Higher Self into the fold. A new triangle emerges, one that serves you rather than crushing you. Centering: Bring your focus inside your body. Allow your muscles to relax, breathe deeply into your belly and chest, and then allow the exhale to center you. Keep repeating this over and over until you get a sense of inner anchoring and calm. Self-Comparison: Stop comparing yourself to the other person in the triangle. Start comparing your Self to your previous Self. What growth have you noticed in your life since a month ago? A year ago? Five years ago? Practice this often. Every time you feel less than or threatened by another person, bring your focus back within and celebrate where you are and where you have come from. Master this, and you learn the essence of spirituality. Die before you die: Reflect on death. What might death be like? How do you feel about it? Does it terrify you? Intrigue you? Inspire you to live fully? Think about it. Read about it. Fall into it. All reaction to triangulation is a desperate clinging to escape death. Death of a relationship. Death of your self-worth. Death of your specialness. Transcend it all by meditating on and accepting death. Outer freedom From Triangulation Everyone has the freedom to do what they want. You cannot control them, and you cannot control every outcome. However, freedom is not free. It comes with consequences. A person can find someone ‘more special.’ A significant other can cheat and leave you behind. But if you drop the triangle regardless, at least you are living on your own terms. You are no longer being crippled and controlled by another person’s manipulation or circumstances. With triangulation, you can only choose from two options: Codependency You control each other, and limit your relationships to each other. You use triangulation to control each other. You lash out whenever you feel insecure or threatened. You panic, feel anxious, and try to snoop around on the other person to feel secure again. You feel terrible about yourself and fight for the acceptance and approval of the other person. OR Freedom There is another way: You let go and accept the consequences. Your parent might yap on about how great your sibling or cousin are. Your partner might have feelings for others, or have awkward close encounters before deciding to pull away. They might cheat. You might be left for someone else. Ultimatley, you need to accept that you have no control over that. We all have the freedom to live on our terms, and connect with whoever we want while agreeing to a set of boundaries and rules. Your significant other might slip. You might slip. You have to live with this possibility while trusting and hoping for the best. This is the nature of freedom. Putting A Stop To Narcissistic Triangulation Until you have done the inner work, you are always susceptible to narcissistic triangulation. When it is done with malignant intent, you will be bashed around by it. When it is situational, your sense of security, serenity and agency will be gradually worn down. When a narcissist triangulates and you have done the work, it will cease to impact you. You can speak out firmly when they introduce another person into the relationship dynamic in a hurtful or manipualtive way. You can simply bask in your own serenity and peace while they focus on the other person. From this place of power, you can then look over what is left of the relationship with calmness, and simply act, rather than react. Walking away is always an option. Either the person is being grossly manipulative, or the situation is simply untenable for you, especially when it comes to exes. But first ask yourself: What inner work can you do? Can you truly be fine with yourself alone? If not, then why? Can you tolerate the presence of ‘outsiders’ in your intimate space? If not, then why? Can you feel sufficient and worthy even when others enter your relationship dynamic? Why do you need to be ‘special’, i.e. the only one? Is it possible that such a rigid two-person world can grow stale, and become a breeding ground for resentment? Do you not think that some flexibility and trust in yourself and others can create space for wonder and growth to enter into your life? Regardless of whether you are alone or in a relationship, in a narcissistic triangle or out of it, ask yourself: Can I direct my focus within every single day, and grow from that place? Do that, and you will discover the greatest triangle of all: You, your True Self, and your Higher Self. There is no more empowering dynamic that you can develop. And there is no other way to stop narcissistic triangulation. If you have just started your narcissistic abuse recovery journey, check out How To Kill A Narcissist. Or if you wish to immunise yourself against narcissists and move on for good, take a look at How To Bury A Narcissist. Upvote · 999 302 99 33 99 13 Sadaf Shoukat Chemical Engineer. · Author has 560 answers and 924.7K answer views ·7y In ideal gas law PV, product of pressure and volume, is constant if the temperature is constant i.e isothermal condition. Upvote · Akshay Devikar Knows Marathi · Author has 395 answers and 614.4K answer views ·8y Related Why is PV^ (Cp/Cv) = constant for Idiabatic process? Universal Gas Eqn: PV=nRT Diff. both sides: P.dV + V.dP = n.R.dT…. (1) For adiabatic process, the expansion or compression of a gas occurs at a very fast rate which make no heat transfer. Therefore, dQ= 0 A/c to First Law of TD: dQ=dU+ dW, implies dU= -dW=-P.dV ….. (2) But dU= n.Cv. dT …. (3) Combining 2 and 3, we get: n.dT= -P.dV/Cv …..(3) Using R= Cp-Cv, the gas law gives: from 1 and 3. n.dT= (P.dV + V.dP)/(Cp-Cv) = -PdV/Cv On rearranging we get, (dP/P)+((Cp/Cv)(dV/V))=0 We know that Cp and Cv are constants for a particular gas, hence integrating: Ln(P)+ r.Ln(V)=constant (replacing Cp/Cv = r or Gamm Continue Reading Universal Gas Eqn: PV=nRT Diff. both sides: P.dV + V.dP = n.R.dT…. (1) For adiabatic process, the expansion or compression of a gas occurs at a very fast rate which make no heat transfer. Therefore, dQ= 0 A/c to First Law of TD: dQ=dU+ dW, implies dU= -dW=-P.dV ….. (2) But dU= n.Cv. dT …. (3) Combining 2 and 3, we get: n.dT= -P.dV/Cv …..(3) Using R= Cp-Cv, the gas law gives: from 1 and 3. n.dT= (P.dV + V.dP)/(Cp-Cv) = -PdV/Cv On rearranging we get, (dP/P)+((Cp/Cv)(dV/V))=0 We know that Cp and Cv are constants for a particular gas, hence integrating: Ln(P)+ r.Ln(V)=constant (replacing Cp/Cv = r or Gamma) therefore, P.V(^r)= constant is termed as adiabatic equation. Upvote · 99 13 Sponsored by SpeechText.AI AI-powered legal transcription service. Convert speech to text from any audio/video format, including TRM files. Fully compliant with GDPR. Start Now 9 3 Kunal Batra Just a student · Author has 54 answers and 135.9K answer views ·8y Related Why is “PV^ (Cp/Cv) =constant” used to represent an adiabatic process? Its not easy to explain this by a text ..so read the img carefully..any doubt do reply.. From (2) and (3) ..its clear that alpha =Cp/Cv Hence verified.. Continue Reading Its not easy to explain this by a text ..so read the img carefully..any doubt do reply.. From (2) and (3) ..its clear that alpha =Cp/Cv Hence verified.. Upvote · 9 2 Dennis Leppin B.E.and M.E. in Chemical Engineering, City College of New York · Author has 2.2K answers and 1.6M answer views ·5y Related Is the equation PV/T=constant applicable for an isothermal or adaibatic process or both? The equation you state is essentially the perfect gas law provided the constant is nR. In any case, if that equation does in fact represent the state of the gas throughout the process, then it is applicable whether the process is adiabatic or isothermal. In reality it deviates unacceptably for higher pressures especially near the critical point. Certainly, for low pressures and temperatures not to Continue Reading The equation you state is essentially the perfect gas law provided the constant is nR. In any case, if that equation does in fact represent the state of the gas throughout the process, then it is applicable whether the process is adiabatic or isothermal. In reality it deviates unacceptably for higher pressures especially near the critical point. Certainly, for low pressures and temperatures not too far from room temperature it would be ok. Even then some gases would deviate badly. Any Equation of State that accurately fits the PVT behavior of the gas or gas mixture in question would also be applicable in that sense. This is nearly a tautology by saying is a good EOS that fits the data applicable as an EOS in the range where the data fits. Of course it is. Of course the equation itself doesn’t tell you anything about the work required in an isothermal process, or adiabatic process, but there do exist closed form solutions for the work of compression or expansion for reversible adiabatic or isothermal processes for perfect gases. For more complex EOS you can do some integrations or numerical solutions to find the enthal... Upvote · Ashish Kushwaha Physicist by heart, engineer otherwise · Author has 479 answers and 4M answer views ·7y Related In a PV diagram, how do you know the process is isothermal or adiabatic? On a P V P V diagram, only adiabatic and isotherm processes have asymptotic shapes. An example case for the curves for the two processes is shown here: It is not possible to just look at the shape of one curve and predict if it adiabatic or isotherm, because they look very similar. However, Isotherms are lot steeper, so of they are next to each other, you can make a prediction by comparing the slope. If there is just one curve present, you can just quickly pick two points (i.e. two pairs of (P 1,V 1)(P 1,V 1) and (P 2,V 2)(P 2,V 2)) on the curve and see if P 1 V 1=P 2 V 2 P 1 V 1=P 2 V 2, which is true for an isotherm Continue Reading On a P V P V diagram, only adiabatic and isotherm processes have asymptotic shapes. An example case for the curves for the two processes is shown here: It is not possible to just look at the shape of one curve and predict if it adiabatic or isotherm, because they look very similar. However, Isotherms are lot steeper, so of they are next to each other, you can make a prediction by comparing the slope. If there is just one curve present, you can just quickly pick two points (i.e. two pairs of (P 1,V 1)(P 1,V 1) and (P 2,V 2)(P 2,V 2)) on the curve and see if P 1 V 1=P 2 V 2 P 1 V 1=P 2 V 2, which is true for an isotherm. If not, it’s an adiabatic, because that is the only option left. Upvote · 99 11 9 2 Mark Roseman Biochemistry Professor, Emeritus (2020–present) · Author has 9.1K answers and 19.6M answer views ·4y Related Isothermal processes are evaporation and melting of a substance, which are performed at constant pressure. How can an isothermal process be performed at P = const, when the isothermal process is opposite to the isobaric one? You seem to be thinking about the isothermal expansion of an ideal gas, a reversible process in which the pressure drops continuously so that PV = nRT is satisfied throughout the process of volume expansion. In this case, the amount of gas, n, is constant. But evaporation of a liquid gives another variable, namely conversion of liquid to vapor (i.e., more gas) as heat is added. In this way, the pressure can be kept constant during the isothermal phase change. In other words, if the vapor obeys the ideal gas law, both n and V increase while P and T remain constant. In the formalism of thermodynam Continue Reading You seem to be thinking about the isothermal expansion of an ideal gas, a reversible process in which the pressure drops continuously so that PV = nRT is satisfied throughout the process of volume expansion. In this case, the amount of gas, n, is constant. But evaporation of a liquid gives another variable, namely conversion of liquid to vapor (i.e., more gas) as heat is added. In this way, the pressure can be kept constant during the isothermal phase change. In other words, if the vapor obeys the ideal gas law, both n and V increase while P and T remain constant. In the formalism of thermodynamics, each phase is an open system since matter can be transferred in and out of them. (If both phases are in a closed container, the two systems combined form a closed system.) Upvote · John Leylegian Former Associate Professor of Mechanical Engineering at Manhattan College (2008–2024) · Author has 435 answers and 182.5K answer views ·4y Related Isothermal processes are evaporation and melting of a substance, which are performed at constant pressure. How can an isothermal process be performed at P = const, when the isothermal process is opposite to the isobaric one? If you look at the saturation curve of any pure substance, the temperature and pressure are not independent of one another. This phenomenon is the reason why water will boil at a lower temperature at high altitudes (lower pressure) Therefore, when you undergo a phase change at constant temperature, it is also at constant pressure. When you are not on the saturation curve, i.e., not undergoing a phase change, then yes, constant pressure and constant temperature processes will not be the same. Upvote · Related questions Why is the PV constant in the adiabatic process? Is the equation PV/T=constant applicable for an isothermal or adaibatic process or both? Why temperature is constant in isothermal process? What is an adiabatic process, an isothermal process, and a cyclic process? In a PV diagram, how do you know the process is isothermal or adiabatic? How can one draw the curve of an isothermal process on a PV diagram? In an isothermal process, is the net heat change zero? Why isothermal process is a reversible process? What is isothermal process? How is the temperature practically constant in an isothermal process? What are the requirements for a process to be isothermal? Why is the isothermal process not possible? What is the difference between isothermal and quasistatic process? What is constant in an isothermal process? Which process has more work done, isothermal or adiabatic? Related questions Why is the PV constant in the adiabatic process? Is the equation PV/T=constant applicable for an isothermal or adaibatic process or both? Why temperature is constant in isothermal process? What is an adiabatic process, an isothermal process, and a cyclic process? In a PV diagram, how do you know the process is isothermal or adiabatic? How can one draw the curve of an isothermal process on a PV diagram? Advertisement About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
10320	https://www.studyguidezone.com/identifying-sentence-errors.htm	Published Time: 2016-07-21T00:04:33+00:00 Identifying Sentence Errors Practice Questions Skip to content Menu Test List College Prep Study Tips Practice Questions Study Resources Menu Test List College Prep Study Tips Practice Questions Study Resources Identifying Sentence Errors Practice Questions Each consecutive underlined section corresponds with an answer choice. Select the choice that has an error, or select choice E for no error. 1. A Because of the Internet, B working at jobs C at home D have become much more common. E No error. A B C D E 2. “Pull it out A by B its plug, not by the C cord,” said D dad. E No error. A B C D E 3. Symptoms of this illness A that warrant a doctor visit B includes fever, C vomiting, and diarrhea, as well as the D loss of appetite. E No error. A B C D E 4. A Either Lisa or Karen B will always volunteer C their valuable D time to serve on our board. ENo error. A B C D E 5. The conversation with her A mother had a more profound B affect on her C than D she expected. E No error. A B C D E 6. The A President and the B Speaker of the House found the C Congressional Republicans’ filibusters to be D all together specious. E No error. A B C D E 7. A Professor Lane, our B Computer Science teacher, was excited when he had the opportunity to meet C Bill Gates, the D president of Microsoft, Inc.E No error. A B C D E 8. Do you think A they B will C except our plan D without an argument? E No error. A B C D E 9. “They A had went to the lake B without me C by the time D I got there,” said Jacques. E No error. A B C D E 10. Work A as quick B as you C can but D as carefully as possible when you take the test. E No error. A B C D E Answers – Identifying Sentence Errors 1. D: The error is “have become,” which should be “has become.” The plural form of the auxiliary verb “have” is incorrect because the subject of the independent clause is “working,” which is singular and thus takes a singular verb—i.e. “working has…” The other underlined sections are correct. 2. D: “Dad” is a name or proper noun and should be capitalized. “By” is a correct preposition to use here. “Its” is the correct use of the third-person singular possessive impersonal pronoun. The punctuation at the end of the quotation is correct. 3. B: “Includes” is incorrect because it is the singular form of the verb, but the subject, “symptoms,” is plural. The verb should be “include.” The subordinating conjunction “that” introducing the dependent clause, and its plural verb “warrant,” (A) are both correct. “…vomiting, and” (C) is punctuated correctly as the last in a series of three or more words. The singular noun “loss” (D) is correct. 4. C: “Their” is incorrect because it is a plural third-person possessive pronoun, but the use of “either (A)…or” indicates a singular form. It should be “her.” “Will always volunteer” (B) is a singular verb phrase and is correct. “Time” (D) is correct regardless of whether it is modified by a singular (“her”) or plural (“their”) possessive pronoun, e.g. “They both volunteered their valuable time.” 5. B: The correct noun for this meaning is spelled “effect.” “Affect” when it is a noun means mood or emotional state, e.g. “The patient presented with a depressed affect.” When it is a verb, the meaning of “affect” is related to the meaning of the noun “effect;” e.g. “The experience had a harmful effect on her, but it did not affect her brother the same way.” “…her mother” (A) is correctly not capitalized as it is a noun, not a name/proper noun (e.g. “Hello, Mother.”) The other underlined parts are correct. 6. D: This is incorrectly spelled as two words. In this sentence, it should be “altogether,” a one-word adverb modifying the adjective “specious” and meaning “entirely” or “completely.” “All together” would be used for a different meaning, e.g. “The family members were all together at the reunion.” The President (A), Speaker of the House (B), and Republicans’ (C) are all correctly capitalized as they are titles. The adjective congressional (C) refers to Congress (a proper name and thus capitalized), but as an adjective it is lower-case unless part of a proper name. 7. E: There is no error in this sentence. Titles and proper names (A) are capitalized. Academic subjects or departments are lower-case (e.g. department of computer science) unless they are adjectives (Computer Science teacher) (B) or proper nouns (English, French, etc.) Bill Gates (C) is a proper noun, i.e. a name, and is always capitalized. “…the president” (D) is correctly lower-case both because it is used after a name, and because it is a used as a description rather than a title here. 8. C: The verb is misspelled here. For the correct meaning, it should be “accept,” i.e. to consent or agree to our plan. “Except” means other than, besides, but, etc. Used as a verb as in this sentence, it would mean to make an exception of our plan, which is incorrect as it contradicts the rest of the sentence (without an argument). “They” (A) and “will” (B) are used correctly as subject pronoun and auxiliary verb. “Without” (D) is correctly used as a preposition. 9. A: The correct form for the past perfect tense of the irregular verb “to go” is “had gone,” not “had went.” “Went” is only used as the past tense, without the auxiliary “had.” The prepositional phrase (B) is correct. The two parts of the dependent clause (C and D) “by the time I got there” are correct. 10. A: The error is “quick,” which is an adjective; here it should be the adverb “quickly” instead, describing manner (how) to modify the verb “Work.” “…as carefully” (D) is an example of the correct usage. The other underlined sections are correct. Last Updated: June 4, 2019 Search for: Search for: Home College Prep Study Resources Privacy Policy © 2025 Study Guide Zone. All rights reserved. All information on this site is provided as is, without warranty. Since we are giving information away, we cannot be held liable for incidental mistakes. Test names and other trademarks are the property of the respective trademark holds. 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Using your mobile phone camera - scan the code below and download the Kindle app. Image Unavailable Image not available forColor: To view this video download Flash Player VIDEOS 360° VIEW IMAGES Follow the author Bryan A. Garner Follow Something went wrong. Please try your request again later. OK Garner's Modern English Usage Hardcover – November 17, 2022 by Bryan A. Garner (Author) 4.6 4.6 out of 5 stars 154 ratings Sorry, there was a problem loading this page. Try again. See all formats and editions {"desktop_buybox_group_1":[{"displayPrice":"$27.14","priceAmount":27.14,"currencySymbol":"$","integerValue":"27","decimalSeparator":".","fractionalValue":"14","symbolPosition":"left","hasSpace":false,"showFractionalPartIfEmpty":true,"offerListingId":"Dyx3cjK4RsIHKNivDLy%2BpBmv%2FH5Fn10rN49E3Lo1NYDDaUymw8ZUpGmbwJOlMq6z2oPu8ZSoNfj4FqH%2F3t%2BlQjULAAhCPOtPZZs5Cq8f4OF45FJZe7bbiPZr%2Fo6l8E4IUoiCTqyiMG9MieODMW7L8Q%3D%3D","locale":"en-US","buyingOptionType":"NEW","aapiBuyingOptionIndex":0}, {"displayPrice":"$19.05","priceAmount":19.05,"currencySymbol":"$","integerValue":"19","decimalSeparator":".","fractionalValue":"05","symbolPosition":"left","hasSpace":false,"showFractionalPartIfEmpty":true,"offerListingId":"Dyx3cjK4RsIHKNivDLy%2BpBmv%2FH5Fn10rZcuL6q6yPQ0G1StHXevajv7vqt846bqIT%2BdId6u4e0Yg7R3B3hYKXpnZi6ct2WbptZPDYacXj3o6zaYwHHmcFBhN%2FfNuqqod1m9AfPvOYndinJ%2B1lQ%2BjAkrkr0VVH3mnpOxXewdSwtmL66BfrknBwlo8no7o%2Fj3y","locale":"en-US","buyingOptionType":"USED","aapiBuyingOptionIndex":1}]} Purchase options and add-ons The most original and authoritative voice of today's English lexicography presents a fully revised new edition of his beloved usage dictionaryWhen Bryan Garner published the first edition of A Dictionary of Modern American Usage in 1999, the book quickly became one of the most influential style guides ever written for the English language. After four previous editions and over twenty years, our language has evolved in many ways, and the powerful tool of big data has revolutionized lexicography. This extensively revised new edition fully captures these changes, featuring a thousand new entries and over two hundred replacement entries, thoroughly updated usage data and ratios on word frequency based on the Google Ngram Viewer, a more balanced coverage of World Englishes, not just American and British, and the inclusion of gender-neutral language. However, one thing has not changed: in no sense is this a "regular" dictionary but a masterpiece of lexicography written with wit and personality by one of the preeminent authorities on the English language. To put it in David Foster Wallace's words, Garner's discussion of rhetoric and style still "borders on genius."From the (lost) battle between self-deprecating and self-depreciating to the misuse of it's for its, from the variant spelling patty-cake taking over pat-a-cake in American English to the singular uses of they, Garner explains the nuances of grammar and vocabulary and the linguistic blunders to which modern writers and speakers are prone, whether in word choice, syntax, phrasing, punctuation, or pronunciation. His empirical approach liberates English from two extremes: from the "purists" who maintain that split infinitives and sentence-ending prepositions are malfeasances and from the linguistic relativists who believe that whatever people say or write must necessarily be accepted. The purpose of Garner's dictionary is to help writers, editors, and speakers use the language effectively. And it does so in a playful and persuasive way that will help you sound "grammatical but relaxed, refined but natural, correct but unpedantic.". Read more Report an issue with this product or seller Previous slide of product details Print length 1312 pages 2. Language English 3. Publisher Oxford University Press 4. Publication date November 17, 2022 5. Dimensions 7.33 x 2.19 x 10.06 inches 6. ISBN-10 0197599028 7. ISBN-13 978-0197599020 Next slide of product details See all details The Amazon Book Review Book recommendations, author interviews, editors' picks, and more. 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He has written more than 25 books, many of them award-winning, including The Chicago Guide to Grammar, Usage, and Punctuation; Garner on Language and Writing; Reading Law: The Interpretation of Legal Texts and Making Your Case: The Art of Persuading Judges (both with Justice Antonin Scalia); Garner's Dictionary of Legal Usage; Legal Writing in Plain English; The Elements of Legal Style; The Winning Brief; and Ethical Communications for Lawyers. As editor in chief of Black's Law Dictionary and author of the grammar and usage chapter in The Chicago Manual of Style, he has become one of the most authoritative and widely cited English lexicographers. He is Distinguished Research Professor of Law at Southern Methodist University and president of LawProse Inc. Product details Publisher ‏ : ‎ Oxford University Press Publication date ‏ : ‎ November 17, 2022 Edition ‏ : ‎ 5th Language ‏ : ‎ English Print length ‏ : ‎ 1312 pages ISBN-10 ‏ : ‎ 0197599028 ISBN-13 ‏ : ‎ 978-0197599020 Item Weight ‏ : ‎ 2.31 pounds Dimensions ‏ : ‎ 7.33 x 2.19 x 10.06 inches Best Sellers Rank: #37,389 in Books (See Top 100 in Books) 19 in English Dictionaries & Thesauruses 31 in Linguistics Reference 59 in Dictionaries (Books) Customer Reviews: 4.6 4.6 out of 5 stars 154 ratings Brief content visible, double tap to read full content. Full content visible, double tap to read brief content. Videos Help others learn more about this product by uploading a video! Upload your video About the author Follow authors to get new release updates, plus improved recommendations. ## Bryan A. Garner Brief content visible, double tap to read full content. Full content visible, double tap to read brief content. Bryan A. Garner (born Nov. 17, 1958) is an American lawyer, grammarian, and lexicographer. He also writes on jurisprudence (and occasionally golf). He is the author of over 25 books, the best-known of which are Garner’s Modern English Usage (4th ed. 2016) and Reading Law: The Interpretation of Legal Texts (2012—coauthored with Justice Antonin Scalia), as well as four unabridged editions of Black’s Law Dictionary. He serves as Distinguished Research Professor of Law at Southern Methodist University. He also teaches from time to time at the University of Texas School of Law, Texas A&M School of Law, and Texas Tech School of Law. In 2009, he was named Legal-Writing and Reference-Book Author of the Decade at a Burton Awards ceremony at the Library of Congress. He has received many other awards, including the Benjamin Franklin Book Award, the Scribes Book Award, the Bernie Siegan Award, and a Lifetime Achievement Award from the Center for Plain Language. His work has played a central role in our understanding of modern judging, advocacy, grammar, English usage, legal lexicography, and the common-law system of precedent. His books are frequently cited by American courts of all levels, including the United States Supreme Court. His friendship with the novelist David Foster Wallace is memorialized in Quack This Way: David Foster Wallace and Bryan A. Garner Talk Language and Writing (2013). His friendship and writing partnership with Justice Antonin Scalia is depicted in the memoir Nino and Me: My Unusual Friendship with Justice Antonin Scalia (2018). Read more about this authorRead less about this author Related books Page 1 of 1Start Over Sponsored Previous page Shop the Store on Amazon › Learn English for Adult Beginners: 7 ESL Books in 1: 7 Steps to Master Grammar, Achieve Fluency, and Expand Your Vocabulary with 1,000 New Words & Phrases 4.44.4 out of 5 stars86 $35.97$35.97 2. 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Learn more how customers reviews work on Amazon Customers say Customers find this book to be a great reference guide that's fun to read, with one customer noting it's particularly useful for working writers and editors. Moreover, they appreciate its true love of language, with one review highlighting how it rewards and enriches readers' language appreciation. However, several customers report issues with missing and duplicated pages. Additionally, the book receives mixed feedback regarding its grammar content. AI Generated from the text of customer reviews Select to learn more Value for moneyUseWriting resourceFun to readLanguage contentGrammarMissing pagesPage duplicates 8 customers mention "Value for money"7 positive1 negative Customers find the book to be a great reference that offers good value for money. "Great book." Read more "What is there to say about this extraordinary work? If you are an admirer of limpid, elegant prose, you will cherish this book...." Read more "Superb and updated edition of B. Garner’s usage guide. This do teach they too rite goud. Seriously, buy this book. Garner is a genius." Read more "GARNER'S MODERN ENGLISH USAGE is an elegant and invaluable reference book of astonishing depth and precision, and it grows stronger with each edition..." Read more 6 customers mention "Use"6 positive0 negative Customers find the book inexhaustibly useful, with one customer noting it provides extensive guidance on common usage errors and another mentioning its practical value in an editing context. "...The hard copy is more fun and more practical in an editing context, but the Kindle version is more convenient and portable. You'll want both." Read more "Bryan Garner has become the greatest researcher of writing, grammar, usage, and language in North America...." Read more "...Of its countless virtues I'll single out only one: the astounding variety of advice...." Read more "...This is one of the most useful guides I've encountered as a Tech Editor...." Read more 5 customers mention "Writing resource"5 positive0 negative Customers find the book to be a valuable writing resource, with one customer noting it is indispensable as a reference and another mentioning it serves as a guide for working writers and editors. "...While it definitely is a resource for working writers and editors, it's also a book that amateurs and word lovers can read for fun...." Read more "Bryan Garner has become the greatest researcher of writing, grammar, usage, and language in North America...." Read more "...Manual of Style is fine for what it is but this is truly as fine a writing resource you could want as an editor...." Read more "Superb and updated edition of B. Garner’s usage guide. This do teach they too rite goud. Seriously, buy this book. Garner is a genius." Read more 3 customers mention "Fun to read"3 positive0 negative Customers find the book fun to read, with one noting it's particularly enjoyable for word lovers, while another describes it as highly entertaining. "...writers and editors, it's also a book that amateurs and word lovers can read for fun...." Read more "...It is also highly entertaining: witty, generous, and often playful, it rewards and enriches its reader's love of language, both written and spoken...." Read more "...The physical book itself is more fun." Read more 3 customers mention "Language content"3 positive0 negative Customers appreciate the language content of the book, with one mentioning how it rewards and enriches readers' love of language, while another notes its focus on language in North America. "...Garner has heart and a true love of language, and it shines through on every page...." Read more "...the greatest researcher of writing, grammar, usage, and language in North America...." Read more "...: witty, generous, and often playful, it rewards and enriches its reader's love of language, both written and spoken...." Read more 3 customers mention "Grammar"2 positive1 negative Customers have mixed opinions about the grammar in the book. "Bryan Garner has become the greatest researcher of writing, grammar, usage, and language in North America...." Read more "You need this if you write. Usage is not grammar...." Read more "...If you are an admirer of limpid, elegant prose, you will cherish this book...." Read more 7 customers mention "Missing pages"1 positive6 negative Customers report issues with missing pages in the book, noting that 50 pages are absent. "Pages are missing..." Read more "...to find that several of the pages duplicated and about 40 pages are therefore missing!" Read more "...Both are missing pages 501 to 548, and both have two sets of pages 549 to 596...." Read more "My copy (ordered on August 2, 2024) is missing pages 501-548, and pages 549-596 are duplicated...." Read more 3 customers mention "Page duplicates"0 positive3 negative Customers report issues with duplicate pages in the book. "...helpful and informative, but was surprised to find that several of the pages duplicated and about 40 pages are therefore missing!" Read more "...is missing pages 501-548, and pages 549-596 are duplicated...." Read more "Disappointed. 50 pages are missing and 50 other pages are duplicated. Had to return." Read more View Image Gallery Amazon Customer 5.0 out of 5 stars Images in this review Reviews with images See all photos Previous page Next page Amazon Customer 1 out of 5 stars This book is missing almost 50 pages I bought TWO copies of this reference; one for my home office and one for my office at work. Both are missing pages 501 to 548, and both have two sets of pages 549 to 596. Of course I'm discovering this almost a year after I bought the first copy, and six months after the second. Horribly disappointing considering Garner's reputation and that I'm out the cost of both books and still don't have an answer for the term I was looking for. MoreHide Thank you for your feedback Sorry, there was an error Sorry we couldn't load the review Top reviews from the United States There was a problem filtering reviews. Please reload the page. 5.0 out of 5 stars Amazing book Reviewed in the United States on April 13, 2023 Format: HardcoverVerified Purchase This amazing book is a bit difficult to pin down. It isn't a dictionary, though it has many thousands of words and definitions. It isn't a grammar book, though most elements of grammar appear on its pages. While it definitely is a resource for working writers and editors, it's also a book that amateurs and word lovers can read for fun.Basically, it's a big, thick, ambitious, almost comically comprehensive book that's best understood as a journey through our language organized in a dictionaryish way. You will find things in alphabetical order, but not every word is present, and entries can range from a sentence to several pages, depending on what the author found interesting.Garner has heart and a true love of language, and it shines through on every page. If he were more pedantic, or more practical, the book would lack its shimmer. This is not really a book to help you clean up your emails or decide whether to end a sentence in a preposition. There are plain-language guides and grammar books that'll help you out with that sort of thing much more efficiently.In the end, it's basically a book for language lovers to savor. Some people love words. I'm one of them. Spending an hour reading this book is a treat. If you love words and how theyexpress ideas, you will probably feel the same.If you're an editor, as I am, you really do need this book. Just hit the "Buy It Now" button.Buy the hard copy, then get the Kindle version. The hard copy is more fun and more practical in an editing context, but the Kindle version is more convenient and portable. You'll want both. Read more 15 people found this helpful Report Ross Bennett ##### 5.0 out of 5 stars The definitive usage reference in 2024 Reviewed in the United States on November 8, 2024 Format: HardcoverVerified Purchase Bryan Garner has become the greatest researcher of writing, grammar, usage, and language in North America. He writes the usage and grammar entries in The Chicago Manual of Style and has completely transformed the practice of professional writing in the legal and publishing professions. This is his master work in the general use of the English language.Unlike the exceptionally good Fowler's Oxford Dictionary of English Usage, Garner offers considered comparisons of how English is used in popular and exceptional publishing and communication. His use of the Internet's collection of the whole of English language trends and usage—notably including Ngram analysis of modern and contemporary English usage—offers usage guidance that has never been as credible or authentic.In short, Bryan Garner is the language teacher and usage specialist who has become the most respected in the world because he backs up his research with hard evidence and data—not just throwing opinions out there "because he said so."If you're a writer or editor in the United States, the first book you should buy is the Chicago Manual of Style. This one is the second. Read more One person found this helpful Helpful Report Walter G ##### 5.0 out of 5 stars Inexhaustibly Useful Reviewed in the United States on November 4, 2024 Format: HardcoverVerified Purchase What is there to say about this extraordinary work? If you are an admirer of limpid, elegant prose, you will cherish this book. Of its countless virtues I'll single out only one: the astounding variety of advice. One entry will instruct you in common usage errors (the difference, say, between adduce, educe, and deduce—heavy weather stuff) and the next will school you in the ways of parallelism (and many other tips for making your sentences sing). There are also charming rants along the way about AIRLINESE, for instance. So enamored am I of this book that I don't even mind its arriving a little beaten up. I imagine I'll be beating it up myself over many years—a lifetime really—of making sure I am not committing grammatical howlers. Read more 4 people found this helpful Helpful Report Amazon Customer ##### 5.0 out of 5 stars Great Usage and Style Manual Reviewed in the United States on February 22, 2024 Format: HardcoverVerified Purchase I just recently became aware of this style & usage guide. I don't usually write reviews but had to in this case. This is one of the most useful guides I've encountered as a Tech Editor. Chicago Manual of Style is fine for what it is but this is truly as fine a writing resource you could want as an editor. Since I received it, hardly a workday goes by in which I do not refer to it at least once - often many times per day. It is really true that you can just start on a page and read some entries just for the heck of it to learn something. Wonderful realistic examples for what to do and what not to do. I highly recommend it, I can't believe that I had not heard of this before - it is well worth the money. Oh and by the way, the book I received does not any missing pages as some of the other reviewers mentioned. Read more 5 people found this helpful Helpful Report ##### 4.0 out of 5 stars Great reference book but Kindle edition does not support sticky notes (for Kindle Scribe) Reviewed in the United States on April 11, 2023 Format: KindleVerified Purchase I have the hardbound volume as well as the Kindle edition, but I am mystified as to why the Kindle edition does not support sticky notes for use on the Kindle Scribe. This volume begs to be annotated! Read more 4 people found this helpful Helpful Report D. Kriner ##### 5.0 out of 5 stars Great book, thoughtless shipping Reviewed in the United States on January 1, 2025 Format: HardcoverVerified Purchase Will give the book itself a 5-star, as the product is fine, but for some reason someone thought it was a good idea to stick the clear packing list envelope directly on the front of the dust jacket, even though the book was not wrapped in plastic and THE PRODUCT WAS SENT IN A BOX. This is just stupid. I had to peel it off and use an adhesive-removing product to get the impossible glue off of the book. While ultimately, with considerable care and effort, I was able to restore the jacket to its original condition, no one should have to go through all of that for a new purchase. Maybe put the sticker on the box, like everyone else? Read more 5 people found this helpful Helpful Report Zoyd Wheeler ##### 5.0 out of 5 stars Superb Reviewed in the United States on December 1, 2024 Format: HardcoverVerified Purchase Superb and updated edition of B. Garner’s usage guide. This do teach they too rite goud. Seriously, buy this book. Garner is a genius. Read more One person found this helpful Helpful Report ##### 5.0 out of 5 stars You need this if you write. Usage is not grammar. Reviewed in the United States on February 1, 2023 Format: HardcoverVerified Purchase This guide is nearly indispensable as a writing reference. Since grammatically correct sentences can make no sense, it is helpful to have a guide to usage. For example, should you use “which” or “that” in your clause? Both are grammatically correct, but they have different meanings. What are you trying to express? I use this and my old copy of Fowler’s for such questions. This guide is very, very useful; I highly recommend it. Read more 12 people found this helpful Helpful Report See more reviews Top reviews from other countries Translate all reviews to English Amazon Customer ##### 1.0 out of 5 stars Pages are missing Reviewed in Singapore on July 16, 2023 Format: HardcoverVerified Purchase After page number 500, it starts from page number 549. Read more Amazon Customer 1.0 out of 5 stars ##### Pages are missing Reviewed in Singapore on July 16, 2023 After page number 500, it starts from page number 549. ###### Images in this review Ian Chadwick ##### 5.0 out of 5 stars How serious are you about the serial comma? Reviewed in Canada on February 17, 2024 Format: HardcoverVerified Purchase Do you wade into discussions on language forums and social media brandishing citations from your favourite authorities? Do you dismiss dissenting authorities as heretics? Are there style and usage guides on your bookshelf with sticky notes and bookmarks in them so you can immediately find your references should anyone post a contrary opinion? Do you haughtily refer to it as the Oxford comma instead of the serial — or, the gods of language forbid, the Harvard — comma?If you answered yes, I have a book for you. Well, for that apparently rare minority who read about language, grammar, style, and usage: Garner's Modern English Usage.For most of the past fifty years, I've had an edition of Fowler’s Modern English Usage on my bookshelves. In fact, today there are three of them: the second, third, and fourth print editions. I’ve been aware of Garner’s work since the late 1990s, when he published A Dictionary of Modern American Usage. In 2022, the fifth edition was published as Garner’s Modern English Usage: The Authority on Grammar, Usage, and Style. How could I resist that siren call?To my delight, Garner dedicates four whole pages to the serial comma, citing numerous authorities and examples both for and against its use. Four pages! Oh, be still my beating heart!In most entries, Garner provides examples, with information about how frequently variants appear in print, and how a change in usage or definition has progressed (his Language-Change Index), from level one (new or rejected) to five (Borg-like: resistance is futile because it has been fully accepted). And even more appealling to my senses: Garner writes with a wry sense of humour about many of the issues he covers.I highly recommend Garner’s Modern English Usage, fifth edition, to anyone who loves to read about language and might also have a Fowler’s or Dreyer's on their bookshelf. I only wish I could convince our local media to get — and read — a copy, too. Read more Report ##### 5.0 out of 5 stars They が単数だって？　ほんまでっか？ Reviewed in Japan on January 22, 2025 Format: HardcoverVerified Purchase 2019年、メリアム・ウェブスター社は「今年の言葉」に “they” を選んだ。その理由は、An employee with a grievance can file a complaint if they need to のように they が単数語の人称代名詞として使われるケース急増したからである。英語には person, anyone, everyone, no one など通性の一般語があるが、通性の三人称代名詞がない。そこで he が男女双方を示す代名詞として長らく使われてきた。つまり、Every-one can think for himself. の himself は herself を含むと解されてきたが、二十世紀後半になると、この用法は男女差別だとして社会的に許されなくなった。そこで　Everyone canthink for himself or herself を使う人が増えたが、この言い方はいかにもぎこちない。米国では 2015年に「同性愛者は同性婚の権利がある」とする最高裁判決が出た。 LGBTQ問題が脚光を浴びるなか、he または she にはトランスジェンダーが含まれていないとして、Everyone can think for themselves のように、通性の複数語人称代名詞 they が単数語の代名詞としても使われるようになった。とはいえ、they の単数語代名詞化には大きな問題がある。例えば、　A person is guilty of conspiracy if, together with one or others, they…の they は a person を示すのか、それとも共謀者全員を示すのか不明瞭になる。they の新用法を認めれば、多数の刑法条文を書き改める必要が生じるだろう。GMEUは、ある語句とその異形（variant）の使用頻度を Google Ngram Viewer を使って調べ、その語句の変化度指数 (Language Change Index) を下記の５段階で示した。　ステージ１ (認められない)　ステージ２ (あまり使われていない)　ステージ３ (広く使われているが. . .)　ステージ４ (あまねく使われているが. . .)　ステージ５ (みんなに認められている)They の新用法は標準文章英語においてどのような地位を占めるのか？ステージ５と主張する向きもあるが、大方はステージ３ないし４と見るのではなかろうか。世代交代が完了しなければステージ５にはならないだろう。本書は文法や語彙の変遷だけでなく、類語の微妙な違いについても明快な記述がある。語法辞典の最高峰といってよいだろう。 Read more Report Translate review to English See more reviews \| \| \| \| Your recently viewed items and featured recommendations › View or edit your browsing history After viewing product detail pages, look here to find an easy way to navigate back to pages you are interested in. 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10322	https://www.youtube.com/watch?v=fO2kUA4YPJY	Precalculus 1.1.7 - Introduction to {Sets} Learning with Gideon 718 subscribers 2 likes Description 64 views Posted: 21 Jul 2025 Introduction to Mathematical Sets In this video, we explore the foundational concept of sets in mathematics. Sets are simply collections of distinct elements, and understanding them is key to many areas of math. We start by defining what a set is and how to represent it using curly braces. Then, we introduce set builder notation, a powerful way to describe large or infinite sets using rules instead of listing every element. You’ll also learn about important concepts like the empty set, and how sets can be combined or compared using operations such as union and intersection. We use simple examples and visual aids like Venn diagrams to help you grasp these ideas clearly. What You'll Learn What a set is and how to define it Using set builder notation to describe sets with patterns or infinite elements The concept of the empty set and its notation Set operations: union and intersection Visualizing sets and operations with Venn diagrams Logical interpretations of union (or) and intersection (and) Why Sets Matter Sets are a fundamental building block in mathematics and help organize information clearly. Learning about sets prepares you for more advanced topics such as functions, probability, and logic. Mathematics Sets SetTheory MathBasics MathEducation MathConcepts UnionAndIntersection VennDiagrams SetBuilderNotation EmptySet LogicalOperators AlgebraFoundations MathForBeginners Feel free to pause and try out the examples yourself! Leave your questions in the comments, and don’t forget to subscribe for more math lessons. Transcript: Okay, today we are looking at sets. These are going to be very useful mathematical constructions. Uh let's look at see what a set actually is. Okay, so a set is simply a collection of distinct elements. And let me show you how we can define a set here. I'm going to define the set F. Okay, we like to use single uh letters to name sets. Uh F is going to be for fruits. And I use a curly brace here. And then I just list the elements of this set. And we're going to have in here we're going to have a apple, banana, coconut, and date. Okay. And then we finish it off here with another curly brace. So this defines what elements are inside my set f. Okay. Okay, so we can say that this here that's going to be the name of my set and then all of these things in here, these are all of the elements, right? That's what's inside of set F. So, it's pretty straightforward. We can define uh many different kinds of sets in many different ways, but this is the basic construction. Uh there's a couple of other important symbols we want to know. uh we're going to look at uh let's say we have apple. Okay, so apple turns out is an element of the set f. And so what we have here is a new symbol that uh describes uh this is the element symbol. So it says that apple is an element of f. What if something's not an element of this set? For example, oak. Okay, we say that oak is not an element of f. Very simple, right? So the idea here this symbol represents when something is not an element. Okay. So that's just a basic introduction to how we can define a set. But what if we want to define a set um that has more than four, five or six elements where it becomes very tedious to write down every single element. In fact, maybe you have an infinite number of elements. Well, then we're going to want to use a better way to describe what's inside of a set. We're going to use set builder notation. All right. So, let me first define a new set using the original method. I'm going to define the set A. And this is a set that consists of the elements 2, 3, four, and five. Okay. So, there we go. A set of four elements. Now, notice that there's a clear pattern here. This is sort of a set that consists of all of the integers from 2 to 5. So I can describe this set using set builder notation that basically encodes a pattern. And if you follow that pattern, you wind up with this set of four elements, right? Um so let's do that. You're also going to use a curly brace for set builder notation. But first we define a variable. In this case, we'll say if uh this set consists of all x such that x is an element of the set of integers and 1 is less than x is less than six and then we'll close out the uh the set here. Okay. So let's try and understand what we're writing down here. First we define a variable, right? And then this bar we can sort of read this as such that. And then here we have what we call a rule or sometimes also known as a predicate. Same idea. Okay. So we've got a variable x such that x is an element of the integers and one is less than x is less than six. And so if you follow this rule, you're going to take all of the integers so that are between 1 and six, not inclusive. Well, that's going to be the integers 2 3 4 and five. Right? So there we go. And you can see now this is sort of like uh writing a little computer program to define what your set is. It's just a rule with some variable that you're going to be using to plug in uh or follow the rule, right? Um, and very clearly we can modify the rule to create very large sets. This one's only four elements, but imagine if we made this, you know, uh, a 16. Well, now you've got 10 more elements, right? You can make it 116. You have a 100 more elements. You could even make infinitely large sets this way. Maybe you just have it so that it's x greater than one, right? That would be an infinitely large set. Now, so set builder notation is very powerful. It allows you to define very clearly uh extremely large sets following this kind of rule. There's another way we can also use set builder notation. So this is one way. Um but this part here when we're saying that x is an element of the set of integers, well that's uh what we call a domain, right? And so we can use that domain instead of just a variable here. So let me define the same exact set right here. We can say that a is the set of all x that is an element of the set of integers such that 1 is less than x is less than six. Okay. So in this case what we're doing is we are adding in our domain here first and then we're going to have our rule. So this is another valid use of the set builder notation. Okay let's take a look at the next very important point here. We're going to look at the idea of an empty set. All right. So, an empty set is an interesting concept. It's literally a set with no elements in it. And we can represent this either using the curly braces that we had before that are, you know, empty. There's nothing inside. Or we can use this with sort of a symbol uh with a zero with a with a stroke through it. Okay. So, both of these symbols represent the empty set. Okay. So for example, let's say I had the set P and I'm going to define this set P as an empty set. Okay. So what does this mean? Well, it means that P is literally an empty set that contains no elements. Now why is this important? Because this P is the empty set. This is not the same as zero. We cannot say that P is zero because zero is an integer, right? or a real number, it is an element. So having um P equal Z is not the same as having an empty set. All right, so that's an important distinction. Zero is actually an element that you could put into a set. All right, so now let's take a look at some important uses and uh operations essentially that we can use with sets. So the first operation we're looking at is the union operation and that is represented with this symbol here. It looks like a U and you can say this is a union B and that represents the set of all elements that are in A or that are in B. It's basically the combination of all the elements in set A plus all of the elements in set B. Right? So that's what we're doing here. Let me uh let me illustrate this for you. So, how about we define uh our set A here? And this is going to include oh, how about uh our apple and our banana. And then I'm also going to define our set B. And that's going to include how about our coconut and our date. Okay. So, here we have uh two different sets. And what we can do now is we can put them together, right? So we're going to have a union b. Now what does this represent? Well, this is going to be a new set that contains apple banana, coconut, and date. There you go. Okay, just the combination of all of the distinct elements that were in set A and uh all of the other elements that were in set B. Okay. Now, logically, this is similar to the logical or operator. And so, that's one way you might use this in the future. Let me show you another way we can use this. So, uh how about uh we define two new sets here. Okay. So, I'm going to define my first set here. How about the set C? That's going to equal uh the numbers 1 2 3. And then I will also define the set D. And that's going to equal the numbers uh how about 3 6 and 12. Okay, so there we go. Um three elements in each set. And now what we're going to do is we're going to have C union D. And we want to know what is this new set going to be? Well, you're going to have to combine all of these elements basically. So it'll be the numbers 1 2 3 6 and 12. Okay? So this is C union D. Now notice even though the number three is present in both of these sets when you combine them a set only includes distinct elements. So you can't have the number three represented twice right? That element is only going to be in the set one time. Okay. So this is the union operator. Let's look at the next operator intersection. So here we see the new notation. It's an upside down u. Uh sometimes we call uh this uh a cap right so sort of like a cap. Sometimes we call the u a cup. Um those are alternate notations but uh this we can read as a intersection b. So this is going to result in a new set that is the intersection here. So it's the set of all elements that are in a and also in b. Okay, this is basically the logical and operator. So let's uh let's look at an example here. Just as before, we're going to have our set C. And that was the elements 1 2 3. And then we also had our set D. And those consisted of the elements of 3 6 and 12. Okay. So now we want to use the intersection operator. So we're going to have C intersection D. And this is going to give us a new set. Now what do you think are going to be the elements in this set? Well, it's going to be all of the elements that are in C and also in B. So, uh, this element one is in C, but it's not also in D, so we can't include it. Same thing with the number two. It's not also in D. The number three is in C and also in D. So, we do include the element three. And in fact, that's the only element because 6 and 12 are not also in C. So this intersection between C and D results in a set with only a single element in this case the number three. What if we try this with two other sets? Okay, what if we try this with uh how about the set uh E? Let's call this uh 1 2 3. And also how about the set F. Okay, let's define this as the elements four five and six. Okay, so now we want to do this again. Let's get the intersection. Oh, I guess we're intersecting E and F. Okay, so what is this going to be? Well, if you notice now, there's there's no overlap. There are no elements that are shared in E and also in F. And so we're going to get is an empty set, also known as the empty set here. Okay, so the intersection is basically the logical and operator. Let me see if I can show you a graphical way to interpret these operations. So it's very common to illustrate these kinds of union operators using these kinds of vin diagrams. We kind of see how things overlap or don't what the result is going to be. Okay, so first let's say that this circle represents set A and this circle represents all of the elements in set B. and we're going to do a union b. We want to know what do we get as a result. So what we're going to get is we're going to look and get all of the elements here just like this. So it's all of the elements. So if an element is in set A or is in set B then it will be in the new set A union B. And so that means you grab all of the elements that are present here. Let's take a look at the intersection now. Okay. So again, this is set A and this is set B. So we're going to have A uh intersection B. Let me draw that a little bit better. Intersection B. So what's going to be in this new set? Well, the elements in this new set must be in set A and also in set B. And so in this case, we are only going to grab this part where they overlap because only in this central region do we have elements that are in set A and also in set B. And so that's how we can visualize the union and intersection operation. Okay, hopefully this was a good overview of sets and we will be using this as a tool in the future.
10323	https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:foundation-algebra/x2f8bb11595b61c86:substitute-evaluate-expression/v/evaluating-expressions-in-two-variables-with-decimals-and-fractions	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.
10324	https://www.theserverside.com/blog/Coffee-Talk-Java-News-Stories-and-Opinions/println-vs-print-difference-printf-Java-newline-when-to-use	The difference between print vs println in Java Search the TechTarget Network Sign-up now. Start my free, unlimited access. LoginRegister Explore the Network TechTarget Network App Architecture Software Quality Cloud Computing Security AWS RSS Core Java DevOps More Topics Development Frameworks Skills and career Tools Other Content Videos Definitions Webinars 2024 IT Salary Survey Results Sponsored Sites More Answers Features Opinions Quizzes Tech Accelerators Tips Tutorials Follow: Home Core Java APIs and programming techniques Coffee Talk: Java, News, Stories and Opinions BLOG The difference between print vs println in Java Cameron McKenzie TechTarget 11 Jul 2025 About This Blog Community driven content discussing all aspects of software development from DevOps to design patterns. Latest Blog Posts How to git cherry-pick a commit example What are the advantages of Java? Create your first Python AWS Lambda function in minutes See More Share this item with your network: Related Content System.out in Java explained– TheServerSide.com String to long in Java– TheServerSide.com Find duplicates in a List in Java– TheServerSide.com Sponsored News Top 3 Strategies for Solving Out-of-Network Challenges for Members–Zelis Healthcare See More Vendor Resources Quocirca: Selecting a cloud printing platform–TechTarget ComputerWeekly.com Print security: An imperative in the IoT era–TechTarget ComputerWeekly.com A Computer Weekly E-Guide on Wired and Wireless LAN–TechTarget ComputerWeekly.com Java’s print vs println methods: What’s the difference? The key difference between Java’s print and println methods is that println appends a newline character (‘\n’) to output, while Java’s print method does not. Unlike println, the print method can be called multiple times and all text sent to the console will appear on the same line. In contrast to print, all subsequent calls println place subsequent output on new lines. print and println examples For example, the following code snippet uses Java’s println method to output the words “Vibe” and “Coding” to two separate lines: System.out.println("Vibe"); System.out.println(" coding!"); Execution of this code results in the following output: Vibe coding! Java print method example The next example is similar to the one above, with the only difference that it uses println vs print. System.out.print("Vibe "); System.out.print("coding!"); Execution of this code results in the text appearing all on one line: Vibe Coding! Java’s newline character Java’s print method can be made to behave like the println method. Simply append the newline escape sequence (‘\n’) to the text String passed into the method, like so: System.out.print("Hello " + "\n"); // newline added with + operator System.out.print("world!\n"); // newline included in text String This newline escape sequence example generates the following output when executed: Hello world! When to use print vs println? The choice between print and println hinges largely on what the developer hopes to achieve. For example, a developer should use print methods if they want to describe the value of two variables with output like this: The value of x is 50, and the value of y is 100. A simple Java print method example The code to achieve the above output is as follows: long x = 50; long y = 100; System.out.print("The value of x is "); System.out.print(x); System.out.print(", and the value of y is "); System.out.print(y); System.out.print("."); A simple Java println method example In contrast to the above example, imagine the developer wants to display output in the following way: The value of x is currently 50. The value of y is currently 100. The code to achieve this output relies on Java println method calls: long x = 50; long y = 100; System.out.println("The value of x is currently " + x + "."); System.out.println("The value of y is currently " + y + "."); Java println vs printf methods For advanced formatting of data and variables, Java provides a printf method that gives much more control to the developer. Here is an example of how to use the Java printf method to format output that uses two text variables and adds a newline character: String name = "Cameron"; String site = "TechTarget"; System.out.printf("I like the stuff %s writes on %S. %n", name, site); / Printf output: I like the stuff Cameron writes on TECHTARGET. / Java methods such as print, println and printf all offer the developer different ways to control output that gets sent to the console or streamed to the log files. However, for application logging, one should use a properly framework such as log4J. Application logging with Java’s print and println methods is not recommended. The printf method provides extra functionality not found in Java print and println methods. Cameron McKenzie is an AWS Certified AI Practitioner, Machine Learning Engineer, Solutions Architect and author of many popular books in the software development and Cloud Computing space. 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10325	https://www.vedantu.com/question-answer/the-diagonals-of-a-rectangle-are-perpendicular-class-10-maths-cbse-5eff26c94060654129168e4f	The diagonals of a rectangle are perpendicular to each other.(a). True(b). False(c). Only when the rectangle is also a rhombus.(d). None of these. 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Please get in touch with us Vedantu Store Question Answer Class 10 Maths The diagonals of a rectangle a... Answer Question Answers for Class 12 Class 12 Biology Class 12 Chemistry Class 12 English Class 12 Maths Class 12 Physics Class 12 Social Science Class 12 Business Studies Class 12 Economics Question Answers for Class 11 Class 11 Economics Class 11 Computer Science Class 11 Biology Class 11 Chemistry Class 11 English Class 11 Maths Class 11 Physics Class 11 Social Science Class 11 Accountancy Class 11 Business Studies Question Answers for Class 10 Class 10 Science Class 10 English Class 10 Maths Class 10 Social Science Class 10 General Knowledge Question Answers for Class 9 Class 9 General Knowledge Class 9 Science Class 9 English Class 9 Maths Class 9 Social Science Question Answers for Class 8 Class 8 Science Class 8 English Class 8 Maths Class 8 Social Science Question Answers for Class 7 Class 7 Science Class 7 English Class 7 Maths Class 7 Social Science Question Answers for Class 6 Class 6 Science Class 6 English Class 6 Maths Class 6 Social Science Question Answers for Class 5 Class 5 Science Class 5 English Class 5 Maths Class 5 Social Science Question Answers for Class 4 Class 4 Science Class 4 English Class 4 Maths The diagonals of a rectangle are perpendicular to each other. (a). True (b). False (c). Only when the rectangle is also a rhombus. (d). None of these. Answer Verified 575.1k+ views 1 likes Hint: Draw a rectangle, with diagonals. From the figure, find out if the diagonals are perpendicular to each other or not, if they are then the statement is true or else the statement is false. Complete step-by-step answer: - Consider the rectangle ABCD drawn below. We know that a rectangle has 2 diagonals. From the figure we can see that AC and BD are the diagonals of the rectangle ABCD. From the figure we can understand that each one is a line segment drawn between the opposite corners of the rectangle. The diagonals of the rectangle are equal i.e. AC = BD and they bisect each other. But the diagonals are not perpendicular to each other. We know that if diagonals are perpendicular then they cut at 90∘. But in the rectangle the diagonals don’t cut at 90∘. Thus the statement given is false. If in case of square and rhombus, the diagonals are perpendicular to each other. But for rectangles, parallelograms, trapeziums the diagonals are not perpendicular. ∴ The diagonals of a rectangle are not perpendicular to each other. The given statement is false. ∴ Option (b) is the correct answer. Note: If you are having doubt about the diagonals if they are perpendicular or not always draw a figure and confirm it. If we draw a rectangle with diagonals, we can see that they are not perpendicular. If we draw a square, their diagonals are always perpendicular. 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10326	https://study.com/learn/lesson/what-is-the-opposite-of-a-number.html	Opposite Numbers \| Definition & Facts - 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 Copyright Math Courses / Math for Kids Course Opposite Numbers \| Definition & Facts Lesson Transcript Yuanxin (Amy) Yang Alcocer, Rayna Cummings Author Yuanxin (Amy) Yang Alcocer Amy has a master's degree in secondary education and has been teaching math for over 9 years. Amy has worked with students at all levels from those with special needs to those that are gifted. View bio Instructor Rayna Cummings Rayna has taught Elementary Education for 12 years (in both 1st, 2nd, and 3rd grades) and holds a M.Ed in Early Childhood Education from The Ohio State University View bio Learn how opposite numbers are defined, understand how to find the opposite of a number, and discover the opposite of a number that is not an integer. See various opposite numbers and learn important facts. Updated: 11/21/2023 Table of Contents Number Line Opposite Numbers Definition Facts Related to Opposite Number Other Opposite Numbers Lesson Summary Show Frequently Asked Questions What is the opposite of 12? The opposite of 12 is -12. This method uses the changing the sign method where positive signs or no signs are changed to negative signs. What is the opposite of 9? The opposite of 9 is -9. The easiest method to find an opposite number is to simply change the sign of a number. What are opposite numbers called? Opposite numbers are those numbers that are on opposite sides of the number line. The 0 on the number acts as a mirror and opposite numbers are mirror images of each other. What is an opposite number in math? An opposite number in math are numbers that are on opposite sides of the number line. The number line switches sides at point 0. What is the opposite number of 80? The opposite to the number 80 is -80. The sign in front of the 80 is a plus, so the opposite number is -80 by changing the positive to a negative. Create an account Table of Contents Number Line Opposite Numbers Definition Facts Related to Opposite Number Other Opposite Numbers Lesson Summary Show Number Line ----------- A ruler is a number line that helps people make measurements. A number line is a line of equally spaced tick marks with evenly spaced numbers written in order from least to greatest going from left to right. Number lines can be as large or as small as needed. The first number line that people are exposed to in school is the simple 0 to 10 number line. Number line 0 to 10 This number line is spaced so that each tick mark is 1 unit long. Number lines can be spaced in any manner that makes it easy to use. For example, when working with larger numbers up to 100, a number line where each tick mark is 10 units long may work better. Number 0 to 100 This number line has 10 units for each tick mark. It makes it easy to see larger numbers and see how they compare to other numbers. Number lines can be drawn to fit any kind of number need. Number lines can contain just decimals or just fractions. Number lines are used to help with addition, subtraction, and even multiplication. Another math concept that number lines can help with is that of opposite numbers. To unlock this lesson you must be a Study.com Member. Create your account Click for sound 2:53 You must c C reate an account to continue watching Register to view this lesson Are you a student or a teacher? I am a student I am a teacher Create Your Account To Continue Watching As a member, you'll also get unlimited access to over 88,000 lessons in math, English, science, history, and more. Plus, get practice tests, quizzes, and personalized coaching to help you succeed. Get unlimited access to over 88,000 lessons. Try it now Try it now. Already registered? Log in here for access Back Resources created by teachers for teachers Over 30,000 video lessons& teaching resources‐all in one place. Video lessons Quizzes & 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 Back Coming up next: Rectangular Numbers \| Definition, Properties & Types You're on a roll. Keep up the good work! Take QuizWatch Next Lesson Replay Just checking in. Are you still watching? Yes! Keep playing. Your next lesson will play in 10 seconds 0:03 The Same, But Different 1:22 It Is All in the Signs 1:58 Other Types of… 2:31 Lesson Summary View Video Only Save Timeline 59K views Video Quiz Course Video Only 59K views Opposite Numbers Definition --------------------------- The opposite numbers definition is that they are the numbers that are on the opposite side of the number line. An opposite number is the mirror image of the number across the number line if the mirror is located at the 0 point. For example, the opposite number to the number 5 is -5 because 5 is five tick marks to the right of the 0 on the number line. If that is the mirror, then going 5 spaces to the left to find its mirror image and opposite number, the result is -5 since going to the left of 0 results in negative numbers. The opposite of 5 is -5 Performing the same steps yields the following opposites on the above number line. The opposite of 4 is -4 The opposite of 3 is -3 The opposite of 2 is -2 The opposite of 1 is -1 0 has no opposite since that is the location of the mirror or the point of reference. When a number is paired with its opposite number, they form an additive inverse pair since adding them together results in 0. For example, adding a 5 with its opposite -5 results in 0. The same is true for all the other above examples. 5+(−5)=0 4+(−4)=0 3+(−3)=0 2+(−2)=0 1+(−1)=0 How to Get the Opposite of a Number The easiest way to find the opposite number of a number is to simply change the sign in front of it. If it is positive or no sign is present, then add a minus or negative sign. If there is a negative or minus sign, then get rid of it or change it to a positive. For example, to change the 5 to its opposite, add a negative since there is no sign in front of it. For a +5, change the plus sign in front of it to a negative. For a -5, change the minus to a plus or remove the sign. This gives the following results. \| Number \| Opposite Number \| --- \| \| 5 \| -5 \| \| +5 \| -5 \| \| -5 \| 5 \| \| -5 \| +5 \| An alternate way to do this is to multiply the number by a -1. 5×−1=−5 5∗(−1)=−5 (−5)(−1)=5 Opposite Numbers Table This changing of the sign from negative to positive and vice versa works for all kinds of real numbers: integers, fractions, decimals, irrationals, rational numbers, and even imaginary numbers. Again, to find the opposite, simply change the sign in front of the number. \| Type of Number \| Number \| Opposite \| --- \| Integer \| 11 \| -11 \| \| Fraction \| -½ \| ½ \| \| Decimal \| 9.34 \| -9.34 \| \| Irrational \| π \| -π \| \| Rational \| -⅛ \| ⅛ \| This method is a quick and easy way to find out what the opposite of a number is. To unlock this lesson you must be a Study.com Member. Create your account Facts Related to Opposite Number -------------------------------- Being able to spot the opposite of a number comes in handy when evaluating mathematical expressions. Spotting such a pair means those numbers add up to 0. When there's a pair, it makes the pair an additive inverse. It is very important to note here that just being a negative number does not mean the number is an additive inverse. A number is only an additive inverse when its opposite is present. And additive inverses can be positive. For example, the additive inverse to a -5 is a positive 5. The additive inverse to any negative number is a positive number. To unlock this lesson you must be a Study.com Member. Create your account Other Opposite Numbers ---------------------- Opposite numbers work the same way for fractions and decimals as seen in the table above. If a fraction or decimal is positive, change it to a negative. If a fraction or decimal is negative, then change it to a positive. The opposite of 7 8 is −7 8 The opposite of −4.3 is 4.3 This same method also holds true for imaginary numbers. The important point to keep in mind here is that both terms of an imaginary number changes sign. For example, the imaginary number 14−4 i has an opposite imaginary number of −14+4 i. Each term of the imaginary number changes sign in the opposite number. The opposite of 14−4 i is −14+4 i To unlock this lesson you must be a Study.com Member. Create your account Lesson Summary -------------- In review, a number line is a line of equally spaced tick marks with evenly spaced numbers written in order from least to greatest going from left to right. The opposite numbers definition is that they are the numbers that are on the opposite side of the number line. The easiest way to find the opposite number of a number is to simply change the sign in front of it. If it is positive or no sign is present, then add a minus or negative sign. If there is a negative or minus sign, then get rid of it or change it to a positive. This changing of the sign from negative to positive and vice versa works for all kinds of real numbers: integers, fractions, decimals, irrationals, rational numbers, and even imaginary numbers. A number paired with its opposite number makes up an additive inverse because when this pair is added together, it will always equal 0. It is very important to note here that just being a negative number does not mean the number is an additive inverse. A number is only an additive inverse when its opposite is present. To unlock this lesson you must be a Study.com Member. Create your account Video Transcript The Same, But Different Quick! What is the opposite of hot? Cold! How about the opposite of slow? Fast! OK. So, as you know, opposite words have different meanings. Well, opposite numbers have different values, but they look very similar. Let's learn more about how to identify opposite numbers. First, let's review our information about a number line. A number line is a horizontal line with numbers that are placed equal distance apart and that are sequentially numbered, meaning the numbers increase or decrease by the same amount. For example, your number line could count by 1s, by 5s, or even by 100s. Opposite numbers are numbers that, when placed on a number line, are the exact same distance away from the 0, but on opposite sides, or in opposite directions. They look the same except for their signs. To the left of the 0 on the number line are negative numbers, with a - sign, and to the right of the zero on the number line are positive numbers which although they have a positive, + , sign, this sign is normally not written. Opposite numbers number line Examples of opposite numbers on our 'Opposite numbers number line' would be -1 and +1, -2 and +2, and so on. Again, opposite number look exactly the same except for their sign, which is based on whether the number is to the left or the right of the 0 on the number line. It Is All in the Signs Good news! You don't have to draw a number line every time you want to find out what the opposite of one number is. Now that you know that opposite numbers are the same number, one with a positive sign (+) and one with a negative sign (-), you can easily find the opposite of any number. So, even if I asked you what the opposite of 454,685,184 was, you would immediately say - 454,685,184! Both of these numbers are the exact same distance away from the 0 on the number line, just in opposite directions! Other Types of Opposite Numbers The easiest way to learn about opposite numbers is to build them and see firsthand how these numbers are opposite. Any numbers can be opposite, such as this example, showing the opposite for the fractions 1/4 and 2/5. Examples of Fraction Opposite Numbers As long as one number has a positive (+) sign and the other number has the negative (-) sign, you have opposites. Opposite numbers can also be decimal numbers. Look at this example: Example of opposite decimal numbers in a number line The opposite of 0.2 is - 0.2 on the number line. Can you name the other opposite decimal numbers shown in the decimal number line picture? Lesson Summary As you just learned from this lesson, opposite numbers are special numbers that can be shown on both sides of a number line with different signs, one with a + and one with a -. You have learned different ways to make opposite numbers from the example pictures, using a number line, and with creating opposite numbers of your own. 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. 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Try it now Math for Kids 23 chapters \| 325 lessons Ch 1. Numbers for Elementary School Ch 2. Geometry for Elementary School Ch 3. Fractions for Elementary School Ch 4. Math Basics for Elementary... Ch 5. Statistics for Elementary School Ch 6. Number Properties for Elementary... Ch 7. Algebra for Elementary School Ch 8. Math Patterns for Elementary... Ch 9. History of Math for Elementary... Ch 10. Math Terms for Elementary School Ch 11. Working with Numbers for Elementary... Ch 12. Types of Numbers for Elementary School What is an Abundant Number? What is a Base Number? 2:32 Cardinal Numbers \| Overview, Definition & Examples 3:06 Consecutive Numbers \| Definition, Example & Formulas 3:22 Finding the Sum of Consecutive Numbers 4:10 What is a Cubed Number? 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10327	https://math.stackexchange.com/questions/661433/how-can-we-rigorously-prove-that-a-rectangle-is-a-parallelogram	geometry - how can we rigorously prove that a rectangle is a parallelogram? - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more how can we rigorously prove that a rectangle is a parallelogram? Ask Question Asked 11 years, 8 months ago Modified11 years, 7 months ago Viewed 3k times This question shows research effort; it is useful and clear 0 Save this question. Show activity on this post. I know that a rectangle has 2 opposite equal and parallel sides but is that enough to say that a rectangle is a parallelogram , i mean how can someone prove it in geometry, any help appreciated geometry Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Feb 4, 2014 at 8:50 Gerry Myerson 187k 13 13 gold badges 235 235 silver badges 408 408 bronze badges asked Feb 3, 2014 at 0:36 kabarykabary 281 5 5 silver badges 10 10 bronze badges 9 In Euclidean geometry, a parallelogram is a simple (non self-intersecting) quadrilateral with two pairs of parallel sides. A rectangle has two pairs of parallel sides, thus a rectangle is a parallelogram.Zafer Cesur –Zafer Cesur 2014-02-03 00:42:12 +00:00 Commented Feb 3, 2014 at 0:42 It seems like you've already proved it.qwr –qwr 2014-02-03 00:42:27 +00:00 Commented Feb 3, 2014 at 0:42 If you are looking to be rigorous, you should prove that a rectangle (a simple quadrilateral with four pairs of perpendicular adjacent sides) has two pairs of parallel sides.David H –David H 2014-02-03 00:52:48 +00:00 Commented Feb 3, 2014 at 0:52 What is the definition of a parallelogram? That will make it much clearer.Sawarnik –Sawarnik 2014-02-03 14:45:35 +00:00 Commented Feb 3, 2014 at 14:45 @Sawarnik A parallelogram is a quadrilateral with two pairs of parallel sides.David H –David H 2014-02-03 15:10:28 +00:00 Commented Feb 3, 2014 at 15:10 \|Show 4 more comments 1 Answer 1 Sorted by: Reset to default This answer is useful 2 Save this answer. Show activity on this post. Ultimately, it boils down to definitions. Any level of rigor can only be established if you can define the object you are talking about. Essentially, what I am asking is what do you mean by this thing, parallelogram? Before we work, we need to pinpoint that what it is exactly. So using Wikipedia's definition of parallelograms: In Euclidean geometry, a parallelogram is a simple (non self-intersecting) quadrilateral with two pairs of parallel sides. And, for rectangle: In Euclidean geometry, a rectangle is any quadrilateral with four right angles. The above two definitions pinpoint what exactly a parallelogram and rectangle, so we can start working without any ambiguity or problems. So what does it mean that a rectangle is a parallelogram? We can say that it means the definition of rectangle fits in the definition of parallelogram. So, the question is, that is any quadrilateral with four right angles [in short, rectangles] has two pairs of parallel sides and is simple (non self-intersecting)? So what are parallel sides, then? Obviously, they can be defined as any two lines in a plane that never meet. But in Euclidean geometry, the one we are working in, we have an axiom (they are statements that cannot really be proven but have to assumed to move ahead) to work with, called the parallel postulate (Google if you want to know more). An equivalent version states: Lines m and l are both intersected by a third straight line (a transversal) in the same plane, and the corresponding angles of intersection with the transversal are congruent, then m and l are parallel. We can see that what we want is an immediate corollary of this. Let the lines m and l be the two [extended] non adjacent sides of the rectangle and we see that the corresponding angles of intersection are 90∘ and hence congruent, the exact thing can be done for the other pair of sides. So, we have proved that the rectangle has two pairs of parallel sides. Now, since the adjacent sides are parallel, they do not intersect [remember the definition], and hence is simple. Thus, every condition is met with, and hence a rectangle is a parallelogram. If you want to make it feel more rigorous, you could rewrite the stuff in formal logic, use stricter definitions or whatever, but I think it is useless to do so, because it does not help us anyway.. I have introduced and specifically wrote out the definitions [and axioms, another important concept, both of which have to be assumed], because I feel that is where you are going into problems. You wrote the properties of the rectangle, that would easily fit within the parallelogram definition, but you did not know when or how to use it. So, I started with definitions and then manipulated to get from one end to the other, so you get a better feel of what is happening. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Feb 17, 2014 at 18:35 answered Feb 3, 2014 at 15:35 SawarnikSawarnik 7,434 7 7 gold badges 37 37 silver badges 79 79 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 geometry See similar questions with these tags. 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10328	https://catdir.loc.gov/catdir/toc/ecip053/2004026995.html	Table of contents for Brock biology of microorganisms Table of contents for Brock biology of microorganisms / Michael T. Madigan, John M. Martinko. Bibliographic record and links to related information available from the Library of Congress catalog. Note: Contents data are machine generated based on pre-publication provided by the publisher. Contents may have variations from the printed book or be incomplete or contain other coding. I CONTENTS UNIT I PRINCIPLES OF MICROBIOLOGY CHAPTER 1 MICROORGANISMS AND MICROBIOLOGY I INTRODUCTION TO MICROBIOLOGY 1 1.1 Microbiology 1.2 Microorganisms as Cells 1.3 Microorganisms and Their Natural Environments 1.4 The Impact of Microorganisms on Humans II PATHWAYS OF DISCOVERY IN MICROBIOLOGY 9 1.5 The Historical Roots of Microbiology: van Leeuwenhoek and Cohn 1.6 Pasteur, Koch, and Pure Cultures 1.7 Microbial Diversity and the Rise of General Microbiology 1.8 The Modern Era of Microbiology CHAPTER 2 AN OVERVIEW OF MICROBIAL LIFE I CELL STRUCTURE AND EVOLUTIONARY HISTORY 22 2.1 Elements of Cell and Viral Structure 2.2 Arrangement of DNA in Microbial Cells 2.3 The Tree of Life II MICROBIAL DIVERSITY 27 2.4 Physiological Diversity of Microorganisms 2.5 Prokaryotic Diversity 2.6 Eukaryotic Microorganisms CHAPTER 3 MACROMOLECULES I CHEMICAL BONDING AND WATER IN LIVING SYSTEMS 39 3.1 Strong and Weak Chemical Bonds 3.2 An Overview of Macromolecules and Water as the Solvent of Life II NONINFORMATIONAL MACROMOLECULES 43 3.3 Polysaccharides 3.4 Lipids III INFORMATIONAL MACROMOLECULES 3.5 Nucleic Acids 3.6 Amino Acids and the Peptide Bond 3.7 Proteins: Primary and Secondary Structure 3.8 Proteins: Higher Order Structure and Denaturation CHAPTER 4 CELL STRUCTURE/FUNCTION I I MICROSCOPY AND CELL MORPHOLOGY 4.1 Light Microscopy 4.2 Three-Dimensional Imaging: Interference Contrast, Atomic Force, and Confocal Scanning Laser Microscopy 4.3 Electron Microscopy 4.4 Cell Morphology and the Significance of Being Small II CELL MEMBRANES AND WALLS 4.5 Cytoplasmic Membrane: Structure 4.6 Cytoplasmic Membrane: Function 4.7 Membrane Transport Systems 4.8 The Cell Wall of Prokaryotes: Peptidoglycan and Related Molecules 4.9 The Outer Membrane of Gram-Negative Bacteria III SURFACE STRUCTURES AND INCLUSIONS OF PROKARYOTES 4.10 Bacterial Cell Surface Structures 4.11 Cell Inclusions 4.12 Gas Vesicles 4.13 Endospores IV MICROBIAL LOCOMOTION 4.14 Flagella and Motility 4.15 Gliding Motility 4.16 Cell Motion as a Behavioral Response: Chemotaxis and Phototaxis CHAPTER 5 NUTRITION, LABORATORY CULTURE, AND METABOLISM OF MICROORGANISMS I I NUTRITION AND CULTURE OF MICROORGANISMS 5.1 Microbial Nutrition 5.2 Culture Media 5.3 Laboratory Culture of Microorganisms II ENERGETICS AND ENZYMES 5.4 Bioenergetics 5.5 Catalysis and Enzymes III OXIDATION-REDUCTION AND ENERGY-RICH COMPOUNDS 5.6 Oxidation-Reduction 5.7 NAD as a Redox Electron Carrier 5.8 Energy-Rich Compounds and Energy Storage IV MAJOR CATABOLIC PATHWAYS, ELECTRON TRANSPORT, AND THE PROTON MOTIVE FORCE 5.9 Energy Conservation: Options 5.10 Glycolysis as an Example of Fermentation 5.11 Respiration and Membrane-Associated Electron Carriers 5.12 Energy Conservation from the Proton Motive Force V CARBON FLOW IN RESPIRATION AND CATABOLIC ALTERNATIVES 5.13 Carbon Flow in Respiration: The Citric Acid Cycle 5.14 Catabolic Alternatives VI BIOSYNTHESIS 5.15 Biosynthesis of Sugars and Polysaccharides 5.16 Biosynthesis of Amino Acids and Nucleotides 5.17 Biosynthesis of Fatty Acids and Lipids CHAPTER 6 MICROBIAL GROWTH I BACTERIAL CELL DIVISION 6.1 Cell Growth and Binary Fission 6.2 Fts Proteins, the Cell Division Plane, and Cell Morphology 6.3 Peptidoglycan Synthesis and Cell Division II GROWTH OF BACTERIAL POPULATIONS 6.4 Growth Terminology and the Concept of Exponential Growth 6.5 The Mathematics of Bacterial Growth 6.6 The Growth Cycle III MEASURING MICROBIAL GROWTH 6.7 Direct Measurements of Microbial Growth: Total and Viable Counts 6.8 Indirect Measurements of Microbial Growth: Turbidity 6.9 Continuous Culture: The Chemostat IV ENVIRONMENTAL EFFECTS ON MICROBIAL GROWTH: TEMPERATURE 6.10 Effect of Temperature on Growth 6.11 Microbial Growth at Cold Temperatures 6.12 Microbial Growth at High Temperatures V ENVIRONMENTAL EFFECTS ON MICROBIAL GROWTH: pH, OSMOLARITY, AND OXYGEN 6.13 Microbial Growth at Low or High pH 6.14 Osmotic Effects on Microbial Growth 6.15 Oxygen and Microbial Growth 6.16 Toxic Forms of Oxygen I CHAPTER 7 ESSENTIALS OF MOLECULAR BIOLOGY I GENES AND GENE EXPRESSION 7.1 Macromolecules and Genetic Information II II DNA STRUCTURE 7.2 DNA Structure: The Double Helix 7.3 DNA Structure: Supercoiling 7.4 Chromosomes and Other Genetic Elements III III DNA REPLICATION 7.5 DNA Replication: Templates and Primers 7.6 DNA Replication: The Replication Fork IV TOOLS FOR MANIPULATING DNA 7.7 Restriction Enzymes and Hybridization 7.8 Sequencing and Synthesizing DNA 7.9 Amplifying DNA: The Polymerase Chain Reaction V V RNA SYNTHESIS: TRANSCRIPTION 7.10 Overview of Transcription 7.11 Diversity of Sigma Factors, Consensus Sequences, and Other RNA Polymerases 7.12 Transcription Terminators 7.13 The Unit of Transcription VI VI PROTEIN SYNTHESIS 7.14 The Genetic Code 7.15 Transfer RNA 7.16 Translation: The Process of Protein Synthesis 7.17 Folding and Secreting Proteins CHAPTER 8 METABOLIC REGULATION I OVERVIEW OF REGULATION 8.1 Major Modes of Regulation II II REGULATION OF ENZYME ACTIVITY 8.2 Inhibiting Enzyme Activity 8.3 Covalent Modification of Enzymes III DNA-BINDING PROTEINS AND REGULATION OF TRANSCRIPTION BY NEGATIVE AND POSITIVE CONTROL 8.4 DNA Binding Proteins 8.5 Negative Control of Transcription: Repression and Induction 8.6 Positive Control of Transcription IV GLOBAL REGULTORY MECHANISMS 8.7 Global Control and the lac Operon 8.8 The Stringent Response 8.9 Other Global Control Networks 8.10 Quorum Sensing IV V OTHER MECHANISMS OF REGULATION 8.11 Attenuation 8.12 Signal Transduction and Two-Component Regulatory System 8.13 Regulation of Chemotaxis 8.14 RNA Regulation and Riboswitches CHAPTER 9 ESSENTIALS OF VIROLOGY I VIRUS AND VIRION 9.1 General Properties of Viruses 9.2 Nature of the Virion II GROWTH AND QUANTIFICATION 9.3 The Virus Host 9.4 Quantification of Viruses III VIRAL REPLICATION 9.5 General Features of Virus Replication 9.6 Virus Multiplication: Attachment and Penetration 9.7 Virus Multiplication: Production of Viral Nucleic Acid and Protein IV IV VIRAL DIVERSITY 9.8 Overview of Bacterial Viruses 9.9 Virulent Bacteriophages and T4 9.10 Temperate Bacteriophages 9.11 Bacteriophage Lambda 9.12 Overview of Animal Viruses 9.13 Retroviruses V SUBVIRAL PARTICLES 9.14 Viroids and Prions I CHAPTER 10 BACTERIAL GENETICS I MUTATION AND RECOMBINATION 10.1 Mutations and Mutants 10.2 Molecular Basis of Mutation 10.3 Mutation Rates 10.4 Mutagenesis 10.5 Mutagenesis and Carcinogenesis: The Ames Test 10.6 Genetic Recombination II II GENETIC EXCHANGE IN PROKARYOTES 10.7 Transformation 10.8 Transduction 10.9 Plasmids: General Principles 10.10 Types of Plasmids and Their Biological Significance 10.11 Conjugation: Essential Features 10.12 The Formation of Hfr Strains and Chromosome Mobilization 10.13 Complementation 10.14 Transposons and Insertion Sequences III III BACTERIAL GENETICS AND GENE CLONING 10.15 Essentials of Molecular Cloning 10.16 Plasmids as Cloning Vectors 10.17 Bacteriophage Lambda as a Cloning Vector 10.18 In Vitro and Site-Directed Mutagenesis IV IV THE BACTERIAL CHROMOSOME 10.19 Genetic Map of the Escherichia coli Chromosome UNIT II EVOLUTIONARY MICROBIOLOGY AND MICROBIAL DIVERSITY CHAPTER 11 MICROBIAL EVOLUTION AND SYSTEMATICS I EARLY EARTH, THE ORIGIN OF LIFE, AND MICROBIAL DIVERSIFICATION 11.1 Evolution of Earth and Earliest Life Forms 11.2 Primitive Life: The RNA World and Molecular Coding 11.3 Primitive Life: Energy and Carbon Metabolism 11.4 Eukaryotes and Organelles: Endosymbiosis II II METHODS FOR DETERMINING EVOLUTIONARY RELATIONSHIPS 11.5 Evolutionary Chronometers 11.6 Ribosomal RNA Sequences and Cellular Evolution 11.7 Signature Sequences, Phylogenetic Probes, and Microbial Community Analyses III MICROBIAL EVOLUTION 11.8 Microbial Phylogeny Derived from Ribosomal RNA Sequences 11.9 Characteristics of the Primary Domains of Life IV MICROBIAL TAXONOMY AND ITS RELATIONSHIP TO PHYLOGENY 11.10 Classical Taxonomy 11.11 Molecular Taxonomy 11.12 The Species Concept in Microbiology 11.13 Nomenclature and Bergey's Manual CHAPTER 12 PROKARYOTIC DIVERSITY: BACTERIA I THE PHYLOGENY OF BACTERIA 12.1 Phylogenetic Overview of Bacteria II II PHYLUM 1: PROTEOBACTERIA 12.2 Purple Phototrophic Bacteria 12.3 The Nitrifying Bacteria 12.4 Sulfur- and Iron-Oxidizing Bacteria 12.5 Hydrogen-Oxidizing Bacteria 12.6 Methanotrophs and Methylotrophs 12.7 Pseudomonas and the Pseudomonads 12.8 Acetic Acid Bacteria 12.9 Free-Living Aerobic Nitrogen-Fixing Bacteria 12.10 Neisseria, Chromobacterium, and Relatives 12.11 Enteric Bacteria 12.12 Vibrio and Photobacterium 12.13 Rickettsias 12.14 Spirilla 12.15 Sheathed Proteobacteria: Sphaerotilus and Leptothrix 12.16 Budding and Prosthecate/Stalked Bacteria 12.17 Gliding Myxobacteria 12.18 Sulfate- and Sulfur-Reducing Proteobacteria III III PHYLUM 2 AND 3: GRAM-POSITIVE BACTERIA AND ACTINOBACTERIA 12.19 Nonsporulating, Low GC, Gram-Positive Bacteria: Lactic Acid Bacteria and Relatives 12.20 Endospore-Forming, Low GC, Gram-Positive Bacteria: Bacillus, Clostridium, and Relatives 12.21 Cell Wall-Less, Low GC, Gram-Positive Bacteria: The Mycoplasmas 12.22 High GC, Gram-Positive Bacteria (Actinobacteria): Coryneform and Propionic Acid Bacteria 12.23 Actinobacteria: Mycobacterium 12.24 Filamentous Actinobacteria: Streptomyces and other Actinomycetes IV IV PHYLUM 4: CYANOBACTERIA AND PROCHLOROPHYTES 12.25 Cyanobacteria 12.26 Prochlorophytes and Chloroplasts V V PHYLUM 5: CHLAMYDIA 12.27 The Chlamydia VI VI PHYLUM 6: PLANCTOMYCES/PIRELLULA 12.28 Planctomyces: A Phylogenetically Unique Stalked Bacterium VII PHYLUM 7: THE VERRUCOMICROBIA 12.29 Verrucomicrobium and Prosthecobacter VIII PHYLUM 8: THE FLAVOBACTERIA 12.30 Bacteroides and Flavobacterium IX IX PHYLUM 9: THE CYTOPHAGA GROUP 12.31 Cytophaga and Relatives X X PHYLUM 10: GREEN SULFUR BACTERIA 12.32 Chlorobium and Other Green Sulfur Bacteria XI XI PHYLUM 11: THE SPIROCHETES 12.33 Spirochetes XII XII PHYLUM 12: DEINOCOCCI 12.34 Deinococcus/Thermus XIII XIII PHYLUM 13: THE GREEN NONSULFUR BACTERIA 12.35 Chloroflexus and Relatives XIV XIV PHYLUM 14(16: DEEPLY BRANCHING HYPERTHERMOPHILIC BACTERIA 12.36 Thermotoga and Thermodesulfobacterium 12.37 Aquifex, Thermocrinis, and Relatives XV XV PHYLUM 17 AND 18: NITROSPIRA AND DEFERRIBACTER 12.38 Nitrospira, Deferribacter, and Relatives CHAPTER 13 PROKARYOTIC DIVERSITY: THE ARCHAEA I PHYLOGENY AND GENERAL METABOLISM 13.1 Phylogenetic Overview of the Archaea 13.2 Energy Conservation and Autotrophy in Archaea II II PHYLUM EURYARCHAEOTA 13.3 Extremely Halophilic Archaea 13.4 Methane-Producing Archaea: Methanogens 13.5 Thermoplasmatales: Thermoplasma, Ferroplasma, and Picrophilus 13.6 Hyperthermophilic Euryarchaeota: Thermococcales and Methanopyrus 13.7 Hyperthermophilic Euryarchaeota: The Archaeoglobales III III PHYLUM CRENARCHAEOTA 13.8 Habitats and Energy Metabolism of Crenarchaeotes 13.9 Hyperthermophiles from Terrestrial Volcanic Habitats: Sulfolobales and Thermoproteales 13.10 Hyperthermophiles from Submarine Volcanic Habitats: Desulfurococcales IV IV PHYLUM NANOARCHAEOTA 13.11 Nanoarchaeum V EVOLUTION AND LIFE AT HIGH TEMPERATURES 13.12 Heat Stability of Biomolecules 13.13 Hyperthermophilic Archaea, H2, and Microbial Evolution CHAPTER 14 EUKARYOTIC CELL BIOLOGY AND EUKARYOTIC MICROORGANISMS I EUKARYOTIC CELL STRUCTURE/FUNCTION 14.1 Eukaryotic Cell Structure and the Nucleus 14.2 Respiratory and Fermentative Organelles: The Mitochondrion and the Hydrogenosome 14.3 Photosynthetic Organelle: The Chloroplast 14.4 Endosymbiosis: Relationships of Mitochondria and Chloroplasts to Bacteria 14.5 Other Organelles and Eukaryotic Cell Structures II EUKARYOTIC GENETICS AND MOLECULAR BIOLOGY 14.6 Replication of Linear DNA 14.7 Overview of Eukaryotic Genetics 14.8 RNA Processing and Ribozymes III EUKARYOTIC MICROBIAL DIVERSITY 14.9 Phylogeny of the Eukarya 14.10 Protozoa 14.11 Slime Molds 14.12 Fungi 14.13 Algae CHAPTER 15 MICROBIAL GENOMICS I GENOMIC CLONING TECHNIQUES 15.1 Vectors for Genomic Cloning and Sequencing 15.2 Sequencing the Genome 15.3 Annotating the Genome II II MICROBIAL GENOMES 15.4 Prokaryotic Genomes: Sizes and ORF Contents 15.5 Prokaryotic Genomes: Bioinformatic Analyses and Gene Distributions 15.6 Eukaryotic Microbial Genomes III OTHER GENOMES AND THE EVOLUTION OF GENOMES 15.7 Genomes of Organelles 15.8 Evolution and Gene Families 15.9 Genomic Mining IV IV GENE FUNCTION AND REGULATION 15.10 Proteomics 15.11 Microarrays and the Transcriptome CHAPTER 16 VIRAL DIVERSITY I VIRUSES OF PROKARYOTES 16.1 RNA Bacteriophages 16.2 Icosahedral Single-Stranded DNA Bacteriophages 16.3 Filamentous Single-Stranded DNA Bacteriophages 16.4 Double-Stranded DNA Bacteriophages: T7 16.5 Mu: A Double-Stranded Transposable DNA Bacteriophage 16.6 Viruses of Archaea II II VIRUSES OF EUKARYOTES 16.7 Plant Viruses 16.8 Positive-Strand RNA Viruses of Animals: Poliovirus and Coronaviruses 16.9 Negative-Strand RNA Viruses of Animals: Rabies, Influenza, and Related Viruses 16.10 Double-Stranded RNA Viruses: Reoviruses 16.11 Replication of Double-Stranded DNA Viruses of Animals 16.12 Double-Stranded DNA Viruses: Herpesviruses 16.13 Double-Stranded DNA Viruses: Pox Viruses 16.14 Double-Stranded DNA Viruses: Adenoviruses 16.15 Viruses Using Reverse Transcriptase: Retroviruses and Hepadnavirus UNIT III METABOLIC DIVERSITY AND MICROBIAL ECOLOGY CHAPTER 17 METABOLIC DIVERSITY I THE PHOTOTROPHIC WAY OF LIFE 17.1 Photosynthesis 17.2 Photosynthetic Pigments and Their Location Within the Cell 17.3 Carotenoids and Phycobilins 17.4 Anoxygenic Photosynthesis 17.5 Oxygenic Photosynthesis 17.6 Autotrophic CO2 Fixation: The Calvin Cycle 17.7 Autotrophic CO2 Fixation: Reverse Citric Acid Cycle and the Hydroxypropionate Cycle II II CHEMOLITHOTROPHY: ENERGY FROM THE OXIDATION OF INORGANIC ELECTRON DONORS 17.8 Inorganic Electron Donors and Energetics 17.9 Hydrogen Oxidation 17.10 Oxidation of Reduced Sulfur Compounds 17.11 Iron Oxidation 17.12 Nitrification and Anammox III III THE ANAEROBIC WAY OF LIFE: ANAEROBIC RESPIRATIONS 17.13 Anaerobic Respiration 17.14 Nitrate Reduction and the Denitrification Process 17.15 Sulfate Reduction 17.16 Acetogenesis 17.17 Methanogenesis 17.18 Ferric Iron, Manganese, Chlorate, and Organic Electron Acceptors IV THE ANAEROBIC WAY OF LIFE: FERMENTATIONS AND SYNTROPHY 17.19 Fermentations: Energetic and Redox Considerations 17.20 Fermentative Diversity 17.21 Syntrophy V HYDROCARBON OXIDATION AND THE ROLE OF O2 IN THE CATABOLISM OF ORGANIC COMPOUNDS 17.22 Molecular Oxygen (O2) as a Reactant in Biochemical Processes 17.23 Hydrocarbon Oxidation 17.24 Methanotrophy and Methylotrophy 17.25 Hexose, Pentose, and Polysaccharide Metabolism 17.26 Organic Acid Metabolism 17.27 Lipids as Microbial Nutrients V VI NITROGEN FIXATION 17.28 Nitrogenase and the Process of Nitrogen Fixation 17.29 Genetics and Regulation of N2 Fixation CHAPTER 18 METHODS IN MICROBIAL ECOLOGY I CULTURE-DEPENDENT ANALYSES OF MICROBIAL COMMUNITIES 18.1 Enrichment and Isolation 18.2 Isolation in Pure Culture II MOLECULAR (CULTURE-INDEPENDENT) ANALYSES OF MICROBIAL COMMUNITIES 18.3 Viability and Quantification Using Staining Techniques 18.4 Genetic Stains 18.5 Linking Specific Genes to Specific Organisms Using PCR 18.6 Environmental Genomics III MEASURING MICROBIAL ACTIVITIES IN NATURE 18.7 Radioisotopes and Microelectrodes 18.8 Stable Isotopes CHAPTER 19 MICROBIAL ECOLOGY I MICROBIAL ECOSYSTEMS 19.1 Populations, Guilds, and Communities 19.2 Environments and Microenvironments 19.3 Microbial Growth on Surfaces and Biofilmshwater Microbial Habitats II SOIL AND FRESHWATER MICROBIAL HABITATS 19.4 Terrestrial Environments 19.5 Freshwater Environments III III MARINE MICROBIOLOGY 19.6 Marine Habitats and Microbial Distribution 19.7 Deep-Sea Microbiology 19.8 Hydrothermal Vents IV IV THE CARBON AND OXYGEN CYCLES 19.9 The Carbon Cycle 19.10 Syntrophy and Methanogenesis 19.11 Carbon Cycling in Ruminant Animals V V OTHER KEY NUTRIENT CYCLES 19.12 The Nitrogen Cycle 19.13 The Sulfur Cycle 19.14 The Iron Cycle VI VI MICROBIAL BIOREMEDIATION 19.15 Microbial Leaching of Ores 19.16 Mercury and Heavy Metal Transformations 19.17 Petroleum Biodegradation 19.18 Biodegradation of Xenobiotics VII VII MICROBIAL INTERACTIONS WITH PLANTS 19.19 The Plant Environment 19.20 Lichens and Mycorrhizae 19.21 Agrobacterium and Crown Gall Disease 19.22 Root Nodule Bacteria and Symbiosis with Legumes UNIT IV IMMUNOLOGY, PATHOGENICITY, AND HOST RESPONSES CHAPTER 20 MICROBIAL GROWTH CONTROL Iii I PHYSICAL ANTIMICROBIAL CONTROL 697 20.1 Heat Sterilization 20.2 Radiation Sterilization 20.3 Filter Sterilization II CHEMICAL ANTIMICROBIAL CONTROL 703 20.4 Chemical Growth Control 20.5 Chemical Antimicrobial Agents for External Use III ANTIMICROBIAL AGENTS USED IN VIVO 707 20.6 Synthetic Antimicrobial Drugs 20.7 Naturally Occurring Antimicrobial Drugs: Antibiotics 20.8 (-Lactam Antibiotics: Penicillins and Cephalosporins 20.9 Antibiotics from Prokaryotes IV CONTROL OF VIRUSES AND EUKARYOTIC PATHOGENS 716 20.10 Antiviral Drugs 20.11 Antifungal Drugs V ANTIMICROBIAL DRUG RESISTANCE AND DRUG DISCOVERY 20.12 Antimicrobial Drug Resistance 20.13 The Search for New Antimicrobial Drugs CHAPTER 21 MICROBIAL INTERACTIONS WITH HUMANS I I BENEFICIAL MICROBIAL INTERACTIONS WITH HUMANS 21.1 Overview of Human-Microbe Interactions 21.2 Normal Microbial Flora of the Skin 21.3 Normal Microbial Flora of the Oral Cavity 21.4 Normal Microbial Flora of the Gastrointestinal Tract 21.5 Normal Microbial Flora of Other Body Regions II HARMFUL MICROBIAL INTERACTIONS WITH HUMANS 737 21.6 Entry of the Pathogen into the Host 21.7 Colonization and Growth 21.8 Virulence III VIRULENCE FACTORS AND TOXINS 742 21.9 Virulence Factors 21.10 Exotoxins 21.11 Enterotoxins 21.12 Endotoxins IV HOST FACTORS IN INFECTION 749 21.13 Host Risk Factors 21.14 Innate Resistance to Infection CHAPTER 22 ESSENTIALS OF IMMUNOLOGY I I OVERVIEW OF THE IMMUNE RESPONSE 757 22.1 Cells and Organs of the Immune System 22.2 The Innate Immune Response 22.3 Inflammation, Fever, and Septic Shock 22.4 The Adaptive Immune Response III II ANTIGENS, T CELLS, AND CELLULAR IMMUNITY 764 22.5 Immunogens and Antigens 22.6 Presentation of Antigen to T Lymphocytes 22.7 T-Cytotoxic Cells and Natural Killer Cells 22.8 T-Helper Cells: Activating the Immune Response III ANTIBODIES AND IMMUNITY 771 22.9 Antibodies (Immunoglobulins) 22.10 Antibody Production 22.11 Complement, Antibodies, and Pathogen Destruction IV IV IMMUNITY AND PREVENTION OF INFECTIOUS DISEASE 22.12 Natural Immunity 22.13 Artificial Immunity 22.14 New Immunization Strategies V IMMUNE RESPONSE DISEASES 781 22.15 Allergy, Hypersensitivity, and Autoimmunity 22.16 Superantigens CHAPTER 23 MOLECULAR IMMUNOLOGY I I RECEPTORS AND IMMUNITY 788 23.1 Innate Immunity and Pattern Recognition 23.2 Adaptive Immunity and the Immunoglobulin Superfamily II THE MAJOR HISTOCOMPATIBILITY COMPLEX (MHC) 789 23.3 MHC Protein Structure 23.4 MHC Genes and Polymorphism III ANTIBODIES 792 23.5 Antibody Proteins and Antigen Binding 23.6 Antibody Genes and Diversity IV T-CELL RECEPTORS 795 23.7 TCR Proteins and Antigen Binding 23.8 TCR Genes and Diversity V MOLECULAR SIGNALS IN IMMUNITY 797 23.9 Clonal Selection and Tolerance 23.10 Second Signals 23.11 Cytokines and Chemokines CHAPTER 24 DIAGNOSTIC MICROBIOLOGY AND IMMUNOLOGY I I GROWTH-DEPENDENT DIAGNOSTIC METHODS 24.1 Isolation of Pathogens from Clinical Specimens 24.2 Growth-Dependent Identification Methods 24.3 Antimicrobial Drug Susceptibility Testing 24.4 Safety in the Microbiology Laboratory II IMMUNOLOGY AND DIAGNOSTIC METHODS 818 24.5 Immunoassays for Infectious Disease 24.6 Polyclonal and Monoclonal Antibodies 24.7 In Vitro Antigen-Antibody Reactions: Serology 24.8 Agglutination 24.9 Fluorescent Antibodies 24.10 Enzyme-Linked Immunosorbent Assay and Radioimmunoassay 24.11 Immunoblot Procedures III III MOLECULAR AND VISUAL CLINICAL DIAGNOSTIC METHODS 24.12 Nucleic Acid Methods 24.13 Diagnostic Virology UNIT V MICROBIAL DISEASES CHAPTER 25 EPIDEMIOLOGY I I PRINCIPLES OF EPIDEMIOLOGY 847 25.1 The Science of Epidemiology 25.2 The Vocabulary of Epidemiology 25.3 Disease Reservoirs and Epidemics 25.4 Infectious Disease Transmission 25.5 The Host Community II CURRENT EPIDEMICS 858 25.6 The AIDS Pandemic 25.7 Hospital-Acquired (Nosocomial) Infections 25.8 Severe Acute Respiratory Syndrome III EPIDEMIOLOGY AND PUBLIC HEALTH 862 25.9 Public Health Measures for the Control of Disease 25.10 Global Health Considerations 25.11 Emerging and Reemerging Infectious Diseases 25.12 Biological Warfare and Biological Weapons 25.13 Anthrax as a Biological Weapon CHAPTER 26 PERSON-TO-PERSON MICROBIAL DISEASES I I AIRBORNE TRANSMISSION OF DISEASES 876 26.1 Airborne Pathogens 26.2 Streptococcal Diseases 26.3 Corynebacterium and Diphtheria 26.4 Bordetella and Whooping Cough 26.5 Mycobacterium, Tuberculosis, and Leprosy 26.6 Neisseria meningitidis, Meningitis, and Meningococcemia 26.7 Viruses and Respiratory Infections 26.8 Colds and Influenza II DIRECT CONTACT TRANSMISSION OF DISEASES 893 26.9 Staphylococcus 26.10 Helicobacter pylori and Gastric Ulcers 26.11 Hepatitis Viruses III SEXUALLY TRANSMITTED DISEASES 897 26.12 Gonorrhea and Syphilis 26.13 Chlamydia, Herpes, and Trichomoniasis 26.14 Acquired Immunodeficiency Syndrome: AIDS and HIV CHAPTER 27 ANIMAL-TRANSMITTED, ARTHROPOD-TRANSMITTED, AND SOILBORNE MICROBIAL DISEASES I I ANIMAL-TRANSMITTED DISEASES 914 27.1 Rabies 27.2 Hantavirus Pulmonary Syndrome II II ARTHROPOD-TRANSMITTED DISEASES 917 27.3 Rickettsial Diseases 27.4 Lyme Disease 27.5 Malaria 27.6 West Nile Virus 27.7 Plague II III SOILBORNE DISEASES 929 27.8 The Pathogenic Fungi 27.9 Tetanus CHAPTER 28 WASTEWATER TREATMENT, WATER PURIFICATION, AND WATERBORNE MICROBIAL DISEASES I I WASTEWATER MICROBIOLOGY AND WATER PURIFICATION 935 28.1 Public Health and Water Quality 28.2 Wastewater and Sewage Treatment 28.3 Drinking Water Purification II WATERBORNE MICROBIAL DISEASES 942 28.4 Sources of Waterborne Infection 28.5 Cholera 28.6 Giardiasis and Cryptosporidiosis 28.7 Legionellosis (Legionnaire's Disease) 28.8 Typhoid Fever and Other Waterborne Diseases CHAPTER 29 FOOD PRESERVATION AND FOODBORNE MICROBIAL DISEASES I I FOOD PRESERVATION AND MICROBIAL GROWTH 951 29.1 Microbial Growth and Food Spoilage 29.2 Food Preservation 29.3 Fermented Foods II II MICROBIAL SAMPLING AND FOOD POISONING 954 29.4 Foodborne Diseases and Microbial Sampling 29.5 Staphylococcal Food Poisoning 29.6 Clostridial Food Poisoning III FOOD INFECTION 29.7 Salmonellosis 29.8 Pathogenic Escherichia coli 29.9 Campylobacter 29.10 Listeriosis 29.11 Other Foodborne Infectious Diseases UNIT VI MICROORGANISMS AS TOOLS FOR INDUSTRY AND RESEARCH CHAPTER 30 INDUSTRIAL MICROBIOLOGY I I INDUSTRIAL MICROOGANISMS AND PRODUCT FORMATION 30.1 Microorganisms and Their Products 30.2 Primary and Secondary Metabolites 30.3 Characteristics of Large-Scale Fermentations 30.4 Fermentation Scale-Up II MAJOR INDUSTRIAL PRODUCTS FOR THE HEALTH INDUSTRY 30.5 Antibiotics: Isolation and Characterization 30.6 Industrial Production of Penicillins and Tetracyclines 30.7 Vitamins and Amino Acids 30.8 Steroids and the Biotransformation Process 30.9 Enzymes as Industrial Products III MAJOR INDUSTRIAL PRODUCTS FOR THE FOOD AND BEVERAGE INDUSTRIES 30.10 Alcohol and Alcoholic Beverages 30.11 Vinegar Production 30.12 Citric Acid and Other Organic Compounds 30.13 Yeast as a Food and Food Supplement 30.14 Mushrooms as a Food Source CHAPTER 31 GENETIC ENGINEERING AND BIOTECHNOLOGY I I THE TECHNIQUES OF GENETIC ENGINEERING 31.1 Review of Principles Underlying Genetic Engineering 31.2 Hosts for Cloning Vectors 31.3 Finding the Right Clone 31.4 Specialized Vectors 31.5 Expression of Mammalian Genes in Bacteria II PRACTICAL APPLICATIONS OF GENETIC ENGINEERING 31.6 Production of Insulin: The Beginnings of Commercial Biotechnology 31.7 Other Mammalian Proteins and Products 31.8 Genetically Engineered Vaccines 31.9 Genetic Engineering in Animal and Human Genetics 31.10 Genetic Engineering in Plant Agriculture: Transgenic Plants Library of Congress Subject Headings for this publication: Microbiology. Microbiology.
10329	https://www.sciencedirect.com/topics/medicine-and-dentistry/kochs-postulates	Skip to Main content Sign in Chapters and Articles You might find these chapters and articles relevant to this topic. Review article Coral Disease Causes, Consequences, and Risk within Coral Restoration Koch's postulates have been widely taken up across microbiology for the study of pathogenic diseases, but the application of these criteria to marine ecosystems has proved challenging due to the connectivity of the marine environment; difficulty in culturing marine microbes; and the inherent difficulties of characterizing consistent and comparable disease signs in marine invertebrates . Koch's postulates include four criteria for identifying the correct pathogen: (i) the bacteria must be present in abundance in every case of the disease and must not be present in a healthy organism; (ii) the bacteria need to be extracted from the host and grown in pure culture and identified; (iii) the bacteria are inoculated back into a healthy host and must then cause the onset of the disease; (iv) the bacteria must be extracted from the inoculated host and grown in pure culture to be identified as the original causative agent. To date, only five marine disease pathogens have satisfied Koch’s postulates . Satisfying Koch’s postulates for coral diseases has been challenging due to several key limiting factors: (i) an inability to accurately replicate all factors within the reef environment in aquarium-based studies; (ii) insufficient data available regarding the mode of infection and disease initiation in the coral host; (iii) the process of inoculating the coral with the potential pathogen may not accurately replicate disease initiation within the reef environment; (iv) proving the onset of disease in aquaria is difficult as contamination of aquarium water may compromise the experiment ; (v) the criteria do not apply to other microbes which may be involved in disease progression. View article Read full article URL: Journal2020, Trends in MicrobiologyT. Moriarty, ... T.D. Ainsworth Chapter The potential role of infectious agents in diseases of unknown etiology 2010, Infectious Diseases (Third Edition)Steven M. Opal KOCH'S POSTULATES REDEFINED A new set of more practical guidelines to assign causation of potential pathogens to diseases of unknown cause is needed in the genomic era of the 21st century. Numerous diagnostic schemes have been proposed to complement or replace Koch's postulates for defining pathogens. A modification of the Hill39 and Evans40 criteria to supplement Koch's postulates is a reasonable framework upon which to build a more logical and practical diagnostic system to define disease causation. This set of criteria can readily be modified to include genomic evidence for noncultivable pathogens (Table 64.3). Each criterion need not be met to determine disease causation of each suspect human pathogen. The more criteria that are satisfied, the more convincing the link becomes between the suspected pathogen as the etiologic agent responsible for the disease process. A more integrated system that takes into account variations in the host response along with the microbial agent responsible for disease causation may be needed in the future.6 View chapterExplore book Read full chapter URL: Book2010, Infectious Diseases (Third Edition)Steven M. Opal Mini review A spirit of scientific rigour: Koch's postulates in twentieth-century medicine 2014, Microbes and InfectionChristoph Gradmann 6 Memories of the bacteriological revolution So what is it that the popularity of Koch's postulates can teach us about twentieth-century medicine? In their numerous variations they represent a trivial ontology of disease, although they are generally to be found in research papers that are anything but trivial in all other respects. The customary practice for invoking the postulates usually entails a discussion of proper methods undertaken in more or less strict accordance with the ideas that Koch is presumed to have held. As a rhetorical tool, invoking Koch's postulates functions in the same way as an anecdote, which presents a plausible story about an important object in a way that reduces complexity. Regardless of their factual content, anecdotes are supposed to contain truth. In the case here, the result is that the authors state with some regularity that although the content of the postulates is out-dated, their spirit is still cutting-edge. Historians from Plutarch to Jacob Burckhardt have emphasized that the value of anecdotes lies in their typical rather than factual representations of histories . Writing about the history of science in ancient Greece, Burckhardt proffered his ideas on the peculiarities of a historiography composed of orally disseminated anecdotes: Oral tradition, however, does not adhere to literal meaning. Instead it becomes typical [...]. Ultimately what remains from a long chain of personalities, events, and circumstances is nothing but an anecdote. In the meantime [...] those who pass the story on from mouth to mouth have made additions to that history, not just from additional knowledge but also by relating it to general features of the subject in question. They have embroidered; they have added to the story. Most of all they have piled all of this onto the name of just one, well-known protagonist [:188/9]. If we apply Burckhardt's argument, discussions of Koch's postulates usually include an anecdote about the historical origins of experimental medicine. At the centre of modern experimental medicine we find a fairy tale that is told and retold in many variations, yet remains essentially the same. It is a type of historiography that we associate more with oral societies and one that academic historians jettisoned long ago. And yet it appears ubiquitous to a style of medical history at home between the lines of scientific papers. When this kind of history-telling encounters conflicting facts, their fate is absorption rather than discussion. For example, when some authors realized that Koch's own methodology was not always in line with his postulates, they devised the explanation that Koch had modified his postulates later on in his career in order to accommodate the concept of carrier-state epidemiology, of which he was the inventor [:774/5], [:2141]. Talking about Koch's postulates is to invoke a tradition of experimental pathology. No matter how many or which postulates are mentioned, it serves to reiterate and confirm certain basic assumptions that render the experimental reconstruction of infectious diseases plausible. In this sense, the countless variations of Koch's postulates represent not change, but stability. Underpinning the bustle of technological invention that has powered twentieth-century medicine we find almost stable assumptions; what we might describe as experimental medicine's world view has changed but very little in more than a century. Discussing the postulates usually involves updating, not replacing them. When the technology of sequencing radically transformed microbiology in the 1990s, two researchers set out to compare their machinery with that which Koch would have had at his disposal when he supposedly laid down his postulates. Unsurprisingly, they stated that the postulates were outdated in their literal form, yet they reminded their readers that ‘the principles behind Koch's postulates still hold’ [:31]. Adding to this, they explicitly cautioned their readers against an understanding that was too simple: Unfortunately, Koch's postulates have frequently been applied to issues of causation with a mathematical zeal that is not warranted in the biological world. […] The power of Koch's postulates comes not from their rigid application, but from the spirit of scientific rigour that they foster [:20]. That leaves one final question unanswered. If talking about Koch's postulates invokes a tradition of experimental medicine, why point to the German physician rather than to Claude Bernard, Almroth Wright, Louis Pasteur, or some other hero of experimental medicine? One possible answer is that the continuous rephrasing of this topic in particular is bound up with the assumed historical relevance of medical bacteriology. Koch was not the only one to champion the idea that infectious diseases may be explained by employing microbial aetiologies. And the idea did not become popular overnight. Nevertheless, it meant that miasmas became a case for medical historians and experimental procedures acquired the status of a cultural foundation of modern medicine. In this sense, Koch's postulates do not stand for any particular methodology. They depict instead a world view that lends itself to be represented in numerous methodologies. It is no contradiction that Koch's postulates undergo continuous modification whilst their invocation simultaneously enacts the rhetorical obligation of twentieth- and twenty-first-century experimental researchers to nineteenth-century medical bacteriology. Critical medical historians have repeatedly tried to bury the notion of a bacteriological revolution.3 However, this bacteriological revolution is still alive and kicking in the memory of medical researchers. In a brilliant paper from 1992 Andrew Cunningham reminded his readers that, whilst nineteenth-century medical bacteriology may not have transformed medicine overnight, this transformation was radical none the less. Once infection had become microbial, explaining it without referring to the involvement of microorganisms became almost unimaginable: ‘To oppose the claims of bacteriology is now not a rival view, nor an alternative view, nor even a dissident view. It is now a lunatic view.’ [:239]. The continued and unshakable popularity of Robert Koch's ill-defined postulates reminds us that this world view is still dominant in medicine today. View article Read full article URL: Journal2014, Microbes and InfectionChristoph Gradmann Chapter The Evolution of Koch's Postulates 2017, Infectious Diseases (Fourth Edition)Jonathan Cohen Conclusions – and a Note of Caution Koch's postulates were invaluable at the time they were developed and remain largely valid for a relatively small number of defined circumstances in which bacteria can be precisely tied to the cause of a particular clinical syndrome. But in a world in which viruses cause cancer and noncultivable bacteria can be demonstrated by molecular probes, Koch's postulates are no longer fit for purpose. What is more, used uncritically they have the potential to mislead.16 Their main purpose now is to provide a framework to ensure that scientific rigor is applied when proposing an organism as the cause of a disease – exactly as Koch intended when he first conceived them. References available online at expertconsult.com. View chapterExplore book Read full chapter URL: Book2017, Infectious Diseases (Fourth Edition)Jonathan Cohen Review article Koch's postulates, microbial dysbiosis and inflammatory bowel disease 2016, Clinical Microbiology and InfectionV.P. Singh, ... B.P. Willing Introduction The Koch's postulates set the standard for proving the role of an organism in a disease. For the four postulates to be fulfilled in their current form, the identified organism must (1) be present in all cases of the disease; (2) be isolated from diseased patients; (3) cause disease when reintroduced to a healthy susceptible animal model; and (4) then be isolated again from the new host. Polymicrobial diseases or diseases linked to microbial dysbiosis, such as inflammatory bowel disease (IBD), do not fulfil the Koch's postulates in their current form. Dysbiosis has been characterized as an alteration in microbial composition or activity that results in an aberrant host response to the microbiota and can be represented by a single or many organisms . Since they were first devised a century ago, the Koch's postulates have been modified in order to accommodate new understandings of microorganisms, such as the inability to culture viruses in the absence of host cells . Therefore, consideration should be given to modify the postulates so as to fit polymicrobial diseases or diseases associated with microbial dysbiosis. As depicted in Fig. 1, we propose that the Koch's postulates can be fulfilled for IBD with some modifications. In both clinical and experimental settings, there is evidence that support an important role of microbes in both Crohn's disease (CD) and ulcerative colitis (UC), the main forms of IBD. Long-standing examples include the ability of antibiotics to improve disease symptoms (in some patients); colostomy results in cessation of symptoms in patients; and with few exceptions, germ-free mice do not develop colitis . Both CD and UC present as an unbalanced immune response to the microbiota, although the nature of the imbalance differs between diseases . While both CD and UC are chronic and relapsing inflammatory conditions of the gastrointestinal tract, they are distinguishable in many ways. For CD, inflammation can occur along the entire gastrointestinal tract, with segmental inflammation localized most frequently in the terminal ileum or large intestine. For UC, inflammation is limited to the mucosal layer in large intestine . CD can also be further defined on the basis of the anatomic location, which can be associated with different genetic predispositions. These unique distinctions suggest that we should not assume that each disease subtype will conform to the Koch's postulates in a similar manner. View article Read full article URL: Journal2016, Clinical Microbiology and InfectionV.P. Singh, ... B.P. Willing Mini review A spirit of scientific rigour: Koch's postulates in twentieth-century medicine 2014, Microbes and InfectionChristoph Gradmann Abstract This article explores one of a citation classics in medical literature—Koch's postulates. It analyses their creation in the nineteenth century and their popularity in the twentieth century. As a genre of historiography, references to the postulates are anecdotes. In referring to a historical event that never happened, such references serve to remind their audiences of a tradition of experimental medicine that supposedly originated with Robert Koch. View article Read full article URL: Journal2014, Microbes and InfectionChristoph Gradmann Chapter Changing how we think about infectious diseases 2019, Taxonomic Guide to Infectious Diseases (Second Edition)Jules J. Berman Section 8.1 Abandoning Koch's postulates It isn't that they can't see the solution. It's that they can't see the problem. G.K. Chesterton Robert Koch's postulates, published in 1890, are a set of criteria that establish whether a particular organism is the cause of a particular disease. Today, Koch's postulates are taught in high school and college classrooms as a demonstration of the rigor and legitimacy of clinical microbiology. To review, the four postulates of Koch are as follows: –1. : The microorganism must be found in the diseased animal, and not found in healthy animals. –2. : The microorganism must be extracted and isolated from the diseased animal and subsequently grown in culture. –3. : The microorganism must cause disease when introduced to a healthy experimental animal. –4. : The microorganism must be extracted from the diseased experimental animal and demonstrated to be the same microorganism that was originally isolated from the first diseased animal. Let's go over these four postulates once more, this time explaining how they ignore or contradict what we now know about infectious diseases. –1. : The microorganism must be found in the diseased animal, and not found in healthy animals. As previously discussed, lots of pathogenic organisms are found in healthy animals, producing disease in only a tiny fraction of the individuals who are infected. For example, Bartonella species can live in blood without causing disease, producing an asymptomatic bacteremia in the wide assortment of animals that they may infect. Hence, we can no longer assume that blood samples from healthy animals are sterile. The mechanism of Bartonella transmission from animal to animal is not fully understood, but arthropod vectors (ticks, fleas, and lice) are suspected, as well as scratches and bites from infected animals (e.g., cats and rats) [Glossary Vector]. There are now about eight species of Bartonella that are known or suspected to be human pathogens. Until just a few decades ago, only two such species were known. Today, the species of Bartonella, which are ubiquitous among mammals, are known or suspected to cause a variety of phenotypically dissimilar diseases : – : Bartonella bacilliformis → Carrion disease – : Bartonella quintana → Bacillary angiomatosis, trench fever, endocarditis – : Bartonella henselae → Bacillary angiomatosis, cat-scratch disease, peliosis hepatis B. henselae – : Bartonella clarridgeiae → Cat-scratch disease – : Bartonella elizabethae → Endocarditis – : Bartonella vinsonii var berkhoffii → endocarditis – : Bartonella vinsonii var arupensis → fever and a valvulopathy – : Bartonella grahamii → uveitis The precise diagnosis of Bartonella species in human blood and lesions has provided us with the names of infectious organism associated with a number of diseases, but this new knowledge has not shed much light on why Bartonella can circulate in the blood without causing any reaction, for indefinite periods of time, or why any given Bartonella species may be associated with any of several diverse clinical manifestations. Furthermore, Koch's third postulate fails miserably for genus Bartonella; injecting any of these Bartonella species into experimental animals, will more than likely produce no symptoms. –2. : The microorganism must be extracted and isolated from the diseased animal (and grown in culture). Many pathogens do not grow in nutrient medium culture. This applies generally to common Mollicute bacteria, including Erysipelothrix, Mycoplasma, and Ureoplasma. This would also apply to viruses, none of which grow in cell-free media. Paradoxically, some of the organisms known to produce bacteremias in human blood grow very poorly in blood cultures, and this would include the aforementioned Bartonella species and the HACEK organisms [1, 3]. The HACEK organisms are a group of proteobacteria, found in otherwise healthy individuals, that are known to cause some cases of endocarditis, especially in children, and which do not grow well in culture. The term HACEK is created from the initials of the organisms of the group: Haemophilus, particularly Haemophilus parainfluenzae; Aggregatibacter, including Aggregatibacter actinomycetemcomitans and Aggregatibacter aphrophilus; Cardiobacterium hominis; Eikenella corrodens; and Kingella, particularly Kingella kingae. –3. : The microorganism must cause disease when introduced to a healthy experimental animal. Again, some of the worst microorganisms will not produce disease in healthy animals. To confuse matters further, we now have examples of nonliving agents that will produce transmissible disease in healthy animals (prions). This third postulate of Koch presumes that each occurrence of an infectious disease has a particular organism that is “the cause” of the disease. We must return here to our often-repeated theme that diseases do not have “a cause,” and infectious diseases are no exception to the rule that pathogenesis is a multistep process. We have already seen that myocardial infarction results from a multitude of conditions that occur through time. In some cases, the last event is infectious, wherein a focal bacterial endocarditis precipitates a thrombus that blocks a narrowed coronary artery. It would be folly to believe that the sequence of events that lead to a myocardial infarction can be precipitated simply by injecting an organism into an animal. Later in this chapter, we will see two examples of rare infections for which several conditions must prevail before a disease emerges [4, 5]. –4. : The microorganism must be extracted from the infected experimental animal and demonstrated to be the same microorganism that was originally isolated from the original diseased animal. Many infections, considered the underlying cause of a disease, are absent from the lesions that ultimately develop. For example, Group A streptococcus infection is considered to be the underlying cause of rheumatic fever. The infection is long gone prior to the appearance of the valvular and endocardial lesions of rheumatic fever. As another example, several species of human papillomavirus are considered to be the underlying cause of nearly all cases of squamous carcinoma of the uterine cervix. Morphologic cytopathic effects are visible in the earliest precancers that precede the development of invasive carcinoma. The cancers, which may occur years following the early papillomavirus infections, may lack recoverable virus. Let's look at an example of an infectious disease that violates every one of Koch's postulates. Whipple disease, previously a disease of unknown etiology, is characterized by organ infiltration with foamy macrophages (i.e., specialized reticuloendothelial cells that “eat” bacteria and debris). The organ most often compromised in Whipple disease is the small intestine, where infiltration of infected macrophages in the lamina propria (i.e., a strip of loose connective tissue subjacent to the epithelial lining of the small intestine) causes malabsorption. Whipple disease is rare. It occurs most often in farmers and gardeners who work with soil. Whipple disease was first described in 1907 , but its cause was unknown until 1992, when researchers isolated and amplified, from Whipple disease tissues, a 16s ribosomal RNA sequence that could only have a bacterial origin . Based on molecular features of the ribosomal RNA molecule, the researchers assigned it to Class Cellulomonadacea, and named the species Tropheryma whipplei, after the man who first described the disease, George Hoyt Whipple. Particularly noteworthy, in the case of Whipple disease, is that Koch's postulates never came close to being satisfied. For the experimentalist, the most important of Koch's postulates require the extraction of the organism from a lesion (i.e., from diseased, infected tissue), the isolation and culture of the organism in the laboratory, and the consistent reproduction of the lesion in an animal injected with the organism. In the case of Whipple disease, none of these criteria were satisfied. The consistent identification in Whipple disease tissue of a particular molecule, characteristic of a particular species of bacteria, was deemed sufficient to establish the infectious origin of the disease. In the general scheme of events, bacteria in the human body are eaten by macrophages, wherein they are degraded. In the case of T. whipplei, only a small population of susceptible individuals lack the ability to destroy T. whipplei organisms. In susceptible individuals, the organisms multiply within macrophages. When organisms are released from dying macrophages, additional macrophages arrive to feed, but this only result in the local accumulation of macrophages bloated by bacteria. Whipple disease is a good example of a disease caused by an organism but dependent on a genetic predisposition, expressed as a defect in innate immunity; specifically, a reduction of macrophages expressing CD11b (also known as macrophage-1 antigen) . Whipple disease cannot be consistently reproduced in humans or any other animals, because it can only infect and grow in a small portion of the human population. As we learn more and more about the complexity of disease causation, formerly useful paradigms, such as Koch's postulates, seem burdensome and useless. When we encounter rare diseases of infectious cause, we might expect to find that the pathogenesis of disease (i.e., the biological steps that lead to a clinical phenotype) may require several independent causal events to occur in sequence. In the case of Whipple disease, the infected individual must be exposed to a soil organism, limiting the disease to farmers and gardeners. The organism, residing in the soil, must be ingested, perhaps by the inhalation of dust. The organism must evade degradation by gut macrophages, limiting disease to individuals with a specific type of defect in cell-mediated immunity, and the individual must have disease that is sufficiently active to produce clinical symptoms. It is unlikely that we could reproduce a complex sequence of steps, leading to a disease, by simply inoculating an organism into an experimental animal [Glossary Underlying cause, Proximate cause, Root cause]. Side-stepping Koch's postulates has become de rigueur in the practice of modern medicine. For example, the United States has experienced a recent increase in cases of acute flaccid myelitis, a rare disease of children . Diagnosis is based on a metagenomic analysis (i.e., culture-independent sequence searches conducted on an assemblage of microbial gene sequences in a biologic sample) of DNA obtained from nasopharyngeal swabs. The organism that is present in most of the examined cases is enterovirus-D68, and this virus is the presumed causal organism of acute flaccid myelitis, until proven otherwise. Genotyping species of organisms has become quite easy, but there are many millions of microorganism species, and it may never be feasible to complete a database of genome sequences of all living species. Though the number of individual species is too large to sequence, we can do a fair job at sequencing most of the different genera of living species. We now have a fairly accurate way of identifying the genus of any organism found within a tissue sample, by sequencing its ribosomal RNA and comparing the sequence against references sequences in public databases [10–13]. There are limitations to this technique, but when we combine our analysis of ribosomal RNA with our accumulated knowledge of clinical features of the infection, we can often arrive at candidate pathogen [11, 14, 15]. Modern medicine has changed the vocabulary of infection. Familiar terms such as primary pathogen, opportunistic infection, and immunocompetent patient need to be reexamined in light of what we have come to know. Even a fundamental concept, such as “the organism causing the disease” should probably be abandoned in light of the multistep pathogenesis of all diseases. Because a microorganism may contribute to the pathogenesis of a disease at a single moment of time, long before the disease becomes clinically manifest, we can expect to see cases in which screening tests for a putative causal organism will be negative in affected patients . Koch, in his own time, understood the practical limitations of his postulates. Maybe it's time to reconsider Koch's postulates in light of the analytic methods now available that assign a taxonomic class to an infective organism, without isolating or characterizing the agent . Read full chapterView PDFExplore book Read full chapter URL: Book2019, Taxonomic Guide to Infectious Diseases (Second Edition)Jules J. Berman Review article The search for disease-associated compositional shifts in bowel bacterial communities of humans 2008, Trends in MicrobiologyGerald W. Tannock Koch's postulates are still relevant Whatever the community compositional shifts that are detected in association with diseases, there is the need to demonstrate the medical consequences of such changes. Population shifts such as the reported reduction in biodiversity of the bowel community of Crohn's disease and ulcerative colitis patients might reflect changes in the ecosystem as the result of disease, not that the microbial changes are the cause of the disease [41,63]. The faecal community of patients with diarrhoea, for example, is greatly simplified compared with that of normal stool because of the wash-out effect produced by the frequent passage of watery digesta that purges the large bowel as a consequence of impaired absorption of water and electrolytes . Fulfillment of Koch's postulates (Box 4) is still required to prove the aetiological role of a specific bacterial species. Postulate 3 tends to be modified de facto if antimicrobial therapy eliminates a putative pathogen and effects a cure. In the case of shifts in community composition, it will presumably be necessary to restore the normal microbial ecology of the bowel to demonstrate cause. Without demonstration of cause, reported shifts in the composition of bowel communities in patients are meaningless from a medical perspective. Box 4 Koch's postulates Postulate 1 The microbe must be present in all people with the disease and should be associated with the lesions of the disease. Postulate 2 The microbe must be isolated in pure culture from a person who has the disease. Postulate 3 The isolated microbe, when administered to humans or animals must cause disease. Postulate 4 The microbe must be isolated in pure culture from the human or animal infected to satisfy postulate 3. View article Read full article URL: Journal2008, Trends in MicrobiologyGerald W. Tannock Review article Amending Koch's postulates for viral disease: When “growth in pure culture” leads to a loss of virulence 2017, Antiviral ResearchJoseph Prescott, ... David Safronetz 3 Koch's postulates Based in part on the earlier perceptions of Jakob Henle, and in consultation with Friedrich Loeffler, Robert Koch devised guidelines to demonstrate that certain human diseases were caused by specific micro-organisms (Table 1). As applied to viral agents, “Koch's Postulates” for establishing causation require virus isolation from a diseased organism, growth of the agent in pure culture, and the development of disease when the virus is re-introduced into a healthy organism (Koch, 1884; Rivers, 1937). This approach has been applied to microbes for over a century and is a current practice not only for identifying pathogenic viruses in diseased organisms, but for the isolation of viruses from their natural reservoirs and vectors that harbor them (see Table 2). Table 1. Koch's postulates to identify the causative agent of an infectious disease. \| \| \| • The microorganism must be found in abundance in all organisms suffering from the disease, but should not be found in healthy organismsa \| \| • The microorganism must be isolated from a diseased organism and grown in pure culture \| \| • The microorganism (from the pure culture) should cause disease when inoculated into a healthy organism \| \| • The microorganism must be re-isolated from the inoculated, diseased experimental host and identified as being identical to the original specific causative agent \| a : Koch dismissed the universal requirement of the first postulate following the discovery of asymptomatic carriers of diseases such as cholera. Table 2. Examples of the alteration of viral virulence upon propagation in cell culture. \| Virus \| Outcome of cell culture passage \| --- \| \| Sin Nombre hantavirus \| Vero-passaged virus is completely attenuated in NHPs, whereas virus propagated in deer mice causes severe disease (Safronetz et al., 2014). \| \| Puumala hantavirus \| Virus passaged in the reservoir (bank vole) causes disease in NHPs, but virus passaged in Vero cells does not (Klingstrom et al., 2002). \| \| Ebola virus \| Accumulation of adenosine residues in the GP gene editing site upon passage in Vero cells leads to attenuation in guinea pigs (Volchkova et al., 2011). \| \| Measles virus \| Cell culture adapted viruses lose pathogenicity in vivo due to a loss in interferon antagonism (Bankamp et al., 2008). \| \| Passage in Vero cells results in a change in entry receptor usage and a decrease in pathogenicity in vivo (Dörig et al., 1993). \| \| Foot and mouth disease virus \| Passage in culture results in a receptor switch between αvβ3 integrin and heparan sulfate (Martínez et al., 1997). \| \| Sindbis virus \| Virus grown on mosquito cells demonstrated increased infectiousness for human dendritic cells when compared to virus grown on Chinese hamster cells (Klimstra et al., 2003). \| \| Rift Valley fever virus \| Virus passaged on mosquito cells retains virulence, whereas when the virus is passaged on Vero cells, in vivo virulence is lost (Weingartl et al., 2014b). \| Although Koch was also instrumental in the birth of the field of virology, at the time he proposed his postulates, knowledge regarding viruses was in its infancy. As obligate intracellular organisms, the procedure of ‘growth in pure culture’ in virology differs substantially from the solid phase media cultures described by Koch for bacteriology. Multiple steps are required for a virus to replicate in cell culture, and each step may impose selective pressure on the population. Host cells are required for the propagation of viruses. This propagation inevitably results in a mixed population of viruses. For the purpose of this article we propose that ‘pure culture’ for virus isolation means propagating viruses using in vitro preparations, such as mammalian cell culture. Historically, viruses were isolated by inoculating susceptible laboratory animals or embryonated eggs with small quantities of homogenized tissues or fluids obtained from biological specimens. Utilizing modern in vitro culture techniques, most viruses are now isolated by inoculating susceptible, generally immortalized, cells with biological material containing the desired agent. Accordingly, virus preparations are obtained by collecting supernatants or lysed cell homogenates. These methods facilitate obtaining high-titer virus stocks which can be concentrated and purified (e.g., using a sucrose gradient), and allow for the serial propagation and molecular characterization of viruses that can readily be grown in culture. While it is often assumed that the starting and final virus populations are the same, in fact there is always some degree of genetic change resulting from “adaptation” to the cultured cells. There are therefore several limitations for many viruses generated in this fashion, potentially reducing the biological relevance of in vivo studies performed with cell culture-derived viruses. Read full articleView PDF Read full article URL: Journal2017, Antiviral ResearchJoseph Prescott, ... David Safronetz Related terms: Enterotoxin Pathogen Diarrhea Virus Infection Pylorus Pathogenicity Rabies Severe Acute Respiratory Syndrome Atherosclerosis Autoimmune Disease View all Topics
10330	https://www.youtube.com/watch?v=JSBUnq535vM	PreCalculus - Trigonometry: Trig Identities (41 of 57) Prove cos^4(x)-sin^4(x)=cos2x Michel van Biezen 1140000 subscribers 235 likes Description 26473 views Posted: 24 Aug 2014 Visit for more math and science lectures! In this video I will prove cos^4(x)-sin^4(x)=cos2x. 28 comments Transcript: welcome to electron line and now that we've got a bunch of these identities under control at least we think we do let's go ahead and try our hand at this one let's see let's show that the left side equals the right side okay here we have something that looks a lot like the difference of squares if we rewrite it like this it becomes more obvious what if we have the cosine square of X quantity squared minus the sine square of X quantity squared equals yeah I can go ahead put the equal sign there alright so now you can see it's the difference of squares which can be factored and it'll look like this will be the cosine square of X plus the sine square of X multiplied times the cosine squared of X minus the sine square of X and that equals well we'll see what it equals and now you can see that this is equal to one so I'd like to write the equal sign in the front right in the back let me do that so equals equals it's a little easier to work with so this becomes one times the cosine squared of X minus the sine square of X and maybe some of you might already recognize that this is equal to an identity but we'll leave that till later all right now what I'm going to do here is go back and remember this one so if we have the cosine of a plus B that is equal to the cosine of a times the cosine of B and if this is plus it becomes minus sine of a sine of B and then if I make that a plus a let's see what happens so the cosine of a plus a that is equal to the cosine of a times the cosine of a minus the sine of a times the sine of a if I simplify that this is equal to the cosine squared of a minus a sine squared of a which is what I have over here so that means that this can now be written as so this is equal to and of course one times that would be the cosine of X plus X which of course is equal to the cosine of 2x which is what I have over there so that other words the cosine squared of X minus the sine squared of X is indeed equal to the cosine of 2x which is equal to what we have over there so we've proven this identity that's how we do that
10331	https://www.britannica.com/science/acid-rain	SUBSCRIBE Ask the Chatbot Games & Quizzes History & Society Science & Tech Biographies Animals & Nature Geography & Travel Arts & Culture ProCon Money Videos acid rain pollution Also known as: acid deposition, acid precipitation Written by Gene E. Likens Distinguished Senior Scientist, Cary Institute of Ecosystem Studies, Millbrook, New York. Gene E. Likens , Thomas J. Butler Site Manager, Cary Institute Cary Institute of Ecosystem Studies, Millbrook, New York, and Visiting Fellow, Cornell University, Ithaca, New York. Thomas J. Butler •All Fact-checked by The Editors of Encyclopaedia Britannica Encyclopaedia Britannica's editors oversee subject areas in which they have extensive knowledge, whether from years of experience gained by working on that content or via study for an advanced degree.... The Editors of Encyclopaedia Britannica Article History Also called: : acid precipitation or acid deposition Related Topics: : What Happened to Acid Rain? : nitrogen dioxide : rain : What Causes Acid Rain? : nitrogen oxide See all related content Top Questions What is acid rain and how is it formed? What chemicals are responsible for acid rain? How does acid rain affect the environment? What are the sources of the pollutants that cause acid rain? How does acid rain impact aquatic life and ecosystems? What are the effects of acid rain on human-made structures? How can acid rain affect human health? What regions of the world are most affected by acid rain? acid rain, precipitation possessing a pH of about 5.2 or below primarily produced from the emission of sulfur dioxide (SO2) and nitrogen oxides (NOx; the combination of NO and NO2) from human activities, mostly the combustion of fossil fuels. In acid-sensitive landscapes, acid deposition can reduce the pH of surface waters and lower biodiversity. It weakens trees and increases their susceptibility to damage from other stressors, such as drought, extreme cold, and pests. In acid-sensitive areas, acid rain also depletes soil of important plant nutrients and buffers, such as calcium and magnesium, and can release aluminum, bound to soil particles and rock, in its toxic dissolved form. Acid rain contributes to the corrosion of surfaces exposed to air pollution and is responsible for the deterioration of limestone and marble buildings and monuments. The phrase acid rain was first used in 1852 by Scottish chemist Robert Angus Smith during his investigation of rainwater chemistry near industrial cities in England and Scotland. The phenomenon became an important part of his book Air and Rain: The Beginnings of a Chemical Climatology (1872). It was not until the late 1960s and early 1970s, however, that acid rain was recognized as a regional environmental issue affecting large areas of western Europe and eastern North America. Acid rain also occurs in Asia and parts of Africa, South America, and Australia. As a global environmental issue, it is frequently overshadowed by climate change. Although the problem of acid rain has been significantly reduced in some areas, it remains an important environmental issue within and downwind from major industrial and industrial agricultural regions worldwide. Chemistry of acid deposition Acid rain is a popular expression for the more scientific term acid deposition, which refers to the many ways in which acidity can move from the atmosphere to Earth’s surface. Acid deposition includes acidic rain as well as other forms of acidic wet deposition—such as snow, sleet, hail, and fog (or cloud water). Acid deposition also includes the dry deposition of acidic particles and gases, which can affect landscapes during dry periods. Thus, acid deposition is capable of affecting landscapes and the living things that reside within them even when precipitation is not occurring. Acidity is a measure of the concentration of hydrogen ions (H+) in a solution. The pH scale measures whether a solution is acidic or basic. Substances are considered acidic below a pH of 7, and each unit of pH below 7 is 10 times more acidic, or has 10 times more H+, than the unit above it. For example, rainwater with a pH of 5.0 has a concentration of 10 microequivalents of H+ per litre, whereas rainwater with a pH of 4.0 has a concentration of 100 microequivalents of H+ per litre. Normal rainwater is weakly acidic because of the absorption of carbon dioxide (CO2) from the atmosphere—a process that produces carbonic acid—and from organic acids generated from biological activity. In addition, volcanic activity can produce sulfuric acid (H2SO4), nitric acid (HNO3), and hydrochloric acid (HCl) depending on the emissions associated with specific volcanoes. Other natural sources of acidification include the production of nitrogen oxides from the conversion of atmospheric molecular nitrogen (N2) by lightning and the conversion of organic nitrogen by wildfires. However, the geographic extent of any given natural source of acidification is small, and in most cases it lowers the pH of precipitation to no more than about 5.2. Anthropogenic activities, particularly the burning of fossil fuels (coal, oil, natural gas) and the smelting of metal ores, are the major causes of acid deposition. In the United States, electric utilities produce nearly 70 percent of SO2 and about 20 percent of NOx emissions. Fossil fuels burned by vehicles account for nearly 60 percent of NOx emissions in the United States. In the atmosphere, sulfuric and nitric acids are generated when SO2 and NOx, respectively, react with water. The simplest reactions are: SO2 + H2O → H2SO4 ←→ H+ + HSO4 ←→ 2H+ + SO42 NO2 + H2O → HNO3 ←→ H+ + NO3 These reactions in the aqueous phase (for example, in cloud water) create wet deposition products. In the gaseous phase they can produce acidic dry deposition. Acid formation can also occur on particles in the atmosphere. Where fossil fuel consumption is large and emission controls are not in place to reduce SO2 and NOx emissions, acid deposition will occur in areas downwind of emission sources, often hundreds to thousands of kilometres away. In such areas the pH of precipitation can average 4.0 to 4.5 annually, and the pH of individual rain events can sometimes drop below 3.0. In addition, cloud water and fog in polluted areas may be many times more acidic than rain falling over the same region. Many air pollution and atmospheric deposition problems are intertwined with one another, and these problems are often derived from the same cause, namely the burning of fossil fuels. In addition to acid deposition, NOx emissions along with hydrocarbon emissions are key ingredients in ground-level ozone (photochemical smog) formation, which is one of the most widespread forms of air pollution. The SO2 and NOx emissions can generate fine particulates, which are harmful to human respiratory systems. Coal combustion is the leading source of atmospheric mercury, which also enters ecosystems by wet and dry deposition. (A number of other heavy metals, such as lead and cadmium, and various particulates are also products of unregulated fossil fuel combustion.) Acid deposition of nitrogen derived from NOx emissions creates additional environmental problems. For example, many lake, estuarine, and coastal marine systems receive too much nitrogen from atmospheric deposition and terrestrial runoff. This eutrophication (or over-enrichment) causes the overgrowth of plants and algae. When these organisms die and decompose, they deplete the dissolved oxygen supply necessary for most aquatic life in water bodies. Eutrophication is considered to be a major environmental problem in lake, coastal marine, and estuarine ecosystems worldwide. Ecological effects of acid deposition Feedback Thank you for your feedback Our editors will review what you’ve submitted and determine whether to revise the article. print Print Please select which sections you would like to print: verifiedCite While every effort has been made to follow citation style rules, there may be some discrepancies. Please refer to the appropriate style manual or other sources if you have any questions. Select Citation Style Likens, Gene E., Butler, Thomas J.. "acid rain". Encyclopedia Britannica, 20 Mar. 2025, Accessed 1 September 2025. Share Share to social media Facebook X External Websites LiveScience - Acid Rain: Causes, Effects and Solutions National Center for Biotechnology Information - PubMed Central - Global Trends of Acidity in Rainfall and Its Impact on Plants and Soil United States Environment Protection Agency - What is Acid Rain? International Journal of Innovative Research in Science, Engineering and Technology - A Study on Acid Rain: Effects and Control Measures (PDF) Cary Institute of Ecosystem Studies - Acid Rain Energy Education - Acid rain American Association for the Advancement of Science - Science - Is Acid Rain a Thing of the Past? USGS - Water Science School - Acid Rain and Water Asia Center for Air Pollution Research - Acid rain Chemistry LibreTexts - The Chemistry of Acid Rain New York State - Department of Environmental Conservation - Acid Rain WebMD - What is Acid Rain? Britannica Websites Articles from Britannica Encyclopedias for elementary and high school students. acid rain - Children's Encyclopedia (Ages 8-11) acid rain - Student Encyclopedia (Ages 11 and up)
10332	https://www.youtube.com/watch?v=e3Y2a6Ysp-w	Common Mistakes with the Pythagorean Theorem \| BHNmath BHNmath 2060 subscribers 2 likes Description 864 views Posted: 11 Dec 2020 In this video we look at common reasons why students get wrong answers when using the Pythagorean Theorem. Transcript: Introduction in this video we'll take a quick look at why you might be arriving at incorrect answers when using the pythagorean theorem now if you find you're getting the wrong answers try asking yourself these questions am i identifying the hypotenuse am i calculating the correct side length am i squaring the side lengths and am i remembering to square root so let's work through an example and see where these issues can creep up Example in this problem we need to determine the unknown side length in the given triangle and first we'll consider the diagram method and after that we'll take a look at the algebra method regardless of which method you use i recommend that the first thing you do is identify the hypotenuse of the triangle and here it's 6.1 centimeters it's the side across from the right angle identifying the hypotenuse right away is a good idea because it'll play a significant role in how our solution unfolds so we've taken care of the first check mark on the left side of the screen we've identified the hypotenuse now notice that the side that we're going to be finding is not the hypotenuse so the second check mark on the left side of the screen is am i calculating the correct side length it's really important to remember the side length we are calculating here is not the hypotenuse so let's do what we normally do with the diagram method and draw squares on each side of our triangle now hopefully drawing the squares is enough to remind you that you need to find areas because the pythagorean theorem is really all about the areas of these squares and that's the third check mark on the left am i squaring the side lengths so let's do that right now this square has a side length of 6.1 centimeters so 6.1 times 6.1 gives us an area of 37.21 centimeters squared and similarly the bottom square has a side length of 2.8 centimeters so we can do 2.8 times 2.8 to get an area of 7.84 centimeters squared so we've taken care of squaring the side lengths now we need to find the area of this square and so often i see students take these two values and add them together to get the area of this square well that would be true if this square was based on the hypotenuse of the triangle but it's not and this goes back to the second check mark on the left side of the screen am i calculating the correct side length remember the two small areas add up to the hypotenuse square area so what we really need to find here is what number plus 7.84 gives us 37.21 and we can calculate that by doing 37.21 minus 7.84 so 37.21 minus 7.84 and that gives us 29.37 so always ask yourself am i calculating the correct side length if we're finding the hypotenuse we'll be adding two areas but if we're finding a side that's not the hypotenuse we'll be subtracting last but not least we have to remember that this is not our final answer that number 29.37 is the area of this square we were asked to find the unknown side length here and in order to do that we need to find which number multiplied by itself gives us 29.37 and we do that by taking the square root so if we take the square root of 29.37 we get approximately 5.4 and that is the unknown side length now let's take a look at the algebra method Algebra Method with the algebra method i once again recommend that you start by making a note of where the hypotenuse is in our triangle and there it is again there 6.1 centimeters it's the side across from the right angle and we usually start the algebra solution by writing out the pythagorean theorem equation a squared plus b squared equals c squared from that point we usually start substituting values for a b and c and this is where the second question on the left comes into play am i calculating the correct side length remember we are not calculating the hypotenuse here so we should not call that side c because side c in our equation over here represents the hypotenuse so if we call this a we can substitute a squared plus b squared which is 2.8 squared equals the hypotenuse squared which is 6.1 squared this is a critical step remember that c has to be the hypotenuse the third question on the left says am i squaring the side lengths this is usually not a big concern with the algebra method because the squaring is visible right there in the equation but make sure you remember to square so we can write a squared plus 2.8 squared which is 7.84 equals 6.1 squared which is 37.21 and we can go and isolate a we'll start by subtracting 7.84 from both sides of our equation the last question on the left says am i remembering to square root it's easy to forget to square root and think that this is the final answer here but remember that's the value of a squared we need to find the length of side a in order to do that we can use the inverse operation of squaring which is square rooting and the square root of 29.37 is approximately 5.4 and there you go so remember to always identify the hypotenuse and pay close attention to which side you're being asked to calculate make sure that you're squaring the side lengths to get the areas and don't forget to square root at the end to get the desired side length
10333	https://arxiv.org/pdf/cond-mat/0411340	-1- Electron effective mobility in strained Si/Si 1-xGe x MOS devices using Monte Carlo simulation V. Aubry-Fortuna, P. Dollfus, S. Galdin-Retailleau Institut d'Electronique Fondamentale, CNRS UMR 8622, Bât. 220, Université Paris-Sud, 91405 Orsay cedex, France. E-mail : valerie.fortuna@ief.u-psud.fr Abstract Based on Monte Carlo simulation, we report the study of the inversion layer mobility in n-channel strained Si/ Si 1-xGe x MOS structures. The influence of the strain in the Si layer and of the doping level is studied. Universal mobility curves μeff as a function of the effective vertical field Eeff are obtained for various state of strain, as well as a fall-off of the mobility in weak inversion regime, which reproduces correctly the experimental trends. We also observe a mobility enhancement up to 120 % for strained Si/ Si 0.70 Ge 0.30 , in accordance with best experimental data. The effect of the strained Si channel thickness is also investigated: when decreasing the thickness, a mobility degradation is observed under low effective field only. The role of the different scattering mechanisms involved in the strained Si/ Si 1-xGe x MOS structures is explained. In addition, comparison with experimental results is discussed in terms of SiO 2 / Si interface roughness, as well as surface roughness of the SiGe substrate on which strained Si is grown. Keywords strained Si, effective mobility, Monte Carlo simulation, MOSFET PACS codes 72.20.Dp, 72.20.Fr, 85.30.Tv -2- 1. Introduction The use of strained-Si channel pseudomorphically grown on a SiGe virtual substrate is becoming a promising way to accelerate the improvement of CMOS performance. In strained-Si surface channel configuration with gate oxide, both n-MOS and p-MOS transistors should strongly benefit from strain-induced enhancement of carrier transport. The biaxial tensile strain introduces splitting of degenerate bands which results, for both electrons and holes, in smaller in-plane conduction mass and reduced intervalley scattering thereby yielding improved carrier velocity. In the case of electrons this effect has been clearly shown from mobility measurement and calculation in SiGe-Si-SiGe quantum wells [2-4] and it has been used for designing high-performance MODFET with low noise figure and high cut-off and maximum oscillation frequencies [5-6]. Now efforts are made to transfer this advantage in CMOS technology on either bulk or insulating substrate [7-10]. It has been demonstrated that the effective electron mobility in MOS structures can be substantially enhanced using tensile strained-Si channel grown on SiGe buffer layer [7, 11, 12, 13]. Compared to unstrained Si structures, a 120% improvement has been recently reported . The aim of this article is to carefully analyse the effect of strain on electron transport in the Si inversion layer for MOS structures designed on bulk substrate. This theoretical study is based on the particle Monte Carlo method for solving the Boltzmann transport equation. Many works have been dedicated to the study of transport in bulk Si MOSFET inversion layer, both experimentally [14-16] and theoretically [17-20]. Since pioneering work of Sabnis and Clemens this type of transport is commonly characterized by the curves of carrier mobility plotted as a function of effective vertical field defined by b ieff Si Q QE η ε+= (1) where bQ and iQ are the bulk depletion and inversion charges per unit area, respectively, Si ε is the Si dielectric constant and η is an empirical parameter. For state-of-the-art bulk-Si MOSFET it has been clearly established that by taking η equal to 0.5 for electrons the mobility-field curves obey a "universal mobility" law independently of substrate doping [15, 16]. More recently, a similar universal behaviour has been observed for strained-Si MOSFET . Similarly, our present work is based on the calculation of mobility versus electric field in MOS structures. Our device Monte Carlo code includes a 3D Poisson solver self-consistently coupled to the Monte Carlo algorithm . For the present work limited to effective mobility extraction, the Poisson coupling is disabled. According to the gate bias, the 1D Poisson's equation is solved in the MOS capacitor prior to the Monte Carlo computation which is then made under frozen vertical field and uniform lateral driving field. This non-self-consistent approach may include quantization effects through coupling to 1D Schrödinger's equation . Such -3- quantization effects are expected to be strong in future bulk devices because the necessary use of high doping to reduce short channel effects may induce high transverse electric field. In this regard device simulation should include these effects. However, device simulators able to accurately describe scattering mechanisms (MC simulators) together with quantization effects through self-consistent coupling of the Poisson and Schrödinger equations have not yet reach a high level of maturity. Thus, despite a loss of accuracy, it may be considered that a 3D description of the carrier gas in the channel is still useful and relevant provided it is able to correctly reproduce experimental results. That is why quantization is not taken into account in this work. The empirical roughness scattering model presented below has been developed for a 3D gas in such a way that it can be used for self-consistent device simulation which does not include quantization yet. To correctly model the transport in inversion layer, a careful description of both roughness scattering and impurity scattering is required. The former dominates the transport at high effective field eff E and the latter strongly varies as a function of eff E as a result of the carrier density dependence of the screening effect. The impurity scattering model has been modified to include the local carrier density in the evaluation of the screening function. Surface roughness scattering is treated with the widely used technique for 3D electron gas which consists of an empirical combination of diffusive and specular reflections at oxide interface [22-24]. However, to recover the universal mobility on a large range of substrate doping and effective field, we show that the fraction of diffusive scattering should not be a constant but a function of eff E .The theoretical background of the Monte Carlo model is presented in Sec. 2. The results of electron transport in bulk-Si MOS structures are described in Sec. 3 and compared with experimental results. The transport in strained Si structures is then carefully analyzed in Sec. 4 as a function of strain and Si thickness. 2. Monte Carlo Model for electrons in inversion layer Here in Sect. 2 we give details of the Monte Carlo model used to describe the electron transport in Si, SiGe and strained Si in the case of MOSFET inversion layer. 2.1. Conduction-band structure of strained Si/ Si 1-x Ge x In this work, the conduction band of unstrained Si consists of six non parabolic ∆ valleys located along the directions at 85% of the Brillouin zone edge. The equi-energy surfaces are ellipsoidal with the transverse effective mass mt = 0.1905 m0 and the longitudinal effective mass ml = 0.9163 m0 (m0 being the free electron mass). The non parabolicity coefficient α is assumed to be 0.5. For Si(100) inversion layer, it is convenient to denote normal (or ∆2 ) valleys the two valleys that have the longitudinal axis normal to the plane of -4-growth and to denote parallel (or ∆4 ) valleys the four valleys that have the longitudinal axis in this plane. For strained Si inversion layers on a (100)-Si 1-x Ge x pseudosubstrate, the normal valleys shift down with respect to the parallel ones, therefore favoring electron transport with a transverse mass in the plane of growth. In bulk Si 1-x Ge x (for 0.3 x ≤ ), the conduction-band structure remains Si-like with six ∆ valleys assumed to be undistorted in the presence of Ge. We assume here that, at first order, transverse and longitudinal masses in tensile-strained Si and bulk Si 1-x Ge x remain unchanged by strain or in the presence of Ge [25, 26]. In addition, the effect of strain in Si is included in the splitting energy ∆Es between the two lowered normal valleys and the four raised parallel valleys. ∆Es is given as a function of the Ge content x in Si 1-x Ge x by ∆Es = 0.68 x (eV) [27, 28]. The presence of strain also reduces the Si band-gap, which leads to a conduction band-offset ∆Ec at the Si/ Si 1-xGe x interface given by ∆Ec = 0.55 x + 0.1 x2 (eV) [27, 28]. 2.2. Scattering mechanisms The scattering mechanisms included in the Monte Carlo model of electron transport in Si are acoustic intravalley scattering, intervalley scattering via three f and three g phonons, ionized impurity scattering, surface-roughness scattering at SiO 2 /Si interface and, for SiGe only, alloy scattering. 2.2.1. Phonon scattering The acoustic intravalley phonon scattering is treated as an elastic process. By neglecting anisotropic effects the scattering rate in ∆ valleys is given by ( ) ( ) ( ) 3/2 2BDac ac 2 2 1 1 2 k T m DE E E Eu λααπ ρ= + + = (2) where E is the electron kinetic energy, ( )1/3 2D t lm m m= is the density of states effective mass, ( )l t2 3u u u= + is the average sound velocity, kB the Boltzmann constant, T the lattice temperature, ρ is the material density and ac D is the acoustic deformation potential. The ∆-∆ intervalley transitions are generally treated via the usual zeroth-order transition matrix, which yields to the following expression of scattering rate ( )( ) ( ) 03/2 2iv D0iv piv iv 2iv iv iv iv iv 12 221 1 2 Z m DE N E EE E E E σλσ ω ∆ρ ωπα σ ω ∆α σ ω ∆⎡ ⎤= + − + + ×⎢ ⎥⎣ ⎦⎡ ⎤× + + + + + +⎣ ⎦ == == = (3) where iv ω= is the phonon energy, D0 is the corresponding deformation potential, Np is the phonon number, ∆Eiv is the intervalley energy transition, Ziv is the number of possible final valleys, σ is equal to –1 in the case of phonon absorption and to +1 in the case of phonon emission. -5-However, according to selection rules, the transitions with low energy phonons, i.e. f1(TA), g 1 (TA), and g 2 (LA) are forbidden via zero order process. Ferry proposed they should be considered by expanding the transition matrix to the first-order in the phonon wave vector without transgressing the selection rules . The deformation potential D1 is then expressed in eV instead of eV/cm for the coupling constant D0 of usual zero order process. The wave vector dependence of the scattering rate becomes a simple energy dependence using the isotropic approximation ( ) 2 2D2E E E k m α+ = = . If E′ stands for the final energy iv iv E E E σ ω ∆′ = + += the complete expression of the first-order scattering rate is finally ( )( ) ( ) ( ) ( ) 15/2 2D1iv p4iv 2 12 21 2 1 1 1 iv Z m D iE NE E E E E E E λ π ρ ωαααα⎡ ⎤= + − ×⎢ ⎥⎣ ⎦′ ′ ′ ′ ′⎡ ⎤× + + + + +⎣ ⎦ = = (4) We have adopted this approach to treat ∆-∆ intervalley scattering through f 1 , g 1 and g 2phonons while scattering through f 2 , f 3 and g 3 phonons are considered via the usual zeroth-order process [4, 31]. A similar approach is used by other authors [32-34]. The material and phonon parameters used in the calculation are listed in Table I. 2.2.2 . Impurity scattering Within the Born approximation, impurity scattering is treated via the screened Coulomb potential using the momentum-dependent screening length L defined by ( ) ( ) 22D2SB ( ) 1 e n F q Fk TL ξ ξε= = r (5) where Dq is the inverse of the Debye-Hückel screening length, ( )n r is the local carrier density, S ε is the dielectric permittivity and the normalised variable ξ is defined by 22 2D B8 qe m k T ξ = = (6) where kkq ′−= is the wave vector involved in scattering event from an initial state k to a final state ′k . The screening function ( )F ξ has been derived for nondegenerate semiconductors [35, 36] and may be conveniently rewritten as ( ) ( ) ( )2 201 exp exp F x dx ξ ξξξ= − ∫ (7) which is a tabulated function . Considering the scattering angle θ , the impurity scattering rate for an electron in state k of energy E is given by ( ) ( ) ( ) ( )( )( ) 1243 / 2 imp Dimp 24222S1D cos 1 2 12 dZ ek m N E E Eq q F θλααπ εξ−= + ++∫= (8) -6-where imp N is the impurity concentration and Z is the number of charge units of the impurity. In the integral, the terms q and ξ depend on the angle θ through ( )2 22 1 cos q k θ= − .Again, the isotropic approximation of the dispersion relation may be conveniently used to make ( )imp k λ an energy-dependent function ( )imp E λ . The electron-impurity scattering is intrinsically an anisotropic mechanism but for mobility calculation it may be treated as an isotropic process by replacing the actual scattering rate by the reciprocal of the momentum relaxation time 1imp τ − obtained by introducing the factor θcos 1 − in the integration of Eq. 8 . For transport calculation in uniform material the density n entering the screening length of Eq. 4 is usually chosen equal to the impurity concentration imp N . It is not valid in the case of transport in MOS inversion layer where the local carrier density is strongly non-uniform and dependent on the applied bias. In our model the screening length is thus calculated from the local density ( ) n r , which is important to obtain good results for the effective mobility on the full range of effective field at any impurity concentration. 2.2.3. Surface-roughness scattering In inversion layers, carrier transport can be strongly affected by irregularities at the SiO 2 /Si interface and by charge distributions in SiO 2 . Surface-roughness scatterings can be treated as described in Refs [18, 40] using two technology-dependent parameters Λ and ∆, the correlation length and amplitude of the interface roughness, respectively. Coulomb scattering with fixed charges located in the oxide and/ or at the interface can also be taken into account [17, 18, 41]. This approach is appropriate to a two-dimensional electron gas, but there is not any standard model for the case of three-dimensional gas. In the present Monte-Carlo simulations, only surface-roughness scattering due to the deviation of the SiO 2 /Si interface from an ideal plane is considered and is treated with an empirical combination of specular and diffusive scattering for carriers that hit the interface . Usually, the fraction of diffusive scattering Ndiff is considered as a constant value which may vary significantly, however, according to the authors: for example, Ndiff can be equal to 6% , 8.5% , 15% or 50% . In the present work, the fraction Ndiff has been chosen to vary with the effective vertical field Eeff and the determination of Ndiff as a unique function of Eeff is developed in Sec. 3.1. 2.2.4 . Alloy scattering in Si 1-xGe x In the simulated structures with thin strained-Si channel some electrons may enter the Si 1-x Ge x virtual substrate where alloy scattering is to be considered. Alloy scattering is treated within the classical model of the "square well" perturbation potential of height all U in a sphere of volume V centred on each alloy site. The radius of this sphere is arbitrarily chosen as the nearest-neighbour distance 43 00 ar = where a0 is the lattice parameter . The alloy potential all U is considered equal to 0.8 eV for electrons in SiGe . The isotropic alloy scattering relaxation time is given by -7- ( ) ( ) ( ) ( ) 33 / 2 20Dall 4all 3 21 1 1 2 164 a m x x U E E EE παατ = − + + = (9) 2.3. Simulated structures and simulation procedure The epilayer stack of the simulated MOS structures consists of tensile-strained Si pseudomorphically grown on thick Si 1-x Ge x (001) pseudo-substrate. The thickness of the top strained Si layer is either 2 nm, 5 nm, 8 nm or 420 nm. The latter thickness is unrealistic for practical applications and is just considered for the purposes of comparison. The Ge content x in the pseudo-substrate varies from 0 to 30%. The doping levels of Si and Si 1-x Ge x are identical and vary from 1 ×10 16 to 1 ×10 18 cm -3 .Prior to Monte Carlo simulation of transport, the 1D Poisson's equation is solved in the MOS capacitor according to the gate bias VGS to obtain the vertical field and carrier density profiles. The corresponding effective field is deduced from Eq. 1. The Monte Carlo calculation is then performed using this vertical profile of electric field and considering a uniform parallel driving field. As mentioned in the introduction, quantization effects are not taken into account. All simulations are performed at room-temperature. 3. Effective mobility in bulk Si MOS structures 3.1. Determination of Ndiff The fraction of diffusive scattering has been determined as a function of Eeff from an adjustment to experimental mobilities obtained for an inversion Si layer in a lightly doped n-MOSFET. Indeed, for small doping levels the effects of impurity Coulomb scattering on carrier transport are not strong and the Eeff dependence of μeff is mainly due to roughness scattering. For a given VGS (i.e. for a given Eeff ), some simulations have been performed by varying Ndiff . The obtained effective mobilities ( )eff diff N μ are then compared to the experimental μeff , which allows us to deduce the relevant Ndiff value. Repeating this procedure for different Eeff leads finally to the following expression of Ndiff as a function of Eeff (with Eeff in kV/cm) : 2 3 44 7 10 14 diff eff eff eff eff 0.176 2.29 10 3.1 10 1.69 10 2.84 10 N E E E E− − − −= − × + × − × + × (10) We have then assumed that this expression can be applied for any other doping levels and whatever the strain in the Si layer, which is discussed in Sec. 4.1. 3.2. Universal mobility curves On Fig. 1 we have plotted, for various impurity concentrations, the ( )eff eff E μ curves obtained by Monte Carlo simulation (lines) as well as some experimental curves (lines with symbols) [11, 15, 16, 45]. As experimental mobilities, the calculated mobilities are in good agreement with the “universal mobility curve”, as defined by Takagi et al. (dashed line). -8-In addition, the ( )eff eff E μ curves mirror the fall-off of the experimental mobilities in weak inversion regime, which results from reduced screening of impurities at low electron density (see Eqs. 5 and 8). The respective influence of impurity and surface-roughness scatterings is clearly illustrated in Fig. 2 where we plot the average rate of scattering for different scattering mechanisms. These rates are obtained by counting the scattering events experienced by all simulated particles (50000) for a given period of time (270 ps) . Under high Eeff , electrons are distributed close to the SiO 2 /Si interface, and the probability for an electron to undergo a diffusive scattering event increases for a given period of time. Indeed, as it can be seen on Fig. 2, for high Eeff values, surface-roughness scatterings (circles) are dominant and independent of the doping level, leading to the decrease of the mobility and to the “universal curve”. At low VGS , the deviation from the “universal mobility curve” is larger as the Si doping level increases and is directly related to impurity Coulomb scattering. Indeed, Fig. 2 shows that these impurity scatterings (triangles) are obviously more frequent when the doping level is high (see the arrows on Fig. 1 and 2, indicating the points corresponding to VGS = 0 V). For 10 18 cm -3 , this type of scattering dominates under low VGS bias, while its influence decreases when increasing VGS , as a consequence of enhanced electron density and higher screening effects. Nevertheless, for a given doping level, we can notice that the experimental curves and those obtained by simulation are not systematically superimposed at low VGS bias. Two possible reasons can be put forward: there is either a difference between real and announced experimental doping levels, or/and a poor estimation of screening effects for the computation of impurity scattering rate. The first explanation can be illustrated with the help of the experimental curves of Fig. 1 corresponding to doping levels of 10 16 , 10 17 and 1.4 ×10 17 cm -3 (triangles and reverse triangles). On one hand, for similar announced doping levels (full triangles and reverse triangles), a significant discrepancy of the μeff values is observed, and, on the other hand, identical μeff values are obtained for two different doping levels (full and open triangles). Second, each experimental curve may be fitted by calculation after adjustment of the screening length entering Eq. 8. A single approach of screening does not allow us to describe exactly each different technology, i.e. all experimental spread data found in the literature . Nevertheless, the experimental trends are correctly reproduced and in addition to previous validation of phonon scattering model [4, 31], these first results validate our combined approach of Coulomb and surface-roughness scatterings. It should be mentioned again that this approach has been developed in such a way that it is easily applicable to self-consistent device Monte Carlo simulation. 4. Effective mobility in strained Si inversion layers 4.1 Effect of strain on electron mobility The simulations have been extended to MOS structures with 8 nm-thick strained Si as surface channel on relaxed Si 1-x Ge x virtual substrate for various values of x (0.05, 0.10, 0.15, -9-0.20 and 0.30) and of doping level (10 16 , 1.4 ×10 17 , 3 ×10 17 and 10 18 cm -3 ). On Fig. 3, we have reported the effective mobility as a function of Eeff for x = 0.15 and bulk Si. The curves ( )eff eff E μ obtained for the other x values are omitted here for clarity. Whatever x, these curves mirror the behaviour of μeff versus Eeff obtained for bulk Si. Indeed, we observe a universal character of μeff for high effective field, as well as a drop of the mobility in weak inversion regime. We can also see the expected mobility enhancement when compared to the case of bulk Si. This effect is more clearly evidenced on Fig. 4, where we have plotted ( )eff eff E μ by varying x from 0 to 0.30, the doping level being equal to 10 16 cm -3 . We observe an increase of the mobility with the Ge content in the pseudosubstrate, i.e. with strain in the Si inversion layer. For x ≤ 0.10 the mobility enhancement is nearly constant over the full range of effective field. For x ≥ 0.15, this mobility enhancement depends on Eeff : the maximum of μeff is reached for x = 0.30 under low Eeff (≤ 300 kV/ cm) and as soon as x = 0.15 under higher effective fields (> 300 kV/ cm). The mobility enhancement is also observed for the other doping levels studied in this work and can be explained in terms of effective mass and of average rate of the different scattering mechanisms . Indeed, when increasing x, more and more electrons reside in the ∆2valleys of the strained Si inversion layer, and then exhibit a transverse effective mass (i.e. a low mass) in the direction of transport and a longitudinal effective mass (i.e. a high mass) in the vertical direction. Concurrently, we observe that the average rate of surface-roughness and intervalley phonon scatterings diminishes (see Fig. 5), which also contributes to the improvement of the mobility. The number of acoustic phonon scatterings per unit of time (not shown in Fig. 5) remains independent of Eeff and strain, and is equal to about 1.93 ×10 12 s -1 . The rate of intervalley phonon scatterings diminishes continuously with increasing x, mainly due to the increase of the splitting energy ∆Ε s between the ∆2 and ∆4valleys. The decrease of the average rate of surface-roughness scattering with x is related to the increase of electron population in the ∆2 valleys. Indeed, the electrons of the ∆2 valleys that hit the SiO 2 / Si interface have a low vertical component of velocity due to the longitudinal effective mass, and then this scattering mechanism happens less frequently when x increases. As soon as x reaches 0.15, about 96% of electrons in the strained Si are in the ∆2valleys, and therefore increasing x beyond 0.15 does not almost change the number of scatterings, as it can be seen on Fig. 5 (full symbols). Under low Eeff , both intervalleys phonon and surface-roughness scattering decrease contribute to the mobility enhancement with x. Nevertheless, under high effective field, surface-roughness scatterings clearly dominate and the maximum of mobility is reached for x = 0.15, due to the saturation of the number of this scattering. We have then compared the ( )eff eff E μ curves of Fig. 4 to experimental mobilities [7, 11-13, 45, 47]. The experimental electron mobilities versus Eeff are generally extracted from the linear regime of drain current ID (VDS ) for long surface-channel strained Si n--10-MOSFETs. Experimentally, the mobility enhancement is nearly constant over the full range of effective field, whatever x [11, 47]: see for example the results of Currie et al. in the insert of Fig. 4 . Rim et al . have shown, with the help of a simple curve-fitting analysis, that both phonon scattering rate and surface-roughness scattering rate should be reduced to correctly describe the experimental mobility in strained Si on Si 1-xGe x for x ≥ 0.15 . This observation suggests that strain may influence surface-roughness scattering: indeed, Boriçi et al. have evidenced that a SiO 2 / strained Si interface is less rough than a SiO 2 / Si one. From the point of view of our simulations, it means that considering ( )diff eff N E as unchanged by strain leads to under-estimated mobility at high effective fields. Further experimental data would be necessary to check this assumption and to quantify the effect of strain on Ndiff . Considering mobility values under low effective fields, typically 0.3 MV/cm, we have reported in Fig. 6 the electron mobility enhancement relative to bulk Si as a function of x for our results (open symbols) and for experimental data (full symbols). The saturation of the mobility when x reaches 0.15 is also evidenced experimentally (see data from Refs [11, 47]), and increasing the Ge content above 0.20 has no more effect on mobility . Nevertheless, the plot of Fig. 6 shows a discrepancy between the various experimental results. For instance, for x = 0.30, the electron mobility can be enhanced by ≈ 80% or by 120% [12, 13]. These differences can be explained in terms of surface roughness of the strained Si/Si 1-x Ge x . Sugii et al. investigated the relationship between Si surface morphology (periods and edge shapes of undulations) and electron mobility . They obtained the best enhancement of mobility when the strained Si was grown on a pseudosubstrate consisting on a Si 1-y Ge y (0 < y < x) graded buffer layer (GBL) followed by a Si 1-x Ge x buffer layer, and they concluded that surface roughness must be controlled to increase mobility in strained Si layer. Recently, an approach with Monte Carlo simulations has confirmed the strong dependence of SiGe pseudosubstrate roughness (correlation length and amplitude) on mobility . Other investigations showed that adding chemical-mechanical-polishing (CMP) of the pseudosubstrate improves the Si surface, whose roughness becomes of about 0.4 nm compared to about 10 nm for strained Si layer grown on GBL without CMP [12, 34]. When a thick metastable strained Si layer (≈ 25 nm) is then grown directly on the pseudosubstrate with CMP, mobilities are significantly enhanced when compared to devices without CMP (see full triangle on Fig. 6). On the other hand, this enhancement is not observed for thin strained Si layer (8 nm in Ref. ) or when a regrowth of SiGe is carried out after CMP . In the former case, carriers must be close to contaminants resulting from the CMP, and in the latter case the roughness has again increased after regrowth of SiGe. According to Fig. 6, a large enhancement of μeff is also observed for Refs and (full triangle and square, respectively), despite no CMP has been used. Olsen et al. obtained low surface roughness by growing a Si/Si 0.7 Ge 0.3 structure on a relaxed constant composition Si 0.85 Ge 0.15 virtual substrate and the same mobility enhancement as Sugii et al . was -11-achieved . Concerning the result of Rim et al. , the lack of experimental details does not allow us to explain this mobility enhancement . For the other data plotted on this figure, the technological processes for NMOS fabrication including a GBL were similar and lead to close mobility values [11, 45, 47]. Our results obtained from Monte Carlo simulation are in accordance with the best experimental mobility enhancements [7, 12, 13]. Other modelling approaches have considered the quantized nature of the inversion layer for transport calculation in strained Si [40, 41, 50, 51]. Using the standard Ando's model of roughness scattering it has been shown that the scattering rate in strained Si is strongly overestimated . The model has been improved by Gámiz et al. [50,51] to include the effect of the buried layer (SiGe or SiO 2 ) on the perturbation Hamiltonian, which yields satisfactory mobility behaviour versus transverse electric field . It should be noted that if plotted on a linear scale, the results obtained from our simple empirical approach are in good agreement with that of Ref. 51. A similar method for the case of a 3D electron gas has been developed by including the correlation length Λ and the amplitude ∆ of surface fluctuations in the determination of the specular scattering probability which, however, is independent of transverse field . Using this approach, a correct agreement with experimental high-field mobility in strained Si is found only if Λ and ∆ are significantly longer and smaller, respectively, than in relaxed Si, which is an additional indication that the strain in Si reduces the roughness of the SiO 2 interface. 4.2 Effect of strained Si thickness on electron mobility We have also investigated the effect of the strained Si thickness on electron mobility. Keeping the Ge content and the doping level constant (0.15 and 10 16 cm -3 , respectively), simulations with four different Si thicknesses have been performed. The corresponding ( )eff eff E μ curves are plotted on Fig. 7. An influence of the strained Si thickness is obvious under low effective field. Indeed, μeff falls off when decreasing the Si thickness down to 5 nm. Between 5 nm and 420 nm, the values of mobility are quite close to each other. Under high effective fields, the mobility remains unchanged whatever the strained Si thickness. In these structures, the distribution of electrons in the different layers as a function of Eeff (see full symbols of Fig. 8) show that under low effective field, the percentage of electrons present in the SiGe pseudosubstrate increases with decreasing the strained Si thickness. This leads to parasitic electron conduction through the low-mobility SiGe underlayer, which degrades the overall mobility. Additionally, the number of electrons close to the SiO 2 /Si interface is higher for thinner strained Si layers, which is then accompanied by an increase of the average rate of surface-roughness scattering events (Fig. 8, open symbols). These facts explain why, under low effective field, mobility decreases with decreasing the Si thickness. Under high effective fields, the electrons are all located in the strained Si layer and the average rate of surface-roughness scattering is identical from a structure to another. Whatever the strained Si -12-thickness, 96% of the electrons located in the Si exhibit a transverse mass, and therefore the Si thickness has no influence on the mobility, as it can be seen on Fig. 7. Currie et al . have studied the effect of channel thickness on mobility enhancement for NMOS devices . They observed large mobility degradations for the thinnest channels, and as soon as the channel thickness increases beyond 5 nm, the devices exhibit the expected mobility enhancements we have reported on Fig. 6 (full circles). This trend is in accordance with our observations. Conclusion In this work, we have carefully analysed the effect of strain in Si, doping level and Si thickness on electron transport in the inversion layer of a strained Si/ Si 1-xGe x MOS structure. Our investigations were based on the calculation of effective mobility versus effective field using Monte Carlo simulations. As for bulk Si inversion layers, a universal relationship between μeff and Eeff is found when Si is strained. In addition, we observe the deviation from these universal curves in weak inversion regime, due to Coulomb scatterings with impurities. The experimental trends are correctly reproduced by these results, which validates our approach of phonon, Coulomb and surface-roughness scatterings. Regarding the strain-induced mobility enhancement factor (maximum of 120 % for strained Si/ Si 0.70 Ge 0.30 ), our simulation results match well with best available experimental data. The role of Si thickness is evidenced under low effective field and for thickness lower than 5 nm. A degradation of the mobility is then observed, due to electron conduction through the SiGe underlayer and increase of surface-roughness scatterings. Acknowledgments This work was partially supported by the European Community 6th FP (Network of Excellence SINANO, contract no 506844). -13- References Van de Walle CG. Strain effect on the valence-band structure of SiGe. In: Kasper E, editor. Properties of strained and relaxed silicon germanium, London: INSPEC, EMIS Datareviews Series 12; 1995. p. 94-102. Ismail K, Nelson SF, Chu JO, Meyerson BS. Electron transport properties of Si/SiGe heterostructures: measurements and device implications. Appl. Phys. Lett. 1993;63:660-2. Hackbarth T, Hoeck G, Herzog HJ, and Zeuner M. Strain relieved SiGe buffers for Si-based heterostructure field-effect transistors. J. Cryst. Growth 1999;201/202:734-38. 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Olsen SH, O'Neill AG, Chattopadhyay S, Kwa KSK, Driscoll LS, Norris DJ, Cullis AG, Robbins DJ, Zhang J. Evaluation of strained Si/ SiGe material for high performance CMOS. Semicond. Sci. Technol. 2004;19:707-14. Kim K, Chuang C-T, Rim K, Joshi RV. Performance assessment of scaled strained-Si channel-on-insulator (SSOI) CMOS. Solid-State Electron. 2004;48:239-43. Langdo TA, Currie MT, Cheng Z-Y, Fiorenza JG, Erdtmann M, Braithwaite G, Leitz CW, Vineis CJ, Carlin JA, Lochtefeld A, Bulsara MT, Lauer I, Antoniadis DA, Somerville M. Strained Si on insulator technology: from materials to devices. Solid-State Elect. 2004;48:1357-67. Currie MT,. Leitz CW, Langdo TA, Taraschi G, Fitzgerald EA, Antoniadis DA. Carrier mobilities and process stability of strained Si n- and p-MPOSFETs on SiGe virtual substrate. J. Vac. Sci. Technol. B 2001;19:2268-79. Sugii N, Hisamoto D, Washio K, Yokoyama N, Kimura S. Performance enhancement of strained-Si MOSFETs fabricated on a chemical-mechanical-polished SiGe substrate. IEEE Trans. Electron Dev. 2002;49:2237-43. Olsen SH, O'Neill AG, Driscoll LS, Kwa KSK, Chattopadhyay S, Waite AM, Tang YT, Evans AGR, Norris DJ, Cullis AG, Paul DJ, Robbins DJ. High-performance nMOSFETs -14-using a novel strained Si/SiGe CMOS architecture. IEEE Trans. Electron Dev. 2002;50:1961-8. Sabnis AG, Clemens JT. Characterization of the electron mobility in the inverted <100> Si surface. In: IEDM Tech. Dig. 1979, p.18-21 Sun SC, Plummer JD. Electron mobility in inversion and accumulation layers on thermally oxidized silicon surfaces. IEEE Trans. Electron Dev. 1980;27:1497-508. Takagi S, Toriumi A, Iwase M, Tango H. On the universality of inversion layer mobility in Si MOSFET's: Part I-Effects of substrate impurity concentration. IEEE Trans. Electron Dev. 1994;41:2357-62. Gámiz F, Melchor I, Palma A, Cartujo P, López-Villanueva JA. 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Sangiorgi E, Pinto MR. A semi-empirical model of surface scattering for Monte Carlo simulation of silicon n-MOSFET's. IEEE Trans. Electron Dev. 1992;39:356-61. Bufler FM, Fichtner W. Scaling of strained-Si n-MOSFETs into the ballistic regime and associated anisotropic effects. IEEE Trans. Electron Dev. 2003;50:278-84. Richard S, Cavassilas N, Aniel F, Fishman G. Strained silicon on SiGe: temperature dependence of carrier effective masses. J. Appl. Phys. 2003;94:5088-94. Fischetti MV, Laux SE. Band structure, deformation potentials, and carrier mobility in strained Si, Ge and SiGe alloys. J. Appl. Phys. 1996;80:2234-52. Galdin S, Dollfus P, Aubry-Fortuna V, Hesto P, Osten HJ. Band offsets predictions for strained group IV alloys: Si 1-x-y Ge x Cy on Si (001) and Si 1-x Ge x on Si 1-z Ge z (001). Semicond. Sci. Technol. 2000;15:565-72. Dollfus P, Galdin S, Osten HJ, Hesto P. Band offsets and electron transport calculation for strained Si 1-x-y Ge x Cy/Si heterostructures. 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Phys. 2002;92:7320-4. Fiegna C, Sangiorgi E. Modeling of high-energy electrons in MOS devices at the microscopic level. IEEE Trans. Electron Dev. 1993;40:619-27. Harrison JW. Alloy scattering in ternary III-V compounds. Phys. Rev. B1976;13:5347-50. V. Venkataraman, C. W. Liu, and J. C. Sturm. Alloy scattering limited transport of two-dimensional carriers in strained Si 1-x Ge x quantum wells. Appl. Phys. Lett. 1993;63: 2795-7. Rim K, Hoyt JL, Gibbons JF. Fabrication and analysis of deep submicron strained-Si N-MOSFET's. IEEE Trans. Electron Dev. 2000;47:1406-15. -16- Roldán JB, Gámiz F. Simulation and modelling of transport properties in strained-Si and strained-Si/SiGe-on-insulator MOSFETs. Solid-State Electron. 2004;48:1347-55. Rim K, Anderson R, Boyd D, Cardone F, Chan K, Chen H, Christansen S, Chu J, Jenkins K, Kanarsky T, Koester S, Lee BH, Lee K, Mazzeo V, Mocuta A, Mocuta D, Mooney PM, Oldiges P, Ott J, Ronsheim P, Roy R, Steegen A, Yang M, Zhu H, Ieong M, Wong HSP. 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Watling JR, Yang L, Boriçi M, Wilkins RCW, Asenov A, Barker JR, Roy S, Solid-State Electron. 2004, 48:1337-46. -17- Figure captions Fig. 1: Calculated and experimental [11, 15, 16, 45] μeff (Eeff ) curves for electrons in bulk Si inversion layers for various doping levels (in cm -3 ). Arrows indicate the bias point V GS = 0. Fig. 2: Average rate of surface-roughness scattering (circles), intervalleys phonon scattering (rhombus) and Coulomb scattering with doping impurities (triangles) as a function of Eeff in two different doped bulk Si: 1 ×10 16 cm -3 (open symbols) and 1 ×10 18 cm -3 (full symbols). These rates are obtained by counting the scattering events experienced by all simulated particles (50000) for a given period of time (270 ps). Fig. 3: Calculated μeff (Eeff ) curves for electrons in 8 nm-thick strained Si/ Si 0.85 Ge 0.15 and bulk Si inversion layers for various doping levels (cm -3 ). Fig. 4: Calculated μeff (Eeff ) curves as a function of x for electrons in 8 nm-thick strained Si/Si 1-xGe x inversion layers doped at 1 ×10 16 cm -3 . For comparison, experimental results of Currie et al. are reported in the insert. Fig. 5: Average rate of surface-roughness scattering (full symbols) and intervalleys phonon scattering (open symbols) as a function of Eeff and for various x values in 8 nm-thick strained Si/Si 1-xGe x inversion layers doped at 1 ×10 16 cm -3 .Fig. 6: Electron mobility enhancement deduced from our simulations (open circles) and from experimental results (full symbols) [7, 11-13, 45, 47]. Fig. 7: Calculated μeff (Eeff ) curves as a function of Si thickness for electrons in strained Si/Si 0.85 Ge 0.15 inversion layers doped at 1 ×10 16 cm -3 .Fig. 8: Average rate of surface-roughness scatterings as a function of Eeff ans Si thickness in strained Si/Si 0.85 Ge 0.15 inversion layers doped at 1×10 16 cm -3 (open symbols). The percentage of electrons in the Si 0.85 Ge 0.15 underlayer is also reported (full symbols). -18- Tables Table I. Material and phonon scattering parameter for electrons in Si Lattice constant a0 (Å) 5.431 Density ρ (g/cm 3 ) 2.329 Longitudinal sound velocity ul (cm/s) 9.0 × 10 5Transverse sound velocity ut (cm/s) 5.4 × 10 5Dielectric constant εs 11.7 Intravalley acoustic phonon scattering: Deformation Potential Dac (eV) 6.6 Intervalley phonon scattering: phonon energy Deformation Potential Scattering iv ω= (meV) D0 (eV/cm) D1 (eV) mode (type) 11.4 3.0 TA (g1) 18.8 3.0 LA (g2) 63.2 3.4 ×10 8 LO (g3) 21.9 3.0 TA (f1) 46.3 3.4 ×10 8 LA (f2) 59.1 3.4 ×10 8 TO (f3) -19-Aubry-Fortuna et al. Figure 1 10 210 3100 1000 simulations Ref. -2×10 16 Ref. -3×10 17 Ref. -2.4×10 18 Ref. -1.4×10 17 Ref. -1×10 16 Ref. -1×10 17 Ref. -universal μ μ eff (cm 2V-1 s-1 )E eff (kV/cm) 2×10 16 1.4×10 17 3×10 17 1×10 18 1×10 16 VGS = 0 V -20-Aubry-Fortuna et al. Figure 2 10 10 10 11 10 12 10 13 10 14 100 1000 roughness scattering intervalley phonon scattering impurity scattering Eeff (kV/cm) open symbols : 10 16 cm -3 full symbols : 10 18 cm -3 Average rate of scattering (s -1 )VGS = 0 V -21-Aubry-Fortuna et al. Figure 3 10 210 3100 1000 1×10 16 1.4×10 17 3×10 17 1×10 18 μ eff (cm 2V-1 s-1 )E eff (kV/ cm) 8 nm-thick strained Si on Si 0.85 Ge 0.15 Si bulk -22-Aubry-Fortuna et al. Figure 4 10 210 3100 1000 bulk Si x = 0.05 x = 0.10 x = 0.15 x = 0.20 x = 0.30 μ eff (cm 2V-1 s-1 )Eeff (kV/ cm) Exp. results of Ref. 300 400 500 600 800 1000 100 200 300 400 600 800 -23-Aubry-Fortuna et al. Figure 5 10 11 10 12 10 13 100 1000 bulk Si x = 0.05 x = 0.15 x = 0.20 x = 0.30 E eff (kV/cm) open symbols: IV phonon scattering full symbols: roughness scattering Average rate of scattering (s -1 )-24-Aubry-Fortuna et al. Figure 6 020 40 60 80 100 120 140 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 this work Ref. Ref. Ref. Ref. Ref. [12,13] Mobility enhancement (%) Ge content -25-Aubry-Fortuna et al. Figure 7 10 210 3100 1000 2 nm 5 nm 8 nm 420 nm μ eff (cm 2V-1 s-1 )E eff (kV/ cm) strained Si on Si 0.85 Ge 0.15 doped at 10 16 cm -3 Si thickness: -26-Aubry-Fortuna et al. Figure 8 10 11 10 12 10 13 0510 15 20 25 30 100 1000 2 nm 5 nm 8 nm 420 nm E eff (kV/cm) % of electrons in SiGe (%) Average rate of scattering (s -1 )Si thickness: roughness scattering
10334	https://math.stackexchange.com/questions/18024/set-vs-multiset	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 Set vs Multiset Ask Question Asked Modified 13 years, 4 months ago Viewed 2k times 2 $\begingroup$ A set is a collection of distinct objects. Suppose I sample some parameter of some population and the results are ${35,35,36,36,37 }$. Do I need to say (i) the multiset of the samples is ${35,35, \dots }$ instead of (ii) the set of samples is ${35,35, \dots }$? statistics multisets Share edited Jun 1, 2012 at 2:54 William 20.3k22 gold badges3636 silver badges6464 bronze badges asked Jan 18, 2011 at 21:00 Lior KoganLior Kogan 21911 silver badge1010 bronze badges $\endgroup$ 1 $\begingroup$ It seems worth mentioning that informal representations of sets such as {a,b,c} where a, b, and c are letters of the alphabet are often used. I say informal because a, b, and c interpreted as letters are not themselves sets while in formal set theory, every element of a set is a set itself. However, we can replace something such as {a, b, c} with a more formal {$0, 1, 2$}, and $0 = \emptyset$, $1 = {0}$, and $2 = {0, 1}$ so each member is indeed a set itself. The point is that how formal one should be really seems to depend on the context as suggested by the answers. $\endgroup$ Jason – Jason 2011-01-18 22:27:00 +00:00 Commented Jan 18, 2011 at 22:27 Add a comment \| 3 Answers 3 Reset to default 3 $\begingroup$ Yes, I think it would be better to use the word "multi-set" here. Formally, say, ${35,35,7}$ and ${35,7}$ are the same set, but in your case the number of times each element is mentioned actually matters. One caveat: Not everybody is careful, and some see this distinction as pedantic, so you may find in books or websites mention of the set ${35,35,7}$ when really, it is the multi-set that is meant. Also, sometimes people use ordered tuples $(35,35,7)$ to avoid this issue. Share answered Jan 18, 2011 at 21:05 Andrés E. CaicedoAndrés E. Caicedo 81.9k1010 gold badges232232 silver badges364364 bronze badges $\endgroup$ Add a comment \| 3 $\begingroup$ Two sets $A$ and $B$ are the same if and only if they have the same elements; that is, if and only if for every $x$, $x\in A$ is true if and only if $x\in B$ is true. That means that the set $A={1,1}$ and the set $B={1}$ are really the same set, because they have the exact same elements. Sets cannot "tell" how many times an element is listed in it. Your set, $A={35, 35, 36, 36, 37}$, and the set $B={35, 36, 37}$ are the same set, for every $x$, $x$ is in $A$ if and only if $x$ is in $B$. They are both also the same as the set $C={37, 37, 37, 37, 35, 36}$, and the set $D={37,36,35,35,35,35,35,35,35,35,35,35}$. Multisets are slightly different; in a multiset, the number of times that an element appears does matter. Two multisets $A$ and $B$ are equal if and only if for every $x$, the number of times that $x$ occurs in $A$ is the same as the number of times that $x$ occurs in $B$. However, multisets don't care about order, so the multiset $A={35, 35, 36, 36, 37}$ and the multiset $B={35, 36, 37, 35, 36}$ are equal. If you want to keep track of both the number of times and the order, the simplest thing is to consider ordered pairs; for example, instead of the set you attempt to give, you could have a set in which the $n$th reading is represented by the ordered pair $(n,a)$, where $a$ is the reading itself. So you could take $A={(1,35), (2,35), (3,36), (4,36), (5,37)}$; this would be different from the set in which the readings are, in order, $35$, $36$, $37$, $36$, and $35$. So you need to figure out exactly what is it that matter and what does not matter. If only what actually occurs matters, use sets. If what matters is what occurs and the number of times it occurs, use multisets. If what matters is what occurs, the order, and the number of times it occurs, use sets of ordered pairs or some other modification. Either way, to avoid confusion, use the corresponding nomenclature: if you want multisets, say "multiset"; if you want sets, say "set". Share answered Jan 18, 2011 at 21:07 Arturo MagidinArturo Magidin 419k6060 gold badges864864 silver badges1.2k1.2k bronze badges $\endgroup$ Add a comment \| 1 $\begingroup$ I think it would be proper to use "multiset". But most statisticians probably just say "set". Share answered Jan 18, 2011 at 21:03 NebulousRevealNebulousReveal 1 $\endgroup$ 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 statistics multisets See similar questions with these tags. 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10335	https://pmc.ncbi.nlm.nih.gov/articles/PMC1489655/	Energetic Consequences of Nitrite Stress in Desulfovibrio vulgaris Hildenborough, Inferred from Global Transcriptional Analysis - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. PMC Search Update PMC Beta search will replace the current PMC search the week of September 7, 2025. Try out PMC Beta search now and give us your feedback. 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Learn more: PMC Disclaimer \| PMC Copyright Notice Appl Environ Microbiol . 2006 Jun;72(6):4370–4381. doi: 10.1128/AEM.02609-05 Search in PMC Search in PubMed View in NLM Catalog Add to search Energetic Consequences of Nitrite Stress in Desulfovibrio vulgaris Hildenborough, Inferred from Global Transcriptional Analysis† Qiang He Qiang He Virtual Institute for Microbial Stress and Survival, Berkeley, California 94720‡,1 Environmental Sciences Division, Oak Ridge National Laboratory, Oak Ridge, Tennessee 37831,2 Department of Civil and Environmental Engineering, Temple University, Philadelphia, Pennsylvania 19122,3 Physical Biosciences Division, Lawrence Berkeley National Laboratory, Berkeley, California 94720,4 Department of Microbiology, Miami University, Oxford, Ohio 45056,5 Earth Sciences Division, Lawrence Berkeley National Laboratory, Berkeley, California 94720,6 Department of Bioengineering, University of California, Berkeley, California 94720,7 Howard Hughes Medical Institute, Chevy Chase, Maryland 20815,8 Departments of Biochemistry and Molecular Microbiology and Immunology, University of Missouri—Columbia, Columbia, Missouri 65211,9 Institute for Environmental Genomics, Department of Botany and Microbiology, University of Oklahoma, Norman, Oklahoma 73019 10 Find articles by Qiang He 1,2,3, Katherine H Huang Katherine H Huang Virtual Institute for Microbial Stress and Survival, Berkeley, California 94720‡,1 Environmental Sciences Division, Oak Ridge National Laboratory, Oak Ridge, Tennessee 37831,2 Department of Civil and Environmental Engineering, Temple University, Philadelphia, Pennsylvania 19122,3 Physical Biosciences Division, Lawrence Berkeley National Laboratory, Berkeley, California 94720,4 Department of Microbiology, Miami University, Oxford, Ohio 45056,5 Earth Sciences Division, Lawrence Berkeley National Laboratory, Berkeley, California 94720,6 Department of Bioengineering, University of California, Berkeley, California 94720,7 Howard Hughes 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Howard Hughes Medical Institute, Chevy Chase, Maryland 20815,8 Departments of Biochemistry and Molecular Microbiology and Immunology, University of Missouri—Columbia, Columbia, Missouri 65211,9 Institute for Environmental Genomics, Department of Botany and Microbiology, University of Oklahoma, Norman, Oklahoma 73019 10 Find articles by Eric J Alm 1,4, Matthew W Fields Matthew W Fields Virtual Institute for Microbial Stress and Survival, Berkeley, California 94720‡,1 Environmental Sciences Division, Oak Ridge National Laboratory, Oak Ridge, Tennessee 37831,2 Department of Civil and Environmental Engineering, Temple University, Philadelphia, Pennsylvania 19122,3 Physical Biosciences Division, Lawrence Berkeley National Laboratory, Berkeley, California 94720,4 Department of Microbiology, Miami University, Oxford, Ohio 45056,5 Earth Sciences Division, Lawrence Berkeley National Laboratory, Berkeley, California 94720,6 Department of Bioengineering, University of California, Berkeley, California 94720,7 Howard Hughes Medical Institute, Chevy Chase, Maryland 20815,8 Departments of Biochemistry and Molecular Microbiology and Immunology, University of Missouri—Columbia, Columbia, Missouri 65211,9 Institute for Environmental Genomics, Department of Botany and Microbiology, University of Oklahoma, Norman, Oklahoma 73019 10 Find articles by Matthew W Fields 1,5, Terry C Hazen Terry C Hazen Virtual Institute for Microbial Stress and Survival, Berkeley, California 94720‡,1 Environmental Sciences Division, Oak Ridge National Laboratory, Oak Ridge, Tennessee 37831,2 Department of Civil and Environmental Engineering, Temple University, Philadelphia, Pennsylvania 19122,3 Physical Biosciences Division, Lawrence Berkeley National Laboratory, Berkeley, California 94720,4 Department of Microbiology, Miami University, Oxford, Ohio 45056,5 Earth Sciences Division, Lawrence Berkeley National Laboratory, Berkeley, California 94720,6 Department of Bioengineering, University of California, Berkeley, California 94720,7 Howard Hughes Medical Institute, Chevy Chase, Maryland 20815,8 Departments of Biochemistry and Molecular Microbiology and Immunology, University of Missouri—Columbia, Columbia, Missouri 65211,9 Institute for Environmental Genomics, Department of Botany and Microbiology, University of Oklahoma, Norman, Oklahoma 73019 10 Find articles by Terry C Hazen 1,6, Adam P Arkin Adam P Arkin Virtual Institute for Microbial Stress and Survival, Berkeley, California 94720‡,1 Environmental Sciences Division, Oak Ridge National Laboratory, Oak Ridge, Tennessee 37831,2 Department of Civil and Environmental Engineering, Temple University, Philadelphia, Pennsylvania 19122,3 Physical Biosciences Division, Lawrence Berkeley National Laboratory, Berkeley, California 94720,4 Department of Microbiology, Miami University, Oxford, Ohio 45056,5 Earth Sciences Division, Lawrence Berkeley National Laboratory, Berkeley, California 94720,6 Department of Bioengineering, University of California, Berkeley, California 94720,7 Howard Hughes Medical Institute, Chevy Chase, Maryland 20815,8 Departments of Biochemistry and Molecular Microbiology and Immunology, University of Missouri—Columbia, Columbia, Missouri 65211,9 Institute for Environmental Genomics, Department of Botany and Microbiology, University of Oklahoma, Norman, Oklahoma 73019 10 Find articles by Adam P Arkin 1,4,7,8, Judy D Wall Judy D Wall Virtual Institute for Microbial Stress and Survival, Berkeley, California 94720‡,1 Environmental Sciences Division, Oak Ridge National Laboratory, Oak Ridge, Tennessee 37831,2 Department of Civil and Environmental Engineering, Temple University, Philadelphia, Pennsylvania 19122,3 Physical Biosciences Division, Lawrence Berkeley National Laboratory, Berkeley, California 94720,4 Department of Microbiology, Miami University, Oxford, Ohio 45056,5 Earth Sciences Division, Lawrence Berkeley National Laboratory, Berkeley, California 94720,6 Department of Bioengineering, University of California, Berkeley, California 94720,7 Howard Hughes Medical Institute, Chevy Chase, Maryland 20815,8 Departments of Biochemistry and Molecular Microbiology and Immunology, University of Missouri—Columbia, Columbia, Missouri 65211,9 Institute for Environmental Genomics, Department of Botany and Microbiology, University of Oklahoma, Norman, Oklahoma 73019 10 Find articles by Judy D Wall 1,9, Jizhong Zhou Jizhong Zhou Virtual Institute for Microbial Stress and Survival, Berkeley, California 94720‡,1 Environmental Sciences Division, Oak Ridge National Laboratory, Oak Ridge, Tennessee 37831,2 Department of Civil and Environmental Engineering, Temple University, Philadelphia, Pennsylvania 19122,3 Physical Biosciences Division, Lawrence Berkeley National Laboratory, Berkeley, California 94720,4 Department of Microbiology, Miami University, Oxford, Ohio 45056,5 Earth Sciences Division, Lawrence Berkeley National Laboratory, Berkeley, California 94720,6 Department of Bioengineering, University of California, Berkeley, California 94720,7 Howard Hughes Medical Institute, Chevy Chase, Maryland 20815,8 Departments of Biochemistry and Molecular Microbiology and Immunology, University of Missouri—Columbia, Columbia, Missouri 65211,9 Institute for Environmental Genomics, Department of Botany and Microbiology, University of Oklahoma, Norman, Oklahoma 73019 10 Find articles by Jizhong Zhou 1,2,10, Author information Article notes Copyright and License information Virtual Institute for Microbial Stress and Survival, Berkeley, California 94720‡,1 Environmental Sciences Division, Oak Ridge National Laboratory, Oak Ridge, Tennessee 37831,2 Department of Civil and Environmental Engineering, Temple University, Philadelphia, Pennsylvania 19122,3 Physical Biosciences Division, Lawrence Berkeley National Laboratory, Berkeley, California 94720,4 Department of Microbiology, Miami University, Oxford, Ohio 45056,5 Earth Sciences Division, Lawrence Berkeley National Laboratory, Berkeley, California 94720,6 Department of Bioengineering, University of California, Berkeley, California 94720,7 Howard Hughes Medical Institute, Chevy Chase, Maryland 20815,8 Departments of Biochemistry and Molecular Microbiology and Immunology, University of Missouri—Columbia, Columbia, Missouri 65211,9 Institute for Environmental Genomics, Department of Botany and Microbiology, University of Oklahoma, Norman, Oklahoma 73019 10 Corresponding author. Mailing address: Institute for Environmental Genomics and Department of Botany and Microbiology, University of Oklahoma, Norman, OK 73019. Phone: (405) 325-6073. Fax: (405) 325-3442. E-mail: jzhou@ou.edu. ‡ Received 2005 Nov 4; Accepted 2006 Mar 18. Copyright © 2006, American Society for Microbiology PMC Copyright notice PMCID: PMC1489655 PMID: 16751553 Abstract Many of the proteins that are candidates for bioenergetic pathways involved with sulfate respiration in Desulfovibrio spp. have been studied, but complete pathways and overall cell physiology remain to be resolved for many environmentally relevant conditions. In order to understand the metabolism of these microorganisms under adverse environmental conditions for improved bioremediation efforts, Desulfovibrio vulgaris Hildenborough was used as a model organism to study stress response to nitrite, an important intermediate in the nitrogen cycle. Previous physiological studies demonstrated that growth was inhibited by nitrite and that nitrite reduction was observed to be the primary mechanism of detoxification. Global transcriptional profiling with whole-genome microarrays revealed coordinated cascades of responses to nitrite in pathways of energy metabolism, nitrogen metabolism, oxidative stress response, and iron homeostasis. In agreement with previous observations, nitrite-stressed cells showed a decrease in the expression of genes encoding sulfate reduction functions in addition to respiratory oxidative phosphorylation and ATP synthase activity. Consequently, the stressed cells had decreased expression of the genes encoding ATP-dependent amino acid transporters and proteins involved in translation. Other genes up-regulated in response to nitrite include the genes in the Fur regulon, which is suggested to be involved in iron homeostasis, and genes in the Per regulon, which is predicted to be responsible for oxidative stress response. The sulfate-reducing bacteria represent a group of microorganisms characterized by the ability to use sulfate as an electron acceptor in anaerobic respiration (47). Microbial sulfate reduction by these microorganisms is recognized as a widely distributed process of great ecological importance (29, 54). Historical interest in sulfate-reducing bacteria has been focused on their involvement in biocorrosion of ferrous metals in the petroleum industry and of concrete structures in wastewater collection systems (15, 24). More recent studies (7, 25, 27, 28) have documented the ability of a number of sulfate-reducing bacteria to reduce soluble metal oxyanions to insoluble forms, a process of great potential in the bioremediation of toxic heavy metals and radionuclides such as chromium and uranium (11, 56). To effectively immobilize heavy metals and radionuclides using sulfate-reducing bacteria, it is important to understand the microbial response to adverse environmental factors common in contaminated subsurface environments. One such factor is the high nitrate concentration of many contaminated sites at the nuclear weapon complexes in the United States managed by the Department of Energy (39, 49). The presence of nitrate may pose a specific stress to sulfate-reducing bacteria as nitrate has been observed to suppress sulfate reduction activity in situ (9, 21). However, it has been suggested that nitrite, an intermediate that transiently accumulates during nitrate reduction (3, 23, 58), is directly responsible for the inhibition of sulfate reduction activity (26, 38). Furthermore, the dissimilatory sulfite reductase, which is a key enzyme in the sulfate reduction pathway, has been implicated in previous reports as the target of nitrite inhibition (13, 16). Consequently, one would predict that energy generation pathways in sulfate-reducing bacteria would be altered upon exposure to nitrite. Thus, it is important to examine the global transcription profiles following nitrite exposure to understand how the complex energy generation pathways in sulfate-reducing bacteria respond to nitrite and to predict the performance of these bacteria for bioremediation. In this report we used Desulfovibrio vulgaris Hildenborough as a model organism to investigate the gene expression profile during the inhibition of sulfate reduction by nitrite. The microbial stress responses at the transcriptional level were studied with whole-genome microarrays. Our results indicate that D. vulgaris is capable of rapid nitrite reduction and that the exposure to nitrite triggers a well-coordinated response in pathways of energy metabolism, nitrogen metabolism, oxidative stress response, and iron homeostasis. MATERIALS AND METHODS Organism and growth conditions. D. vulgaris strain Hildenborough (ATCC 29579) was grown in an anaerobic medium containing lactate plus sulfate (LS medium) of the following composition (per liter): 6.72 g of sodium lactate, 7.10 g of Na 2 SO 4, 0.963g of anhydrous MgSO 4, 1.07 g of NH 4 Cl, 0.383 g of K 2 HPO 4, 0.088 g of CaCl 2 · 2H 2 O, 0.620 mg of resazurin, 0.600 g of Na 2 CO 3, 25.0 ml of 1 M HEPES [4-(2-hydroxyethyl)-1-piperazineethanesulfonic acid] (pH 7.0), 12.5 ml of a trace mineral solution, 0.75 g of l-cysteine, and 1.0 g of yeast extract. The trace mineral solution contained the following (per liter): 1.0 g of FeCl 2 · 4H 2 O, 0.3 g of CoCl 2 · 4H 2 O, 0.5 g of MnCl 2 · 4H 2 O, 0.2 g of ZnCl 2, 20 mg of H 3 BO 3, 50 mg of Na 2 MoO 4 · 4H 2 O, and 0.1 g of NiCl 2 · 6H 2 O. Cysteine was added as a reductant after the medium had been boiled and cooled to room temperature. The headspace of the medium container was continuously flushed with oxygen-free nitrogen gas, and the pH was adjusted to 7.2 ± 0.1 with 5 M NaOH. A vitamin solution (10 ml per liter) (5) was added to the autoclaved medium from filter-sterilized anaerobic stock solutions. Cultures were incubated in the dark at 37°C in stoppered 160-ml serum bottles with 100 ml of LS medium or in 30-ml anaerobic culture tubes with 10 ml of medium and sealed with butyl rubber stoppers and aluminum seals. Strictly anaerobic techniques were used throughout all experimental manipulations. Oligonucleotide probe design and microarray construction. DNA microarrays covering 3,482 of the 3,531 annotated protein-coding sequences of the D. vulgaris genome were constructed with 70-mer oligonucleotide probes (18). Oligonucleotide probes (3,574) were designed to cover all open reading frames (ORFs) for the genome of D. vulgaris Hildenborough, using the computer software tool ArrayOligoSelector (4), based on an early version (June 2003) of the gene model with 3,584 ORFs. The specificity of all designed oligonucleotide probes was examined with two criteria, as follows. From a BLAST analysis (1), 496 oligonucleotide probes were considered nonspecific to individual genes if they showed a >85% sequence identity or a >18-base continuous homologous stretch with other ORFs in the genome (18). These nonspecific oligonucleotide probes were redesigned against the genome using two other programs, OligoPicker (60) and OligoArray (51), with the same design parameters. Following the examination of the entire probe set according to the oligonucleotide probe design criteria (18), 3,471 (97.1%) specific oligonucleotide probes were obtained, and 103 (2.9%) remained nonspecific. When this early version gene model was mapped to the published version of the D. vulgaris gene model (19), 3,482 of the 3,531 protein-coding sequences were covered with 3,439 specific and 43 nonspecific oligonucleotide probes (see Table S1 in the supplemental material). All designed oligonucleotides were commercially synthesized without modification by MWG Biotech Inc. (High Point, NC). The concentration of oligonucleotides was adjusted to 100 pmol/μl. Oligonucleotide probes were prepared in 50% (vol/vol) dimethyl sulfoxide (Sigma-Aldrich, St. Louis, MO) and spotted onto UltraGAPS glass slides (Corning Life Sciences, Corning, NY) using a BioRobotics Microgrid II microarrayer (Genomic Solutions, Ann Arbor, MI). Each oligonucleotide probe had two replicates on a single slide. Additionally, six different concentrations (5, 25, 50, 100, 200, and 300 ng/μl) of genomic DNA were also spotted (eight duplicates of each of the six concentrations on a single slide) as additional positive controls. After the oligonucleotide probes were printed, they were fixed onto the slides by UV cross-linking (600 mJ of energy), according to the protocol of the manufacturer of the UltraGAPS glass slides (Corning Life Science, Corning. NY). Exposure to nitrite stress. In experiments for nitrite stress microarray analysis, D. vulgaris cultures were grown to exponential phase (optical density at 600 nm [OD 600] of ca. 0.4) in LS medium. To triplicate cultures, nitrite from anaerobic stock solutions was added to a final concentration of 2.5 mM. Cell samples of each culture were harvested immediately after the addition of nitrite, and after 30, 60, 90, 150, and 240 min by centrifugation (5,000 × g) for 5 min at 4°C. Cell samples from triplicate control cultures without the addition of nitrite were collected simultaneously at the same time points. Cell pellets were then immediately frozen in liquid N 2 and stored at −80°C prior to RNA isolation. To perform appropriate statistical analysis, cell samples were not pooled in subsequent processing. To prevent organic contamination, all glassware was acid washed and baked at 300°C overnight. Analytical methods. Growth of cultures was monitored spectrophotometrically (OD 600). Nitrite was analyzed on a Dionex DX-120 ion chromatograph apparatus with a PeakNet analysis software package and a Dionex IonPak Anion Exchange column as previously described (17). The mobile phase contained 1.8 mM Na 2 CO 3 and 1.7 mM NaHCO 3. Peaks were quantified via a conductivity detector, and concentrations were determined using known standards. Samples from the cultures were filtered through 0.20-μm-pore-size filters prior to ion chromatographic analysis. Total RNA extraction, purification, and labeling. RNA extraction, purification, and labeling were performed independently on each cell sample. Total cellular RNA was isolated using TRIzol Reagent (Invitrogen, Carlsbad, CA) following the manufacturer's protocol. RNA extracts were purified according to the RNeasy Mini Kit (QIAGEN, Valencia, CA) instructions, and on-column DNase digestion was performed with an RNase-free DNase set (QIAGEN, Valencia, CA) to remove genomic DNA contamination, according to the manufacturer's procedure. To generate cDNA targets with reverse transcriptase, 10 μg of purified total RNA was used for each labeling reaction using a previously described protocol (55). Briefly, random hexamers (Invitrogen) were used for priming, and the fluorophore Cy3-dUTP or Cy5-dUTP (Amersham Biosciences, Piscataway, NJ) was used for labeling. After the labeling, RNA was removed by NaOH treatment, and cDNA was immediately purified with a QIAGEN PCR mini kit. The efficiency of labeling was routinely monitored by measuring the absorbance at 260 nm (for DNA concentration), 550 nm (for Cy3), or 650 nm (for Cy5). Two samples of each total RNA preparation were labeled, one with Cy3-dUTP and another with Cy5-dUTP, for microarray hybridization. Microarray hybridization, washing, and scanning. To hybridize microarray glass slides, the Cy5-dUTP-labeled cDNA targets from one nitrite-treated culture were mixed with the Cy3-dUTP-labeled cDNA targets from one untreated control culture and vice versa (dye swap). As a result, each biological sample was hybridized to two microarray slides. Equal amounts of Cy3- or Cy5-labeled probes were mixed and resuspended in 35 to 40 μl of hybridization solution that contained 50% (vol/vol) formamide, 5× saline-sodium citrate (5× SSC; 1× SSC is 150 mM NaCl plus 15 mM sodium citrate, pH 7.0), 0.1% (wt/vol) sodium dodecyl sulfate (SDS), and 0.1 mg of herring sperm DNA/ml (Invitrogen). The hybridization solution was incubated at 95 to 98°C for 5 min, centrifuged to collect condensation, kept at 50°C, and applied onto microarray slides. Hybridization was carried out in hybridization chambers (Corning Life Sciences, Corning, NY) at 45°C overnight (16 to 20 h). A total of 10 μl of 3× SSC solution was added to the wells at both ends of the microarray slides to maintain proper chamber humidity and probe hydration around the edges of the coverslips. Microarray slides were washed according to the instructions for spotted oligonucleotide microarrays on UltraGAPS slides by the manufacturer (Corning) in the following steps: two washes in a solution containing 2× SSC and 0.1% (wt/vol) SDS at 42°C at 5-min intervals, two washes in a solution containing 0.1× SSC and 0.1% (wt/vol) SDS at room temperature at 10-min intervals, and two washes in 0.1× SSC at room temperature at 2-min intervals. After being blown dry by a stream of N 2, the slides were scanned for the fluorescence intensity of both the Cy5 and Cy3 fluorophores using a ScanArray Express microarray analysis system (Perkin Elmer, Boston, MA). Image quantification and data analysis. To determine signal fluorescence intensities for each spot, 16-bit TIFF scanned images were analyzed by application of the software ImaGene, version 6.0 (Biodiscovery, Marina Del Rey, Calif.) to quantify spot signal, spot quality, and background fluorescence intensities. Empty spots, poor spots, and negative spots were flagged according to the instruction of the software and removed in subsequent analysis (12). The resulting data files were subjected to Lowess intensity-based normalization and further analyzed using GeneSpring, version 5.1 (Silicon Genetics, Redwood City, Calif.). Lowess normalization was performed on each microarray slide, and results of the triplicate cultures from the same time point were used for statistical analysis. To assess the statistical significance of individual data points, a Student t test was used to calculate a P value to test the null hypothesis that the expression level was unchanged. A statistical model incorporating both per gene variance and operon structure was further used to compute the posterior probability that each gene changed its expression level in the direction indicated by its mean value (48). An average linkage hierarchical clustering analysis of the time course transcriptional response to nitrite stress with the Euclidean distance as the similarity metric was performed and visualized with Hierarchical Clustering Explorer, version 3.0 (53). The expression of genes encoding iron-containing proteins was evaluated by comparing the expression levels of transcripts coding for iron-containing proteins to the average values for all genes in the cell. To identify genes encoding iron-containing proteins, we examined domain structure as predicted by InterPro, version 9.0 (37), and included all genes with domains annotated as binding iron. Real-time quantitative RT-PCR analysis. To independently validate gene expression results from the microarray analysis, eight ORFs exhibiting a range of expression levels from low to high (as identified by microarray analysis) were chosen for analysis using real-time reverse-transcription PCR (RT-PCR). Primer pairs (in parentheses, forward and reverse, respectively) were designed for the following genes to yield products of ∼100 bp (52): DVU0625, encoding a cytochrome c nitrite reductase (5′-AGAACCTCTGGCTCGGCTAT and 5′-CGATTGATACGGTCGATGTG); DVU0942, encoding a Fur family transcriptional regulator (5′-CATCGCCGTATTTCAGGATT and 5′-GAGATGCCCGCCTACTTTC); DVU1290, encoding the gamma subunit of a putative nitrate reductase (5′-TTTCCGGCTTTCAGTACGTT and 5′-AGACTTGGCCCAATCCACTA); DVU1574, encoding a ribosomal protein L25 (5′-GGTGGCAAGCTCGAAGTCTA and 5′-GATGTCGAGTTCGGTCAGGT); DVU2247, encoding an AhpC/Tsa family antioxidant (5′-TCTATCCGCTGGACTTCACC and 5′-ACACCGATGACCTC GACATT); DVU2543, encoding a hybrid cluster protein (5′-ACCTCACCATCTACGCCTTG and 5′-GCTTTGGCCGTGTATTCATC); DVU2571, encoding a ferrous iron transport protein B (5′-GAAGGAGGTCATCGTCTCCA and 5′-GGGGTCGTTCCTGATCTGT); and DVU2680, encoding a flavodoxin (5′-CTTCAT and 5′-CCCGCAGAAGTACTCGTAGG). The RT-PCR analysis was carried out using a previously described protocol (59). Briefly, the cDNA template for real-time RT-PCR was synthesized from 5 μg of total RNA using the reverse transcriptase reaction with random hexamer priming (Invitrogen). The quantitative PCR was carried out in an iCycler thermal cycler (Bio-Rad, Hercules, Calif.) that measured the increases in fluorescence resulting from the incorporation of SYBR green dye (Molecular Probes, Eugene, Oreg.) into double-strand DNA. Real-time data acquisition and analysis were performed with the software iCycle 2.3, version B, according to the manufacturer's instructions. Standards for each gene of interest were obtained by serial dilutions of PCR amplification product from D. vulgaris genomic DNA using the procedure described above but without SYBR green dye. The standards were used to establish a standard curve consisting of seven points serially diluted from 10 7 to 10 1 copies. Copy numbers of the target gene transcripts were determined by comparison with the standard curves, and then gene expression differences between the treatment and control samples were determined. RESULTS Growth inhibition of D. vulgaris by nitrite. The inhibitory effect of nitrite on the growth of D. vulgaris was evaluated by adding different concentrations of nitrite to actively growing cultures (OD 600 of ca. 0.4). No significant growth inhibition was observed with nitrite concentrations below 1 mM, as the growth curves overlapped between nitrite-treated cultures and control cultures (Fig. 1). When 2.5 mM nitrite was added to the medium, a slower growth rate was observed. When the nitrite concentration was increased to 5 mM, the growth of D. vulgaris was significantly decreased, and no change in OD was observed over the monitored time (Fig. 1). FIG. 1. Open in a new tab Impact of nitrite on the growth of D. vulgaris. Nitrite of different concentrations was added to sulfate-reducing bacterial cultures in mid-log phase, and growth was subsequently monitored as the OD 600. Data are averaged from triplicate cultures, with error bars indicating standard deviations. Reduction of nitrite by D. vulgaris. To further determine the connection between nitrite and the growth inhibition of D. vulgaris, nitrite levels were monitored in the D. vulgaris cultures (Fig. 2). Abiotic reduction of nitrite by the medium or sulfide was excluded as nitrite was stable in the presence of 40 mM sulfide but no cells in the controls. On the other hand, with initial concentrations lower than 2.5 mM, nitrite decreased rapidly in active D. vulgaris cultures, indicating that nitrite was reduced by D. vulgaris cells. In contrast, D. vulgaris cultures were unable to reduce 5 mM nitrite, which coincided with nearly complete inhibition of growth, as shown in Fig. 1. FIG. 2. Open in a new tab Nitrite reduction by D. vulgaris cultures. Following the addition of various initial concentrations of nitrite into mid-log-phase cultures (OD 600, 0.4), reduction of nitrite was monitored over time. (Inset) Changes in nitrite concentration in the presence of 40 mM sulfide but no cells (a) and mid-log cells (OD 600, 0.4) only (b). Axis labels in the inset are the same as in the main figure. Results shown are averages of triplicates, with error bars indicating standard deviation. Transcriptome analysis of nitrite stress. To examine the nitrite stress response in D. vulgaris, microarray experiments were carried out to compare global gene expression profiles between nitrite-stressed D. vulgaris cultures and control cultures without nitrite exposure (see Table S2 in the supplemental material). In order to achieve an optimal stress response, D. vulgaris cells were challenged by a sublethal nitrite level (2.5 mM), which effectively inhibited cell growth but still allowed active reduction of nitrite (Fig. 2). Because tolerance to nitrite was found to be dependent on the biomass concentration at the time of nitrite addition (16), cultures of similar optical densities (OD 600 of ca. 0.4) were used throughout the study. Significant changes in gene expression profiles occurred within 30 min following nitrite addition and peaked at 60 min, with 330 genes up-regulated and 273 genes down-regulated (Fig. 3) more than twofold. Subsequently, transcriptional responses rapidly diminished with only 82 genes still up-regulated and 86 genes down-regulated more than twofold 4 h after nitrite addition (Fig. 3). Concurrently, the initial 2.5 mM nitrite concentration dropped below 0.5 mM. The steady decline in transcriptional response subsequent to its peaking at 60 min mirrored the time course of reduction of nitrite by D. vulgaris. These results indicated a correlation between the dynamics of transcriptional response and the reduction of nitrite between 60 min to 240 min. FIG. 3. Open in a new tab Temporal profiling of the transcriptional response to sodium nitrite by D. vulgaris. Each column represents the number of genes showing significant changes (P< 0.05) in gene expression level versus time elapsed following the addition of nitrite. Positive and negative values indicate up- and down-regulation, respectively. To illustrate the gene functions involved in the nitrite stress response, genes with altered expression levels after 1 h of nitrite exposure were grouped into functional role categories according to the annotation of The Institute for Genomic Research (TIGR) of the D. vulgaris Hildenborough genome sequence (19, 45), as shown in Fig. 4. Notably, a large portion of the highly up-regulated genes were grouped into cellular roles involved in regulatory functions, signal transduction, and organic acid oxidation including genes encoding the l-lactate dehydrogenase, formate dehydrogenase, and pyruvate ferredoxin oxidoreductase, suggesting a shift in energy flow through complex regulatory pathways when D. vulgaris cells sense the presence of nitrite. On the other hand, many of the highly down-regulated genes have functions in protein biosynthesis and encode transport and binding proteins, perhaps indicating a slowdown in normal cellular biosynthetic activities when challenged by nitrite stress. FIG. 4. Open in a new tab Functional profiling of the transcriptional response by D. vulgaris 1 h following 2.5 mM sodium nitrite addition. The functional role category annotation is that provided by TIGR (www.tigr.org). Each column represents the number of genes in a selected functional category showing significant changes in mRNA abundance in response to nitrite. Positive and negative values indicate up-and down-regulation, respectively. Columns: 1, amino acid biosynthesis; 2, biosynthesis of cofactors, prosthetic groups, and carriers; 3, cell envelope; 4, cellular processes; 5, energy metabolism; 6, protein synthesis; 7, regulatory functions; 8, signal transduction; and 9, transport and binding proteins. Shown are selected role categories with highly differentially expressed genes (change of more than threefold). Validation of microarray results. To validate transcriptional results generated by microarray hybridization, real-time RT-PCR analysis was conducted to quantify the expression levels of eight genes (see Fig. S1 in the supplemental material). A high degree of correlation was observed between results from microarray analysis and RT-PCR (r 2 = 0.93), as reported in previous studies that used this microarray procedure (12, 59), thus affirming the accuracy of this approach in the current study. In addition to real-time RT-PCR analysis, expression differences for gene pairs within the same predicted operon or gene pairs selected at random were compared to determine whether changes observed in microarray experiments were authentic (12). Consistent with our expectation (Fig. 5A), genes within the same operon responded more similarly than genes randomly selected from the genome. As shown in Fig. 5A, the within-operon pairs had higher probabilities to exhibit much smaller log ratio differences than gene pairs chosen at random, thus confirming the agreement between microarray results and operon prediction and the high quality of the expression data. Furthermore, a second operon-based computational method was also used to test the validation of the microarray results through evaluation of the confidence levels of gene expression (48). Consistently, genes identified as having confident changes in expression were in agreement with other genes found in the same operon (Fig. 5B). Thus, the comparison with operon structure confirmed both the high quality of the expression data and our ability to identify reliable data points. FIG. 5. Open in a new tab Validation of microarray results by computational approaches. (A) Log ratio expression difference of gene pairs within the same operon versus gene pairs selected at random. The normalized frequency was plotted against the ratio expression difference between the treatments and control. Genes within the same operon responded more similarly than genes randomly selected from the genome under sodium nitrite exposure. (B) Agreement within predicted operons at the 90-min time point. All genes were divided into eight groups based on the confidence level of the measured change computed by the OpWise program ( A confidence of 0.5 indicates complete uncertainty as to whether the gene was up- or down-regulated, while a value of 1 indicates certainty that the measured change in mean reflects the actual direction of change. The y axis shows the fraction of genes (above that expected by chance) in each group that changed in the same direction as adjacent genes predicted to be in the same operon, together with 95% confidence intervals for the estimate. Values near 1 indicate perfect agreement with all co-operonic genes changing in the same direction, while values near 0 indicate the level of agreement expected by chance (i.e., 50%). Hierarchical clustering analysis of temporal gene expression profile. To identify differential gene expression patterns in the nitrite stress response by D. vulgaris, a hierarchical clustering analysis was conducted on the transcriptional profile (Fig. 6). Cluster A consists of genes highly induced throughout the duration of the experiment, including the operon encoding the nitrite reductase and several redox-active proteins. Notably, the gene for the hybrid cluster protein was highly up-regulated in response to the presence of nitrite, providing more evidence for its suggested role in nitrogen metabolism (57, 63). Cluster B represents a group of genes highly induced within 1.5 h of the addition of nitrite but for which the induction diminished or even reversed subsequently. Interestingly, genes included in this cluster encode ferrous iron transport proteins, suggesting a role of iron homeostasis in the initial stress response to nitrite. Cluster C covers a large number of genes repressed during the early response to nitrite that were later restored to expression levels observed in the control cultures. Within Cluster C, genes for ribosomal proteins were particularly in abundance, consistent with prior findings of an early slowdown in protein synthesis (Fig. 4). Cluster D is comprised of genes down-regulated throughout the experiment's duration, including genes that have functions in energy metabolism. From the concurrent induction of genes in nitrite reduction (cluster A) and the repression of genes encoding ATP synthase subunits (cluster D), we hypothesize that electron flow was likely shifted from respiratory phosphorylation to the reduction of nitrite. FIG. 6. Open in a new tab Hierarchical clustering of selected genes with significant changes (P< 0.05 and a change of more than twofold at least at one time point) in expression in response to 2.5 mM nitrite. Red indicates up-regulation, whereas green represents repression. Each row represents the expression of a single gene, and each column represents an individual time point following nitrite addition: T1, 0.5 h; T2, 1.0 h; T3, 1.5 h; T4, 2.5 h; and T5, 4.0 h. Listed genes are examples from each cluster. Cluster A consists of genes highly induced throughout the duration of the experiment; cluster B consists of genes highly induced within 1.5 h of the addition of nitrite but for which induction diminished or even reversed subsequently; cluster C consists of genes repressed during the early response to nitrite but for which repression was later alleviated; and cluster D consists of genes down-regulated throughout the experiment. Genes involved in energy metabolism. Genes having functions in energy metabolism exhibited considerable divergence in their transcriptional response to nitrite (Table 1). In contrast to the severe repression of ATP synthase genes, genes encoding the membrane-bound lactate dehydrogenase (ldh) and pyruvate ferredoxin oxidoreductase (porAB) were induced, pointing to potential increases in substrate level phosphorylation and electron flow. In parallel, a number of operons encoding redox proteins participating in periplasmic electron transfer were also up-regulated, including the gene encoding the periplasmic [NiFe] hydrogenase (hynBA-2) and its putatively associated tetraheme cytochrome (DVU2524-2526) and one encoding a formate dehydrogenase (DVU2810-2812). Thus, the possibility exists that these redox protein complexes are involved in electron transfer to nitrite or in the turnover of reduced electron carriers, allowing a greater rate of substrate oxidation for substrate level phosphorylation. Additionally, the fumarate reductase operon (frdABC) was induced, possibly allowing fumarate to serve as an alternative electron acceptor (Table 1). TABLE 1. Effect of nitrite exposure on the transcriptional responses of D. vulgaris genes involved in energy metabolism \| Gene identifier and response \| Annotation \| Expression ratio (treatment/control) at time pointa: \| :---: \| 0.5 h \| 1.0 h \| 1.5 h \| 2.5 h \| 4.0 h \| \| Up-regulation \| \| \| \| \| \| \| \| DVU0172 \| Iron-sulfur cluster-binding protein \| 3.0 \| 2.8 \| 1.8 \| 0.50 \| 0.31 \| \| DVU0173 \| Thiosulfate reductase; putative \| 2.5 \| 1.9 \| 1.5 \| 0.55 \| 0.45 \| \| DVU0600 \| l-Lactate dehydrogenase (Ldh) \| 2.1 \| 1.8 \| 1.9 \| \| \| \| DVU0624 \| NapC/NirT cytochrome c family protein \| 13.5 \| 11.4 \| 12.2 \| 9.4 \| 11.4 \| \| DVU0625 \| Cytochrome c nitrite reductase; catalytic subunit NfrA; putative \| 18.5 \| 12.8 \| 14.9 \| 7.5 \| 5.7 \| \| DVU1080 \| Iron-sulfur cluster-binding protein \| 54.2 \| 81.8 \| 58.6 \| 28.7 \| 27.2 \| \| DVU1081 \| Iron-sulfur cluster-binding protein \| 12.6 \| 10.5 \| 12.5 \| 6.8 \| 7.2 \| \| DVU1569 \| Pyruvate ferredoxin oxidoreductase; alpha subunit (PorA) \| — \| 2.0 \| 2.3 \| — \| — \| \| DVU1570 \| Pyruvate ferredoxin oxidoreductase; beta subunit (PorB) \| — \| 2.2 \| 2.3 \| — \| — \| \| DVU2524 \| Cytochrome c 3; putative \| 2.2 \| 3.4 \| 5.0 \| — \| — \| \| DVU2525 \| Periplasmic [NiFe] hydrogenase; small subunit; isozyme 2 (HynB-2) \| — \| 1.5 \| 1.6 \| — \| — \| \| DVU2526 \| Periplasmic [NiFe] hydrogenase; large subunit; isozyme 2 (HynA-2) \| 8.1 \| 9.1 \| 9.5 \| 4.7 \| 3.2 \| \| DVU2543 \| Hybrid cluster protein \| 27.5 \| 34.5 \| 65.1 \| 82.7 \| 37.3 \| \| DVU2544 \| Iron-sulfur cluster-binding protein \| 39.8 \| 50.7 \| 82.0 \| 233.0 \| 30.3 \| \| DVU2810 \| Formate dehydrogenase formation protein FdhE; putative \| 1.5 \| — \| 1.5 \| 1.8 \| — \| \| DVU2811 \| Formate dehydrogenase; beta subunit; putative \| — \| — \| 2.1 \| 2.8 \| — \| \| DVU2812 \| Formate dehydrogenase; alpha subunit; selenocysteine-containing (FdnG-3) \| — \| — \| 1.8 \| — \| — \| \| DVU3261 \| Fumarate reductase; cytochrome b subunit (FrdC) \| 2.0 \| 3.5 \| 3.2 \| — \| — \| \| DVU3262 \| Fumarate reductase; flavoprotein subunit (FrdA) \| 1.8 \| 2.9 \| 2.6 \| — \| 1.4 \| \| DVU3263 \| Fumarate reductase; iron-sulfur protein (FrdB) \| 1.9 \| 3.7 \| 3.1 \| — \| — \| \| Down-regulation \| \| \| \| \| \| \| \| DVU0431 \| Ech hydrogenase; subunit EchD; putative \| 0.37 \| 0.53 \| — \| — \| — \| \| DVU0432 \| Ech hydrogenase; subunit EchC; putative \| 0.41 \| 0.51 \| 0.59 \| — \| — \| \| DVU0433 \| Ech hydrogenase, subunit EchB; putative \| 0.49 \| 0.59 \| — \| — \| — \| \| DVU0434 \| Ech hydrogenase, subunit EchA; putative \| 0.57 \| 0.59 \| — \| — \| — \| \| DVU0774 \| ATP synthase; F 1 epsilon subunit (AtpC) \| — \| 0.42 \| 0.39 \| 0.46 \| — \| \| DVU0775 \| ATP synthase; F 1 beta subunit (AtpD) \| — \| 0.31 \| 0.32 \| 0.30 \| — \| \| DVU0776 \| ATP synthase; F 1 gamma subunit (AtpG) \| — \| 0.30 \| 0.42 \| 0.18 \| — \| \| DVU0777 \| ATP synthase, F 1 alpha subunit (AtpA) \| 0.40 \| 0.31 \| 0.23 \| 0.24 \| — \| \| DVU0778 \| ATP synthase; F 1 delta subunit (AtpH) \| 0.48 \| 0.31 \| 0.38 \| 0.25 \| — \| \| DVU0779 \| ATP synthase; F 0 B subunit; putative (AtpF2) \| 0.57 \| 0.38 \| 0.36 \| 0.40 \| — \| \| DVU0780 \| ATP synthase; F 0 B subunit; putative (AtpF1) \| 0.44 \| 0.40 \| 0.40 \| — \| — \| \| DVU0917 \| ATP synthase; F 0 C subunit (AtpE) \| — \| 0.37 \| 0.29 \| 0.31 \| 0.59 \| \| DVU0918 \| ATP synthase; F 0 A subunit (AtpB) \| 0.46 \| 0.37 \| 0.29 \| 0.42 \| 0.67 \| \| DVU1286 \| Reductase; transmembrane subunit; putative (DsrP) \| 0.34 \| 0.33 \| 0.24 \| 0.56 \| — \| \| DVU1287 \| Reductase; iron-sulfur binding subunit; putative (DsrO) \| 0.19 \| 0.19 \| 0.17 \| 0.54 \| — \| \| DVU1288 \| Cytochrome c family protein (DsrJ) \| 0.19 \| 0.21 \| 0.23 \| 0.43 \| — \| \| DVU1289 \| Reductase; iron-sulfur binding subunit; putative (DsrK) \| 0.15 \| 0.19 \| 0.22 \| 0.47 \| — \| \| DVU1290 \| Nitrate reductase; gamma subunit; putative (DsrM) \| 0.16 \| 0.17 \| 0.17 \| 0.53 \| — \| \| DVU1769 \| Periplasmic [Fe] hydrogenase; large subunit (HydA) \| — \| 0.59 \| 0.38 \| — \| 0.38 \| \| DVU1770 \| Periplasmic [Fe] hydrogenase; small subunit (HydB) \| 0.55 \| 0.42 \| 0.24 \| 0.25 \| 0.21 \| \| DVU2792 \| Electron transport complex protein RnfC; putative \| 0.43 \| 0.33 \| 0.42 \| 0.47 \| 0.50 \| \| DVU2793 \| Electron transport complex protein RnfD; putative \| 0.58 \| 0.36 \| 0.59 \| 0.47 \| 0.54 \| \| DVU2794 \| Electron transport complex protein RnfG; putative \| 0.51 \| 0.36 \| 0.49 \| 0.43 \| 0.51 \| \| DVU2795 \| Electron transport complex protein RnfE; putative \| — \| 0.40 \| — \| 0.48 \| — \| \| DVU2796 \| Electron transport complex protein RnfA; putative \| — \| 0.47 \| 0.61 \| 0.56 \| 0.61 \| \| DVU2797 \| Iron-sulfur cluster-binding protein \| — \| 0.42 \| — \| 0.62 \| — \| \| DVU2798 \| ApbE family protein \| 0.58 \| 0.41 \| 0.53 \| 0.62 \| — \| Open in a new tab a Changes of gene expression level at different time points following addition of 2.5 mM sodium nitrite to cultures compared to controls without nitrite addition. Expression levels were obtained at the same time points from both the treatment and control cultures for the calculation of the expression changes resulting from the stressor. Expression ratio values greater than 1 denote increases in gene expression and values between 0 and 1 indicate decreases in gene expression (P< 0.05). Dashes indicate that no significant change in gene expression was observed. Among genes involved in the sulfate reduction pathway, the operon for the triheme cytochrome c-containing membrane-bound oxidoreductase (dsrMKJOP), which has been implicated as having a role in electron transfer for sulfite reduction (16), was considerably down-regulated in response to nitrite. Two other operons with unknown functions, one encoding the membrane-bound decaheme electron transport complex (rnfABCDEFG) and the other for the membrane-bound cytoplasmically oriented Ech hydrogenase (echABCD), were also significantly down-regulated. Genes with functions in nitrogen metabolism. One important observation in response to nitrite was the down-regulation of multiple genes encoding ATP-binding ABC-transporters for amino acids and polyamines (Table 2). With the down-regulation of genes in the sulfate reduction pathway and oxidative phosphorylation (Table 1), the decrease in the expression of energy-dependent transport systems could be linked to the lowered expression of genes involved in energy production in the cells under nitrite stress. The reduced expression of genes for amino acid transporters could also be a result of the down-regulation of genes encoding the protein biosynthetic machinery. TABLE 2. Effect of nitrite exposure on the transcriptional responses of D. vulgaris genes involved in nitrogen metabolism \| Gene identifier \| TIGR annotation \| Expression ratio (treatment/control) at time pointa: \| :---: \| 0.5 h \| 1.0 h \| 1.5 h \| 2.5 h \| 4.0 h \| \| DVU0095 \| Polyamine ABC transporter, periplasmic polyamine-binding protein \| 0.24 \| 0.19 \| 0.21 \| — \| — \| \| DVU0105 \| Glutamine ABC transporter; ATP-binding protein \| 0.30 \| 0.47 \| 0.59 \| — \| — \| \| DVU0106 \| Glutamine ABC transporter; permease protein \| 0.64 \| 0.64 \| 0.62 \| — \| — \| \| DVU0107 \| Glutamine ABC transporter; periplasmic glutamine-binding protein \| 0.40 \| 0.30 \| 0.39 \| — \| — \| \| DVU0388 \| Amino acid ABC transporter; ATP-binding protein \| 0.25 \| 0.19 \| 0.28 \| — \| — \| \| DVU0624 \| NapC/NirT cytochrome c family protein \| 13.5 \| 11.4 \| 12.2 \| 9.4 \| 11.4 \| \| DVU0625 \| Cytochrome c nitrite reductase; catalytic subunit NfrA; putative \| 18.5 \| 12.8 \| 14.9 \| 7.5 \| 5.7 \| \| DVU0751 \| Amino acid ABC transporter; permease protein; His/Glu/Gln/Arg/opine family \| 0.50 \| 0.39 \| 0.42 \| 0.53 \| 0.45 \| \| DVU0752 \| Amino acid ABC transporter; amino acid-binding protein \| 0.49 \| 0.25 \| 0.33 \| 0.28 \| 0.27 \| \| DVU0966 \| Amino acid ABC transporter, periplasmic amino acid-binding protein \| 0.40 \| 0.17 \| 0.17 \| 0.38 \| 0.27 \| \| DVU0967 \| Amino acid ABC transporter; permease protein; His/Glu/Gln/Arg/opine family \| 0.47 \| 0.40 \| 0.56 \| — \| — \| \| DVU0968 \| Amino acid ABC transporter, ATP-binding protein \| 0.32 \| 0.38 \| 0.36 \| — \| — \| \| DVU1026 \| Uracil permease \| 0.14 \| 0.21 \| 0.30 \| — \| 0.47 \| \| DVU1237 \| Amino acid ABC transporter; permease protein; His/Glu/Gln/Arg/opine family \| 0.28 \| 0.61 \| 0.41 \| — \| — \| \| DVU1238 \| Amino acid ABC transporter; periplasmic amino acid-binding protein \| 0.39 \| 0.62 \| 0.32 \| — \| 0.47 \| \| DVU1766 \| Aspartate ammonia-lyase; putative \| 0.53 \| 0.49 \| 0.36 \| 0.51 \| 0.53 \| \| DVU2113 \| Xanthine/uracil permease family protein \| 0.23 \| 0.28 \| 0.31 \| — \| 0.54 \| \| DVU2242 \| Asparaginase family protein \| 0.64 \| 0.58 \| 0.50 \| — \| — \| \| DVU2543 \| Hybrid cluster protein \| 27.5 \| 34.5 \| 65.1 \| 82.7 \| 37.3 \| \| DVU3392 \| Glutamine synthetase type I \| 1.5 \| 2.2 \| 2.8 \| 2.2 \| 1.6 \| Open in a new tab a See footnote to Table 1. Paradoxically, the glutamine synthetase gene (DVU3392), which participates in nitrogen assimilation (31), was induced during nitrite stress (Table 2). This might result from the increased flux of carbon needed to support substrate level phosphorylation that could spill over into the tricarboxylic acid cycle, increasing the ratio of α-ketoglutarate to glutamine, a possible signal for nitrogen limitation. Additionally, genes encoding the aspartate ammonia-lyase (DVU1766) and asparaginase (DVU2242), which are responsible for the catabolism of amino acids acquired from the medium, were down-regulated (Table 2). Thus, the transition from respiration of sulfate to alternative energy sources could possibly influence the expression profile of genes that participate in the overall carbon and nitrogen metabolism of the cells. Genes in the predicted Fur regulon. Among the highly induced genes during nitrite stress are ferrous iron transporter genes (Fig. 6), which were predicted to be controlled by the ferric uptake regulator (Fur) at the transcriptional level (50). Interestingly, all genes in the predicted Fur regulon (50) were highly up-regulated for 1.5 h following the onset of nitrite stress (Table 3). Since the Fur regulons of many bacteria are known to be derepressed by iron deficiency (10, 14), induction of the Fur regulon in nitrite stress implies a link between this stress and iron depletion. Indeed, genes encoding iron-containing proteins, including nitrite reductase, were on average up-regulated in response to nitrite (Fig. 7), potentially resulting in a higher demand for iron. It is thus suggested that the highly induced Fur-regulated genes, which include ferrous iron transporters, served as a response to the higher expression of genes of iron-containing proteins. TABLE 3. Effect of nitrite exposure on the transcriptional responses of D. vulgaris genes in the predicted Fur regulona \| Gene identifier \| TIGR annotation \| Expression ratio (treatment/control) at time pointb: \| :---: \| 0.5 h \| 1.0 h \| 1.5 h \| 2.5 h \| 4.0 h \| \| DVU0763 \| GGDEF domain protein \| 11.9 \| 2.1 \| — \| — \| — \| \| DVU2378 \| Transcriptional regulator; AraC family \| 4.3 \| 4.1 \| 2.4 \| — \| — \| \| DVU2574 \| Ferrous iron transport protein; putative FeoA \| 3.5 \| 5.0 \| 3.9 \| — \| — \| \| DVU2680 \| Flavodoxin \| 27.6 \| 22.6 \| 4.9 \| — \| — \| \| DVU3330 \| Conserved hypothetical protein \| 2.3 \| 5.7 \| 2.3 \| — \| — \| \| DVU0273 \| Conserved hypothetical protein \| 15.3 \| 5.2 \| 1.8 \| — \| 0.46 \| \| DVU0304 \| Hypothetical protein \| 34.0 \| 10.1 \| 3.7 \| — \| — \| Open in a new tab a Predicted Fur-binding sites (50). b See footnote to Table 1. FIG. 7. Open in a new tab Average changes in expression levels of selected D. vulgaris gene groups following 2.5 mM sodium nitrite addition, with “all” representing all genes covered by the microarray, “iron-binding” representing genes encoding iron-containing proteins, and “fur-regulated” representing genes belonging to the predicted Fur regulon (50). Genes belonging to the predicted PerR regulon. Because reactive nitrogen species have been shown to trigger oxidative stress responses (36, 41), expression levels of genes predicted to be regulated by the oxidative stress regulator PerR (50) were examined (Table 4). All genes in the PerR regulon were moderately up-regulated at one or more sampling points during the experiment, suggesting that the oxidative stress response is a derivative from nitrite stress. TABLE 4. Effect of nitrite exposure on the transcriptional responses of D. vulgaris genes in the predicted PerR regulona \| Gene identifier \| TIGR annotation \| Expression ratio (treatment/control) at time pointb: \| :---: \| 0.5 h \| 1.0 h \| 1.5 h \| 2.5 h \| 4.0 h \| \| DVU0772 \| Hypothetical protein \| 1.8 \| 2.4 \| 2.6 \| 2.1 \| — \| \| DVU2247 \| Antioxidant; AhpC/Tsa family \| 3.0 \| 3.1 \| 2.1 \| 1.8 \| — \| \| DVU2318 \| Rubrerythrin; putative \| — \| — \| 1.5 \| — \| 0.53 \| \| DVU3095 \| Transcriptional regulator; Fur family; PerR \| — \| — \| — \| 2.2 \| — \| \| DVU3096 \| Hypothetical protein \| — \| 1.8 \| — \| — \| — \| Open in a new tab a Predicted PerR-binding sites are from reference 50. b See footnote to Table 1. DISCUSSION The sulfate-reducing bacteria are of great potential in the bioremediation of heavy metals and radionuclides in anaerobic environments (56, 61). Therefore, considerable research efforts using physiological and genetic approaches have been made to understand the response of these bacteria to unfavorable environmental factors. One significant finding from previous studies is the inhibition of sulfate reduction by nitrite, an important intermediate during microbial nitrate reduction (20, 40, 44). The availability of the genome sequence of D. vulgaris makes it possible to study stresses at the whole-genome level (19). Using global transcriptional analysis, this work revealed that D. vulgaris cells responded to the presence of nitrite with a series of well-coordinated regulatory pathways linking energy metabolism, nitrogen metabolism, iron homeostasis, and oxidative stress response. Physiological and transcriptional analyses demonstrated that nitrite reduction was the primary mechanism for detoxification by D. vulgaris (Fig. 2 and Table 2), which is consistent with previous observations (13). A significant increase in the expression of the nitrite reductase genes reported here (Table 2) was also seen by Haveman et al. (16). However, earlier reports measuring the specific activity of nitrite reductase indicated that the enzyme was essentially constitutive (32) or that activity was actually less in cells exposed to nitrite (44). This disparity has not yet been resolved here and requires further examination. Major effects on energy generation pathways were expected from the biochemical studies that demonstrated nitrite inhibition of sulfite reductase (16, 62) and the periplasmic [Fe] hydrogenase of D. vulgaris (44). Furthermore, as nitrite reduction in the periplasm consumes electrons and protons that are central to respiration, one would expect changes in energy metabolism. Indeed, a number of genes with important roles in energy metabolism were differentially expressed, suggesting the extensive response to nitrite stress in the energy metabolism pathways at the transcriptional level (Table 1), which is illustrated in the proposed conceptual model of the transcriptional responses in the energetics of nitrite reduction (Fig. 8). D. vulgaris cells could respond to this energy requirement by the up-regulation of ldh and porAB, thus increasing the electron flow and the opportunity for substrate level phosphorylation. Simultaneously, the triheme cytochrome c (dsrMKJOP) operon, which has been suggested to transfer electrons to the sulfite reductase (16), was significantly down-regulated. These results are in good agreement with the earlier work of Haveman et al. (16), who showed the repressive effects of nitrite on sulfate reduction including sulfate adenylyl transferase and pyrophosphatase. FIG. 8. Open in a new tab Conceptual model of the transcriptional responses in the energy metabolism pathways to nitrite stress (2.5 mM NaNO 2) by D. vulgaris based on the transcriptional profile obtained 60 min after stress exposure. The repression of genes encoding the dsrMKJOP triheme transmembrane complex by nitrite suggests that reducing equivalents derived from lactate oxidation were shifted to nitrite reduction. Red designates up-regulation and blue designates down-regulation; changes in the intensity of the red or blue represent the extent of the up- or down-regulation, respectively; white indicates no change detected in expression level. Interestingly, earlier work on the effects of nitrite on the growth of D. vulgaris on lactate- or sulfate-containing medium reported a large accumulation of hydrogen (44). Thus, some of the excess reductant may be channeled to hydrogen production, consistent with the up-regulation of the [NiFe] hydrogenase (isozyme-2) gene. Additionally, the up-regulation of the genes in the fumarate reductase operon (frdBAC) (Table 1) could signal the use of fumarate as a terminal electron acceptor in the absence of sulfate/sulfite reduction. Taken as a whole, the repression of sulfate reduction, the increase in nitrite reduction, and the inhibition of [Fe] hydrogenase likely contribute to a diminished proton motive force, which in turn may be responsible for the repression of genes encoding the ATP synthase subunits (Table 1). The resulting slowdown in growth might also reflect the down-regulation of genes for ribosomal proteins and those for the biosynthesis of amino acids during nitrite stress (Fig. 6). Notably, while the [NiFe] hydrogenase isozyme-2 gene was up-regulated under nitrite stress, the [Fe] hydrogenase gene was down-regulated (Table 1). Currently, the physiological roles of the various hydrogenases in D. vulgaris are still not clear. However, the [NiFe] hydrogenase is apparently more suited to functioning in the presence of nitrite, based on prior reports that nitrite strongly inhibits the periplasmic [Fe] hydrogenase but has no impact on [NiFe] hydrogenase (2). Thus, the redundancy in periplasmic hydrogenases may allow for functional compensation under stress conditions (19). Furthermore, the up-regulation of periplasmic formate dehydrogenase points to the possibility that formate oxidation acts as another mechanism to supply electrons and protons for nitrite reduction, but the source of formate in the periplasm needs to be resolved. Interestingly, it is proposed that the “hydrogen cycling” model (43) for contributing to a proton gradient could potentially be one example of a more general phenomenon termed redox cycling (22). In fact, consideration of the D. vulgaris genome sequence reveals the potential for production of formate in the cytoplasm. Movement of the protonated, uncharged species through the cytoplasmic membrane and its oxidation in the periplasm could contribute to the electron and proton flows across the membrane (19). Thus, it is possible that uncharged formate generated during pyruvate oxidation diffuses across the membrane and then is oxidized by the formate dehydrogenase to contribute to the electrons and protons used in nitrite reduction or hydrogen generation. The transcriptional response to nitrite stress in energy metabolism pathways also appeared to affect the expression of genes involved in nitrogen metabolism in D. vulgaris (Table 2). The down-regulation of multiple genes encoding ATP-requiring ABC amino acid and polyamine transporters may reflect the inhibition of ATP generation from sulfate respiration and/or the decreased demand for amino acids for protein biosynthesis. In contrast, the up-regulation of the glutamine synthetase gene (glnA) would appear to signal nitrogen-limiting conditions (31, 64). It is possible that an increased flux of carbon to support substrate level phosphorylation might overflow into the tricarboxylic acid cycle, altering the α-ketoglutarate/glutamine ratio controlling glnA expression. Another strategy to preserve amino acids for biosynthetic demand was the repression of genes encoding aspartate ammonia-lyase and asparaginase, which catabolize amino acids when they are present in excess. Interestingly, results from this study show that genes in the Fur regulon were among the most highly up-regulated genes in response to nitrite stress (Table 3). As Fur is known as the primary regulator of iron homeostasis in many other microorganisms (30, 34), this observation raises a question about the connection between nitrite stress and iron homeostasis. The Fur family metalloregulatory proteins are typically dimeric DNA-binding transcriptional factors that also bind Fe 2+ as a corepressor in order to repress downstream genes. It is therefore proposed that derepression of Fur-regulated genes could be attributed to interactions of nitrite, directly or indirectly, with either the Fur protein, Fe 2+, or both. Derepression of the Fur regulon could be effected by iron deficiency resulting from consumption of cytoplasmic Fe(II), since genes encoding many iron-containing proteins, including the nitrite reductase, were up-regulated in response to nitrite. Compounding this demand for iron, chemical oxidation of Fe 2+ by NO 2− has been readily observed (6, 42), and Fe 3+ is generally unavailable for biosynthesis or signaling. A less likely mechanism is that nitrite might react directly with the protein-bound Fe 2+ corepressor, generating Fe 3+, leading to dissociation and concomitant derepression. The intracellular concentrations of NO 2− are likely to be small because of the rapid reduction of nitrite in the periplasm and the apparent absence of a specific transport system for this ion. However, given the complex chemistry of reactive nitrogen species (46), it is still possible that reactive nitrogen species generated from nitrite reduction could enter the cytoplasm to react with Fe 2+ (8). However, the contribution of each mechanism to the relief from Fur repression during nitrite exposure is not clear, and further biochemical study is needed to address the mechanism and importance of Fur regulation in this stress response in D. vulgaris. Since both Fur and PerR respond to oxidative stress and belong to the same superfamily of metalloregulatory proteins that respond to metal ions (33), it is possible that the same mechanism derepresses both regulons. Studies on other microorganisms have shown that reactive nitrogen species, including nitrite, incidentally induce genes responsive to oxidative stress, in addition to genes specifically designed to protect cells from nitrosative stress (35, 36). Whether proteins encoded in the PerR regulon confer protection against nitrite or are adventitiously derepressed remains to be determined. In summary, the results reveal that D. vulgaris cells initiate a coordination of transcriptional regulations allowing the alleviation of nitrite toxicity via nitrite reduction. The down-regulation of genes in the energy metabolism pathways suggests a shift in the flow of reducing equivalents from oxidative phosphorylation to nitrite reduction. Based on the transcriptional response to nitrite stress, it is also proposed that substrate level phosphorylation becomes prominent and that the excess reductant generated may be disposed of as succinate or hydrogen. It is further suggested that increased demand for iron resulting from these regulatory events likely contributes to iron depletion along with the chemical oxidation of available Fe 2+, derepressing the Fur regulon. However, further biochemical study is needed to elucidate the regulatory mechanisms and importance of transcriptional regulators in D. vulgaris during stress responses. Supplementary Material [Supplemental material] aem_72_6_4370__index.html (1.2KB, html) Acknowledgments This research was supported by the U.S. Department of Energy under the Genomics: GTL Program through the Virtual Institute for Microbial Stress and Survival and Microbial Genome Program of the Office of Biological and Environmental Research, Office of Science. Oak Ridge National Laboratory is managed by University of Tennessee-Battelle LLC for the Department of Energy under contract DEAC05-00OR22725. Footnotes † Supplemental material for this article may be found at REFERENCES 1.Altschul, S., T. Madden, A. 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10336	https://www.oxfordlearnersdictionaries.com/definition/english/covet	Definition of covet verb from the Oxford Advanced Learner's Dictionary covet \| \| \| --- \| \| present simple I / you / we / they covet \| /ˈkʌvət/ /ˈkʌvət/ \| \| he / she / it covets \| /ˈkʌvəts/ /ˈkʌvəts/ \| \| past simple coveted \| /ˈkʌvətɪd/ /ˈkʌvətɪd/ \| \| past participle coveted \| /ˈkʌvətɪd/ /ˈkʌvətɪd/ \| \| -ing form coveting \| /ˈkʌvətɪŋ/ /ˈkʌvətɪŋ/ \| Definitions on the go Look up any word in the dictionary offline, anytime, anywhere with the Oxford Advanced Learner’s Dictionary app. Nearby words Oxford Learner's Dictionaries More from us Who we are Oxford University Press is a department of the University of Oxford. It furthers the University's objective of excellence in research, scholarship, and education by publishing worldwide
10337	https://www.mometrix.com/academy/rates-and-unit-rates/	Unit Rate and Rate (Video & Practice Questions) Skip to content Online Courses Study Guides Flashcards Online Courses Study Guides Flashcards Menu Rates and Unit Rates Unit Rate and Rate (Video & Practice Questions) On this page What is a Rate? How to Find Unit Rate Frequently Asked Questions What is a Unit Rate? PDF Unit Rate Problems [x] Transcript - [x] FAQs - [x] Fact Sheet - [x] Practice Welcome to this video on rates and, more specifically, unit rates. This math concept is practical and useful, so we hope that you come away with a solid understanding and confidence in interpreting the rates that are all around you. Let’s get started! Rates are spoken about and used every day in many aspects of life. The concept is pretty straightforward, as rates are nothing more than ratios of values that represent different units of measure. The values that are being compared are called the terms of the rate. What is a Rate? Let’s consider the treasure trove of rates that may be in your grocery cart the next time you go shopping. In the examples listed here, the rate that you agree to pay is simply the cost related to the quantity of the product. Let’s say we have a 12-ounce box of pasta, which costs $1.49. The rate is then $1.49 per 12 ounces. If we have a pound of deli meat and the cost is $9.99, our rate would then be $9.99 per pound. Lastly, let’s say we have an 8-pack of 20-ounce bottles of soda and the cost is $5.98. Our rate then is $5.98 per pack. \| Item \| Cost \| Rate \| --- \| 12 oz. box of pasta \| $1.49 \| $1.49/12 oz. of pasta \| \| 1 lb. of deli meat \| $9.99 \| $9.99/1 lb. of deli meat \| \| 8-pack of 20 oz. soda \| $5.98 \| $5.98/eight 20 oz. bottles \| If we look at this last example a little closer, we’ll see that there is room to break down the rate even further. The example provides the price of an 8-pack, but what if I want to determine the cost of one 20-ounce bottle? The ratio $5.98 8 bottles$5.98 8 bottles provides that information. This quick calculation tells me that each 20-ounce bottle costs approximately $0.75. This breakdown to determine the cost per bottle may be helpful to determine whether I buy the eight pack of one type of soda, or the individual 20-ounce bottles of another brand on sale for $0.50 each. How to Find Unit Rate The process of breaking down the cost to a smaller unit reveals the unit rate of the product. This is helpful to make informed decisions at the store, as the volumes of product in various packaging are often different. By comparing unit rates, savvy customers are able to make price comparisons based on common units of the product regardless of packaging and advertised “sale” prices. Let’s break down the soda cost per bottle further to determine the cost per ounce. If one 20 ounce bottle costs roughly $0.75, then dividing that cost by 20 ounces reveals the cost per ounce. $0.75 20 oz$0.75 20 oz≈$0.04≈$0.04 As you can see, breaking down costs to the smallest unit reveals the cost savings of the sale. Of course, saving a few cents per ounce of soda may not be the deciding factor of your purchase. Other factors come into play when consumers are shopping, like brand loyalty and personal preference. However, comparing unit costs provides an objective way of making consumer choices based on the price. The important thing to remember when analyzing unit rates is that the units must be the same. Let’s consider another example to illustrate this point. Suppose you are on a road trip in Wyoming and on the first day you covered 300 miles in 4 hours of mostly highway driving. You can quickly determine your average rate of speed as miles per hour with the following calculation: 300 mi 4 hrs 300 mi 4 hrs=75 mph=75 mph Coincidentally, your friend is traveling in Germany, where the standard unit of measure is in kilometers, and she reports that she covered approximately 513 kilometers in 4 hours on her first day of the road trip. Her average speed would be calculated as: 513 km 4 hrs 513 km 4 hrs=128.25 kph=128.25 kph Clearly, this is comparing “apples to oranges” in the sense that the underlying units are not the same. A conversion of either miles to kilometers or kilometers to miles must be made to make a fair comparison of average speed. Keep in mind that a kilometer is a shorter unit of distance than a mile. 1 mi≈1.609 km 1 mi≈1.609 km. To convert your average speed of 75 miles per hour to kilometers, simply multiply 75×1.609 75×1.609: 75 mph×1.609 km/mi 75 mph×1.609 km/mi=120.6 kph=120.6 kph On the other hand, you could convert your friend’s reported kilometers per hour to miles per hour. 1 km≈0.6215 mi 1 km≈0.6215 mi. Multiply this conversion factor by your friend’s daily average speed to convert to miles per hour: 128.25 kph×0.6215 mi/km 128.25 kph×0.6215 mi/km=79.7 mph=79.7 mph The way that you convert does not matter as long as you compare the average speeds of the same unit. Both conversions show that your friend in Europe traveled at a faster rate on the first day of her trip. I hope that you feel more equipped to interpret and solve the math “puzzles” that you may come across. Thanks for watching, and happy studying! Frequently Asked Questions Q How do you find the rate and unit rate? A Since rate is a ratio that compares two different units, find the rate of something by comparing one unit to another. For example, a bag of 8 apples for $3.60, can be described as $3.60 per bag of apples. You are comparing price to a bag of apples. A unit rate is a rate expressed for the quantity of one. In this example, that would be the price in dollars for 1 single apple. To find the unit rate, we will divide $3.60 by the number of apples in the bag, which is 8, to get the unit rate $0.45/apple. Q What are examples of unit rates? A A very common example of a unit rate, which we encounter on a daily basis, is miles per hour. Since this rate is describing the number of miles in one hour, it is a unit rate. Another common unit rate is the price of gas per gallon. You can see this unit rate when passing any gas station sign. When buying meat, the price is displayed in dollars per one pound, which is also a unit rate. The price of most snack foods in the grocery store are also unit rates, such as the price of a bag of chips or a bag of pretzels. Q What does unit rate mean? A A unit rate describes the ratio of two different units for the quantity of one. Some of the unit rates we use every day are miles per hour someone travels, price per pound of meat, and price per day for a rental car. We can find the unit rate when given a rate by dividing the unit in the numerator by the quantity in the denominator. For example, if we are given the rate $4.50/5 gallons of milk, we can divide 4.50 by 5 to find the unit price, which is dollars per one gallon of milk. Q How do I calculate rates? A We do not actually calculate rate; rate is a ratio between two things that have different units of measure. For example, if we know that a pack of 3 Romaine Hearts costs $2.99, then the rate is $2.99/pack of Romaine Hearts. The distance between Chicago and New York is approximately 790 miles. On average a flight from Chicago to New York is 3 hours. We can show the rate of travel for a plane as 790 miles/3 hours. Q What is the difference between a rate and a unit rate? A The difference between a rate and a unit rate is that a rate is the ratio between two different units of measure, while a unit rate is the ratio of between two different units of measure for a single thing. Q Can a unit rate be used to compare two rates? A A unit rate can be used to compare two rates as long as the units of measure are the same. For example, in the US we use $/gallon when describing the price of gas. In Europe they use Euro/liter. When comparing the price of gas in the US and Europe, it would be misleading if we did not convert the units to match because $1 = 0.82 Euro and 1 gallon = 3.785 liters. What is a Unit Rate? PDF Your Unit Rate Review Download Unit Rate Problems Question #1: Calculate the following rate as a unit rate: $36 for 4 movie tickets $9 for one movie ticket $9 for two movie tickets $18 for 2 movie tickets $18 for one movie ticket [x] Show Answer Answer: The correct answer is $9 for one movie ticket. In order to express $36 for one movie ticket as a unit rate, we need to determine the cost for one movie ticket. This requires us to divide $36 by 4, which equals $9. [x] Hide Answer Question #2: Olivia can bake 95 cookies in 5 hours. At this rate, how many cookies can she bake in 11 hours? 100 cookies 90 cookies 475 cookies 209 cookies [x] Show Answer Answer: The correct answer is 209 cookies. Olivia bakes 95 cookies in 5 hours, which can be expressed as the following rate: 95 cookies 5 hours 95 cookies 5 hours. From here we can divide 95 5 95 5 in order to determine how many cookies she bakes per one hour. 95 5=19 95 5=19, so 19 cookies per hour. If Olivia bakes 19 cookies per hour, and we need to know how many cookies she can bake in 11 hours, we can simply multiply 19×11 19×11, which equals 209 cookies. [x] Hide Answer Question #3: Franklin can jog approximately 2,000 meters in 4 minutes. Express this rate as a unit rate. 800 meters per minute 500 meters per minute 4,000 meters per 8 minutes 300 meters per 5 minutes [x] Show Answer Answer: The correct answer is 500 meters per minute. A unit rate describes a ratio that represents “per one”. In order to express 2,000 meters in 4 minutes as a unit rate we need to divide 2,000 by 4. 2,000 4=500 2,000 4=500, which is the unit rate 500 meters per one minute. [x] Hide Answer Question #4: Jamie wants to buy a 6 pack of pineapple juice boxes. The price for a 6 pack is $2.64. Jamie notices that one juice box is leaking, so he only wants to buy five out of the six. How much would five juice boxes cost? $2.35 $2.80 $2.20 $2.99 [x] Show Answer Answer: The correct answer is $2.20. In order to calculate the price for 5 juice boxes, we need to determine the price for one juice box. This can be calculated by dividing $2.64 6$2.64 6, which equals $0.44, or 44 cents. Each juice box costs 44 cents. If Jamie wants to purchase 5 juice boxes at 44 cents each, we simply multiply 5×0.44 5×0.44, which is 2.2, or $2.20. [x] Hide Answer Question #5: Your local grocery stores are having sales on bags of candy. Store A is selling 40 oz. bags of candy for $8.75. Store B is selling 30 oz. bags of candy for $7.49. Which store has the better deal? Store A Store B [x] Show Answer Answer: The correct answer is Store A. In order to identify the better deal, we need to calculate the unit cost (unit rate) for each store. Store A sells candy for $8.75 40 oz$8.75 40 oz, so let’s divide $8.75 by 40 in order to determine how much one ounce of candy will cost. $8.75 40 oz$8.75 40 oz is approximately 22 cents per one ounce. Store B sells candy for $7.49 30 oz$7.49 30 oz so let’s divide $7.49 by 30 to determine its unit cost. $7.49 30 oz$7.49 30 oz is approximately 25 cents per one ounce. Now that we have calculated the unit cost for each store, we can clearly see that Store A has the better deal on bags of candy! Better hurry before they sell out! [x] Hide Answer Return to Pre-Algebra Videos 185363 by Mometrix Test Preparation \| Last Updated: August 8, 2025 On this page What is a Rate? How to Find Unit Rate Frequently Asked Questions What is a Unit Rate? PDF Unit Rate Problems Why you can trust Mometrix Raising test scores for 20 years 150 million test-takers helped Prep for over 1,500 tests 40,000 5-star reviews A+ BBB rating Who we are About Mometrix Test Preparation We believe you can perform better on your exam, so we work hard to provide you with the best study guides, practice questions, and flashcards to empower you to be your best. Learn More... 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10338	https://legal-resources.uslegalforms.com/s/special-agent	We use cookies to improve security, personalize the user experience, enhance our marketing activities (including cooperating with our marketing partners) and for other business use. Click "here" to read our Cookie Policy. By clicking "Accept" you agree to the use of cookies. Read less Read more Accept US Legal Forms Legal Resources Definitions S Special Agent What is a Special Agent? A Comprehensive Legal Overview Definition & meaning A special agent is a type of agent who has limited authority to act on behalf of another person or entity. This authority is confined to specific tasks or transactions as designated by the principal, the person who appoints the agent. For instance, a real estate broker may serve as a special agent for a property owner, tasked solely with finding a buyer for a property. The agent's role is defined and restricted to the particular assignment given to them. Table of content Legal use & context Special agents are commonly referenced in various legal contexts, including real estate, business transactions, and contract law. Their limited authority means they can only act within the scope of their designated tasks. This type of agency is important in ensuring that the principal's interests are protected while allowing the agent to perform specific functions. Users can utilize legal templates from US Legal Forms to create agreements or documents that define the roles and responsibilities of special agents. Key legal elements Real-world examples Here are a couple of examples of abatement: Example 1: A real estate agent is hired by a homeowner to sell their house. The agent's authority is limited to marketing the property and negotiating offers. They cannot make decisions outside this scope, such as altering the sale price without the homeowner's consent. Example 2: A company hires a consultant to negotiate a specific contract with a supplier. The consultant acts as a special agent, and their authority is limited to this negotiation only. (hypothetical example) State-by-state differences \| State \| Special Agent Definition \| --- \| \| California \| Special agents must have a written agreement outlining their authority. \| \| New York \| Special agents can only act within the parameters set by the principal. \| \| Texas \| Special agents' authority is limited to specific transactions as defined in the contract. \| This is not a complete list. State laws vary, and users should consult local rules for specific guidance. Comparison with related terms \| Term \| Definition \| Key Differences \| --- \| General Agent \| An agent with broad authority to act on behalf of the principal. \| General agents can make decisions across a range of activities, unlike special agents. \| \| Power of Attorney \| A legal document granting someone authority to act on another's behalf. \| Power of attorney can be general or specific, while a special agent's authority is always limited to specific tasks. \| Common misunderstandings What to do if this term applies to you If you find yourself needing to appoint a special agent, clearly outline the authority and responsibilities in a written agreement. This document should specify the tasks the agent is authorized to perform. For assistance, consider using legal templates from US Legal Forms to create an appropriate agreement. If the situation is complex, it may be wise to consult a legal professional for tailored advice. Find a legal form that suits your needs Browse our library of 85,000+ state-specific legal templates. This field is required Select state Select state Alabama Alaska Arizona Arkansas California Colorado Connecticut Delaware District of Columbia Florida Georgia Guam Hawaii Idaho Illinois Indiana Iowa Kansas Kentucky Louisiana Maine Maryland Massachusetts Michigan Minnesota Mississippi Missouri Montana Multi-State Nebraska Nevada New Hampshire New Jersey New Mexico New York North Carolina North Dakota Ohio Oklahoma Oregon Pennsylvania Puerto Rico Rhode Island South Carolina South Dakota Tennessee Texas Utah Vermont Virgin Islands Virginia Washington West Virginia Wisconsin Wyoming Quick facts Authority is limited to specific tasks. Commonly used in real estate and business transactions. Requires clear communication of roles. Legal templates are available for creating agreements. Key takeaways FAQs The main role of a special agent is to act on behalf of a principal in a specific transaction or task, with limited authority. No, a special agent cannot make decisions outside the specific tasks they are authorized to perform. You can appoint a special agent by drafting a written agreement that clearly outlines their authority and responsibilities. Definitions in alphabetical order Special Administrator Special Appearance Special Allocatur Special Agricultural Workers (SAW) Special Access Program Special Administration Special Acceptance Special Airworthiness Certificate Special Agents [U.S. Fish and Wildlife Service] Special Agent Spear Phishing Related legal terms
10339	https://opencw.aprende.org/courses/economics/14-123-microeconomic-theory-iii-spring-2005/lecture-notes/	Lecture Notes \| Microeconomic Theory III \| Economics \| MIT OpenCourseWare Subscribe to the OCW Newsletter Help\|Contact Us Find Courses Find courses by: Topic MIT Course Number Department Instructional Approach Teaching Materials View All Courses Collections Audio/Video Lectures Online Textbooks New Courses Most Visited Courses OCW Scholar Courses This Course at MIT Supplemental Resources Translated Courses 繁體字 / Traditional Chinese Español / Spanish Português / Portuguese فارسی / Persian Türkçe / Turkish (비디오)한국 / Korean More... Cross-Disciplinary Topic Lists Energy Entrepreneurship Environment Introductory Programming Life Sciences Transportation About About MIT OpenCourseWare Site Statistics OCW Stories News Donate Make a Donation Why Donate? Our Supporters Other Ways to Contribute Shop OCW Become a Corporate Sponsor Featured Sites OCW Initiatives Highlights for High School OCW Educator MIT Crosslinks and OCW MITx and Related OCW Courses Beyond OCW MIT+K12 Videos Teaching Excellence at MIT Outreach @ MIT Open Education Consortium Advanced Search Home » Courses » Economics » Microeconomic Theory III » Lecture Notes Lecture Notes Course Home Syllabus Calendar Readings Assignments Exams Lecture Notes Download Course Materials Lecture Notes\| LEC# \| TOPICS \| LECTURE NOTES \| --- \| 1 \| Existence and Optimality of General Equilibrium \| AC MC Pricing (PDF) \| \| 2 \| The Core and Convergence Theorem \| \| \| 3 \| Arrow's Impossibility Theorem \| \| \| 4 \| Externalities and Public Goods \| Externalities (PDF) Taxing Externalities with Measurable Pollution (PDF) Taxing Externalities with Uniform Taxes (PDF) \| \| 5 \| Intertemporal Competitive Equilibrium \| \| \| 6 \| Uncertainty with Complete Markets \| Contingent Commodities in an Exchange Economy (PDF) \| \| 7 \| Incomplete Markets \| Stock Market Model (PDF) Contrasting SM & CCC Models (PDF) Inefficiency with Incomplete Markets (PDF) \| Find Courses Find by Topic Find by Course Number Find by Department Instructional Approach Teaching Materials Audio/Video Courses Courses with Subtitles Online Textbooks New Courses Most Visited Courses OCW Scholar Courses This Course at MIT Supplemental Resources Translated Courses View All Courses About About OpenCourseWare Site Statistics OCW Stories News Press Releases Tools Help & FAQs Contact Us Advanced Search Site Map Privacy & Terms of Use RSS Feeds Donate Make a Donation Why Donate? Our Supporters Other Ways to Contribute Shop OCW Become a Corporate Sponsor Featured Sites Highlights for High School OCW Educator MIT Crosslinks and OCW MITx and Related OCW Courses MIT+K12 Videos Teaching Excellence at MIT Outreach@MIT Open Education Consortium Our Corporate Supporters Support for MIT OpenCourseWare's 15th anniversary is provided by About MIT OpenCourseWare OCW is a free and open publication of material from thousands of MIT courses, covering the entire MIT curriculum. Learn more » © 2001–2015 Massachusetts Institute of Technology Your use of the MIT OpenCourseWare site and materials is subject to our Creative Commons License and other terms of use.
10340	https://artofproblemsolving.com/wiki/index.php/Vieta%27s_Formulas?srsltid=AfmBOoo2fMSgOe-sOAlJJhnod0nItGSHX-yQwV_nHGBG32lOE5RlU6Xr	Art of Problem Solving Vieta's Formulas - 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 Vieta's Formulas Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Vieta's Formulas In algebra, Vieta's formulas are a set of results that relate the coefficients of a polynomial to its roots. In particular, it states that the elementary symmetric polynomials of its roots can be easily expressed as a ratio between two of the polynomial's coefficients. It is among the most ubiquitous results to circumvent finding a polynomial's roots in competition math and sees widespread usage in many math contests/tournaments. Contents [hide] 1 Statement 2 Proof 3 Problems 3.1 Introductory 3.2 Intermediate 4 Advanced 5 See also Statement Let be any polynomial with complex coefficients with roots , and let be the elementary symmetric polynomial of the roots. Vieta’s formulas then state that This can be compactly summarized as for some such that . Proof Let all terms be defined as above. By the factor theorem, . We will then prove Vieta’s formulas by expanding this polynomial and comparing the resulting coefficients with the original polynomial’s coefficients. When expanding the factorization of , each term is generated by a series of choices of whether to include or the negative root from every factor . Consider all the expanded terms of the polynomial with degree ; they are formed by multiplying a choice of negative roots, making the remaining choices in the product , and finally multiplying by the constant . Note that adding together every multiplied choice of negative roots yields . Thus, when we expand , the coefficient of is equal to . However, we defined the coefficient of to be . Thus, , or , which completes the proof. Problems Here are some problems with solutions that utilize Vieta's quadratic formulas: Introductory 2005 AMC 12B Problem 12 2007 AMC 12A Problem 21 2010 AMC 10A Problem 21 2003 AMC 10A Problem 18 2021 AMC 12A Problem 12 Intermediate 2017 AMC 12A Problem 23 2003 AIME II Problem 9 2008 AIME II Problem 7 2021 Fall AMC 12A Problem 23 2019 AIME I Problem 10 Advanced 2020 AIME I Problem 14 See also Polynomial Retrieved from " Categories: Algebra Polynomials Theorems 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.
10341	https://www.mathway.com/popular-problems/Trigonometry/316545	Find the Exact Value arctan(tan(-pi/3)) \| Mathway Enter a problem... [x] Trigonometry Examples Popular Problems Trigonometry Find the Exact Value arctan(tan(-pi/3)) arctan(tan(−π 3))arctan(tan(-π 3)) Step 1 Add full rotations of 2 π 2 π until the angle is greater than or equal to 0 0 and less than 2 π 2 π. arctan(tan(5 π 3))arctan(tan(5 π 3)) Step 2 Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because tangent is negative in the fourth quadrant. arctan(−tan(π 3))arctan(-tan(π 3)) Step 3 The exact value of tan(π 3)tan(π 3) is √3 3. arctan(−√3)arctan(-3) Step 4 The exact value of arctan(−√3)arctan(-3) is −π 3-π 3. −π 3-π 3 Step 5 The result can be shown in multiple forms. Exact Form: −π 3-π 3 Decimal Form: −1.04719755…-1.04719755… a r c t a n(t a n(−π 3))a r c t a n⁡(t a n⁡(-π 3)) arctan(tan(−π 3))x arctan⁡(tan⁡(-π 3))x arctan(tan(−π 3))x 2 arctan⁡(tan⁡(-π 3))x 2 arctan(tan(−π 3))x 3 arctan⁡(tan⁡(-π 3))x 3 ( ( ) ) \| \| [ [ ] ] √ √   ≥ ≥   ° °       7 7 8 8 9 9       ≤ ≤   θ θ       4 4 5 5 6 6 / / ^ ^ × ×   π π       1 1 2 2 3 3 - - + + ÷ ÷ < <          , , 0 0 . . % %  = =     Report Ad Report Ad ⎡⎢⎣x 2 1 2√π∫x d x⎤⎥⎦[x 2 1 2 π∫⁡x d x] Please ensure that your password is at least 8 characters and contains each of the following: a number a letter a special character: @$#!%?& 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. 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10342	https://www.gauthmath.com/solution/1833425821446193/Simplify-to-a-single-trig-function-with-no-denominator-frac-1-sin-2xsin-2x-cos-2	Question Solution contact@gauthmath.com
10343	https://www.youtube.com/watch?v=S9DcU2gMtpM	【例題6】等腰三角形面積的計算劉繼文 9370 subscribers 8 likes Description 7694 views Posted: 5 Apr 2021 Transcript:
10344	https://math.stackexchange.com/questions/433367/considering-a-sum-of-a-monotonically-increasing-and-decreasing-sequence	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 Considering a sum of a monotonically increasing and decreasing sequence. Ask Question Asked Modified 12 years, 3 months ago Viewed 3k times 3 $\begingroup$ The following is the problem that I am working on. Let ${z_n} = {x_n}+{y_n}$ be a sequence where ${x_n}$ is monotonically increasing, ${y_n}$ monotonically decreasing, and ${z_n}$ is bounded. Is ${z_n}$ convergent ? What if ${x_n}$ and ${y_n}$ are also bounded ? I can clearly see and prove that in the second case, ${z_n}$ must converge to the sum of each of the limits of the sequences (because they exist). However, this is what I think about the case where only ${z_n}$ is bounded. Intuitively I want to say that it would be nice if ${z_n}$ converges but that sounds too good. So I considered the following case. If ${x_n}$ increases "faster" than ${y_n}$ decreasing, ${z_n}$ will become a monotonically increasing sequence that is bounded, thus it will converge to the sup of ${z_n}$. If ${y_n}$ decreases "faster" then with the similar argument ${z_n}$ will converge to the inf. If their increasing and decreasing rate are equal then it's a constant sequence, so the limit is cogent. But I am not 100% confident that there doesn't exist a "rate of increase and decrease that lies in between" so that ${z_n}$ eventually oscillates or something. Can someone help me out ? analysis limits Share edited Jun 12, 2020 at 10:38 CommunityBot 1 asked Jul 1, 2013 at 1:44 hyg17hyg17 5,28144 gold badges4343 silver badges9090 bronze badges $\endgroup$ 4 $\begingroup$ It is pretty easy to construct a sequence where $z_i = (-1)^i$, which is bounded but doesn't converge. $\endgroup$ Calvin Lin – Calvin Lin 2013-07-01 01:50:51 +00:00 Commented Jul 1, 2013 at 1:50 $\begingroup$ @CalvinLin: From the answers I can see that, thank you. Do you have any more examples so that in the future I can get the idea of what kind of thinking I should be doing ? $\endgroup$ hyg17 – hyg17 2013-07-01 02:30:35 +00:00 Commented Jul 1, 2013 at 2:30 $\begingroup$ Generally, you should know why a counter-example works, and what 'property' it is exploiting. In this case, we want a bounded sequence that is not convergent, and the typical example is $(-1)^i$ (and equivalently $0,1,0,1,0,1, \ldots$). Then, you can think about how to create such a sequence $z_i$. Of course, it is not guaranteed that this approach always works, so understanding more/various counter-examples will be very helpful. $\endgroup$ Calvin Lin – Calvin Lin 2013-07-01 02:33:58 +00:00 Commented Jul 1, 2013 at 2:33 $\begingroup$ You had a good analysis of what happens if $(x_n)$ increases faster than $(y_n)$ decreases, and what happens in the opposite case. All that remained is to realize that, since these are infinite sequences they can switch between the two cases. $(x_n)$ can increase faster than $(y_n)$ decreases for a while, say for $n\leq100$, and then it can be slower for a while, say for $100\leq n\leq200$, and then faster for a while, etc. That way you can manipulate $(z_n)$ to do just about anything you can imagine. $\endgroup$ Andreas Blass – Andreas Blass 2013-07-01 03:02:14 +00:00 Commented Jul 1, 2013 at 3:02 Add a comment \| 2 Answers 2 Reset to default 7 $\begingroup$ What about something like $ x_n = n + .5 \sin (n) $ and $ y_n = -n + .5 \sin(n) $? Share answered Jul 1, 2013 at 1:51 supersnoopersupersnooper 17122 bronze badges $\endgroup$ 1 $\begingroup$ Wow, this is exactly the reason I was not confident with my argument. Is this a rather common technique ? $\endgroup$ hyg17 – hyg17 2013-07-01 02:32:07 +00:00 Commented Jul 1, 2013 at 2:32 Add a comment \| 4 $\begingroup$ Hint: $x_n = \lfloor \frac{n}{2} \rfloor$ and $y_n = - \lfloor \frac{n+1}{2} \rfloor$ Share answered Jul 1, 2013 at 1:52 Calvin LinCalvin Lin 77.6k55 gold badges8686 silver badges170170 bronze badges $\endgroup$ 3 $\begingroup$ Ohhhhhh, that is clever ! I am not very articulate with the use of floor and roof functions, but this gives me great insight. Thanks ! $\endgroup$ hyg17 – hyg17 2013-07-01 02:35:20 +00:00 Commented Jul 1, 2013 at 2:35 $\begingroup$ @hyg17 Note that you could simply have created the sequences explicitly as $x_n = 0,0,1,1, 2, 2, 3, 3 \ldots $ and $y_n = -0, -1, -1, -2, -2, -3, -3, \ldots$. It's clear how to continue these sequences, and why it works. The use of floor/ceiling functions merely give a closed form for the sequence, which is not completely necessary. $\endgroup$ Calvin Lin – Calvin Lin 2013-07-01 02:46:12 +00:00 Commented Jul 1, 2013 at 2:46 $\begingroup$ Ahh, thanks for loosening up some knots in my head. We are virtually at the same age but our brains are at a different level. lol $\endgroup$ hyg17 – hyg17 2013-07-01 02:53:25 +00:00 Commented Jul 1, 2013 at 2:53 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 analysis limits See similar questions with these tags. 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10345	https://www.inchcalculator.com/percentage-of-a-percentage-calculator/	Percent of a Percent Calculator - Inch Calculator Inch CalculatorSkip to Content search for a calculator Popular Searches calculate body fat what is today's date? calculate yards of concrete how long until 3:30? calculate a pay raise 30 minute timer calculate board and batten wall layout how to do long division calculate the best TV size how many days until Christmas? Enable Dark Mode Enable Light Mode Open Navigation Menu Close Navigation Menu search for a calculator Popular Searches calculate body fat what is today's date? calculate yards of concrete how long until 3:30? calculate a pay raise 30 minute timer calculate board and batten wall layout how to do long division calculate the best TV size how many days until Christmas? Calculators Calculators Automotive Calculators Construction Calculators Conversion Calculators Cooking & Baking Calculators Electrical Calculators Financial Calculators Health & Fitness Calculators Math Calculators Pets Calculators Science Calculators Shipping & Logistics Calculators Time & Date Calculators TV & Video Calculators U.S. Patriotic Calculators Enable Dark Mode Enable Light Mode Home Math Percentages Percent of a Percent Calculator Percent of a Percent Calculator Find the percentage of a percentage by entering them below. This is useful to find a compound percentage of a value. Percent 1 % Percent 2 % ×Close dialog Have a Question or Feedback? Name (optional) Email (optional) Question or Feedback Submit Feedback Result: 25% of 60% = 15% Steps to Solve Start with the formula to calculate a percentage of a percentage p=percent 1 / 100×percent 2 / 100×100 Substitute values in the equation p=25 / 100×60 / 100×100 Simplify and solve p=0.25×0.6×100 p=15% Learn how we calculate this below Add this calculator to your site LATEST VIDEOS next stay CC Settings Off Arabic Chinese English French German Hindi Portuguese Spanish Font Color white Font Opacity 100%Font Size 100%Font Family Arial Text Shadow none Background Color black Background Opacity 50%Window Color black Window Opacity 0% White Black Red Green Blue Yellow Magenta Cyan 100%75%50%25% 200%175%150%125%100%75%50% Arial Georgia Garamond Courier New Tahoma Times New Roman Trebuchet MS Verdana None Raised Depressed Uniform Drop Shadow 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% By Joe Sexton Joe is the creator of Inch Calculator and has over 20 years of experience in engineering and construction. He holds several degrees and certifications. Full bio Reviewed by Pateakia Heath, PhD Pateakia has worked in education for 15 years and has three degrees, including a PhD, Master's degree, and Bachelor's degree. She specializes in mathematics. Full bio Copy LinkShare on FacebookShare on XShare on Pinterest Cite ×Close dialog Cite This Page Sexton, J. (n.d.). Percent of a Percent Calculator. Inch Calculator. Retrieved September 29, 2025, from How to Find the Percentage of a Percentage It might sound confusing, but sometimes you need to find the percentage of a percentage, which might seem confusing. You can find the percent of a percent by converting both percentages to decimal form, then multiplying them together. Start by dividing each percentage by 100 to convert them to a decimal form. Then multiply the decimals together to get the result. Multiply the result by 100 to convert the decimal back to a percentage. Percent of a Percent Formula To find a percentage of a percentage, use the following formula: p = pct 1 / 100 × pct 2 / 100 × 100 Thus the percentage of a percentage p is equal to the first percentage divided by 100 times the second percentage divided by 100, times 100. For example, let’s find 25% of 80%. p = (pct 1 ÷ 100) × (pct 2 ÷ 100) × 100 p = (25 ÷ 100) × (80 ÷ 100) × 100 p = .25 × .8 × 100 p = .2 × 100 p = 20% So, 25% of 80% is equal to 20%. Similar Percentage Calculators Percentage Calculator Marks Percentage Calculator Percent Yield Calculator Percent to Fraction Calculator Percent to Ratio Calculator See All Inch Calculator Have Feedback or a Suggestion? 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10346	https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:linear-equations-graphs/x2f8bb11595b61c86:horizontal-vertical-lines/e/horizontal-and-vertical-lines	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
10347	https://www.elainetron.com/apcalc/apcalc.pdf	Copyright 1996,1997 Elaine Cheong All Rights Reserved Study Guide for the Advanced Placement Calculus AB Examination By Elaine Cheong 1 Table of Contents INTRODUCTION 2 TOPICS TO STUDY 3 • Elementary Functions 3 • Limits 5 • Differential Calculus 7 • Integral Calculus 12 SOME USEFUL FORMULAS 16 CALCULATOR TIPS AND PROGRAMS 17 BOOK REVIEW OF AVAILABLE STUDY GUIDES 19 ACKNOWLEDGEMENTS 19 2 Introduction Advanced Placement1 is a program of college-level courses and examinations that gives high school students the opportunity to receive advanced placement and/or credit in college. The Advanced Placement Calculus AB Exam tests students on introductory differential and integral calculus, covering a full-year college mathematics course. There are three sections on the AP Calculus AB Examination: 1. Multiple Choice: Part A (25 questions in 45 minutes) - calculators are not allowed 2. Multiple Choice: Part B (15 questions in 45 minutes) - graphing calculators are required for some questions 3. Free response (6 questions in 45 minutes) - graphing calculators are required for some questions Scoring Both sections (multiple choice and free response) are given equal weight. Grades are reported on a 1 to 5 scale: Examination Grade Extremely well qualified 5 Well qualified 4 Qualified 3 Possibly qualified 2 No recommendation 1 To obtain a grade of 3 or higher, you need to answer about 50 percent of the multiple-choice questions correctly and do acceptable work on the free-response section. In both Parts A and B of the multiple choice section, 1/4 of the number of questions answered incorrectly will be subtracted from the number of questions answered correctly. 1 Advanced Placement Program® and AP® are trademarks of the College Entrance Examination Board. 3 Topics to Study Elementary Functions Properties of Functions A function ƒ is defined as a set of all ordered pairs (x, y), such that for each element x, there corresponds exactly one element y. The domain of ƒ is the set x. The range of ƒ is the set y. Combinations of Functions If ƒ(x) = 3x + 1 and g(x) = x2 - 1 a) the sum ƒ(x) + g(x) = (3x + 1) + (x2 - 1) = x2 + 3x b) the difference ƒ(x) - g(x) = (3x + 1) - (x2 - 1) = -x2 + 3x + 2 c) the product ƒ(x)g(x) = (3x + 1)(x2 - 1) = 3x3 + x2 - 3x - 1 d) the quotient ƒ(x)/g(x) = (3x + 1)/(x2 - 1) e) the composite (ƒ ° g)(x) = ƒ(g(x)) = 3(x2 - 1) + 1 = 3x2 - 2 Inverse Functions Functions ƒ and g are inverses of each other if ƒ(g(x)) = x for each x in the domain of g g(ƒ(x)) = x for each x in the domain of ƒ The inverse of the function ƒ is denoted ƒ-1. To find ƒ-1, switch x and y in the original equation and solve the equation for y in terms of x. Exercise: If ƒ(x) = 3x + 2, then ƒ-1(x) = (A) 1 3 2 x+ (B) x 3 - 2 (C) 3x - 2 (D) 1 2 x + 3 (E) x−2 3 The answer is E. x = 3y + 2 3y = x - 2 y = x−2 3 Even and Odd Functions The function y = ƒ(x) is even if ƒ(-x) = ƒ(x). Even functions are symmetric about the y-axis (e.g. y = x2) The function y = ƒ(x) is odd if ƒ(-x) = -ƒ(x). Odd functions are symmetric about the origin (e.g. y = x3) 4 Exercise: If the graph of y = 3x + 1 is reflected about the y-axis, then an equation of the reflection is y = (A) 3x - 1 (B) log3 (x - 1) (C) log3 (x + 1) (D) 3-x + 1 (E) 1 - 3x The answer is D. The reflection of y = ƒ(x) in the y-axis is y = ƒ(-x) Periodic Functions You should be familiar with the definitions and graphs of these trigonometric functions: sine, cosine, tangent, cotangent, secant, and cosecant Exercise: If ƒ(x) = sin(tan-1 x), what is the range of ƒ? (A) (-π /2,π /2) (B) [-π /2,π /2] (C) (0, 1] (D) (-1, 1) (E) [-1, 1] The answer is D. The range of sin x is (E), but the points at which sin x = ± 1 (π /2 + kπ ), tan-1 x is undefined. Therefore, the endpoints are not included. Note: The range is expressed using interval notation: ( , ) a b a x b ⇔ < < [ , ] a b a x b ⇔ ≤ ≤ Zeros of a Function These occur where the function ƒ(x) crosses the x-axis. These points are also called the roots of a function. Exercise: The zeros of ƒ(x) = x3 - 2x2 + x is (A) 0, -1 (B) 0, 1 (C) -1 (D) 1 (E) -1, 1 The answer is B. ƒ(x) = x(x2 - 2x + 1) = x(x -1)2 5 Properties of Graphs You should review the following topics: a) Intercepts b) Symmetry c) Asymptotes d) Relationships between the graph of y = ƒ(x) and y = kƒ(x) y = ƒ(kx) y - k = ƒ(x - h) y = \|ƒ(x)\| y = ƒ(\|x\|) Limits Properties of Limits If b and c are real numbers, n is a positive integer, and the functions ƒ and g have limits as x c → , then the following properties are true. 1. Scalar multiple: lim x c →[b(ƒ(x))] = b[lim x c →ƒ(x)] 2. Sum or difference: lim x c →[ƒ(x) ± g(x)] = lim x c →ƒ(x) ± lim x c →g(x) 3. Product: lim x c →[ƒ(x)g(x)] = [lim x c →ƒ(x)][lim x c →g(x)] 4. Quotient: lim x c →[ƒ(x)/g(x)] = [lim x c →ƒ(x)]/[lim x c →g(x)], if lim x c →g(x) ≠0 One-Sided Limits lim x a →+ƒ(x) x approaches c from the right lim x a →−ƒ(x) x approaches c from the left Limits at Infinity lim x→+∞ƒ(x) = L or lim x→−∞ƒ(x) = L The value of ƒ(x) approaches L as x increases/decreases without bound. y = L is the horizontal asymptote of the graph of ƒ. Some Nonexistent Limits lim x→0 1 2 x lim x→0 \| \| x x lim x→0 sin1 x Some Infinite Limits lim x→0 1 2 x = ∞ lim x→+ 0 ln x =−∞ 6 Exercise: What is lim x→0 sin x x ? (A) 1 (B) 0 (C) ∞ (D) π 2 (E) The limit does not exist. The answer is A. You should memorize this limit. Continuity Definition A function ƒ is continuous at c if: 1. ƒ(c) is defined 2. lim x c →ƒ(x) exists 3. lim x c →ƒ(x) = ƒ(c) Graphically, the function is continuous at c if a pencil can be moved along the graph of ƒ(x) through (c, ƒ(c)) without lifting it off the graph. Exercise: If f x x x x f k ( ) ( ) = + =      3 2 0 2 , for x ≠0 and if ƒ is continuous at x = 0, then k = (A) -3/2 (B) -1 (C) 0 (D) 1 (E) 3/2 The answer is E. lim x→0 ƒ(x) = 3/2 Intermediate Value Theorem If ƒ is continuous on [a, b] and k is any number between ƒ(a) and ƒ(b), then there is at least one number c between a and b such that ƒ(c) = k. 7 Differential Calculus Definition ƒ'(x) = lim ∆x→0 f x x f x x ( ) ( ) + − ∆ ∆ and if this limit exists ƒ'(c) = lim x c → f x f c x c ( ) ( ) − − If ƒ is differentiable at x = c, then ƒ is continuous at x = c. Differentiation Rules General and Logarithmic Differentiation Rules 1. d dx [cu] = cu' 2. d dx [u ± v] = u' ± v' sum rule 3. d dx [uv] = uv' + vu' product rule 4. d dx [u v ] = vu uv v ' ' − 2 quotient rule 5. d dx [c] = 0 6. d dx [un] = nun-1u' power rule 7. d dx [x] = 1 8. d dx [ln u] = u u ' 9. d dx [eu] = euu' 10. d dx [ƒ(g(x))] = ƒ' (g(x)) g' (x) chain rule Derivatives of the Trigonometric Functions 1. d dx [sin u] = (cos u)u' 2. d dx [csc u] = -(csc u cot u)u' 3. d dx [cos u] = -(sin u)u' 4. d dx [sec u] = (sec u tan u)u' 5. d dx [tan u] = (sec2 u)u' 6. d dx [cot u] = -(csc2 u)u' Derivatives of the Inverse Trigonometric Functions 1. d dx [arcsin u] = u u ' 1 2 − 2. d dx [arccsc u] = − − u u u ' \| \| 2 1 3. d dx [arccos u] = − − u u ' 1 2 4. d dx [arcsec u] = u u u ' \| \| 2 1 − 5. d dx [arctan u] = u u ' 1 2 + 6. d dx [arccot u] = − + u u ' 1 2 Implicit Differentiation Implicit differentiation is useful in cases in which you cannot easily solve for y as a function of x. 8 Exercise: Find dy dx for y3 + xy - 2y - x2 = -2 dy dx [y3 + xy - 2y - x2] = dy dx [-2] 3y2 dy dx + (x dy dx + y) - 2 dy dx - 2x = 0 dy dx (3y2 + x - 2) = 2x - y dy dx = 2 3 2 2 x y y x − + − Higher Order Derivatives These are successive derivatives of ƒ(x). Using prime notation, the second derivative of ƒ(x), ƒ''(x), is the derivative of ƒ'(x). The numerical notation for higher order derivatives is represented by: ƒ(n)(x) = y(n) The second derivative is also indicated by d y dx 2 2 . Exercise: Find the third derivative of y = x5. y' = 5x4 y'' = 20x3 y''' = 60x2 Derivatives of Inverse Functions If y = ƒ(x) and x = ƒ-1(y) are differentiable inverse functions, then their derivatives are reciprocals: dx dy dy dx = 1 Logarithmic Differentiation It is often advantageous to use logarithms to differentiate certain functions. 1. Take ln of both sides 2. Differentiate 3. Solve for y' 4. Substitute for y 5. Simplify Exercise: Find dy dx for y = x x 2 2 1 3 1 1 + −       / ln y = 1 3 [ln(x2 + 1) - ln(x2 - 1)] y y ' = 1 3 1 1 1 1 2 2 x x + − −       9 y' = − + − − +       2 1 1 1 1 2 2 2 2 1 3 ( )( ) / x x x x y' = − + − 2 1 1 2 4 3 2 2 3 ( ) ( ) / / x x Mean Value Theorem If ƒ is continuous on [a, b] and differentiable on (a, b), then there exists a number c in (a, b) such that ƒ'(c) = f b f a b a ( ) ( ) − − L'Hôpital's Rule If lim ƒ(x)/g(x) is an indeterminate of the form 0/0 or ∞∞ / , and if lim ƒ'(x)/g'(x) exists, then lim f x g x ( ) ( ) = lim f x g x '( ) '( ) The indeterminate form 0⋅∞ can be reduced to 0/0 or ∞∞ / so that L'Hôpital's Rule can be applied. Note: L'Hôpital's Rule can be applied to the four different indeterminate forms of ∞∞ / : ∞∞ / , ( ) / −∞ ∞, ∞ −∞ / ( ), and ( ) / ( ) −∞ −∞ Exercise: What is lim sin x x x → + 0 1 ? (A) 2 (B) 1 (C) 0 (D) ∞ (E) The limit does not exist. The answer is B. lim cos x x →0 1 = 1 Tangent and Normal Lines The derivative of a function at a point is the slope of the tangent line. The normal line is the line that is perpendicular to the tangent line at the point of tangency. Exercise: The slope of the normal line to the curve y = 2x2 + 1 at (1, 3) is (A) -1/12 (B) -1/4 (C) 1/12 (D) 1/4 (E) 4 10 The answer is B. y' = 4x y = 4(1) = 4 slope of normal = -1/4 Extreme Value Theorem If a function ƒ(x) is continuous on a closed interval, then ƒ(x) has both a maximum and minimum value in the interval. Curve Sketching Situation Indicates ƒ'(c) > 0 ƒ increasing at c ƒ'(c) < 0 ƒ decreasing at c ƒ'(c) = 0 horizontal tangent at c ƒ'(c) = 0, ƒ'(c-) < 0, ƒ'(c+) > 0 relative minimum at c ƒ'(c) = 0, ƒ'(c-) > 0, ƒ'(c+) < 0 relative maximum at c ƒ'(c) = 0, ƒ''(c) > 0 relative minimum at c ƒ'(c) = 0, ƒ''(c) < 0 relative maximum at c ƒ'(c) = 0, ƒ''(c) = 0 further investigation required ƒ''(c) > 0 concave upward ƒ''(c) < 0 concave downward ƒ''(c) = 0 further investigation required ƒ''(c) = 0, ƒ''(c-) < 0, ƒ''(c+) > 0 point of inflection ƒ''(c) = 0, ƒ''(c-) > 0, ƒ''(c+) < 0 point of inflection ƒ(c) exists, ƒ'(c) does not exist possibly a vertical tangent; possibly an absolute max. or min. Newton's Method for Approximating Zeros of a Function xn + 1 = xn - f x f x n n ( ) '( ) To use Newton's Method, let x1 be a guess for one of the roots. Reiterate the function with the result until the required accuracy is obtained. Optimization Problems Calculus can be used to solve practical problems requiring maximum or minimum values. Exercise: A rectangular box with a square base and no top has a volume of 500 cubic inches. Find the dimensions for the box that require the least amount of material. Let V = volume, S = surface area, x = length of base, and h = height of box V = x2h = 500 S = x2 + 4xh = x2 + 4x(500/x2) = x2 + (2000/x) S' = 2x - (2000/x2) = 0 2x3 = 2000 11 x = 10, h = 5 Dimensions: 10 x 10 x 5 inches Rates-of-Change Problems Distance, Velocity, and Acceleration y = s(t) position of a particle along a line at time t v = s'(t) instantaneous velocity (rate of change) at time t a = v'(t) = s''(t) instantaneous acceleration at time t Related Rates of Change Calculus can be used to find the rate of change of two or more variable that are functions of time t by differentiating with respect to t. Exercise: A boy 5 feet tall walks at a rate of 3 feet/sec toward a streetlamp that is 12 feet above the ground. a) What is the rate of change of the tip of his shadow? b) What is the rate of change of the length of his shadow? 12 5 y x z z 5 12 x z = x y x + = 12 5 z x = 12 5 x y = 5 7 ( ) dx dt dy dt = 5 7 ( ) dz dt dx dt = 12 5 ( ) dx dt = 5 7 3 ( ) dz dt = 12 5 15 7 b) = 15 7 ft/sec a) = 36 7 ft/sec Note: the answers are independent of the distance from the light. Exercise: A conical tank 20 feet in diameter and 30 feet tall (with vertex down) leaks water at a rate of 5 cubic feet per hour. At what rate is the water level dropping when the water is 15 feet deep? V = 1 3 π r2h dv dt = 1 9 π h2 dh dt r h = 10 30 5 = 1 9 π h2 dh dt r h = 1 3 dh dt h = 45 2 π V = 1 27 π h3 dh dt = 1 5π ft/hr 12 Integral Calculus Indefinite Integrals Definition: A function F(x) is the antiderivative of a function ƒ(x) if for all x in the domain of ƒ, F'(x) = ƒ(x) ∫ƒ(x) dx = F(x) + C, where C is a constant. Basic Integration Formulas General and Logarithmic Integrals 1. kƒ(x) dx = k ƒ(x) dx 2. ∫[ƒ(x) ± g(x)] dx = ƒ(x) dx ± g(x) dx 3. ∫k dx = kx + C 4. ∫xn dx = x n n+ + 1 1 + C, n ≠-1 5. ∫ex dx = ex + C 6. ∫ax dx = a a x ln + C, a > 0, a ≠1 7. ∫dx x = ln \|x\| + C Trigonometric Integrals 1. ∫sin x dx = -cos x + C 2. ∫cos x dx = sin x + C 3. ∫sec2 x dx = tan x + C 4. ∫csc2 x dx = -cot x + C 5. ∫sec x tan x dx = sec x + C 6. ∫csc x cot x dx = -csc x + C 7. ∫tan x dx = -ln \|cos x\| + C 8. ∫cot x dx = ln \|sin x\| + C 9. ∫sec x dx = ln \|sec x + tan x\| + C 10. ∫csc x dx = -ln \|csc x + cot x\| + C 11. ∫ dx a x x a C 2 2 − = + arcsin 12. ∫ dx a x a x a C 2 2 1 + = + arctan 13. ∫ dx x x a a x a C 2 2 1 − = + arcsec Integration by Substitution ∫ƒ(g(x))g'(x) dx = F(g(x)) + C If u = g(x), then du = g'(x) dx and ∫ƒ(u) du = F(u) + C Integration by Parts ∫u dv = uv - ∫v du Distance, Velocity, and Acceleration (on Earth) a(t) = s''(t) = -32 ft/sec2 v(t) = s'(t) = ∫s''(t) dt = ∫-32 dt = -32t + C1 at t = 0, v0 = v(0) = (-32)(0) + C1 = C1 s(t) = ∫v(t) dt = ∫(-32t + v0) dt = -16t2 + v0t + C2 13 Separable Differential Equations It is sometimes possible to separate variables and write a differential equation in the form ƒ(y) dy + g(x) dx = 0 by integrating: ∫ƒ(y) dy + ∫g(x) dx = C Exercise: Solve for dy dx x y = −2 2x dx + y dy = 0 x2 + y2 2 = C Applications to Growth and Decay Often, the rate of change or a variable y is proportional to the variable itself. dy dt = ky separate the variables dy y = k dt integrate both sides ln \|y\| = kt + C1 y = Cekt Law of Exponential Growth and Decay Exponential growth when k > 0 Exponential decay when k < 0 Definition of the Definite Integral The definite integral is the limit of the Riemann sum of ƒ on the interval [a, b] lim ( ) ∆ ∆ x i i n f x x → = = ∑ 0 1 ∫a b ƒ(x) dx Properties of Definite Integrals 1. ∫a b [ƒ(x) + g(x)] dx = ∫a b ƒ(x) dx + ∫a b g(x) dx 2. ∫a b kƒ(x) dx + k ∫a b ƒ(x) dx 3. ∫a a ƒ(x) dx = 0 4. ∫a b ƒ(x) dx = - ∫b a ƒ(x) dx 5. ∫b a ƒ(x) dx + ∫b c ƒ(x) dx = ∫a c ƒ(x) dx 6. If ƒ(x) ≤ g(x) on [a, b], then ∫a b ƒ(x) dx ≤ ∫a b g(x) dx 14 Approximations to the Definite Integral Riemann Sums ∫a b ƒ(x)dx = Sn = f x x i i n ( )∆ = ∑ 1 Trapezoidal Rule ∫a b ƒ(x)dx ≈[ 1 2 ƒ(x0) + ƒ(x1) + ƒ(x2) + ... + ƒ(xn-1) + 1 2 ƒ(xn)] b a n − The Fundamental Theorem of Calculus If ƒ is continuous on [a, b] and if F' = ƒ, then ∫a b ƒ(x) dx = F(b) - F(a) The Second Fundamental Theorem of Calculus If ƒ is continuous on an open interval I containing a, then for every x in the interval, d dx ∫a x ƒ(t) dt = ƒ(x) Area Under a Curve If ƒ(x)≥0 on [a, b] A = ∫a b ƒ(x) dx If ƒ(x)≤0 on [a, b] A = - ∫a b ƒ(x) dx If ƒ(x)≥0 on [a, c] and A = ∫a c ƒ(x) dx - ∫c b ƒ(x) dx ƒ(x)≤0 on [c, b] Exercise The area enclosed by the graphs of y = 2x2 and y = 4x + 6 is: (A) 76/3 (B) 32/3 (C) 80/3 (D) 64/3 (E) 68/3 The answer is D. Intersection of graphs: 2x2 = 4x + 6 2x2 - 4x + 6 = 0 x = -1, 3 A = ∫−1 3 4x + 6 - 2x2 = (2x2 + 6x - 2 3 3 x ) −1 3 = 18 + 18 - 18 - (2 - 6 + 2/3) = 64/3 Average Value of a Function on an Interval 1 b a a b − ∫ ƒ(x) dx 15 Volumes of Solids with Known Cross Sections 1. For cross sections of area A(x), taken perpendicular to the x-axis: V = ∫a b A(x) dx 2. For cross sections of area A(y), taken perpendicular to the y-axis: V = ∫a b A(y) dy Volumes of Solids of Revolution: Disk Method V = ∫a b π r2 dx Rotated about the x-axis: V = ∫a b π [ƒ(x)]2 dx Rotated about the y-axis: V = ∫a b π [ƒ(y)]2 dy Volumes of Solids of Revolution: Washer Method V = ∫a b π (ro2 dx - ri2 ) dx Rotated about the x-axis: V = ∫a b π [(ƒ1(x))2 - (ƒ2(x))2] dx Rotated about the y-axis: V = ∫a b π [(ƒ1(y))2 - (ƒ2(y))2] dy Exercise: Find the volume of the region bounded by the y-axis, y = 4, and y = x2 if it is rotated about the line y = 6. π ∫0 2 [(x2 - 6)2 - (4 - 6)2 ]dx = 192 5 π cubic units Volumes of Solids of Revolution: Cylindrical Shell Method V = ∫a b 2π rh dr Rotated about the x-axis: V = 2π ∫a b xƒ(x) dx Rotated about the y-axis: V = 2π ∫a b yƒ(y) dy 16 Some Useful Formulas loga x = log log x a sin2x + cos2x = 1 1 + tan2x = sec2x 1 + cot2x = csc2x sin 2x = 2 sin x cos x cos 2x = cos2x - sin2x sin2x = ½(1- cos 2θ) cos2x = ½(1+ cos 2θ) Volume of a right circular cylinder = π r2h Volume of a cone = 1 3 πr2h Volume of a sphere = 4 3 πr3 17 Calculator Tips and Programs Your calculator will serve as an extremely useful tool if you take advantage of all of its functions. We will base all of the following tips and programs on the TI-82, which most calculus students use today. On the AP Calculus Exam, you will need your calculator for Part B of Section I and for Section II. You will need to know how to do the following: 1. simple calculations 2. find the intersection of two graphs 3. graph a function and be able to find properties listed under Elementary Functions in the Topics to Study section (e.g. domain, range, asymptotes) Here are some of the functions available that you should know how to use: In the CALC menu: 1. calculate the value of a function at x = c 2. calculate the roots of a function 3. find the minimum of a function 4. find the maximum of a function 5. find the point of intersection of two functions 6. find the slope of the tangent at (x, y) 7. find the area under the curve from a to b In the MATH MATH menu: 6. find the minimum of a function fMin(expression, variable, lower, upper) 7. find the maximum of a function fMax(expression, variable, lower, upper) 8. find the numerical derivative at a given value nDeriv(expression, variable, value) 9. find the numerical integral of an expression fnInt(expression, variable, lower, upper) 0. calculate the root of an expression solve(expression, variable, guess, {lower, upper}) Calculator Programs One of the easiest programs to create is one that will solve for f(x). You can also run the program multiple times to find other values for the same function. PROGRAM: SOLVE : Input X : 3x2 + 2 → X [type your function here and place → X at the end] : Disp X 18 Here is a program to solve for a quadratic equation: PROGRAM: QUADRAT :Input "A? ", A :Input "B? ", B :Input "C? ", C :(- B + √ (B2 - 4AC) ) / 2A → D :(- B - √ (B2 - 4AC) ) / 2A → E :B2 - 4AC → F :ClrHome :Disp "+ EQUALS" :Disp D :Disp "- EQUALS" :Disp E :Disp "B2 - 4AC EQUALS" :Disp F To Run: Enter a, b, and c for ax2 + bx + c. "+ EQUALS" and "- EQUALS" give the roots of the equation Here is a program that will use the trapezoidal rule to approximate a definite integral: PROGRAM: TRAP :ClrHome :Input "F(X) IN QUOTES:", Y0 :Input "START(A):", A :Input "END(B):", B :Input "NO. OF DIV. (N):", N :(B - A) / N → D :0 → S :For (X, A, B, D) :S + Y0 → S :End :A → X : Y0 → F :B → X : Y0 → L :D (-F + S - L) → A :ClrHome :Disp "EST AREA=" :Disp A 19 Book Review of Available Study Guides Brook, Donald E., Donna M. Smith, and Tefera Worku. The Best Test Preparation for the Advanced Placement Examination in Calculus AB. Piscataway: Research & Education Association, 1995. This book contains six full-length AP Exams and a short, comprehensive AP course review. This book is for the student who wishes to practice taking the Calculus AB Exam. The topic review is not very clear, and there are several errors in the questions and keys. However, detailed solutions are presented for all problems. Hockett, Shirley O. Barron's How to Prepare for the Advanced Placement Examinations, Mathematics. New York: Barron's Educational Series, Inc., 1987. This book contains a review of calculus topics, practice multiple-choice questions for each unit, and four practice examinations and 3 actual examinations for both the Calculus AB and BC Exams. This book is the most comprehensive study guide that I have found. However, the current edition of this text would be much more useful. Smith, Sanderson M. and Frank W. Griffin. Advanced Placement Examinations in Mathematics. New York: Simon & Schuster, Inc., 1990. This book contains a complete review of all exam topics, including multiple-choice and free-response questions with explanations. It also includes two full-length practice tests with explanatory answers and BASIC computer programs to reinforce calculus concepts. This book is also a good way to prepare for the Calculus Exams AB and BC. Zandy, Bernard V. Cliffs Quick Review Calculus. Lincoln: Cliffs Notes, Inc., 1993. This book is a compact review of all topics covered in a first-year calculus class. Example problems are given in each unit. This book contains the best topical review that I have found. It was not prepared specifically for the AP Exam, so students will need to review the trapezoidal rule, which was not covered in this book. Students will not need to know the arc length formula for the Calculus AB Exam. Acknowledgements I would like to thank these people for their time and support: Dr. Rena Bezilla, Raymond Cheong, Mr. Eric Ebersole, Mr. Charlie Koppelman, and Bonnie Zuckerman.
10348	https://ocw.mit.edu/courses/res-8-009-introduction-to-oscillations-and-waves-summer-2017/mitres_8_009su17_lec8n.pdf	Massachusetts Institute of Technology MITES 2017–Physics III Lecture 08: Traveling Waves and Boundary Interactions In these notes, we solve the wave equation for traveling wave solutions and calculate the transmission and reﬂection coeÿcients characterizing how waves propagate across boundaries. We also discuss waves trav eling through media and the energy dissipation that can result. We end with a short discussion of sound waves. 1 Waves en route In the previous notes, we successfully derived the wave equation describing how strings underwent both longitudinal and transverse1 oscillations. For transverse oscillations deﬁned by a vertical displacement y(x, t), we found the wave equation was given by ∂2 ∂2 2 y(x, t) = v y(x, t), (1) ∂t2 ∂x2 p where v (which is T/µ for the string) was a quantity with units of velocity that we had yet to interpret. In deriving this equation, we considered the string to reside within a speciﬁc domain of space and considered the wave motion propagating within this domain. Such an assumption was a convenient starting point in analyzing waves, but not all waves exist bounded in ﬁxed domains. For example, light and sound waves (both of which are described by wave equations) travel from one place to another interacting with the various media they come in contact with. With regard to light, we are able to see because light waves are reﬂected o various objects in our envi ronment and travel to the rod and cone cells in our eyes. During travel, the intensity of the light can decrease if it has to propagate through fog, or it might vanish entirely if it was incident on an opaque object. With regard to sound, if you can hear cars driving outside or your neighbors sitting rooms away, it is because sound waves are traveling through the air, being transmitted through walls and boundaries sepa rating and ﬁnally hitting your ear drums. Not all of the sound waves emitted by their respective sources reach you. Some of these waves are reﬂected back toward their source and others decay away quickly. Figure 1: We can hear and see objects around us due to traveling sound and light waves. How do we mathematically describe such waves? We would like to be able to describe such propagation. Since we have only so far considered bounded 1As a useful mnemonic to di erentiate the two, remember that longitudinal waves travel ”along” the axis of propagation 1 2 TRAVELING WAVE SOLUTIONS M. WILLIAMS waves, we would need to expand our conception of solutions to the wave equation in order to describe traveling waves. To this end, the framing question for this lesson is Framing Question How do we model the ways waves travel through space and are reﬂected, transmitted, and dispersed through boundaries? 2 Traveling wave solutions Figure 2: The two terms in the general solution Eq.(4) deﬁne wave propagation at the same speed but in di erent directions. Our goal is to solve Eq.(1) such that our solutions describe traveling waves. Our method of solution will parallel our previous solution methods (namely guess and check), but before we make the attempt, let us discuss some properties we want these traveling wave solutions to have. Unlike our standing wave solutions (which were deﬁned only within the domain [0, L]) we want our traveling waves to be deﬁned for a much larger space of x values. For simplicity, we will take this larger space of x values to be the entire real line2, i.e., we want our wave to be deﬁned for x = +∞ and x = −∞. Also, we will take our waves to be propagating with a ﬁnite velocity v. Concerning v, we would want our waves to propagate such that if we translate them by a distance Δx in the opposite direction of the propagating wave front, then this would be tantamount to not moving at all and instead waiting a time Δt = Δx/v for the wave to advance by the position Δx. To satisfy this property, our wave y(x, t) must be a function of x − vt. We can see this by noting that if we take x → x − Δx (i.e., move by Δx in the opposite direction of +v), then the function f1(x − vt) becomes f1(x − Δx − vt) = f1(x − v(t + Δt)) (2) which is equivalent to taking t → t + Δt if Δt = Δx/v. Therefore, as our guess for a solution to Eq.(1), we take y(x, t) = f1(x − vt) where f1 is any suÿciently well-behaved3 function. Checking this solution we have ∂2 ∂2 2 y(x, t) = v y(x, t) ∂t2 ∂x2 ∂2 ∂2 2 f1(x − vt) = v f1(x − vt) ∂t2 ∂x2 (−v)2f1(x − vt) = v 2f1(x − vt) 2In reality waves always interact with something (either the source of the wave or its target) at some point during their propagation so their domain is actually constricted. 3By ”well-behaved” we mean there are no singularities. 2 2 TRAVELING WAVE SOLUTIONS M. WILLIAMS f1(x − vt) = f1(x − vt). (3) Thus any function which is a function of the quantity x − vt is a solution to the wave equation. Through a similar calculation, we can show that the solution y = f2(x + vt) (where f2 is not necessarily equivalent to f1) is also a solution to Eq.(1). These two solutions f1(x − vt) and f2(x + vt) represent wave propagation in two di erent directions: The solution f2(x + vt) represents a wave propagating to the left, as opposed to the rightward propagation of f1(x − vt). Now, given that the general solution to a linear di erential equation is a linear combination of the possible solutions, we ﬁnd that the general solution to Eq.(1) is y(x, t) = f1(x − vt) + f2(x + vt) (4) where the coeÿcients of the linear combination were absorbed into redeﬁnitions of f1 and f2. Although, it does not look like it, Eq.(4) includes the standing wave solutions we found through separation of variables in the previous notes. Thus, Eq.(4) is indeed the most general (although not always the most useful) form of the solutions to Eq.(1). 2.1 The Connection between Standing and Traveling Waves We stated that Eq.(4) is the true general solution for a wave equation. What relationship does it have with the previous solution we found for standing waves? To understand the relationship, we consider Eq.(4) with two sinusoidal waves of equal phase and am plitude but traveling in opposite directions. We deﬁne this wave amplitude as u(x, t): u(x, t) = 1 [A cos (k(x − vt)) − A cos (k(x + vt))] , (5) 2 where k is an as of yet unspeciﬁed wave number. Using the sum of angles formula for cosine functions, we ﬁnd that Eq.(5) becomes u(x, t) = A sin(kx) sin(ωt), (6) where we used the deﬁnition of angular frequency to replace kv with ω. From here, er vsn extend this result by presuming that instead of having two waves traveling on the string, we have inﬁnitely many waves all of the form Eq.(5) except each wave’s wavenumber k depends on an index n = 1, 2, . . . such that kn = nπ/L (where L is some length of interest). The coeÿcient A is then reasonably expected to depend on n, so we would replace A with An. The total wave displacement would then be ∞ X y(x, t) = un(x, t) n=1 ∞ X An = [cos (kn(x − vt)) − cos (kn(x + vt))] 2 n=1 ∞ X = A sin(knx) sin(ωnt), (7) n=1 which is the solution we found for the waves on a string bounded within x = 0 and x = L. Eq.(7) is somewhat less general than the Fourier series solution found before in that it satisﬁes y(x, 0) = 0, and thus an initial condition has (implicitly) already been imposed. To ﬁnd the more general solution, we would have needed to begin with a more general combination of sines and cosines in Eq.(5). In either case, we see that standing waves can be represented by linear combinations of traveling waves moving in opposite directions. 3 2 TRAVELING WAVE SOLUTIONS M. WILLIAMS 2.2 Fourier Integrals and Periodic Motion Given that our focus is on periodic motion, it is natural to ﬁrst explore sinusoidal solutions in Eq.(4). So we will take our function f1 and f2 to be a linear combination of sinusoids: Thus the solution Eq.(4) becomes y(x, t) = A cos [k(x − vt)] + B sin [k(x − vt)] + C cos [k(x + vt)] + D sin [k(x + vt)] (8) where A, B, C, and D are real quantities. For the purposes of calculation, it will prove easier to deal with a more general form of these sinusoidal solutions. We deﬁne the complex variable z(x, t) such that y(x, t) is the real part of z(x, t): y(x, t) = Re [z(x, t)], (9) We will take z(x, t) to satisfy the same wave equation as y(x, t), ∂2 ∂2 2 z(x, t) = v z(x, t), (10) ∂t2 ∂x2 with the same form of the general solution e z(x, t) = f1(x − vt) + f e 2(x + vt) (11) The fact that z(x, t) is complex allows us to consider sinusoidal solutions written as complex exponentials. Namely, the complex analog of Eq.(8) is ik(x−vt) −ik(x−vt) ik(x+vt) −ik(x+vt) z(x, t) = A+e + B+e + A−e + B−e , (12) iku −iku where A± and B± are complex quantities. We include both e and e type of functions in Eq.(12) because complex exponentials with opposite arguments are independent of one another. We can obtain Eq.(8) by applying Eq.(9) and taking Re[A+] = A, Im[A+] = −B, and so on. The way we will use Eq.(12), is that we will take its real part before we physically interpret any result computed from it. Recalling our previous nomenclature, for the forward moving wave solution \|A+\| is the amplitude of the wave, k is the wave number, and kv = ω is the frequency. In the previous lesson, we noted that solutions to the wave equation deﬁned by a single wave number k only represent one solution to the wave equation. For example, we found that one solution to the wave equation in a bounded domain was An sin (knvt) sin (knx) . (13) In order to ﬁnd the general solution, we needed to sum this solution over all possible values of kn. Given that each solution is associated with a speciﬁc coeÿcient, the general solution was then ∞ X y(x, t) = An sin (knvt) sin (knx) (14) n=1 Similarly, to ﬁnd the most general form of Eq.(12) we need to sum over all possible values of k. Here, k is a continuous (rather than discrete) variable so this summation will take the form of an integral. Moreover, the continuous analog of the An in Eq.(14) is A+(k) and A−(k). Thus, the most general form of the sinusoidal solution Eq.(12) is Z ∞ h i ik(x−vt) ik(x+vt) z(x, t) = dk A+(k)e + A−(k)e , (15) −∞ We do not need to include the B coeÿcients from Eq.(12); since our integral is running from −∞ to +∞, the e−ik(x−vt) and e−ik(x+vt) solutions are automatically included in the negative domain of k. In the subsequent analysis, we will mostly be considering solutions of the form Eq.(11). However, Eq.(15) is important because 4 3 WAVES CHANGING MEDIA M. WILLIAMS it is related to ﬁeld of mathematics called Fourier analysis which is built around the following theorem. Fourier’s Theorem: The two forms of the general solution to Eq.(10) (i.e., Eq.(11) and Eq.(15)) are actually equivalent. By a mathematical result called Fourier’s Theorem, a general function f(x) can be represented as f(x) = Z ∞ −∞ dk φ(k)e ikx , (16) where the function φ(k) is in turn given by φ(k) = 1 2π Z ∞ −∞ dx f(x)e −ikx . (17) Using Fourier’s theorem we can express the arbitrary functions e f1 and e f2 used in Eq.(11) as ˜ f1(s) = Z ∞ −∞ dk A+(k)e iks and ˜ f2(s) = Z ∞ −∞ dk A+(k)e iks , (18) where A+ and A− are deﬁned by equations similar to Eq.(17). With Eq.(16), we thus ﬁnd z(x, t) = e f1(x − vt) + e f2(x + vt) = Z ∞ −∞ dk h A+(k)e ik(x−vt) + A−(k)e ik(x+vt) i , (19) which establishes the equivalence between Eq.(11) and Eq.(15). 3 Waves changing media With Eq.(15), we have at last found a general mathematical description of waves traveling in free space, but as mentioned in the introduction, waves rarely travel unimpeded. Rather, waves often interact with their surrounding by changing media or reﬂecting o of surfaces. For example, mirrors work by reﬂecting light directly back at its source, and we can see through to the bottom of pools because light is propagating from the bottom of the pool through the water and then through the air to our eyes. Whenever the medium through which a wave is propagating changes, its properties change. In this section, we study these changes using a simple model of a propagating string. 3.1 Reﬂection and Transmission Figure 3: Two strings of di erent mass densities joined at x = 0. Figure from © AIP Publishing LLC. All rights reserved. This content is excluded from our Creative Commons license. For more information, see 5 3 WAVES CHANGING MEDIA M. WILLIAMS For the string system that we have been considering the ”medium” through which the wave is propagat ing is the string itself. The deﬁning quality of the string which deﬁnes the properties of propagation is its mass density µ. Thus to study how the properties of waves change as they cross media, we have to consider how waves change as they transition between strings with di erent mass densities4. We begin with the system shown in Fig. 3 where a string of mass density µ1 for x < 0 is connected at the point x = 0 to a string of mass density µ2 for x > 0. We assume that for all time, we have a continuos wave f0(x − vt) coming in from the left (and we know it is coming in from the left because f0 is a function of x − vt and not x + vt). We want to know what happens to the wave after it reaches the origin. We presume two things can happen: part of the wave will be transmitted to the string of mass density µ2 and part of the wave will be reﬂected back at its source. Namely, if we were to write the equation representing the total wave for x < 0 and x > 0 we would have ( f0(x − v1t) + fR(x + v1t) for x < 0 y(x, t) = (20) fT (x − v2t) for x > 0 where R stands for the reﬂected wave and T stands for transmitted wave. We use di erent velocities for p x < 0 and x > 0 because the velocity of a wave on a string is deﬁned by v = T/µ and the sections of the string on either side of x = 0 have di erent densities. We note that the transmitted wave is traveling to the right because it is a function of x − vt, but the reﬂected wave is traveling to the left because it is a function of x + vt. 3.2 Continuity and Zero Net-Transverse Force We will determine a relationship between f0, fR, and fT by using physical properties of the string. First, the string is continuous, so the total wave amplitude at x = 0 must be the same on both sides of the point connecting the two parts of the string. This means we must have y(0+, t) = y(0−, t), [Continuity condition] (21) where 0+ deﬁnes x = 0 found by approaching the point from the right and 0− deﬁnes x = 0 found by approaching the point from the left. In terms of our general solution for , this continuity condition becomes f0(−v1t) + fR(v1t) = fT (−v2t). (22) Next, any point on the string is essentially massless, so the total force exerted on that point must be zero. This must be true of the point joining the x < 0 and x > 0 regions of the string as well. For the string with transverse oscillations, the force arises from the tension T and the slope of the string. We can deﬁne this slope by the angle θ the string makes with the horizontal. We will assume θ is small which is true for low amplitude waves. When a section of string is at an angle θ with the horizontal, the force which is being exerted on it to produce this angle has horizontal and vertical components. For a string of tension T , the horizontal component of the force at the point x = 0 coming from left part of the string is Fx(x = 0) = −T cos θ ' −T. (23) We use a negative sign because the tension from the left part of the string is pulling the point at the origin toward the x < 0 direction. Similarly, the vertical component of the force coming from the left part of the string is ∂y Fy(x = 0) = −T sin θ ' Tθ ' T tan θ = −T , (24) ∂x 4More generally, we could also consider di erent tensions, but for simplicity we will take the tensions to be equal on both sides of the string. 6 3 WAVES CHANGING MEDIA M. WILLIAMS where we used the small-angle approximation of sin θ and tan θ. Here, the negative sign arises because if the slope of the string at the origin is positive then the left part of the string must be pulling down on the point at the origin. The requirement that the net-force on the part of the string at x = 0 is zero, amounts to requiring that the sum of F for x < 0 and F for x > 0 at x = 0 is zero. Given Eq.(23), this condition is automatically satisﬁed for the horizontal forces acting within the string. Namely we have Fnet,x(x = 0) = lim Fx + lim Fx ' −T + T = 0. (25) x→0+ x→0− where we use the notation x → 0+ or x → 0− to signify approaching x = 0 from the right or left, respectively. Conversely, we also require that the net-force in the y-direction is zero, and so we have ∂y ∂y Fnet,y (x = 0) = lim Fy + lim Fy ' − lim T + lim T = 0. (26) x→0+ x→0− x→0+ ∂x x→0− ∂x Therefore, for our string of constant tension, this condition of zero net-force in turn implies ∂y ∂y T = T . [Zero net-transverse force condition] (27) ∂x ∂x 0+ 0− Given that the string tension is the same on both sides of the boundary, written in terms of the general solution Eq.(3.1), this result becomes f0 0 (−v1t) + fR 0 (v1t) = fT 0 (−v2t) (28) In summary, the two conditions that the string displacement at an interface (placed at x = 0) must satisfy are 1. Continuity: The string must be continuous across the interface (Eq.(21)). This means the displacement y(x, t) when approaching the interface from the left must be the same as that when approaching the interface from the right. 2. Zero net-transverse force: The net-vertical force at the interface must be zero (Eq.(27)). This is to ensure there is no net-force applied to the inﬁnitesimal mass at the interface; such a net force would yield an inﬁnite acceleration. 3.3 Reﬂection and Transmission Coeÿcients Eq.(22) and Eq.(28) are convenient starting points for deriving properties relating the incident wavefront f0 to the reﬂected and transmitted wavefronts fR and fT . We can go even further by positing a standard form for these waves. For simplicity, we will return to the complex exponential form of the general sinusoidal solution 5. Our traveling wave solutions are then ik1u+ +B0e −ik1u+ ik1 u− +BRe −ik1u− ik2 u+ +AT e −ik2u+ f e 0(u+) = A0e , f e R(u−) = ARe , f e T (u+) = AT e , (29) where u+ = x − vt and u− = x + vt and where we use k1 and k2 to refer to the wave numbers of the µ1 string and µ2 string, respectively. Given the complex exponential analog of Eq.(29), we ﬁnd that the total wave for x < 0 is ik1(x−v1t) −ik1 (x−v1t) ik1(x+v1t) ik1(x+v1t) z(x, t) = A0e + B0e + ARe + BRe [For x < 0] (30) 5To be more precise we should have performed this analysis with the function z(x, t), but this measure of sloppiness does not a ect the ﬁnal results. If we ﬁnd any complex quantities in a ﬁnal result, we will simply take the real part and assume we implemented the procedure correctly from the beginning. 7 3 WAVES CHANGING MEDIA M. WILLIAMS and the total wave for x > 0 is ik2(x+v2t) ik2(x+v2t) z(x, t) = AT e + BT e [For x > 0]. (31) We note that because string 1 and 2 di er in their density, the parameters in the system which are dependent on length scale are di erent for the two sections of the strings. Namely, any quantity which includes units of distance is di erent when it is deﬁned on the leftside versus on the rightside of Fig. 3. However, any quantity with out any units of distance is the same across the string. This means that although k and v (with units of m−1 and m/s, respectively) are not the same across the two sections of the string, ω = kv (with units of rad/sec) is the same. Therefore, Eq.(30) and Eq.(31) becomes i(k1x−ωt) −i(k1x−ωt) i(k1x+ωt) −i(k1x+ωt) z(x, t) = A0e + B0e + ARe + BRe [For x < 0] (32) or i(k2x−ωt) −i(k2x−ωt) z(x, t) = AT e + BT e [For x > 0]. (33) Now, imposing Eq.(21) on these forms of z(x, t), we ﬁnd −iωt iωt −iωt iωt iωt −iωt AT e + BT e = A0e + B0e + ARe + BRe iωt = (A0 + BR)e −iωt + (B0 + AR)e , (34) which given the linear independence of eiωt and e−iωt, leads to the two equations AT = A0 + BR, BT = B0 + AR. (35) Similarly, imposing Eq.(27) on these forms, gives us −iωt − ik2BT e −iωt −iωt − ik1B0e iωt iωt − ik1BRe −iωt ik2AT e = ik1A0e + ik1ARe iωt 0 = (k1A0 − k1BR)e iωt + (−k1B0 + k1AR)e , (36) which then yields the other two equations k2AT = k1(A0 − BR), −k2BT = k1(−B0 + AR). (37) Together Eq.(35) and Eq.(37) give four equations which can be broken up into two systems of two equations. The unknowns in the equations are the amplitudes of the reﬂected and transmitted waves. Solving these systems is a matter of basic algebra and their solutions mirror one another. So we will write the solutions for a single system and ignore the other under the assumption that it can be easily found through a similar procedure. Presuming we know the initial wave amplitude A0, the amplitude of the the reﬂected wave and the transmitted wave are (from solving the system given by the left equations in Eq.(35) and Eq.(37)) 2k1 k1 − k2 AT = A0, BR = A0. (38) k1 + k2 k1 + k2 It will prove more useful to write this result in terms of the mass density µ of each string. Dividing Eq.(38) by ω (which is the same on both sides of the string), and using k/ω = v−1, we ﬁnd −1 −1 −1 v v − v 1 1 2 AT = A0, BR = A0, (39) −1 −1 −1 −1 v + v v + v 1 2 1 2 8 3 WAVES CHANGING MEDIA M. WILLIAMS Figure 4: µ2 = µ1 Figure 5: µ2 →∞ Figure 6: µ2 → 0 p and then using the deﬁnition of velocity as v = T/µ (with µ di erent on both sides of the string, but T the same by imposition) we ﬁnd √ √ √ 2 µ1 µ1 − µ2 AT = √ √ A0, BR = √ √ A0. (40) µ1 + µ2 µ1 + µ2 These are the ﬁnal expressions relating the initial wave amplitude A0 to the reﬂected amplitude BR and the transmitted amplitude AT . We would ﬁnd identical expressions relating B0, AR and BT if we solved the system given by the right equations in Eq.(35) and Eq.(37). Most often, we deﬁne the coeÿcients of Eq.(40) as reﬂection and transmission coeÿcients √ √ √ 2 µ1 µ1 − µ2 T = √ √ , R = √ √ (41) µ1 + µ2 µ1 + µ2 Because these coeÿcients are real, they apply equally well when deﬁning the relationships between the initial, reﬂected, or transmitted amplitude for the real traveling waves of the form Eq.(8). What properties do these equations lend to real traveling waves? We can best understand this through limiting cases concerning how the mass density of the string changes across the origin. – µ2 = µ1 (Uniform String): If we have a uniform string across the origin, then we expect no wave to be reﬂected back and that all of the initial wave is transmitted from the left to the right. With Eq.(40) we ﬁnd exactly this, for with µ1 = µ2, Eq.(41) gives us T = 1 and R = 0, indicating no wave is reﬂected and the transmission amplitude matches the initial amplitude. (Fig. 4) – µ2 →∞ (Inﬁnitely heavy on right): If we have an inﬁnitely heavy string on the right, then this is tantamount to having the string ﬁxed at x = 0. Eq.(41) then tells us that no wave is transmitted and the wave which is reﬂected has negative the amplitude of the initial wave. (Fig. 5) More generally, we see that whenever µ2 > µ1, the amplitude of the reﬂected wave BR always has a negative sign relative to the incident wave A0. This means that if A0 is positive, BR would be negative and vice versa; Summarily, whenever µ2 > µ1 the incident wave experiences a phase change of π upon iπ reﬂection (because e = −1). Moreover, for µ2 > µ1, we \|R\| > \|T \| meaning that when transitioning to a denser string, the amplitude of the reﬂected wave is greater than the amplitude of the transmitted wave. This makes sense seeing as it would be harder for a wave to propagate through a heavier string. – µ2 → 0 (Inﬁnitely light on right): If we have an inﬁnitely light string on the right, then this is tan tamount to having no string after x = 0. Eq.(41) tells us the transmitted amplitude is twice that of the initial amplitude, but this is ﬁctitious because if µ2 = 0 then there is no string through which this amplitude can propagate. Thus all of the (real) wave returns to its source and we have R = 1. (Fig. 6) 9 3 WAVES CHANGING MEDIA M. WILLIAMS More generally, for µ2 < µ1, we ﬁnd \|R\| < \|T \| indicating that when transitioning to a less dense string, the amplitude of the transmitted wave is greater than the amplitude of the reﬂected wave. This makes sense seeing as it would be easier for a wave to propagate through a lighter string. 3.4 Wave Attenuation In our derivation of the wave equation, we analyzed the properties of coupled oscillators under the assump tion that each oscillator experienced the energy-conserving spring force deﬁnitive of Hooke’s law. This was also our basic assumption when we analyzed simple harmonic motion, but to bring our models closer to reality, we later incorporated the e ects of air drag given the knowledge that mechanical energy is rarely conserved in actual systems. Similarly, to describe real waves we would need to incorporate such energy-dissipating e ects into our derivation of the wave equation. For the case of coupled masses oscillating in the transverse direction (i.e., a direction perpendicular to the axis which deﬁned their couplings), our previous equation of motion for the mass in the jth position was my ¨ j = k(yj+1 − yj ) − k(yj − yj−1). [Energy-conserving system]. (42) If we were to incorporate air-drag into this equation, we would need to include the velocity dependent drag force −by ˙j . Doing so, gives us the equation of motion my ¨ j = −by ˙j + k(yj+1 − yj ) − k(yj − yj−1). (43) Mirroring our previous derivation of the wave equation, we can take our lattice spacing a to 0 and deﬁne macroscopic quantities like the linear mass density µ = lima→0 m/a and the average string tension T = lima→0 ka. We would also have to deﬁne a drag coeÿcient for the string itself as β ≡ lim b . (44) a→0 a The resulting wave equation would be ∂2y ∂y ∂2y 2 + 2λ = v , (45) ∂t2 ∂t ∂x2 p where v = T/µ and 2λ = β/µ. To solve this wave equation, we can use the fact that the di erential equation is ”linear and homogeneous with constant coeÿcients” to apply methods similar to those used to solve the other equations of motion with similar properties. First we rewrite this equation in terms of the complex variable z(x, t) (where y(x, t) = Re[z(x, t)]). ∂2z ∂z 2 ∂2z + 2λ = v , (46) ∂t2 ∂t ∂x2 Given its properties, we can assume that the solution to this equation is an exponential (representing a wave traveling to the right) of the form z(x, t) = Aei(kx−ωt), (47) where the relationship between k and ω has yet to be determined. Inserting this solution into Eq.(46), we ﬁnd − ω2 − i2λω = −v 2k2 , (48) which is the desired relationship between k and ω. Given Eq.(48), there can be various types of dissipative behaviors in our system and these behaviors are determined by our system’s boundary conditions. As a speciﬁc example, we follow the case outlined at the end of dj morin/waves/transverse.pdf. Imagine that at x = 0, the string is oscillating with a wave amplitude Ae−iωt which persists for all time. Since there is no decay in the amplitude we know ω must be exclusively real and 10 4 SOUND WAVES M. WILLIAMS Figure 7: Wave amplitude decreasing after entering a dissipative medium. Figure from mpozek/research.html Courtesy of Miroslav Požek. Used with permission. thus Eq.(48) must deﬁne a wave number k which is complex. Solving for this wave number we ﬁnd r ω 2 k = −2iλ ≡ k0 + iK, (49) v where we used the fact that √ a + ib is a complex number to replace it with a real part k0 and an imaginary part K6. We thus ﬁnd that Eq.(47) becomes i(k0x−ωt) z(x, t) = Ae−Kx e . (50) This is a single solution, the other solution could be found through a similar procedure by beginning with y(x, t) = Be−i(kx−ωt). We would then ﬁnd the general solution is i(k0x−ωt) −i(k0 −ωt) z(x, t) = Ae−Kx e + Be−Kx e . (51) Computing the real part of this quantity gives us the physical solution for this situation: y(x, t) = Re[z(x, t)] = A0e −Kx cos(k0x − ωt − φ), (52) where A0 and φ are phases set by the initial conditions of this system. Given what we know about how exponentials change the amplitude of waves, we should easily be able to visualize Eq.(52). It would consist of a wave decaying exponentially as a function of distance from the origin (See Fig. 7). Compared to our previous traveling wave solutions, Eq.(52) better represents how real waves propagate because such waves are usually moving through media which dissipate the wave’s initial energy. 4 Sound Waves We have so far been discussing the properties of waves rather generally using the waves on a string as an example. What about the descriptions of the sound and light waves which served as the introduction to this chapter? In this ﬁnal section, we will brieﬂy discuss sound waves, leaving light (or electromagnetic waves) to the next lesson. Sound waves (like waves on a string and unlike electromagnetic waves) comprise propagation of a dis turbance through a medium. For sound waves the medium consists of a gas, liquid, or a solid, and the disturbances are longitudinal oscillations (i.e., oscillations along the direction of wave propagation) in the positions of the molecules which deﬁne the medium. We present the wave equation for sound waves, with √ 6To solve for it explicitly we could use the identity a + ib = ρ(cosh η + i sinh η) where ρ2 = a and 2η = sinh−1(b/2a2). 11 REFERENCES M. WILLIAMS out derivation to highlight its similarity to the wave equations we’ve studied so far (derivations can be found in most of the references on the website). We will only consider (again for simplicity) sound waves propa gating in a single direction. Say we have a region of space consisting of a gas of molecules of mass density ρ. If the total pressure in that region of space is Ptot, we can separate Ptot into a base pressure P0 and a pressure ﬂuctuation p, the latter of which leads to what we experience as a sound wave: Ptot = P0 + p. (53) Then considering the properties of the longitudinal oscillation and some results from thermal physics, we ﬁnd that the wave equation for p is ∂2p γP0 ∂2p = , (54) ∂t2 ρ ∂x2 where γ is a quantity which depends on the atomic composition of a molecule of the gas; for diatomic gases (like N2 and O2 which constitute much of air), γ = 7/5. Given our previous wave equation, we can infer that the speed of the wave propagation for pressure waves is s c = γP0 . (55) ρ Given that, for air, γ ≈ 7/5, ρ = 1.275 kg/cm3, and P0 = 1 atm, we ﬁnd that the numerical value of the speed of the pressure wave is c ≈ 330 m/s, (56) which is indeed the speed of sound. References H. Georgi and A. P. French, “The physics of waves,” 1993. 12 MIT OpenCourseWare Resource: Introduction to Oscillations and Waves Mobolaji Williams For information about citing these materials or our Terms of Use, visit:
10349	https://www.youtube.com/watch?v=vQjg4y54Oac	Complex Numbers: Argument and Representation \| 23/27 \| UPV Universitat Politècnica de València - UPV 384000 subscribers 1 likes Description 140 views Posted: 7 Dec 2023 Título: Complex Numbers: Argument and Representation Descripción: In this unit, the argument of a complex number is defined, and the representation in the complex plane is reviewed. Moll López, SE. (2013). Complex Numbers: Argument and Representation. Descripción automática: In this video, the presenter introduces a new unit on complex numbers, focusing on their arguments and representation on the complex plane. The prerequisites mentioned for understanding this unit include knowledge of complex number basics, operations, modulus, and trigonometric functions, with a preference for using radians over degrees. The video discusses how to represent complex numbers like "a + bi" on a graph by placing the real part "a" on the x-axis and the imaginary part "b" on the y-axis. The x-axis is deemed the real axis, while the y-axis is the imaginary axis. The term 'affix' is introduced to describe the corresponding point of a complex number on the plane, and the 'modulus' represents the distance from this point to the origin. The concept of the argument of a complex number is then explained as the angle that the complex number (or its affix) forms with the positive real axis. This argument, also referred to as 'alpha' in the presentation, can have infinitely many values by adding multiples of 2π to the initial angle. The formula using arctangent to calculate the argument based on the real and imaginary components is derived using the definitions of the trigonometric functions sine and cosine. The instructor emphasizes understanding which quadrant of the plane the complex number falls in to accurately calculate the argument, noting the signs of the real and imaginary parts for each quadrant. Throughout the discussion, the presenter provides graphical representations and quadrant-wise examples, and closes by mentioning that future units will offer examples and additional tools for calculating and understanding complex numbers. Autor/a: Moll López Santiago Emmanuel Curso: Este vídeo es el 23/27 del curso MOOC Math Fundamentals: Numbers and Terminology. Inscríbete en: Universitat Politècnica de València UPV: Más vídeos en: Accede a nuestros MOOC: Complex Numbers #Argument #APPLIED MATHEMATICS Transcript: hello welcome to a new unit on complex numbers in this unit we are going to try to Define what is the argument of a complex number and the representation of complex numbers in the plane the minimum requirements are a little more extensive than previously we are accumulating Concepts they are to know what a complex number is to operate with complex numbers modulus of a complex number and know how to handle the basic trigonometric functions such as s cine and tangent we will work mainly in radians although we will usually say the equivalence degrees well suppose we want to represent a complex number A plus b i what we are going to do is as a complex number is separated into the real part and the imaginary part we are going to represent the real part in one of the axis and the imaginary part in another of the axes in these axes in particular notice that we have called the x-axis the real axis and the y- axis the imaginary axis how would we represent the complex number very simple we take the real part and represent it on the real axis and the imaginary part vertically parall parallel to the imaginary axis so this is the complex number A plus b i represented on the plane well what things we need to know the previous plane is called the complex plane to each complex number corresponds a point on the plane which we will call an AIX the modulus of a complex number corresponds to the distance from the complex to the origin or from the ax to the origin the obsa axis will be called the real axis and the ordinate axis will be called the imaginary axis the purpose of this unit is that every complex forms an angle with respect to the real axis from the positive part of the real axis to the AIX well this that's going to be the definition of argument we'll see it right away look at the graphical representation here we have the real part the imaginary part and the graphical representation of the complex number A plus b i if you notice above the hypotenuse you have the absolute value of Z that is the distance that goes from the origin to the AIX a + b i and Alpha in green is the angle that the complex number forms with respect to the real axis well the more strict definition would be it is the angle formed by the positive semi-axis of the real axis and the vector defined by the origin of coordinates and the AIX of Z this is exactly what we have seen in the previous representation it is usually denoted as argument of Z or in this case in the presentation we will also call it Alpha well keep in mind that there is not only one argument there are infinitely many values of the argument of Z once we have obtained the angle at forms we can add 2 pi as many times as necessary to obtain the same complex number well the formula to obtain a complex number is the arct tangent of B divided by a where B is the imaginary part of the complex number and a is the real part of the complex number it is also very important to note which quadrant the Z FX is in both are very important to calculate as we will see below notice where does the formula come from a fairly quick explanation if if we have the graphical representation of the complex number A plus b i if we look at the situation of alpha we have that the S of alpha is the opposite leg which is B divided by the hypotenuse which is the modulus of Z the cine of alpha is defined as the adjoining side which would be a divided by the hypotenuse which is the modulus of Z therefore if we apply the definition of the tangent which is the S of alpha divid by the coine of alpha we divide B by the modulus of alpha ID B divided by the modulus of Z ided A divided modulus of Z and we are left with B / a if we apply AR tangent in both parts of the equality or clear Alpha what we have is that Alpha the argument of the Z number is the arctangent of B divided a it is very important to know in which quadrant the complex number is located here we have a representation of the quadrants the first quadrant is when the real part and the imaginary part are positive the second quadrant is when the real part is negative and the imaginary part is positive the third quadrant is when both the real and imaginary parts are negative and the fourth quadrant is when the real part is positive but the imaginary part is negative if we have a complex number that is in the first quadrant that is the real part and the imaginary part are positive its representation is like this as you can see the angle Alpha is an acute angle and is in the first quadrant if the complex number is in the second quadrant that is the real part is negative what we get is a representation of this form you see that the angle is obtuse and comes to meet the AIX which is in the second quadrant if it is in the third quadrant we have exactly the same we measure the angle from the real axis until we come to touch the complex number which in this case has the real part and the imaginary part negative we could also have measured the angle in negative that is from the real axis down and say that the angle would be in this case for example- pi4 or - 5 pi4 this way of measuring it is to try that the angle is always between minus pi and 0 in the fourth quadrant we have have the same thing if we want to measure the angle we measure from the real axis and go through the whole angle until we reach the AIX up to the complex number x - y i there in this case the imaginary part is negative we can also measure the other way around measure from the real axis down and give the negative value in this case for example the representation would be minus Pi quter that would be the negative meter so far we have seen the definition of argument of a complex number and we have tried to represent the complex numbers in the complex explain so we have called it well in the next units we will see examples of how to calculate these complex numbers and so we will have a little bit more tools when it comes to calculating them so far this unit thank you
10350	https://www.hmhco.com/blog/teaching-ratios-and-unit-rates-in-math?srsltid=AfmBOoqtM1VbvR3c20LpjXYRQVVhcV4GzL6XZjzkfBJ6DgYPral4nbjQ	Teaching Ratios & Unit Rates in Math \| HMH Curriculum Literacy Core Curriculum Into Literature, 6-12 Into Reading, K-6 See all Literacy Performance Suite Intervention English 3D, K-12 Read 180, 3-12 See all Reading Intervention Assessment Supplemental A Chance in the World SEL, 8-12 Amira Learning, PreK–8 Classcraft, K-8 JillE Literacy, K-3 Waggle, K-8 Writable, 3-12 See all Reading Supplemental Personalized Path Math Core Curriculum Arriba las Matematicas, K-8 Go Math!, K-6 Into Algebra 1, Geometry, Algebra 2, 8-12 Into Math, K-8 Math in Focus, K-8 See all Math Performance Suite Supplemental Classcraft, K-8 Waggle, K-8 See all Supplemental Math Personalized Path Intervention Math 180, 3-12 See all Math Intervention See all Assessment Science Core Curriculum Into Science, K-5 Into Science, 6-8 Science Dimensions, K-12 See all Science Readers ScienceSaurus, K-8 Social Studies Core Curriculum HMH Social Studies, 6-12 See all Social Studies Supplemental Writable Professional Development For Teachers Coachly Teacher's Corner Live Online Courses Program-Aligned Courses See all Professional Development For Leaders The Center for Model Schools More AI Tools on HMH Ed Assessment Early Learning English Language Development Homeschool Intervention Literacy Mathematics Professional Development Science School Improvement Social Studies Special Education Summer School See all Solutions Resources Browse Resources AI Tools Classroom Activities Customer Resource Hub Customer Success Stories Digital Samples Events Grants & Funding Family & Caregiver Resources International Product Updates Research Library Shaped, HMH Blog Webinars Customer Support Contact Sales Customer Service & Technical Support Portal Platform Login Company Learn about us About Corporate Responsibility Leadership News Announcements Our Culture Our Legacy Join us Careers Educator Input Panel Suppliers Divisions Center for Model Schools Heinemann NWEA Shop Support Platform Login From setup to support, HMH will ensure your back-to-school season runs smoothly. 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The numbers or measurements being compared are sometimes called the terms of the ratio. For example, if a store sells 6 red shirts and 8 green shirts, the ratio of red to green shirts is 6 to 8. You can write this ratio as 6 red/8 green, 6 red:8 green—or when writing fast or trying to make a point—simply 6/8 or 6:8. Both expressions mean that there are 6 red shirts “for every” 8 green shirts. Notice how you can rewrite 6/8 as 3/4, no different from any other time a math concept can appear as a fraction. A rate is a special ratio in which the two terms are in different units. For example, if a 12-ounce can of corn costs 69¢, the rate is 69¢ for 12 ounces. This is not a ratio of two like units, such as shirts. This is a ratio of two unlike units: cents and ounces. The first term of the ratio (69¢) is measured in cents, and the second term (12) in ounces. You can write this rate as 69¢/12 ounces or 69¢:12 ounces. Both expressions mean that you pay 69¢ “for every” 12 ounces of corn, and similar to the shirt ratio, can enter calculations as the fraction 69/12. But notice that this time, a new unit is created: cents per ounce. Rates are used by people every day, such as when they work 40 hours per week or earn interest every year at a bank. When rates are expressed as a quantity of 1, such as 2 feet per second (that is, per 1 second) or 5 miles per hour (that is, per 1 hour), they can be defined as unit rates. You can write any rate as a unit rate by reducing the fraction so it has a 1 as the denominator or second term. As a unit rate example, you can show that the unit rate of 120 students for every 3 buses is 40 students per bus. 120/3 = 40/1 You could also find the unit rate by dividing the first term of the ratio by the second term. 120 ÷ 3 = 40 When a price is expressed as a quantity of 1, such as $25 per ticket or $0.89 per can, it is called a unit price. If you have a non-unit price, such as $5.50 for 5 pounds of potatoes, and want to find the unit price, divide the terms of the ratio. $5.50 ÷ 5 pounds = $1.10 per pound The unit price of potatoes that cost $5.50 for 5 pounds is $1.10 per pound. Rates in the Real World Rate and unit rate are used to solve many real-world problems. Look at the following problem. “Tonya works 60 hours every 3 weeks. At that rate, how many hours will she work in 12 weeks?” The problem tells you that Tonya works at the rate of 60 hours every 3 weeks. To find the number of hours she will work in 12 weeks, write a ratio equal to 60/3 that has a second term of 12. 60/3 = ?/1260/3 = 240/12 Removing the units makes the calculation easier to see. However, it is important to remember the units when interpreting the new ratio. Tonya will work 240 hours in 12 weeks. You could also solve this problem by first finding the unit rate and multiplying it by 12. 60/3 = 20/1 20 × 12 = 240 When you find equal ratios, it is important to remember that if you multiply or divide one term of a ratio by a number, then you need to multiply or divide the other term by that same number. Let's take a look at a problem that involves unit price. “A sign in a store says 3 Pens for $2.70. How much would 10 pens cost?” To solve the problem, find the unit price of the pens, then multiply by 10. $2.70 ÷ 3 pens = $0.90 per pen $0.90×10 pens = $9.00 Finding the cost of one unit enables you to find the cost of any number of units. What Is a Unit Rate in Math? Your students have no doubt encountered rates and ratios before (have they seen a speed limit sign?), but it may help them to review these concepts before solving problems that use them. Standard:Understand the concept of a unit rate a/b associated with a ratio a:b with b ≠ 0. (6.RP.A.2) Prerequisite Skills and Concepts: Students should have a basic understanding of ratios, how to write them, and an ability to simplify a ratio. Students should also have an ability to work with fractions and find equivalent fractions. Say:Today we are going to look at a special type of ratio called a rate. Does anyone know what I mean by a rate? Students may say that a rate is a ratio in which the quantities being compared use different units such as dollars and ounces or miles and hours. Students may use common English synonyms for rate such as speed. If so, point out that speed means calculating how fast something is going by comparing distance to time, such as miles to hours. If necessary, explain what a rate is. Say:Rates are commonly found in everyday life. The prices in grocery stores and department stores are often rates. Rates are also used in pricing gasoline or tickets, measuring speed, or paying hourly wages and monthly fees. Have students think of other examples of rates. In addition to common real-world examples, encourage silly or unusual rates such as hip-hop artists per zip code or diamond collars per chihuahua. Say:Two important ideas are unit rates and unit prices. What is the difference between rate and unit rate? Or price and unit price? Does anybody have any ideas? Students will probably not know what a unit rate is, so provide them with the following explanation to explain rate vs. unit rate. Say:A unit means one of something. A unit rate means a rate for one of something. We write this as a ratio with a denominator of one. For example, if you ran 70 yards in 10 seconds, you ran on average 7 yards in 1 second. Both of the ratios, 70 yards in 10 seconds and 7 yards in 1 second, are rates, but the 7 yards in 1 second is a unit rate. Ask:Now that you know what a unit rate is, what do you think a unit price is? Students will say that it is the price of one item. If they don't, tell them what it is. Ask:What is the unit price of 10 pounds of potatoes that costs $2.80? Help students calculate that the unit price is $0.28 per pound by dividing the price by 10. Share the following problem: “One store has carrots on sale for $1.14 for 3 pounds, while another store has carrots on sale for $0.78 for two pounds. Which store has the better deal?” Ask:What are we trying to find in this problem? Students should say that we are trying to find out which is the better deal for carrots when thinking about the cost of each carrot. Ask:What would help us find the better deal? Students should say that if we find the unit price for the carrots at each store, we would know which was the better deal. Say:Find the unit prices for the carrots at both stores and then we will discuss what you did. Have students independently calculate the unit price and answer which store has the better deal. Compare how different students did the calculations, and have students discuss similarities and differences among the models they used and solutions they found. Allow for varying answers such as “the second store had the better deal for me because I would only want two carrots anyway.” If time permits, have students solve the following problem as well. “One animal can run 60 feet in 4 seconds, while another animal can run 100 feet in 8 seconds. Which animal runs faster?” (The first animal runs faster at 15 feet per second.) Developing the Concept: Rates Now that students know how to find a unit rate, they will learn how to find an equivalent ratio using unit rates. Finding equivalent ratios uses the same thought process as finding equivalent fractions. Standard:Use ratio and rate reasoning to find equivalent ratios and solve real-world problems (6.RP.A.3) Say:Before we learned how to find a unit rate. Now we are going to learn how to use that unit rate to solve problems. Look at this problem. Share the following problem: “Yesterday Ebony ran 18 laps around the track in 12 minutes. If she runs at that rate for 30 laps, how long will it take her?” (Tip: You can substitute the context with anything that would interest your students.) Ask:What are we trying to find in this problem? We are trying to find out how long it takes Ebony to run 30 laps. Ask:What information do we know that will help us solve this problem? We know that Ebony can run 18 laps in 12 minutes. We also know she is going to run at that same rate for 30 laps. Ask:How far does Ebony run in one minute? Have students try to figure it out individually. Compare student solutions, and discuss why Ebony runs 1.5 laps in one minute. Say:Let's make a table to list the information we know. Make the following table but leave "Minutes" blank. Fill it in by soliciting class input. LapsMinutes 1.5 1 3 2 6 4 12 8 18 12 24 16 30 20 Ask:Does anyone see another way we could have found the time of 20 minutes without writing down the whole table? Compare the different ideas students offer. Some students might recognize that if they divided 30 by 1.5, they would get 20 minutes. Discuss this strategy with the class. Write the following problem on the board: “Maria wants to buy a pencil for everyone in her class. It costs $0.78 for 3 pencils. How much would Maria have to spend if she bought a pencil for each of her 24 classmates?” Say:I'd like you to solve this problem on your own, and then we'll discuss what you did. Have students individually share their solutions. Some students may have solved it by finding the unit price or completing a table. Others may have solved it by noticing that 3 pencils cost $0.78, and 24 = 3 × 8. So they multiplied $0.78 by 8 to get $6.24. Looking for more free math lessons and activities for elementary school students? 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10351	https://math.stackexchange.com/questions/4259262/angle-chasing-to-show-three-points-are-collinear	geometry - Angle chasing to show three points are collinear. - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Angle chasing to show three points are collinear. Ask Question Asked 4 years ago Modified4 years ago Viewed 616 times This question shows research effort; it is useful and clear 4 Save this question. Show activity on this post. Let A B C A B C be an acute triangle with circumcenter O O and let K K be such that K A K A is tangent to the circumcircle of △A B C△A B C and ∠K C B=90∘∠K C B=90∘. Point D D lies on B C B C such that K D\|\|A B.K D\|\|A B. Show that D O D O passes through A.A. This is a problem from EGMO, eg.1.32. My approach: I created a dummy point D' such that D'O passes through A. Now I just have to show that BA and D'A are \|\| which will solve the problem. I think I have all the necessary theorems and I think this problem can be solved with angle chasing, but I can't really figure out how to proceed. Can you please guide me? geometry Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications edited Sep 25, 2021 at 14:49 Sathvik 3,826 1 1 gold badge 12 12 silver badges 26 26 bronze badges asked Sep 24, 2021 at 16:21 StackpackedKarStackpackedKar 300 1 1 silver badge 11 11 bronze badges 0 Add a comment\| 2 Answers 2 Sorted by: Reset to default This answer is useful 3 Save this answer. Show activity on this post. We shall try to work backwards in order to reach the desired conclusion. Let △A B C△A B C be an acute triangle with circumcenter O O and let K K be such that K A K A is tangent to the circumcircle of △A B C△A B C and ∠K C B=90∘.∠K C B=90∘. Define L L to be any point on line K A K A such that A A lies between K K and L L. Also, extend segment A O A O to point X X on B C B C. Using Alternate Segment Theorem, ∠B A L=A C B=α⟹∠A O B=2 α⟹∠O A B=∠B A X=90−α.∠B A L=A C B=α⟹∠A O B=2 α⟹∠O A B=∠B A X=90−α. Since, ∠X A L=∠X C K=90∘,A K C X∠X A L=∠X C K=90∘,A K C X is cyclic and hence, ∠A K X=∠A C X=α⟹A B∥K X.∠A K X=∠A C X=α⟹A B∥K X. But there exists a unique point D D on B C B C such that A B∥K D A B∥K D. Therefore, X≡D X≡D and A,O,D A,O,D are collinear. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Sep 24, 2021 at 17:09 SathvikSathvik 3,826 1 1 gold badge 12 12 silver badges 26 26 bronze badges Add a comment\| This answer is useful 3 Save this answer. Show activity on this post. It is sufficient to show that ∠C A O=∠C A D∠C A O=∠C A D. Let the tangent be labelled K A K′K A K′. Then due to the Alternate Segment Theorem, ∠K′A B=∠A C B∠K′A B=∠A C B And since A B∥K C A B∥K C, both of these equal ∠A K D∠A K D. The tangent is perpendicular to the radius so ∠B A O=90 o−∠K′A B∠B A O=90 o−∠K′A B. But from the data given, ∠K C A=90 o−∠A C B⟹∠K C A=∠B A O∠K C A=90 o−∠A C B⟹∠K C A=∠B A O Considering the total angle in triangles A B C A B C and A K C A K C, the remaining angle in each is the same, so ∠O A C=∠D K C∠O A C=∠D K C However, points A K C A K C and D D are concyclic since ∠A K D=∠A C D∠A K D=∠A C D (due to Angles in the Same Segment are Equal). Therefore, for the same reason, ∠D A C=∠D K C∠D A C=∠D K C. Hence, finally, ∠C A O=∠C A D∠C A O=∠C A D QED Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Sep 24, 2021 at 17:28 David QuinnDavid Quinn 35.1k 3 3 gold badges 25 25 silver badges 52 52 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 geometry See similar questions with these tags. 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10352	https://www.soa.org/4ab88e/globalassets/assets/files/resources/research-report/2022/understanding-the-connection.pdf	Understanding the Connection between Real-World and Risk-Neutral Scenario Generators AUGUST \| 2022 2 Copyright © 2022 Society of Actuaries Research Institute Understanding the Connection between Real-World and Risk-Neutral Scenario Generators AUTHOR Stephen J. Strommen, FSA, CERA, MAAA SPONSOR Quantitative Finance and Investment Curriculum Committee Caveat and Disclaimer The opinions expressed and conclusions reached by the authors are their own and do not represent any official position or opinion of the Society of Actuaries Research Institute, the Society of Actuaries or its members. The Society of Actuaries Research Institute makes no representation or warranty to the accuracy of the information. Copyright © 2022 by the Society of Actuaries Research Institute. All rights reserved. 3 Copyright © 2022 Society of Actuaries Research Institute CONTENTS Introduction.............................................................................................................................................................. 4 Section 1: Why are there two kinds of generators? (Fitness for purpose) .................................................................. 5 Section 2: What it means to be market-consistent .................................................................................................... 8 2.1 Reflecting expectations ......................................................................................................................................... 8 2.2 Common mathematical form of scenario generators .......................................................................................... 9 2.3 The common connection between real-world and risk-neutral calibrations ...................................................... 9 Section 3: Interest rate models ............................................................................................................................... 10 3.1 Market behavior in interest rate models ............................................................................................................ 11 3.2 The market price of risk in interest rate models ................................................................................................ 13 Section 4: Equity index models ............................................................................................................................... 16 4.1 Market behavior in equity index models ............................................................................................................ 16 Alternatives for the expected drift term ...................................................................................................... 16 Alternatives for the random shock term ...................................................................................................... 17 4.2 Market price of risk in equity index models ....................................................................................................... 17 Section 5: Calibration of market-consistent and market-coherent models .............................................................. 18 5.1 General approach ................................................................................................................................................ 18 5.2 Risk-neutral calibration ....................................................................................................................................... 18 5.3 Real-world calibration ......................................................................................................................................... 19 5.4 Combined calibration: a market-coherent real-world model ............................................................................ 20 Section 6: Acknowledgments .................................................................................................................................. 23 Appendix A: The P-measure and the Q-measure .................................................................................................... 24 Interest rates ............................................................................................................................................................. 24 Equity returns ............................................................................................................................................................ 27 Appendix B: Some more complex interest rate models .......................................................................................... 28 Double mean reverting models................................................................................................................................. 28 The three-factor Cox-Ingersoll-Ross model .............................................................................................................. 30 Appendix C: The meaning of arbitrage-free ............................................................................................................ 32 Appendix D: Term premiums and the market price of risk ...................................................................................... 33 Notes on calibration of term premiums ................................................................................................................... 35 Appendix E: Fair value, Principle-based Reserves, and the Valuation Manual ......................................................... 37 Obtaining scenario-specific values ............................................................................................................................ 37 Blending scenario-specific values into a single reported value ................................................................................ 38 Selecting a scenario generator that is fit for purpose .............................................................................................. 39 References .............................................................................................................................................................. 40 About The Society of Actuaries Research Institute .................................................................................................. 41 4 Copyright © 2022 Society of Actuaries Research Institute Understanding the Connection between Real-World and Risk-Neutral Scenario Generators Introduction Scenario generators are sometimes described as being as either “real-world” or “risk-neutral”. The difference between real-world and risk-neutral generators is in the treatment of the market price of risk. This document focuses on the difference in treatment of the market price of risk, and helps the reader understand the relationship between real-world and risk-neutral scenarios and when it is appropriate to use each type. Risk-neutral generators are referred to as “market-consistent” because they are calibrated to reproduce market prices of traded financial instruments. It is often assumed that real-world generators cannot reproduce market prices, but in fact the ability to reproduce market prices is a matter of calibration and a real-world generator can be calibrated to reproduce market prices. We introduce the term “market-coherent” to describe a real-world generator that is calibrated to reproduce market prices, explain how such calibration can be done, and discuss why it can be useful. The term “market-coherent” was chosen because such a generator not only reproduces current market prices but also reflects the real-world evolution of those prices into the future by using the P-measure (real-world probabilities) rather than the Q-measure (risk-neutral probabilities). To help illustrate how real-world and risk-neutral methods can be calibrated to produce the same market price, an Excel workbook has been prepared to accompany this paper. The workbook illustrates the valuation of a simple put option on equity stock using real-world and risk-neutral methods. The probability distributions associated with real-world and risk-neutral probabilities are shown side-by-side, and the means by which the methods use different probabilities to arrive at the same value are shown in detail. 5 Copyright © 2022 Society of Actuaries Research Institute Section 1: Why are there two kinds of generators? (Fitness for purpose) Before addressing the connection between real-world and risk-neutral scenario generators, it is important to understand why there are two different kinds of generators in the first place. The two kinds of generators were created to address two fundamentally different problems. That means they are fit for different purposes, and it is important when choosing a generator to choose one that is fit for the purpose at hand. Real-world generators were created to provide realistic stochastic simulation of the path of future economic conditions through time. Each path is a scenario, and such realistic but stochastically generated scenarios can be used in financial models to quantify risk. By running simulations using such scenarios, one can try to answer the questions “How bad could things get” and “How likely is that?” in the context of risk management for a financial institution. Therefore, real-world scenario generators are fit for the purpose of simulation across time. Risk-neutral generators were created for market-consistent valuation of options and other derivatives with an uncertain payoff. A market-consistent valuation method is one which reproduces the market price of market-traded instruments including options and derivatives. Not all such options and derivatives are market-traded and have observable market prices. When buying or selling a derivative that is not traded widely, it is very useful to determine its value by comparison with similar derivatives that are widely traded. It is useful to determine a “market-consistent” price or value. Risk-neutral methods, and generators based upon them, allow determination of a market-consistent value without making any assumption about the market price of risk. That is useful because the price of risk is not directly observable, but the market prices used to calibrate a risk-neutral generator are observable. Therefore, risk-neutral scenario generators are fit for the purpose of valuing an item with an uncertain payoff in a market-consistent way. The world would be much simpler if this distinction in purpose were always clear-cut. It is more complex when the problem at hand is to value an item with an uncertain payoff by means of stochastic simulation across time. In the realm of accounting for insurance contracts, that kind of valuation approach has been widely adopted. In that context the choice as to which kind of scenario generator is fit for the purpose requires understanding the difference between them at a deeper level. One key to understanding this difference is to consider the universe of all possible scenarios. In abstract terms, any scenario generator is a means of picking scenarios from that universe. The parameters of the generator essentially assign a probability of selection to each possible scenario. The probability of selection differs in a fundamental and systematic way between real-world and risk-neutral scenarios. Real-world scenarios use the P-measure, or real-world probabilities. Risk-neutral scenarios use the Q-measure, or risk-neutral probabilities. These different measures assign different probabilities to the same set of events in the future. The chart below illustrates this concept in a very simple case. Suppose the problem at hand involves projecting the future value of a share of equity stock one year from now. The universe of possible scenarios ranges from very low value to very high value. At the starting date of the projection, the real-world probabilities are for a mean expected value near the middle of that range. Real-world scenarios are probability weighted around that mean expected value. 6 Copyright © 2022 Society of Actuaries Research Institute Figure 1 DISTRIBUTION OF REAL-WORLD SCENARIOS What about risk-neutral scenarios? The next chart adds the probability distribution for a set of risk-neutral scenarios. Note that there is a lot of overlap between the risk-neutral and real-world distributions, indicating that many of the same scenarios may appear in both sets. Note also that the risk-neutral distribution is not centered on the same central value: it is centered on a lower value. This means that the average projected value of a risky investment in risk-neutral scenarios will be lower than in real-world scenarios. The probability weighting in risk-neutral scenarios (Q-measure) gives more weight to adverse results (lower projected value in this case) than the P-measure. For example, the central value in the risk-neutral probability weighting is based on the price increasing at the risk-free rate. That rate is typically lower than a realistic market expectation that would include a risk premium. Figure 2 DISTRIBUTION OF REAL-WORLD AND RISK-NEUTRAL SCENARIOS 7 Copyright © 2022 Society of Actuaries Research Institute In these charts the horizontal axis is a projected value. Keep in mind that the starting value is the same in both real-world and risk-neutral scenarios. And, importantly, the expected present value of the stochastic future price is the same in both the real-world and risk-neutral scenario sets. That’s because the discount rate differs between real-world and risk-neutral methods. The risk-neutral discount rate is always the risk-free short-term rate. The real-world discount rate is higher because it includes a risk premium for the equity price risk. That risk premium exists because real-world markets are risk-averse; an uncertain future price is discounted at a rate that includes a risk premium related to the degree of uncertainty in the future price. Therefore, projected values and prices are higher in real-world market-coherent scenarios than in risk-neutral scenarios, but the present values are the same. This means that projections of value based on risk-neutral scenarios diverge more and more from real-world projections the further into the future you go. Risk-neutral methods are clearly meant for market-consistent valuation on the scenario start date, where they match current real-world prices, but not for projections or simulations that involve decision-making based on simulated conditions on dates in the future. In risk-neutral scenarios, simulated future conditions do not have a realistic probability distribution. This has implications when using stochastic scenario methods for valuation of insurance contracts or pension obligations. When the valuation basis is market value, risk-neutral and real-world methods can (in theory) produce the same present value in cases where there are no future decisions to be made based on future market conditions. That would be the case for a block of insurance contracts or pensions backed by a true replicating portfolio. But in many cases simulation of obligations involves decisions to be made by either the administrator or the consumer based on future conditions1. These decisions have financial effects that are not considered in the calibration of risk-neutral scenarios because that calibration is based on other financial instruments. Therefore, use of the risk-neutral probabilities (Q-measure) to probability-weight the financial effects of such decisions is less than ideal because those are not the real probabilities. It would be ideal if the real-world probabilities (P-measure) could be used. Using real-world scenarios and the P-measure for this purpose requires that they be calibrated to reproduce market prices. Such a calibration requires an assumption concerning the market price of risk, which is not directly observable. Because of this, and since risk-neutral methods and the Q-measure do not require an assumption for the market price of risk, risk-neutral methods are commonly used. Nevertheless, the market price of risk can be inferred indirectly using some generally accepted models. With a reasonable assumption for the market price of risk, the P-measure can be calibrated to reproduce market prices. We will call real-world scenarios based on such a calibration “market-coherent” because they not only reproduce current market prices but also employ real-world probabilities (the P-measure) concerning the evolution of financial conditions in the future. Market-coherent scenarios can be used in valuation of the kind of obligations described above. What about real-world scenarios that are not market-coherent? Such generators can be useful for risk management or in a regulatory context where black swan events are an important consideration. The next chart illustrates one possible distribution for such a set of scenarios. In this case the distribution is wider with longer tails and would not reproduce market prices of many derivatives. For this kind of purpose, the tail scenarios are the focus of attention, and the exact center of the distribution is less important. 1 Such future decisions on the part of the insured may include whether to keep a contract in force because guarantees have become valuable or surrender it because guarantees no longer have much value. Decisions on the part of the insurer may include how to invest future renewal premiums. The distribution of future market prices and interest rates that would drive future investment decisions are not realistic in risk-neutral scenarios. 8 Copyright © 2022 Society of Actuaries Research Institute Figure 3 DISTRIBUTION OF REAL-WORLD SCENARIOS WITH EXTRA TAIL RISK Section 2: What it means to be market-consistent 2.1 REFLECTING EXPECTATIONS The term “market-consistent” means a valuation or a set of scenarios must be consistent with current market prices. Market prices for financial instruments reflect expectations of the future. For example, the market price of fixed income investments will be lower if it is expected that interest rates will rise and higher if it is expected that interest rates will fall in the future. The market price of equity investments like corporate stock depends on expectations regarding future earnings. The market price of options depends not only on the strike price but also on the expected volatility of market prices over the term of the contract because the probability of an out-of-the money option having value depends on the volatility. A calibration that reproduces current market prices focuses mainly on two aspects of market expectations: • the central path or direction of future prices • the volatility around that path; how far actual returns might vary from the expected path Calibration of a scenario generator to reproduce current market prices involves using observed market prices to quantify the central path and volatility around that path. Real-world and risk-neutral calibration methods infer different central paths and different volatilities. If real-world and risk-neutral scenarios were used in the same way to calculate a present value, they would produce different values. But real-world valuation methods and risk-neutral valuation methods use the scenarios differently, and that difference allows them to both reproduce the same market price. An Excel workbook accompanying this paper illustrates this idea using an example valuation of a put option on equity stock. 9 Copyright © 2022 Society of Actuaries Research Institute Since both real-world and risk-neutral methods can be calibrated to reproduce market prices, under the definition above they could both be categorized as market-consistent when calibrated that way. Common usage has come to associate the term “market-consistent” only with the risk-neutral approach. To avoid confusion, the term “market-coherent” can be used to refer to a real-world generator calibrated to reproduce market prices. Scenario generators also need to reflect understandings about how markets behave. These include the idea that markets are arbitrage-free and that market participants are risk-averse. Scenario generators can be designed and calibrated to embody those expectations as well. There are some popular forms of generators that are not arbitrage-free. These are commonly used for real-world simulation. Since they are not arbitrage-free they do not reproduce market prices. There is generally no risk-neutral version of such models, so one cannot explain the connection between risk-neutral and real-world calibrations and they will not be discussed further in this document. 2.2 COMMON MATHEMATICAL FORM OF SCENARIO GENERATORS The common mathematical form of a scenario generator is simple: (next period value) = (current value) + (expected drift) + (random shock) The expected drift is calibrated to the central expectation of the change in value. The random shock is calibrated to volatility in the value. 2.3 THE COMMON CONNECTION BETWEEN REAL-WORLD AND RISK-NEUTRAL CALIBRATIONS Consider the common mathematical form shown above. Taken together, the terms for the expected drift and the random shock define a probability distribution for next period’s value. A key to understanding the connection between real-world and risk-neutral models is that the distribution in a real-world calibration is different from the distribution in a risk-neutral calibration. In particular, the drift term is always different between real-world and risk-neutral calibrations. The risk-neutral approach to calibration is based on the idea that the market price of risk is zero and the expected drift in the price of all investments is the short-term risk-free rate. But since the real-world drift in price is higher for riskier investments, the lower expected drift in a risk-neutral calibration amounts to giving more probability weight to adverse events that result in lower future market prices. That is how a risk-neutral calibration reflects risk aversion; it effectively gives more probability weight to adverse future events. This is the key to understanding why the intended usage of risk-neutral scenarios is limited to valuation. Risk-neutral scenarios are unrealistic in that they embed an implicit margin for risk by giving more weight to adverse events – that is, events that lead to lower market prices. But that is exactly why they are useful for valuation purposes. Valuations using risk-neutral techniques include risk margins implicitly and avoid the need to add an explicit risk margin. That is why, when using risk-neutral scenarios, the estimated value is an average across all scenarios with no add-on margin, but when using real-world scenarios some sort of risk margin needs to be included. One common means of adding a risk margin in real-world valuations is to set the value to the contingent 10 Copyright © 2022 Society of Actuaries Research Institute tail expectation (CTE), the average of the subset of scenarios containing the most adverse results rather than the average across all scenarios.2 There are countless different implementations of the common mathematical form shown above. The implementation depends on the kind of value for which scenarios are to be generated, and on the kind of market behavior over time that one wants to simulate. The following sections illustrate some generator formulations for interest rates and for a stock price index that includes dividend re-investment. Differences between various formulations are explained in terms of the market behavior they simulate and the way the market price of risk is embedded. Section 3: Interest rate models Arbitrage-free interest rate models focus on the path of the short-term risk-free interest rate. The remainder of the risk-free yield curve at any time is defined as an expectation based upon the path of the short-term risk-free rate. The expectation operator is applied to the discounted value of a future payment using path-wise discounting. The expected discounted values are treated as spot prices, and those spot prices define spot rates that represent the yield curve. If the short-term risk-free rate at time t is 𝑟𝑡 then the spot price P for maturity t=T is the risk-neutral expectation3: 𝑃𝑇= 𝐸[𝑒𝑥𝑝(−∫𝑟𝑡𝑑𝑡 𝑇 0 )] The expression for the spot price given above applies to risk-neutral calibrations. Real-world calibrations simulate movement of the short-term rate differently and therefore are not consistent with this formula for the spot price. Why is the formula defining movement of the short-term rate different between real-world and risk-neutral models? The reason is the difference in purpose. The purpose of a real-world model is for simulation of the actual path of the short-term risk-free rate. The purpose of a risk-neutral model is valuation, so the model must produce forward paths of short-term rates that reproduce the prices of pure discount bonds based on the formula shown above. The forward path in a risk-neutral model is different from the path in a real-world model because of the risk associated with locking in the fixed interest rate in a long-term pure discount bond. That risk is reflected in the 2 Another common means of adding a risk margin in real-world valuation is to start with the average value across all real-world scenarios and then add an amount that represents the present value of the cost of capital required to manage the risk of the item being valued. The theory in that approach is that the cost of capital represents the market price of risk. 3 The spot price is not the same as the discounted value using the expected path of the short-term rate, which would be expressed as follows: 𝑃𝑇≠[𝑒𝑥𝑝(−∫𝐸(𝑟 𝑡)𝑑𝑡 𝑇 0 )] This expression may have a very similar value, but it is not the same. Since both expressions involve an expectation, they are often confused. The situation here is somewhat analogous to measures of central tendency in a probability distribution, where the mean and the median are both measures of central tendency, but they are not the same. 11 Copyright © 2022 Society of Actuaries Research Institute volatility of the market price of a pure discount bond – volatility that does not exist in the price of a short-term money market fund earning the short-term risk-free rate. Investors in pure discount bonds demand a higher return as a reward for taking that risk, so the forward rate paths consistent with the price of a pure discount bond are higher than the forward paths of the truly risk-free short-term rate. Previously it was said that risk-neutral scenarios give more weight to adverse events, meaning declines in prices. Yet now we say that forward rate paths in a risk-neutral model are higher than in a real-world projection. How is this consistent? It is consistent because an upward movement in the forward rate leads to a decline in the price. Higher interest rates mean lower prices. Giving more weight to adverse events that decrease the market price means giving more weight to increases in interest rates that decrease the market price. Risk-neutral scenarios are not intended for use in simulation where those future higher interest rates might lead to higher returns on future new investments. Risk-neutral scenarios are intended for use in valuation of investments purchased in the past. Any use of risk-neutral scenarios to simulate investment purchases in the future violates the intended purpose of such scenarios for these reasons4: • It results in purchase of future investments at higher yields (i.e., more favorable future conditions) than the market actually expects. • It is inconsistent with the financial theory underlying risk-neutral models which was developed for the purpose of valuation of existing investments, not simulation of future investments. Given a formulation for 𝑟𝑡 it is often possible to derive a closed-form expression for the spot price consistent with the formula above involving the central expectation along possible paths. A closed-form expression is desirable because it allows the shape of the full yield curve to be calculated directly. When such a closed-form expression does not exist, spot prices and the yield curve are approximated using Monte Carlo simulation of many scenarios for 𝑟𝑡 to estimate the central expectation in the formula above. 3.1 MARKET BEHAVIOR IN INTEREST RATE MODELS There are two important and separate characteristics to any arbitrage-free model for interest rates, and both center on the formula defining movements of the short-term risk-free rate. One characteristic is the kind of general market behavior that is simulated. The other characteristic is whether the calibration is real-world or risk-neutral in nature. In what follows we first discuss a few different models of general market behavior. Then we discuss the difference between real-world and risk-neutral versions of each model. This is done to emphasize that when choosing an ESG, one needs to be aware of both characteristics of the model. All market-consistent models are not alike and do not produce identical results, and the differences are not limited to whether they are real-world or risk-neutral; the differences depend also on the underlying model of market behavior. Perhaps the best-known and simplest interest rate model is the Vasicek model. The Vasicek model treats the short-term risk-free interest rate as a mean-reverting random walk. Using the common mathematical form from above, the Vasicek model in discrete time is expressed as follows: 4 The underlying theory does not support this usage in an asset-liability simulation where there is re-investment risk or where there are options in the contracts that create decision points where the decision will be based on future economic conditions. In those situations, use of real-world market-coherent scenarios is more consistent with the underlying theory. 12 Copyright © 2022 Society of Actuaries Research Institute (next period value) = (current value) + (expected drift) + (random shock) 𝑟𝑡+1 = 𝑟𝑡+ 𝜅(𝜃−𝑟𝑡) + 𝜎𝑊 𝑡 𝜅= mean reversion strength or speed 𝜃= mean reversion point 𝜎= volatility 𝑊 𝑡= a random number from a Gaussian distribution with mean zero and variance one To develop a closed-form expression for the spot price and the remainder of the yield curve, the discrete model above is expressed as a stochastic differential equation in continuous time5: 𝑑𝑟= 𝜅(𝜃−𝑟)𝑑𝑡+ 𝜎𝑑𝑊 The Vasicek model assumes a certain kind of market behavior. Aspects of that behavior are that the volatility is constant and that interest rates can become negative if enough negative random shocks accumulate. Other models have been developed assuming different kinds of market behavior. For example, one can observe that volatility is not constant but tends to be higher when interest rates are higher. One can also observe that there is a strong tendency for interest rates not to go below zero. The Cox-Ingersoll-Ross model and the Black-Karasinski model each embed those behaviors in different ways. Here is the Cox-Ingersoll-Ross model: 𝑑𝑟= 𝜅(𝜃−𝑟)𝑑𝑡+ 𝜎√𝑟𝑑𝑊 The only difference from the Vasicek model is the random shock term which is multiplied by the square root of the current rate. This makes the random shocks (i.e. the volatility) depend on the current level of the interest rate. When interest rates approach zero, the random shocks get very small and mean reversion forces the interest rate back up toward the mean reversion point. This prevents interest rates from ever getting to or below zero. Here is the Black-Karasinski model: 𝑑 𝑙𝑛(𝑟) = 𝜅(ln (𝜃) −ln (𝑟))𝑑𝑡+ 𝜎𝑑𝑊 This model is also a mean-reverting random walk, but the simulation is done with the logarithm of the interest rate rather than the interest rate itself. The volatility term is constant in absolute terms, but in proportion to the logarithm of the interest rate the volatility is smaller when interest rates are lower and vice versa, as in the Cox-Ingersoll-Ross model. But when the interest rate approaches zero, the magnitude of its logarithm approaches infinity and that makes the mean reversion term very sensitive to the level of interest rates. In this model the mean reversion term explodes when interest approach zero, providing an even stronger reversion to the mean than in the Cox-Ingersoll-Ross model. And when interest rates are high, the mean reversion term gets substantially smaller and has less effect. 5 Note that the values of the parameters to be used in the discrete time version of the model are not the same as those in the continuous time version. The values depend on the size of the discrete time step in use. 13 Copyright © 2022 Society of Actuaries Research Institute These differences in assumed market behavior are illustrated in the chart below. The chart shows the distribution of the short-term interest rate 15 years in the future based on each of the models, using comparable calibrations. The distributions were simulated using 10,000 scenarios and counting the number of scenarios that fell into each 0.25% bucket. The distribution for the Vasicek model is a bell-shaped curve that includes some negative interest rates. The distribution of the Cox-Ingersoll-Ross model is skewed with a longer tail on the high end and does not include any negative interest rates. The distribution for the Black-Karasinski model is even more skewed. Figure 4 SIMULATED INTEREST RATE DISTRIBUTIONS FROM THREE MODELS All three of these distributions are based on calibrations with the same mean reversion point, the same value for the mean reversion strength parameter, and the same volatility at the mean. The point of showing the difference between them is to motivate the idea that the choice of model affects the valuation because it reflects the model’s assumptions about market behavior and the likelihood of different events in the future. This dependence of any market-consistent valuation on the choice of underlying model should not be overlooked. 3.2 THE MARKET PRICE OF RISK IN INTEREST RATE MODELS The equations above do not explicitly include the market price of risk, which is normally denoted as 𝜆. Let’s briefly review what the market price of risk is and then see how it has been incorporated into these models. To define the market price of risk we start with the idea that risk-averse investors require a higher expected or average return on riskier investments. We measure risk by the volatility of the market price. Greater volatility and greater risk increase the probability that the price may decline causing a loss, but they are associated with higher returns on the average. With that in mind we define the market price of risk as a ratio: Market price of risk = 𝜆= (𝐸𝑥𝑝𝑒𝑐𝑡𝑒𝑑 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑟𝑒𝑡𝑢𝑟𝑛)−𝑟 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑟𝑖𝑐𝑒 14 Copyright © 2022 Society of Actuaries Research Institute The excess average return over the risk-free rate is called a risk premium. The market price of risk is the ratio of the risk premium to the volatility of the price. One might ask why the market price of risk should be incorporated into a formula defining movements of the short-term risk-free rate. After all, since the rate is risk-free, then the market price of risk does not apply. The explanation is the formula given earlier for the spot price of a pure discount bond as an expectation based on the future movement of the short-term rate. Pure discount bonds are not risk-free; their price is volatile because the market interest rates used to discount their fixed future payoff are volatile. Since the price of a pure discount bond is volatile, the market expects a return higher than the risk-free rate and uses that higher return to calculate the market price. The path of the short-term rate used to price pure discount bonds (the risk-neutral path) must be higher than the real-world expected path to a degree that depends on the market price of risk. Therefore, it is common to introduce the market price of risk into these formulas as an upward drift in the risk-free rate, an upward drift equal to the volatility times 𝜆. For example, the Vasicek formula including 𝜆 is this6: 𝑑𝑟= 𝜅(𝜃−𝑟)𝑑𝑡+ 𝜎(𝑑𝑊+ 𝜆) Note that this is equivalent to simply increasing the mean reversion point parameter: 𝑑𝑟= 𝜅(𝜃−𝑟)𝑑𝑡+ 𝜎𝜆+ 𝜎𝑑𝑊 𝑑𝑟= 𝜅((𝜃+ 𝜎𝜆 𝜅) −𝑟) 𝑑𝑡+ 𝜎𝑑𝑊 The last expression above is identical to the original expression for the Vasicek model, but with the mean reversion parameter 𝜃 replaced by (𝜃+ 𝜎𝜆 𝜅). Because of that, sometimes the market price of risk in a model like this is re-defined to be the value added to the mean reversion parameter rather than the more abstract concept of risk premium divided by the volatility. The simple relationships shown above help one understand some of the language commonly used with reference to these models. The P-measure and the Q-measure. A measure refers to the probability distribution of future movements of the variable under consideration, in this case 𝑑𝑟. As shown above, in an arbitrage-free model the spot price is defined (measured) in terms of an expectation based on the path of 𝑟. Both the drift term and the random shock term are part of the measure. As was shown above, the change in measure from real-world (the P-measure) to risk-neutral (the Q-measure) in the Vasicek model is accomplished by modifying the random shock term, replacing 𝑑𝑊 with 𝑑𝑊 ̂ = 𝑑𝑊+ 𝜆. Equivalently, the change in measure can be accomplished by changing the drift term, replacing 𝜃 with 𝜃 ̂ = 𝜃+ 𝜎𝜆 𝜅. This equivalence between changing the random term and the drift term is what makes this and many similar models mathematically tractable so that a fixed-form equation for the spot price at any maturity can be derived. This equivalence is also what makes the P-measure and the Q-measure “equivalent Martingale measures”. An appendix is included to provide further illustrations and explanations of the relationship between the P-measure and the Q-measure. 6 In many references, lambda is introduced with a negative sign rather than a positive sign. When that is done, calibration shows that lambda has a negative value. Here we introduce it with a positive sign and a positive value because the market price of risk is defined in a way that anticipates a positive value. 15 Copyright © 2022 Society of Actuaries Research Institute The real-world parameters vs. the risk-neutral parameters. This expression refers to the values of the parameters, not their definitions. The mathematical definition of the parameters is the same in real-world and risk-neutral versions of the model. The parameters of the Vasicek model are 𝜅, 𝜃, and 𝜎. The difference between real-world and risk-neutral parameter values is in the value of 𝜃. If the real-world value of that parameter is 𝜃= 𝑥 then the risk-neutral value of that parameter is 𝜃 ̂ = 𝑥+ 𝜎𝜆 𝜅 as shown in the formulas above7. Let’s turn now to the Cox-Ingersoll-Ross model. The market price of risk can be introduced into the Cox-Ingersoll-Ross model in at least two different ways. The difference illustrates the complexity of reflecting the market price of risk in other models. The first method is to replace the random term 𝑑𝑊 in the Cox-Ingersoll-Ross model with 𝑑𝑊 ̃𝑡= 𝑑𝑊 𝑡− 𝜆√𝑟 𝜎𝑑𝑡. This is the approach used in the original Cox-Ingersoll-Ross paper8. 𝑑𝑟𝑡= 𝜅(θ −𝑟𝑡)𝑑𝑡+ 𝜎√𝑟𝑡𝑑𝑊 ̃𝑡 𝑑𝑟𝑡= 𝜅(θ −𝑟𝑡)𝑑𝑡+ 𝜎√𝑟𝑡(𝑑𝑊 𝑡−𝜆√𝑟 𝜎𝑑𝑡) 𝑑𝑟𝑡= 𝜅(θ −𝑟𝑡)𝑑𝑡−𝜆𝑟𝑡𝑑𝑡+ 𝜎√𝑟𝑡𝑑𝑊 𝑡 𝑑𝑟𝑡= (𝜅+ 𝜆) ( κθ (𝜅+ 𝜆) −𝑟𝑡) 𝑑𝑡+ 𝜎√𝑟𝑡𝑑𝑊 𝑡 In the last line above we see that the terms containing lambda have been removed from the random term and put equivalently into the drift term. The parameter 𝜅 has been replaced by 𝜅̂ = 𝜅+ 𝜆 and the parameter 𝜃 has been replaced by 𝜃 ̂ = 𝜅𝜃 𝜅+𝜆. One can use this form of the model to derive a formulaic expression for the spot price. The formulaic expression is very complex and is shown in Economic Scenario Generators – A Practical Guide (Pedersen et. al, 2016). Another method for introducing the market price of risk is to notice the following: 𝑑𝑟𝑡= 𝜅(θ −𝑟𝑡)𝑑𝑡+ 𝜎√𝑟𝑡𝑑𝑊 ̃𝑡= 𝜅̂(𝜃 ̂ −𝑟𝑡)𝑑𝑡+ 𝜎√𝑟𝑡𝑑𝑊 𝑡 The second form of the model at the end of the line above can be used to develop a formulaic expression for the spot price that depends on adjusted values of 𝜅 and 𝜃 without any reference to lambda. If the adjustments are 𝛾1 and 𝛾2 then define 𝜅̂ = 𝜅+ 𝛾1 and 𝜃 ̂ = 𝜃+ 𝛾2. The effect of the two adjustments together is assumed to reflect the market price of risk. The value of this approach is that if one is only interested in risk-neutral valuation, then one can use the formula for the spot price based on 𝜅̂ and 𝜃 ̂ to calibrate their values 7 One might be confused by the fact that lambda appears in this formula for the risk-neutral parameter value 𝜃 ̂. Isn’t the market price of risk zero in a risk-neutral calibration? The explanation is that calibration can be done in a way that estimates the value of 𝜃 ̂ directly without the need to adjust for the market price of risk. The market price of risk is not zero, rather it is reflected implicitly so no adjustment for the market price of risk is required. A risk-neutral calibration does not really assume the market price of risk is zero; it could not reproduce market prices if it did. Rather, a risk-neutral calibration provides parameter values that reflect the market price of investment risk implicitly, so no explicit adjustment for investment risk is required in a risk-neutral valuation. 8 This is one of the references in which lambda has a negative value. 16 Copyright © 2022 Society of Actuaries Research Institute directly to market prices and never need to specify the implicit values for 𝛾1 and 𝛾2 or the real-world values 𝜅 and 𝜃. The simple interest rate models discussed above are not commonly used in practice because their simplicity does not reproduce actual market behavior very well. An appendix is included in this document to illustrate the treatment of the market price of risk in some more complex interest rate models. The market price of risk can seem obscure when presented as a factor in a mathematical formula. Understanding can be improved by viewing the effect of the market price of risk on the yield curve. Recall that there is often a closed-form formula for the yield curve based on a definition of the process for 𝑟𝑡. The result of the formula depends on the parameter values used. When there are both real-world and risk-neutral parameter values for the same model, they will produce different yield curves starting from the same short-term rate. The difference between those yield curves is caused by the market price of risk because the market price of risk is the difference in the parameters. The difference between those yield curves is a set of “term premiums” that vary by term to maturity. Term premiums arise because of the risk associated with locking in a fixed rate for a long time in a world where future interest rates are uncertain. Term premiums represent the market price of that risk. One can judge the reasonableness of any formula for the market price of risk based on the term premiums that it implies, that is, based on the difference between yield curves produced using real-world and risk-neutral parameter values. Section 4: Equity index models Arbitrage-free equity index models focus on the total rate of return of investments in the equity index. Here we consider only models that include re-investment of dividends. Such models follow the general form outlined above: (next period value) = (current value) + (expected drift) + (random shock) Generally, these models are simpler than interest rate models. The expected drift for any period is typically the sum of the short-term risk-free rate and a risk premium. The realized returns in any specific scenario are dominated by the random shocks which tend to be much larger in magnitude than the expected drift. In risk-neutral calibrations of these models, the risk premium is zero so the expected drift is always equal to the short-term risk-free interest rate. Differences between risk-neutral versions of such models are limited to the random shock term. 4.1 MARKET BEHAVIOR IN EQUITY INDEX MODELS The design of an equity index model focuses on ways to model the expected drift and the random shock (volatility). When multiple equity indices are included in one ESG, there is focus on the relationships of the drift and volatility of various indices. We will not address relationships between separate equity indices in this document. ALTERNATIVES FOR THE EXPECTED DRIFT TERM There are no alternatives for behavior of the expected drift in a risk-neutral model. The expected drift in a risk-neutral model is always the current short-term risk-free rate. Alternatives for market behavior of the expected drift in a real-world model include: 17 Copyright © 2022 Society of Actuaries Research Institute • Whether the expected drift is the sum of the current short-term risk-free rate and a fixed risk premium or is constant. If the expected drift is a constant, then the implied risk premium changes over time to offset changes in the short-term risk-free rate. • Whether the expected drift is regime-switching. • Whether the expected drift is mean-reverting over time, so that the level of the equity index tends to mean-revert to a level that grows at a more stable compounded rate over long periods. ALTERNATIVES FOR THE RANDOM SHOCK TERM Alternatives for market behavior of the random shock include: • Whether the volatility of random shocks is constant or is stochastic or regime-switching in nature. • Whether the volatility of random shocks is supplemented by a “jump process” that creates large jumps in value at random intervals. Many of these alternatives are discussed in Economic Scenario Generators – A Practical Guide (Pedersen et. al. 2016) and are not elaborated here. 4.2 MARKET PRICE OF RISK IN EQUITY INDEX MODELS The market price of risk in an equity index model is reflected in the excess of the drift term over the short-term risk-free rate. That excess is the equity risk premium. In a risk-neutral model the drift term is equal to the short-term risk-free rate so the market price of risk and the equity risk premium are zero. Risk-neutral equity models are intended for valuation of options and other derivatives. One does not need a model to perform valuation of a stock index – the value is the current level of the index. Since the drift term in a risk-neutral equity model is always the short-term risk-free rate, the valuation of options depends only on the random shock term in the model. Risk-neutral models can vary in the way the random shock term is constructed, and calibration is important. The basic risk-neutral equity model is the Black-Scholes model in which volatility is Gaussian and calibration amounts to determining the implied volatility level that corresponds to an observed market price. Implied volatility in the Black-Scholes model is determined by calibration to the market price of an option. In the Black-Scholes model the drift in price is the short-term risk-free rate and the only variable is the volatility. Implied volatility is the volatility which, when used in the Black-Scholes formula, leads to an option value that equals the market price of that option. When this is done, implied volatility tends to depend on the strike price and tenor of the option used for calibration. The variation by strike price and tenor is called the “volatility surface”. The volatility surface is very useful for valuation. But it presents a problem when the purpose of a simulation is not valuation of an option at a point in time but focuses on the future path of the underlying index. In that context, a choice must be made concerning the level of volatility for the underlying index in the scenarios. Reasonable choices come from the range of volatilities in the volatility surface.9 9 The volatility surface corresponds to the Black-Scholes model, which is a constant volatility model. More complex models involve stochastic volatility, and such models are often calibrated to value options using simulation. In that context, the parameters of a stochastic volatility model are calibrated to market prices of some options, and the result may be several sets of parameters that differ by the strike and tenor of the option. An analogous issue arises there because when the purpose of a simulation is not valuation of an option at a point in time but focuses on the future path of the underlying index, a single set of parameters is needed to generate scenarios for the path of the index. 18 Copyright © 2022 Society of Actuaries Research Institute In a real-world model the equity risk premium tends to be positive. But some of the options for real-world models involve the idea that equity returns are driven by forces other than a risk premium over the risk-free rate. Such forces may include fiscal and monetary policy, GDP growth, unemployment, and others. Because of those other influences, the equity risk premium may not be constant and the drift in equity prices may not be tied to the sum of the risk-free rate and a fixed risk premium. Section 5: Calibration of market-consistent and market-coherent models 5.1 GENERAL APPROACH Economic Scenario Generators – A Practical Guide (Pedersen et. al. 2016) provides a good discussion of calibration procedures in section 7. Importantly, it defines separate calibration approaches for risk-neutral and real-world calibrations. Here we generalize so that the two calibration approaches can be viewed as two versions of the same process. The generalized process involves iterative adjustment of parameter values to improve fit to historical observations. The difference between risk-neutral and real-world calibration is not in the use of iterative fitting, it is in selecting which historical observations are used as targets of the ESG’s fit, and how the goodness of fit is measured. No model yet invented can perfectly fit all historical observations. There will always be some historical data that a model can fit better than other data. Historical data that is not used directly in calibrating the model may not be fit well at all. As noted above, risk-neutral and real-world calibrations focus on fitting different aspects of historical data. Because of this, it should not come as a surprise that sometimes risk-neutral and real-world calibrations are very different from one another and appear inconsistent. The kind of conceptual connection between risk-neutral and real-world calibrations that was presented earlier in this document can be reflected when both kinds of model are calibrated together. But often they are calibrated independently using different samples of historical data. Odd relationships between real-world and risk-neutral parameters can arise from independent calibration to separate data sets. In the sections below we briefly discuss risk-neutral and real-world calibration separately and then explain the kind of combined calibration required for a real-world market-coherent model. 5.2 RISK-NEUTRAL CALIBRATION The purpose of a risk-neutral ESG is valuation. Calibration of a risk-neutral ESG is the process of choosing parameters that reproduce market prices, so that the ESG can be used to determine a market-consistent value for items that are not widely traded and therefore have no observable market price. Typically a risk-neutral ESG is formulated in a way that leads to closed-form expressions for the prices of various kinds of items traded in the market, including both direct investments and derivatives. The process of calibration amounts to iteratively adjusting the values of the parameters in the closed-form expressions for market prices to achieve the best fit to the widest array of observed market prices. These considerations involved in risk-neutral calibration deserve special attention: • Market conditions are constantly changing. Often (but not always) a risk-neutral ESG is re-calibrated on every valuation date to current prices in order to be most consistent with current market prices. Sometimes recalibration is less frequent, and the calibration is intended to be consistent with market prices over a period of time, but the period of time is selected to represent relatively stable market conditions. 19 Copyright © 2022 Society of Actuaries Research Institute The point here is that the time frame of the historical data sample used for risk-neutral calibration is typically shorter than that used for real-world calibration. • Once a formula for a market price based on a set of parameters is developed, the process of fitting parameter values to reproduce market prices often does not associate any meaning to those parameters. They are just numbers to be adjusted in a mathematical optimization. The resulting parameter values may not make much sense in the real-world. This is not viewed as a problem in the context of valuation. But care must be exercised when using such risk-neutral parameter values as a starting point when building a real-world model. • Sometimes the calibration process includes solving for different parameter values for different derivatives of the same investment. This is most common in connection with the volatility parameter. For example, in the Black-Scholes model the “volatility surface” is an array of different values for the volatility parameter depending on the strike prices and tenors of options on some underlying index. By using an array of values like this, market prices can be matched more exactly. But the user should understand that there is only one underlying item with only one underlying volatility, and that the need for an array of different volatilities for derivatives is evidence that the model does not fit the underlying item and its derivatives with the same parameters. The use of a “volatility surface” is just a way to improve the fit of the model to current prices. It is useful in the context of valuation, but not in the context of simulation because there can be only one volatility for the underlying item. In a simulation the volatility of the underlying item governs payoffs of any derivatives of that item. 5.3 REAL-WORLD CALIBRATION Calibration of real-world models is covered well in Economic Scenario Generators – A Practical Guide (Pedersen et. al. 2016). There are some key differences from risk-neutral calibration. • The “goodness of fit” is measured differently. Instead of measuring the fit of formulaic prices to actual market prices, “goodness of fit” is measured with reference to expected movements over time in the generated scenarios. The Practical Guide says: o “Because real-world parameterizations are forward looking, they require explicit views as to how the economy will develop in the future, so they require a significant amount of expert judgment to determine the veracity of the scenarios that result from the parameterization process. In practice, real-world calibrations often are parameterized to be consistent with historical dynamics of economic variables, although the long-term steady-state levels associated with these parameterizations can differ from long-term historical averages in favor of current consensus expectations or even individual viewpoints. • Because goodness of fit is measured in terms of movements over time, parameters like the mean reversion point and volatility have real meaning and the steady-state levels mentioned in the Practical Guide have real meaning. This contrasts with risk-neutral calibration where, as noted above, sometimes the parameter values diverge from their real meaning. • The market price of risk is always non-zero in a real-world model and is always zero in a risk-neutral model. That means that the expected return of any investment that involves risk will be different on average when comparing real-world model and risk-neutral models. That makes real-world and risk-neutral calibrations of the same model different from one another. 20 Copyright © 2022 Society of Actuaries Research Institute 5.4 COMBINED CALIBRATION: A MARKET-COHERENT REAL-WORLD MODEL A real-world model can be calibrated to reproduce market prices, but this is achieved differently in the real-world context than in the risk-neutral context. In both cases, reproducing market prices means • Choosing a probability measure to use in calibration. The real-world measure is called the P-measure. The risk-neutral measure is called the Q-measure. Calibrating the path of future interest rates using observed spot prices of pure discount bonds. • Reflect market volatility around that path through calibration to some observed data. Real-world and risk-neutral calibrations accomplish these things differently. • Choosing a probability measure o Risk-neutral calibration uses the Q-measure o Real-world calibration uses the P-measure • Reflecting expectations for the path of interest rates o Risk-neutral calibration centers the path of the short-term risk-free rate on the forward rate path implied by observed spot prices on pure discount bonds. o Real-world calibration centers the path of the short-term risk-free rate on the forward rate path implied by observed spot prices on pure discount bonds, adjusted downward to remove the term premiums included in observed long term bond rates for the risk of locking in a long-term rate. • Reflecting market volatility o Risk-neutral calibration solves for the “implied volatility”, the volatility parameter value required in a closed-form formula for the price of a security to reproduce the market price. This often results in a “volatility surface” of different implied volatilities for different options on the same underlying security. o Real-world calibration uses historical realized volatility in observed market prices for the underlying security. Before discussing further details of the real-world calibration process, it is important to emphasize that real-world market-coherent scenarios are designed to be used differently than risk-neutral market-consistent scenarios in any stochastic valuation exercise. When using risk-neutral scenarios, the valuation result is the average of the scenario-specific valuations. When using real-world scenarios, the average of the scenario-specific values needs to be adjusted by adding a margin for investment risk, either at the scenario level or in total. There is some debate over whether the margin for risk can be determined in a reliable way. If not, then the real-world valuation cannot reproduce market prices even if it is based on scenarios calibrated in the real-world market- 21 Copyright © 2022 Society of Actuaries Research Institute coherent manner described above. In this document we take the view that risk margins can be determined in a reliable way10. Techniques for calculating risk margins are outside the scope of this document. We will first discuss characteristics of a real-world model that allows these things to be done, and then provide more detail on how they are done. Characteristics of a real-world market-coherent model include: • It must start with an arbitrage-free model framework because a model must be arbitrage-free to be market-coherent. • An arbitrage-free model framework can have both real-world and risk-neutral calibrations. A market-coherent real-world model must have both real-world and risk-neutral calibrations because both are used in scenario generation. • The difference between the parameter values in real-world and risk-neutral calibrations represents a specific assumption about the market price of risk. • The model must have a clear division between “state variables” and “parameters”. There are two sets of parameters (real-world and risk-neutral) but only one set of “state variables”. The statement that such a model must have both real-world and risk-neutral calibrations deserves explanation. Both calibrations are required because they are both used when generating interest rate scenarios. Interest rate scenarios in a market-coherent real-world model are generated using the following general process: 1. The initial values of the state variables are fit to the observed starting yield curve (spot prices) using the risk-neutral parameters. 2. Future scenario paths for the short-term rate are generated using the real-world parameters to create paths for the state variables. 3. The yield curve at any time step in a scenario path is calculated using the state variables and the risk-neutral parameters. This process assumes we have a model that has a closed-form formula or other fast procedure to calculate the yield curve based on the state variables and the parameters. This is the same as the process used by a risk-neutral model, except for step 2 where the real-world parameters are used in place of the risk-neutral parameters for simulating the forward path of the state variables. Steps 1 and 3 are the same as in a risk-neutral model because those steps deal with fitting the state variables to a full yield curve at a point in time. Step 2 is different because it deals with generating a path through time based on the probabilities of different paths. In step 2 a real-world generator uses the real probabilities (P-measure) while a risk-neutral generator uses the risk-neutral probabilities (Q-measure) which implicitly give more weight to adverse events that lead to lower market prices. 10 Valuation of insurance contracts for accounting purposes requires inclusion of margins for both investment risks and insurance risks. Risk-neutral scenarios are often used to provide an implicit market-consistent margin for investment risks. But the margin for insurance risks must then be added separately. If it is possible to determine a separate margin for insurance risks, one may argue that it is also possible to determine a separate margin for investment risks, or for the sum of all risks in a block of insurance contracts. 22 Copyright © 2022 Society of Actuaries Research Institute To say this another way, the future real-world path of the short-term rate is governed by the real-world parameter values, while the forward rates that define the yield curve (and spot prices) are governed by the risk-neutral parameter values. Equity return scenarios in a market-coherent real-world model are generated using the following general process: 1. Generate the risk-free rate for the next time step 2. Add a risk premium (expected equity return spread) based on the market price of risk 3. Add a random shock based on the volatility process This differs from the process in a risk-neutral equity return generator because the risk premium in step 2 is zero in a risk-neutral generator. Also, the volatility in step 3 may differ. In the modeling of interest rates, flexibility in simulating market behavior generally means using a multi-factor model for interest rates. Each factor corresponds to a “state variable” whose value represents the current state of that factor. In a one-factor interest rate model, the state variable is typically the current short-term risk-free interest rate. Multi-factor models are discussed here in the Appendix on “More complex interest rate models”. In a multi-factor model the values of the state variables are shared between real-world and risk-neutral versions of the model, while the values of the parameters differ. The state variables define conditions on the starting date and any simulated date thereafter, while the parameters govern the stochastic paths of the state variables. The parameters remain fixed over time, but the state variables move over time. It is this division between “state variables” and parameters that enables the creation of a market-coherent real-world model. The state variables are shared between real-world and risk-neutral versions of the model. Risk-neutral parameters are calibrated for consistency with market prices, including the spot prices in the yield curve. Real-world parameters are calibrated for consistency with movements of the state variables across points in time. In a real-world market-coherent model, the prices (yield curve) at any point in time are based on the risk-neutral parameters while the movements of the state variables are governed by the real-world parameters, as is most consistent with the way each set of parameters is calibrated. The use of both real-world and risk-neutral parameters in the same model imposes a calibration constraint that does not exist when such parameters are used in separate models. The constraint arises because the difference between the real-world and risk-neutral parameters is the market price of risk. For the two sets of parameters to make sense together, the market price of risk that connects them must be reasonable. That means, at a minimum, that it should generally be positive so that the generated scenarios tend to show higher expected average returns for riskier investments. Calibration of risk-neutral parameters is typically done by measuring “goodness of fit” to market prices. Calibration of real-world parameters is typically done by measuring “goodness of fit” to movements in state variables. The connection in theory is the market price of risk, but when the two calibrations are done independently, the market price of risk implied by the difference between them may not make sense. That’s because, as noted earlier, models do not fit perfectly, and the approach used in risk-neutral calibration can lead to parameter values that lose their real meaning. Therefore, the calibration of the risk-neutral parameters in a real-world market-coherent model may be done differently than for other risk-neutral models. The real-world parameters are calibrated first and taken as given. The risk-neutral parameters are treated as the sum of the real-world parameters and adjustments that reflect the market price of risk, and those adjustments are the subject of the risk-neutral calibration. 23 Copyright © 2022 Society of Actuaries Research Institute One way to calibrate the market price of risk in an interest rate model is to set it so that it produces reasonable term premiums. As noted earlier, term premiums implied in the yield curve are defined by the difference in two yield curves. One curve is generated using the real-world parameters and the other is generated using the risk-neutral parameters which are the real-world parameters adjusted for the market price of risk. The difference between those curves is a set of term premiums by term to maturity. The two yield curves that are compared to get the term premiums are both based on the same values of the initial state variables. Since state variables move over time, one should determine the term premiums implied starting from a range of different values for the state variables. The process just described for calibrating the market price of risk involves a fair amount of judgment along with knowledge of what a reasonable set of term premiums should be. Generally they should be positive if one believes that markets are risk-averse, but historical studies have provided more refined estimates. An appendix to this document discusses term premiums and the market price of risk in more detail. Section 6: Acknowledgments The researchers’ deepest gratitude goes to those without whose efforts this project could not have come to fruition: the Project Oversight Group for their diligent work overseeing, reviewing and editing this report for accuracy and relevance. Project Oversight Group members: Charles V Ford, FSA, MAAA, CFA Gary Hatfield, FSA, CERA, CFA, PhD Jason Kehrberg, FSA, MAAA, PhD James Kosinski, FSA, MAAA, CFA, PhD Steve Marco, ASA, CERA Hal Pedersen, ASA, MAAA, PhD At the Society of Actuaries Research Institute: Doug Chandler, FSA, FCIA, Canadian retirement research actuary Dale Hall, FSA, CERA, CFA, MAAA, SOA managing director of research 24 Copyright © 2022 Society of Actuaries Research Institute Appendix A: The P-measure and the Q-measure This appendix provides graphic depictions of the future distributions of returns created by scenario generators, focusing on the difference between real-world and risk-neutral versions. The real-world distributions are called the P-measure and the risk-neutral distributions are called the Q-measure. INTEREST RATES The chart below illustrates the drift of the future short-term risk-free rate in one calibration of a Cox-Ingersoll-Ross model. Both the real-world (RW) and risk-neutral (RN) drifts are shown. Since the drift is stochastic in nature, three percentile points in the probability distribution are shown: the 10th percentile, the 50th percentile and the 90th percentile. In the scenarios charted here, the starting level of the short-term risk-free rate is 4.00% and the calibrated real-world mean reversion point is also 4.00%. The point of this chart is to show that the risk-neutral drift is higher than the real-world drift. The next chart shows the probability density of the short-term risk-free rate based on the same scenarios as the previous chart. The density is charted based on the number of scenarios in each 0.25% interval out of a set of 10,000 scenarios. The point of this chart is that the risk-neutral probabilities (Q-measure) define a distribution that has a higher mean and is slightly wider than the real-world probabilities (P-measure). 25 Copyright © 2022 Society of Actuaries Research Institute Risk-neutral scenarios have been described as weighting adverse events more heavily and thereby leading to lower future values of current investments. Consider how that is reflected in a simulation of a pure discount bond that matures for $1,000. We know the starting value, and we know the maturity value, and those must be the same in both real-world and risk-neutral scenarios. But what about the value between those two points? Consider a situation where the current risk-free short-term rate is at its mean reversion point at the start of a simulation. In the real world the short-term rate is equally likely to move up or down, so let’s look at a scenario where it remains constant, and the rest of the yield curve remains constant. The chart below shows the yield curve we’ll use for illustrative purposes. The one-year spot rate is 4% and the 10-year spot rate is 6%. The 10-year rate includes term premiums that compensate for the risk of locking in the rate for 10 years. 26 Copyright © 2022 Society of Actuaries Research Institute If we generate both real-world and risk-neutral scenarios with this starting yield curve, we know that the scenarios will differ, with the risk-neutral scenario drifting higher. While the short-term rate in the real-world scenario will be unchanging, the short-term rate in the risk-neutral scenario will follow the forward rate path in the initial yield curve. This will lead to higher interest rates in the risk-neutral scenario. That means lower discounted present values for the bond in the risk-neutral scenario. The chart below illustrates the average path of the value of the bond. At every point between the start date and the maturity date, the value is lower in the risk-neutral scenario. One can calculate the return on the bond for each projection year in each of these scenarios. In the real-world scenario the return “walks down the curve” in the usual way, showing a return equal to long-term forward rates when the maturity date is far in the future, and declining over time to the short-term rate just before maturity. In the risk-neutral scenario, all investments are assumed to earn the same short-term risk-free rate (on average), but the short-term risk-free rate follows the path of the forward rate curve. The forward rate curve starts with the short-term rate and ends with the long-term rate. Basically, the forward rate curve is earned in reverse order. 27 Copyright © 2022 Society of Actuaries Research Institute EQUITY RETURNS For equities, the real-world probabilities define a distribution of future values that is higher than the distribution based on risk-neutral probabilities. The higher values include the market reward for taking risk. The chart below is based on simulation of 10,000 scenarios, counting the number of accumulated values in each $0.05 interval. When percentile points on these distributions are tracked over time one can see how the difference in probabilities affects the drift of accumulated value over time. 28 Copyright © 2022 Society of Actuaries Research Institute Appendix B: Some more complex interest rate models The interest rate models discussed in the body of this document are not used widely in practice for simulation. They were used to explain concepts in a simple context. The reason those simple models are not used in practice is that the market behavior that they capture is unrealistically simple. They are one-factor models where the only factor describing the current state of the economy is the short-term risk-free interest rate. Since those models are mean reverting with a fixed mean reversion point, the market behavior they imply is: • if the short-term rate is above the mean reversion point then then forward rates will trend down • if the short-term rate is below the mean reversion point then forward rates will trend up The only shape of yield curve such models can produce is monotonic, either upward sloping or downward sloping, depending on where the current short-term rate is relative to the mean reversion point. Interest rate models for simulation in practice are multi-factor in nature. Such models can simulate behavior where interest rates will trend in one direction in the short run and in another direction in the long run. That makes it possible to simulate behavior consistent with the shape of market yield curves that are not monotonic; hump-shaped and valley-shaped yield curves are possible. The role of “state variables” is a key to such models. Each “factor” in a one-factor or multi-factor model has a state variable that defines the current state of that factor. In a one-factor model, the state variable is typically the current level of the short-term risk-free interest rate. In multi-factor models, the additional factors have state variables that correspond to the current level of other variables that are included in the model. An important aspect of multi-factor models is that each factor has both a state variable that defines its current value and parameters whose values define future stochastic paths of that state variable. The general idea is that the parameter values are anticipated to be stable over time and the movements of the state variables define the behavior of the yield curve over time. Fitting a current yield curve to such a model means selecting values for the state variables that best fit the model yield curve to the observed yield curve. Here we describe two very different approaches to multi-factor models. In a “double mean reverting” model, each factor governs a different aspect of market behavior. In a “three-factor Cox-Ingersoll-Ross” model, each factor is a separate CIR model of the short-term risk-free rate that has its own parameter values, and the simulated short-term rate is the sum of the three factors. DOUBLE MEAN REVERTING MODELS A double mean reverting model has two state variables: the short-term rate and a short-term mean. The short-term rate is attracted to the short-term mean in a random walk, and the short-term mean itself follows a random walk and is attracted to a long-term mean which remains fixed. Definitions of the state variables: 𝑟𝑡= current short-term risk-free rate at time t 𝑚𝑡= current short-term mean reversion point for the risk-free rate at time t 29 Copyright © 2022 Society of Actuaries Research Institute Definitions of the parameters: 𝜇= long-term mean reversion point 𝜎1 = volatility of the short-term rate 𝜎2 = volatility of the short-term mean reversion point 𝜅1 = mean reversion strength of the short-term rate to the short-term mean 𝜅2 = mean reversion strength of the short-term mean to the long-term mean Definition of the stochastic process in discrete time: 𝑟𝑡+1 = 𝑟𝑡+ 𝜅1(𝑚𝑡−𝑟𝑡) + 𝜎1𝑊 2 𝑚𝑡+1 = 𝑚𝑡+ 𝜅2(𝜇−𝑚𝑡) + 𝜎2𝑊 2 The random shocks 𝑊 1 and 𝑊 2 are random draws from a Gaussian (Normal) distribution with mean 0 and variance 1. There are several variations on this model. • The two random shocks may or may not be correlated. • The two stochastic processes (two factors) shown above are in a form analogous to the Vasicek model. Each could be changed to be analogous to a Cox-Ingersoll-Ross, Black-Karasinski, or other formulation, with suitable changes to values of the parameters. Characteristics of this model type that deserve note: • A variety of yield curve shapes can be generated using a fixed set of parameter values while changing the values of the state variables. For example, a humped yield curve is generated when the short-term mean is greater than either the short-term rate or the long-term mean. • In some scenarios the short-term mean will remain far from the long-term mean for an extended period of time. Since the short-term rate reverts to the short-term mean, this tends to create scenarios that may be characterized as “low-for-long” or “high-for-long” in greater proportion than with a model with a single fixed mean reversion point. Reflecting the market price of risk in a double-mean-reverting model The market price of risk is reflected by the difference in drift due to the difference between real-world and risk-neutral parameter values. The simplest approach is to simply increase the long-term mean reversion point so that the risk-neutral value of the parameter 𝜇 is 𝜇̂ = 𝜇+ 𝜆𝜇. Alternately, one can use the approach discussed earlier where one starts with the formulaic yield curve based on real-world parameters and add a vector of term premiums to reflect the market price of risk. There are, of course, other approaches as well. 30 Copyright © 2022 Society of Actuaries Research Institute THE THREE-FACTOR COX-INGERSOLL-ROSS MODEL A three-factor Cox-Ingersoll-Ross model is simply the sum of three independent one-factor Cox-Ingersoll-Ross models. The three factors each have their own parameters for mean, volatility, and mean reversion strength. The short-term rate is the sum of the state variables of the three factors. Definitions of the state variables: 𝑟 1,𝑡= current short-term risk-free rate in factor 1 𝑟2,𝑡= current short-term risk-free rate in factor 2 𝑟3,𝑡= current short-term risk-free rate in factor 3 Definitions of the parameters: Each of the three factors has its own set of the three parameters in a Cox-Ingersoll-Ross model: 𝜅= mean reversion strength or speed 𝜃= mean reversion point 𝜎= volatility The values of these parameters are different between the factors, so there are nine parameters in total before reflecting the market price of risk. Definition of the stochastic process in discrete time: Each of the factors follows the basic Cox-Ingersoll-Ross process using its own state variable and its own set of parameter values. Here is the process used for each factor expressed in continuous time: 𝑑𝑟= 𝜅(𝜃−𝑟)𝑑𝑡+ 𝜎√𝑟𝑑𝑊 Since the short-term rate is the sum of the state variables, the price of a pure discount bond is 𝑃𝑇= 𝐸[𝑒𝑥𝑝(−∫(𝑟 1,𝑡+ 𝑟2,𝑡+ 𝑟3,𝑡)𝑑𝑡 𝑇 0 )] The mathematical form of the Cox-Ingersoll-Ross model allows derivation of a closed-form expression for this price. That is a significant advantage of this type of model. Characteristics of this model type that deserve note: • The three factors in this model are loosely analogous to the three components of the yield curve that arise from principal component analysis. A calibration of this model generally results in the three factors having very different parameter values for the mean reversion strength or speed. The parameter values are low, medium, and high. These correspond to the principal components as follows: o Low mean reversion speed corresponds to the “level” component o Medium mean reversion speed corresponds to the “slope” component 31 Copyright © 2022 Society of Actuaries Research Institute o High mean reversion speed corresponds to the “curvature” component • The three factors each have a fixed mean reversion point, so in sum they have a fixed mean reversion point. This makes this kind of model less likely to produce scenarios that remain far from the mean for an extended period of time as in “low for long” or “high for long”. • The movements of the state variables in this model are assumed to be independent, and scenarios are generated using independent random shocks to each factor. However, when the state variables are fit to historical yield curves, the historical movements of the state variables are significantly correlated. This indicates that while this model can generate reasonable distributions of future interest rates, the path-wise behavior of scenarios may be less realistic than some other models. Reflecting the market price of risk in a three-factor Cox-Ingersoll-Ross model As was discussed earlier for the one-factor Cox-Ingersoll-Ross model, it is common to reflect the market price of risk by simply defining adjustments to the parameters 𝜃 and 𝜅 in each of the factors. That makes six different adjustments (2 per factor x 3 factors). Combined with the 9 parameters already in the model, that makes 15 total parameters in a combined real-world / risk-neutral model11. As with any mean-reverting model, adding a market price of risk that produces positive term premiums in the generated yield curves is most directly done by an adjustment (increase) in the mean reversion point. In this model each factor has its own mean reversion point and its own mean reversion speed. One can manipulate the shape of the curve of term premiums by deciding how much of the adjustment to the total mean reversion point is allocated to each of the three factors. Putting most of the adjustment in the faster-reverting factors will lead to a curve of term premiums that is steeper on the short end and levels off at the long end, which some actuaries believe is most realistic. 11 That’s before any shift that may be added to allow the model to simulate negative interest rates. The shift would be a 16th parameter. 32 Copyright © 2022 Society of Actuaries Research Institute Appendix C: The meaning of arbitrage-free Arbitrage is the generation of profit with no risk. It occurs when the same item can be purchased somewhere at one price and sold elsewhere at a higher price. Markets are assumed to be arbitrage-free, so any market consistent or market-coherent ESG must be arbitrage-free. In a world where countless kinds of derivative securities exist, the detection of arbitrage in any scenario generator is a complex task. While arbitrage is defined as buying and selling the same “item” at different prices, the “item” could be composed of two or more other items, often including a derivative of the item itself. A test for arbitrage must evaluate all equivalent combinations of items. Complicated mathematical tests have been devised for that purpose. Section 9 of Economic Scenario Generators – A Practical Guide (Pedersen et. al. 2016) describes such tests. A simple concept underlies those complicated tests. The concept is that the generated paths of future returns (i.e., interest rates, stock prices) must be consistent with market expectations on the starting date of the scenarios. One of those expectations is the path of the short-term risk-free rate. To be arbitrage-free, an ESG for investments that involve risk must produce some scenario returns higher and some lower than the short-term risk-free rate. That way there are some scenarios that result in gains and some that result in losses, relative to keeping money risk-free in a money market fund. If all scenarios result in gains, or all result in losses, then arbitrage exists, and the generator is not arbitrage-free. To be arbitrage-free, any possibility of gain (earnings higher than the risk-free rate) must be accompanied by a possibility of loss. This does not mean that the average return must equal the risk-free rate. The average scenario return for a class of investments can be much different from the risk-free rate while still allowing the possibility of both gains and losses and thereby remaining arbitrage-free. This is one of the keys to understanding the connection between real-world and risk-neutral ESGs. In real-world ESGs the average scenario return for an investment depends on the level of risk in the investment. In risk-neutral ESGs the average scenario return for all investments is the same short-term risk-free rate, but that risk-free rate follows an unrealistic average path. Both kinds of ESGs can be arbitrage-free; the difference is that the market price of risk is nonzero in a real-world generator and that leads to average returns on risky investments that differ from the risk-free rate. 33 Copyright © 2022 Society of Actuaries Research Institute Appendix D: Term premiums and the market price of risk Term premiums are the reward investors demand for taking the risk of locking in a fixed interest rate in a world where interest rates change over time. The risk is that interest rates may rise, so a locked-in rate has both an element of opportunity cost and an element of risk that the market value of a fixed-rate investment may decline. Those two risks are flip sides of the same thing – they are 100% correlated. Since term premiums are a reward for taking risk, they are a function of the market price of risk. If you can quantify the market price of risk, you should be able to calculate the corresponding term premiums. This appendix explains how to do that. The market price of risk has been defined as this ratio, where r is the short-term risk-free rate: Market price of risk = 𝜆= (𝐸𝑥𝑝𝑒𝑐𝑡𝑒𝑑 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑟𝑒𝑡𝑢𝑟𝑛)−𝑟 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑟𝑖𝑐𝑒 For default-free fixed-income investments, the numerator of the ratio is the term premium. The denominator is the volatility of the price. Recall that in an arbitrage-free model, if the short-term risk-free rate at time t is 𝑟𝑡 then the spot price P for maturity t=T is the expectation: 𝑃𝑇= 𝐸[𝑒𝑥𝑝(−∫𝑟𝑡𝑑𝑡 𝑇 0 )] The price is a function of the risk-free rate at time 0, which is 𝑟0. One should be able to estimate the sensitivity of the price to a change in 𝑟0, which would be noted as 𝜕𝑃𝑇 𝜕𝑟0. We also can estimate the volatility of 𝑟0 which can be notated as 𝜎𝑟. We can then quantify term premiums using the following algebra. 𝜆= (𝐸𝑥𝑝𝑒𝑐𝑡𝑒𝑑 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑟𝑒𝑡𝑢𝑟𝑛) −𝑟 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑟𝑖𝑐𝑒 𝜆= (𝑇𝑒𝑟𝑚 𝑝𝑟𝑒𝑚𝑖𝑢𝑚) 𝜎𝑟(𝜕𝑃𝑇 𝜕𝑟0 ) 𝜆𝜎𝑟(𝜕𝑃𝑇 𝜕𝑟0 ) = (𝑇𝑒𝑟𝑚 𝑝𝑟𝑒𝑚𝑖𝑢𝑚) This gives an expression for the term premium given three things: • An estimate of the market price of risk 𝜆 • The volatility of the short-term interest rate 𝜎𝑟 • The sensitivity of the price to the short-term rate 𝜕𝑃𝑇 𝜕𝑟0 34 Copyright © 2022 Society of Actuaries Research Institute Consider the situation where the values of 𝜆 and 𝜎𝑟 are constants12. The sensitivity of the price is a function of the term to maturity and is greater for longer maturities. This means that term premiums can be expected to increase with term to maturity and follow a curve that mirrors the shape of 𝜕𝑃𝑇 𝜕𝑟0 which varies by term to maturity. The Cox-Ingersoll-Ross model is one such model and it has a formulaic expression for the spot prices 𝑃𝑇. That makes it possible to calculate the value of 𝜕𝑃𝑇 𝜕𝑟0 and quantify the term premiums resulting from a set of chosen parameter values. The chart below illustrates term premiums from the Cox-Ingersoll-Ross model based on one reasonable calibration. Similar curves arise from other mean-reverting models. When a curve of term premiums is quantified in this way, one needs to be clear about the measure of the observed yield curve to which term premiums correspond. When term premiums are quantified in this way, they are part of the forward rate curve. If they were quantified as part of the spot curve or the par coupon curve, they would typically be lower and less steep on the short end13. The connection between real-world and risk-neutral interest rate models can be seen clearly in this context. Consider the formula for the price of a pure discount bond: 𝑃𝑇= 𝐸[𝑒𝑥𝑝(−∫𝑟𝑡𝑑𝑡 𝑇 0 )] 12 There are of course models where 𝜆 and 𝜎𝑟 are not constants. These include multi-factor models or models with stochastic volatility. The mathematics associated with such models gets very complex and is beyond the scope of this document. Nevertheless, the same conceptual framework applies. One can gain much intuitive insight from the simple analysis presented here. 13 The reader is assumed to know how to transform a yield curve into its three equivalent forms – the par coupon curve, the spot curve, and the forward rate curve. One source is Fabozzi 2021, pages 692-695, 709-713. 0.00% 0.50% 1.00% 1.50% 2.00% 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 Years to maturity Illustrative term premiums from a Cox-Ingersoll-Ross model 𝜆𝜎(𝜕𝑃/𝜕𝑟) 35 Copyright © 2022 Society of Actuaries Research Institute The price is defined based on the path of 𝑟𝑡, the instantaneous forward discount rate. We can break the path of 𝑟𝑡 into two parts: the term premium 𝑝𝑡 and the remainder 𝑠𝑡. The remainder conceptually corresponds to the path of the risk-free short-term rate, which does not include term premiums. We then have: 𝑃𝑇= 𝐸[𝑒𝑥𝑝(−∫(𝑠𝑡+ 𝑝𝑡)𝑑𝑡 𝑇 0 )] The risk-neutral parameters for an interest rate scenario generator define the stochastic path for 𝑟𝑡. The real-world parameters for an interest rate scenario generator define the stochastic path for 𝑠𝑡, which is the path of 𝑟𝑡 with term premiums removed. The difference between those paths is the path of term premiums 𝑝𝑡. While term premiums are not directly observable at a point in time, cross-sectional studies over time can provide estimates of term premiums. Such studies have been done, and they suggest that term premiums follow a pattern by maturity that is much like that shown in the chart above, rising quickly for short maturities and leveling off for longer maturities. NOTES ON CALIBRATION OF TERM PREMIUMS The chart shown above is based on one reasonable calibration of a model. Term premiums are sensitive to the calibration. The chart below illustrates term premiums based on some alternate parameter values for volatility and mean reversion speed, all using the same market price of risk within the same model. In more complex interest rate models, term premiums can also be sensitive to the level and shape of the starting yield curve. One may wish to adapt an ESG so that it can reflect a desired set of term premiums. The exhibit below outlines the steps used to generate scenarios in a market-coherent real-world ESG. The basic process is shown on the left, and the changes needed to reflect pre-specified term premiums are shown on the right. 36 Copyright © 2022 Society of Actuaries Research Institute Step Basic process Revised process (specified term premiums) Initialize state variables Fit state variables to the spot prices in the observed starting curve using the risk-neutral parameters. Subtract the specified term premiums from the observed starting curve and fit the state variables to the resulting curve using real-world parameters. Generate paths of state variables Use the real-world parameters to generate the stochastic paths. Same. Calculate yield curves at each time step based on the state variables at that time Use the risk-neutral parameters to calculate the yield curve based on the state variables. Use the real-world parameters to calculate a yield curve based on the state variables, and then add the specified term premiums to that curve. Note that there is no use of risk-neutral parameters in the revised process. The use of pre-defined term premiums replaces the role of the risk-neutral parameters. This emphasizes the fact that the role of risk-neutral parameters in a real-world market-coherent ESG is simply to define the term premiums. If term premiums are provided explicitly in another way, then there is no role for risk-neutral parameters in generation of real-world market-coherent scenarios. The model under discussion here assumes that the yield curve is governed by two forces – the future expected path of the short-term rate and the term premiums based on the market price of risk. The observed yield curve can also be influenced by other things such as supply and demand factors and inflation expectations. These issues need to be considered when evaluating the fit of a model to historical data. 37 Copyright © 2022 Society of Actuaries Research Institute Appendix E: Fair value, Principle-based Reserves, and the Valuation Manual The valuation of insurance reserve liabilities (“reserves”) for accounting purposes has evolved in the 21st century so that it is often based on stochastic simulation and discounting of future cash flows. In the U.S. such valuation methods have been called “principle-based” (Center for Insurance Policy Research 2021) and are encoded in the National Association of Insurance Commissioner’s (NAIC) Valuation Manual for use in required regulatory financial reporting. Internationally, a different and new standard of accounting for insurance contracts has also employed such methods (IFRS 2021), but with important differences. There is a common principle on which all such methods are based. The principle is that the value of a block of insurance contract reserves is equal to the value of the assets required on the valuation date to pay them off, with some degree of uncertainty or margin for risk. This principle ties the valuation of the insurance contract reserves to the valuation of the assets. The largest difference between reserve valuations in the international accounting standard and the U.S. statutory Valuation Manual is that the valuation of assets is at market in the international standard while it is largely based on amortized cost in U.S. insurance regulatory (“statutory”) reporting. There are two ways to tie the value of the reserves to the value of the assets in this kind of principle-based approach. One way is to directly determine the amount of assets needed on the valuation date to pay off the contracts and set the value of the reserve equal to the value of those assets. The other way is to determine the rate of investment return in each time step of an asset-liability cash flow simulation and discount the liability cash flows at a rate equal to the path of projected investment returns. If the amount of starting assets in the simulation is just sufficient to run out when the last contract payment is made, then these two methods should produce identical results14. Different accounting bases prescribe different asset valuation methods, and different asset valuation leads to the reporting of different investment yields. When the reserve is calculated by discounting cash flows at the projected investment return on the assets, one must use the investment return as constrained by the applicable accounting standard. The reported investment return may differ from the return on market value in cases where assets and reserves are not held at a market value. For example, the investment return as measured under U.S. statutory accounting can be substantially different from current returns based on market value. Stochastic valuation of reserves with an asset-liability simulation model involves calculation of a separate liability value for each stochastic scenario and then using the set of scenario-specific values to determine a single reported value. The next two subsections discuss methods commonly used to get scenario-specific values and to determine the single reported value. OBTAINING SCENARIO-SPECIFIC VALUES When using stochastic scenarios for valuation of reserves in an asset-liability model, the amount of starting assets required to fulfill the contract differs for each scenario, unless the asset portfolio is a true replicating portfolio. True replicating portfolios are rare, so one must deal with the question of how to set the amount of starting assets. As a practical matter, one does not use a different amount of starting assets for each scenario. This creates two issues. First, one must determine what amount of assets to start with, and second, one must have a method to 14 Significant differences in results from these two approaches can arise when the path of investment returns has not been determined correctly based on the accounting standard in use. 38 Copyright © 2022 Society of Actuaries Research Institute calculate the value of the reserve when the assets do not exactly satisfy the contract cash flows in a particular scenario. A generally accepted approach is to start with an amount of assets very close in value to the ending single reported value of the reserve. This approach is somewhat circular and may require a bit of trial and error to arrive at that value. Nevertheless, this approach is required in the Valuation Manual, where the amount of starting assets is required to be within a specified narrow range around the reported reserve amount. Generally, this means that the starting assets will be more than sufficient to fund the reserve in most scenarios, but there many still be some particularly adverse scenarios where the assets run out before all contract obligations are paid. In those scenarios, the asset-liability model simulates borrowing to pay the obligations, and the investment return becomes the rate of interest simulated as paid on the borrowed funds. If the same amount of assets is used at the start of each scenario, then the scenario-specific value of the reserve cannot be set equal to the value of the starting assets. Therefore, the approach of discounting the cash flows at the path of investment returns (as modeled scenario by scenario and as reported under the applicable accounting standard) can be used to determine the scenario-specific value of the liability. In the Valuation Manual, another method is sometimes required; it is the “greatest present value of accumulated deficit” (GPVAD). Under this method, one determines a deficit defined as the excess of the outstanding liability over accumulated assets at each time step in the projection15. Each of those amounts is discounted back to the valuation date, and the GPVAD is the most negative of those discounted values. The scenario-specific reserve value is then defined as the sum of the value of the starting assets and the GPVAD. In a scenario where the starting assets are exactly sufficient to pay off the obligations, the GPVAD is zero and the scenario-specific reserve value is equal to the value of the starting assets. The discount rate used for the GPVAD theoretically reflects the expected return on the extra starting assets that would be required to fully fund the obligation and eliminate any deficits in a particular scenario. For regulatory purposes that could be the same as the projected return on the assets in the asset-liability projection, or it could be set conservatively at a level close to the risk-free rate. In either case, it is important to understand that this discount rate is not used to discount cash flows; it is only used to discount accumulated deficits. The biggest part of the liability value is set equal to the value of the starting assets, with the GPVAD often being a relatively small adjustment that varies by scenario. BLENDING SCENARIO-SPECIFIC VALUES INTO A SINGLE REPORTED VALUE When the accounting basis for asset valuation is market value, then risk-neutral methods can be used, and the single liability value is the simple average of the scenario-specific values. Any valuation of insurance contracts needs to include a margin for risk. Risk-neutral methods include a margin implicitly because risk-neutral calibration gives added probability weight to adverse scenarios. Whenever real-world scenarios are used, the simple average of the scenario-specific values does not include such an implicit margin and a margin must be added in some other way. The two most common methods for adding such a margin in the context of real-world scenarios are these: 15 Since it can be difficult or time-consuming to determine the outstanding liability value at each time step in a scenario, an approximation or “working reserve” may be defined for this purpose. Sometimes the “working reserve” is simply zero and the deficit is simply the negative of the amount of accumulated assets. 39 Copyright © 2022 Society of Actuaries Research Institute • Contingent tail expectation (CTE). Under this method the scenario-specific values are sorted from least to greatest, and only the greatest are included when calculating an average value. The number of scenarios included in the average is a specified percentage of the total. The Valuation Manual specifies the “70 CTE” which means that the smallest 70% of the scenarios are excluded from the average and only the largest 30% are included. • Cost of capital method. Under this method the scenario specific values are each increased by adding an imputed cash flow equal to the cost of capital in each time step of the scenario. The margin is the present value of those imputed cash flows. The cost of capital is intended to represent the market price of risk. In theory this approach may be considered consistent with market pricing but estimating the cost of capital is felt by some actuaries to be problematic. SELECTING A SCENARIO GENERATOR THAT IS FIT FOR PURPOSE If the accounting basis for assets is market value, then the scenarios used for liability valuation should be able to reproduce market prices of traded assets. While risk-neutral scenarios are always calibrated to do that, real-world scenarios may or may not be calibrated that way. When using a real-world ESG it is important that the calibration approach be appropriate for the accounting basis. Real-world scenarios that are not calibrated to reproduce market prices may be appropriate when the accounting basis is not market value. Recall that valuation using real-world scenarios must include an add-on margin for risk. When the scenarios used are not market-coherent, then the margin may arise partly from conservatism in the scenario calibration and partly from an explicit margin. That is the case with the Valuation Manual. U.S. regulators mandate the use of a real-world scenario generator that is not market-coherent and then add a margin using the CTE approach. A special issue arises when hedging is to be simulated within each scenario in the asset-liability model. Hedging depends on market-consistent16 valuation, including “Greeks”. To simulate hedging activity in each time step one needs market-consistent values for both the hedges and the contractual obligations being hedged. Such market-consistent values can be obtained through separate stochastic valuations at each time step. Stochastic valuations at each time step in a long-term simulation are often called “inner loop” valuations to indicate that the scenarios used for valuation are separate and different from the “outer loop” scenario that defines the conditions at each time step. The starting date for a set of “inner loop” scenarios is a valuation date anywhere within time span of the outer loop scenario. Since inner loop valuations are often used to determine a market-consistent value for derivatives used for hedging, they often employ risk-neutral scenarios calibrated on the fly to the yield curve at their starting date. This is true even if the outer loop scenario is real-world and neither market-consistent nor market-coherent in nature. In situations involving hedging, a risk-neutral ESG may be used for the inner loop valuations even though the outer loop is based on scenarios from a real-world ESG. That can be appropriate, for example, when simulating a clearly defined hedging strategy in a valuation that complies with the Valuation Manual where the outer loop scenarios are real-world and neither market-consistent nor market-coherent.17 16 In this paragraph the term “market-consistent” means a valuation technique that reproduces observed market prices. Both “market-consistent” and “market-coherent” valuations satisfy this definition. 17 See Question 12.1 (Slutsker 2019). 40 Copyright © 2022 Society of Actuaries Research Institute References Center for Insurance Policy and Research. Principle-Based Reserving (PBR). National Association of Insurance Commissioners (NAIC.org). Last updated May 28, 2021. Fabozzi, Frank. 2021. Fixed Income Securities. Ninth Edition. McGraw Hill. IFRS. 2021. IFRS 17 Insurance Contracts Standard 2022 Issued. International Financial Reporting Standards Foundation. Pedersen, Hal, Mary Pat Campbell, Stephen L Christiansen, Samuel H. Cox, Daniel Finn, Ken Griffin, Nigel Hooker, Matthew Lightwood, Stephen M Sonlin and Chris Suchar. 2016. Economic Scenarios – A Practical Guide. Chicago: Society of Actuaries. Slutsker, Benjamin and Dylan Strother, Cindy McGovern et al. 2019. Principle-Based Approach Projections Practice Note. Washington. American Academy of Actuaries. 41 Copyright © 2022 Society of Actuaries Research Institute About The Society of Actuaries Research Institute Serving as the research arm of the Society of Actuaries (SOA), the SOA Research Institute provides objective, data-driven research bringing together tried and true practices and future-focused approaches to address societal challenges and your business needs. The Institute provides trusted knowledge, extensive experience and new technologies to help effectively identify, predict and manage risks. Representing the thousands of actuaries who help conduct critical research, the SOA Research Institute provides clarity and solutions on risks and societal challenges. The Institute connects actuaries, academics, employers, the insurance industry, regulators, research partners, foundations and research institutions, sponsors and non-governmental organizations, building an effective network which provides support, knowledge and expertise regarding the management of risk to benefit the industry and the public. Managed by experienced actuaries and research experts from a broad range of industries, the SOA Research Institute creates, funds, develops and distributes research to elevate actuaries as leaders in measuring and managing risk. These efforts include studies, essay collections, webcasts, research papers, survey reports, and original research on topics impacting society. Harnessing its peer-reviewed research, leading-edge technologies, new data tools and innovative practices, the Institute seeks to understand the underlying causes of risk and the possible outcomes. The Institute develops objective research spanning a variety of topics with its strategic research programs: aging and retirement; actuarial innovation and technology; mortality and longevity; diversity, equity and inclusion; health care cost trends; and catastrophe and climate risk. The Institute has a large volume of topical research available, including an expanding collection of international and market-specific research, experience studies, models and timely research. Society of Actuaries Research Institute 475 N. Martingale Road, Suite 600 Schaumburg, Illinois 60173 www.SOA.org
10353	https://www.quora.com/Can-you-explain-why-tan-x-sin-x-cos-x	Something went wrong. Wait a moment and try again. Tangent Function Trigonometric Identities Cosine (math function) Trigonometry Functions Sine and Cosine Trigonometric Expressions Math Identities 5 Can you explain why tan(x) = (sin x) / (cos x)? Jeff Suzuki Mathematician and historian of mathematics. · Upvoted by Michael Jørgensen , PhD in mathematics and Gerhard Heinrichs , Master Mathematics, Ludwig Maximilian University of Munich (1973) · Author has 4.4K answers and 10M answer views · 4y For some reason, when we teach tangent, we never seem to point out an important connection. Behold! The unit circle, from which all trigonometry derives: Let DC be perpendicular to the diameter of a circle at C. You should know: By similar triangles: And since it’s a unit circle, OC = 1, so Now we need a catchy name for the length of DC. We could call it “the line drawn perpendicular to the diameter from an endpoint.” However, this perpendicular line has a rather interesting property…it’s tangent to the circle. (For extra credit: What’s the length of OD?) For some reason, when we teach tangent, we never seem to point out an important connection. Behold! The unit circle, from which all trigonometry derives: Let DC be perpendicular to the diameter of a circle at C. You should know: By similar triangles: And since it’s a unit circle, OC = 1, so Now we need a catchy name for the length of DC. We could call it “the line drawn perpendicular to the diameter from an endpoint.” However, this perpendicular line has a rather interesting property…it’s tangent to the circle. (For extra credit: What’s the length of OD?) Related questions Why does sin x/cos x = tan x? What is the relation between sin x + cos x and tan x? Can you explain why sin (tan x) =cos(x)? Can you explain why tan(x) = sin(x) /cos(x) and not vice versa? Why is tan(x) sin(x) /cos(x) and not cos(x) /sin(x)? Bernard Boucher Studied Electrical and Electronics Engineering & Mathematics at Clark College -- Vancouver, Washington (Graduated 1982) · Author has 385 answers and 529.6K answer views · 4y Originally Answered: Why does sinx / cosx = tan x? · Sinx = opposite/hypotenuse cosx = adjacent/hypotenuse tanx = opposite/adjacent sinx/cosx = (opposite/hypotenuse)/ (adjacent/hypotenuse) or opposite/hypotenuse hypotenuse/adjacent = opposite/adjacent = tanx QED Thank you for your view. If you like my answer, please consider an upvote. Elaine Dawe BMath, in Mathematics & Computer Science, University of Waterloo (Graduated 1985) · Author has 5.4K answers and 6.6M answer views · 5y Originally Answered: Why does sinx / cosx = tan x? · Because of the very definitions of sinx,cosx, and tanx. In a right triangle with acute angle x, we have defined the trig ratios as follows: sinx=oppositehypotenuse cosx=adjacenthypotenuse tanx=oppositeadjacent From this we get the acronym SOH−CAH−TOA Anyway, if we take the expression for tanx and divide numerator and denominator by hypotenuse we get: tanx=opposite/hypotenuseadjacent/hypotenuse=sinxcosx Andrew Coffey B.S. in Pure Mathematics & Physics, University of New Mexico (Graduated 2021) · Author has 270 answers and 253.4K answer views · 6y Originally Answered: Why does sinx / cosx = tan x? · Let’s start with a picture (credit:Right Triangle -- from Wolfram MathWorld) We will focus on the left one, but the right two are very important in trigonometry. I will use the convention that the angle opposite side a is α and the angle opposite the side b is β. Recall: sinα=a√a2+b2 cosα=b√a2+b2 tanα=ab Now, let’s divide sine by cosine: sinαcosα=a√a2+b2b√a2+b2=ab=tanα. We can do the same thing with β. In general, we can do this same tric Let’s start with a picture (credit:Right Triangle -- from Wolfram MathWorld) We will focus on the left one, but the right two are very important in trigonometry. I will use the convention that the angle opposite side a is α and the angle opposite the side b is β. Recall: sinα=a√a2+b2 cosα=b√a2+b2 tanα=ab Now, let’s divide sine by cosine: sinαcosα=a√a2+b2b√a2+b2=ab=tanα. We can do the same thing with β. In general, we can do this same trick with any right triangle, so it must be an intrinsic property of the trigonometric functions. We know what sine and cosine are because of how we defined them, as those particular ratios. Promoted by The Penny Hoarder Lisa Dawson Finance Writer at The Penny Hoarder · Updated Sep 16 What's some brutally honest advice that everyone should know? Here’s the thing: I wish I had known these money secrets sooner. 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Related questions Can you explain why cos(-x) = -cos x? If tan x is positive, what can be said about the values of sin x and cos x? How do we prove that? cos(x) 1 ______ + tan(x) =_____________ sin(x) sin(x).cos(x) cos (x)/sin(x)+ tan(x) = 1/sin(x).cos(x) At what angle does tan(x) = sin(x) /cos(x)? Can we say that tan (sin x) = sec (1/cos x)? Gary Ward MaEd in Education & Mathematics, Austin Peay State University (Graduated 1997) · Author has 4.9K answers and 7.6M answer views · Updated 2y Related Can you explain why tan(x) = sin^2 (x) /cos^2 (x)? Can you explain why tan(x) = sin^2 (x) /cos^2 (x)? I cannot. The red graph is the tan(x); the blue graph is the sin^2(x) / cos^2(x). Since for the most part the above premise is false, the premise is false. Red is tan(x); blue is (sin(x)/cos(x))^2. Can you explain why tan(x) = sin^2 (x) /cos^2 (x)? I cannot. The red graph is the tan(x); the blue graph is the sin^2(x) / cos^2(x). Since for the most part the above premise is false, the premise is false. Red is tan(x); blue is (sin(x)/cos(x))^2. Enrico Gregorio Associate professor in Algebra · Author has 18.4K answers and 16M answer views · 11mo Related How can you prove that tan(x) =sin(x) /cos(x)? It depends on how you define the tangent. Traditionally, considering the angle x=ˆAOB as in the picture below the definitions went as follows: sinx=AB cosx=OB, which is the sine of the complementary angle ˆAOQ tanx=AP secx=OP cotx=AQ cscx=OQ The radius of the circle was taken to be a large number, so as truncating at the integral part gave good approximations when division by the radius was performed. Later it was realized that using a radius of length one largely simplified formulas. The picture should also explain the terminology. Now it just takes a minute with similar t It depends on how you define the tangent. Traditionally, considering the angle x=ˆAOB as in the picture below the definitions went as follows: sinx=AB cosx=OB, which is the sine of the complementary angle ˆAOQ tanx=AP secx=OP cotx=AQ cscx=OQ The radius of the circle was taken to be a large number, so as truncating at the integral part gave good approximations when division by the radius was performed. Later it was realized that using a radius of length one largely simplified formulas. The picture should also explain the terminology. Now it just takes a minute with similar triangles to discover that tanx=sinxcosx secx=1cosx and similarly for cotx and cscx. Nowadays the geometric definition of sine and cosine is based on the unit circle and tanx=sinxcosx by definition. Sponsored by CDW Corporation Want document workflows to be more productive? The new Acrobat Studio turns documents into dynamic workspaces. Adobe and CDW deliver AI for business. Jack Vogt Former Rocket Scientist (1956–1990) · Author has 1.8K answers and 260.4K answer views · 3y Consider a right triangle with sides A, B, and C, where B is the base and C is the hypotenuse. Let the angles opposite the sides be labelled a, b, and c in the obvious manner. Then by definition, the sine of a is A/C and the cosine of a is B/C and the tangent of a is A/B. Isn’t it obvious to you that tan(x) = (sin x) / (cos x) = A/C / B/C ? What is there to explain? Andy Baker Works at University of Glasgow · Author has 7.3K answers and 1.7M answer views · 1y Originally Answered: Why is tanx=sinx/cosx? · The obvious answer is that it is defined to be that! In a right angled triangle with another angle θ , tan θ is of course the slope/gradient of the hypotenuse. Sponsored by Grammarly Is your writing working as hard as your ideas? Grammarly’s AI brings research, clarity, and structure—so your writing gets sharper with every step. Joseph Lau Badminton expert and esotericism in physics and math · Author has 133 answers and 290.8K answer views · 4y Originally Answered: Why does sinx / cosx = tan x? · From basic rules of trigonometry, in a right angle triangle where the opposite is a, the adjacent is b and the hypotenuse is c, sinx is defined by a/c. cosx is defined by b/c and tanx is defined by a/b. sinx / cosx = (a/c) / (b/c) = (a/c) • (c/b). You can see that the c's cancel, resulting in a/b which is exactly tanx. There are obviously numerous other ways to prove this, but this method is relatively simple and succinct. The Chosen One Service Desk Analyst at Samsung SDS (2007–present) · Author has 14.7K answers and 5.6M answer views · 6y Originally Answered: Why does sin x/cos x = tan x? · Taking a right angled triangle, tan x = Opposite/Adjacent cos x = adjacent/hypotenuse sin x = opposite /hypotense opposite/hypotenuse divided by adjacent/hypotenuse = opposite/adjacent, which is tan x. Roger Asdale Author has 2.3K answers and 327.6K answer views · 3y Ratios of angles in a right-angled triangle are derived from 3 sides — adjacent, opposite and hypotenuse. sin(x) = opposite / hypotenuse cos(x) = adjacent / hypotenuse tan(x) = opposite / adjacent sin(x) / cos(x) = sin(x) x 1 / cos(x) = (opposite / hypotenuse) x (hypotenuse / adjacent). Hypotenuse cancels out, so sin(x) / cos(x) =opposite / adjacent = tan(x). Robert Colburn Studied Mathematics at Cabrillo College · Author has 3K answers and 2.4M answer views · 3y Originally Answered: Why does sinx / cosx = tan x? · Why does sinx / cosx = tan x? Definitions of the functions. sin θ = y/r cos θ = x/r tan θ = y/x sin θ / cos θ = (y/r) / (x/r) = (y/r) (r/x) = y/x = tan θ Marco Biagini MSc Mathematics ETH Zurich, Switzerland · Author has 5.4K answers and 5.8M answer views · 2y Related Can you explain why tan(x) = sin^2 (x) /cos^2 (x)? Nobody can, since it’s not true: tan(x) = (sin^2(x))/(cos^2(x)) = tan(x) = tan^2(x)tan(x) = tan^2(x) Solve for x over the integers: tan(x) = tan^2(x) Subtract tan^2(x) from both sides: tan(x) - tan^2(x) = 0 Factor tan(x) and constant terms from the left hand side: -tan(x) (tan(x) - 1) = 0 Multiply both sides by -1: tan(x) (tan(x) - 1) = 0 Split into two equations: tan(x) - 1 = 0 or tan(x) = 0 Add 1 to both sides: tan(x) = 1 or tan(x) = 0 Take the inverse tangent of both sides: x = π n_1 + π/4 for x element Z and n_1 element Z or tan(x) = 0 The roots x = π n_1 + π/4 violate x element Z for all n_1 element Z: Nobody can, since it’s not true: tan(x) = (sin^2(x))/(cos^2(x)) = tan(x) = tan^2(x)tan(x) = tan^2(x) Solve for x over the integers: tan(x) = tan^2(x) Subtract tan^2(x) from both sides: tan(x) - tan^2(x) = 0 Factor tan(x) and constant terms from the left hand side: -tan(x) (tan(x) - 1) = 0 Multiply both sides by -1: tan(x) (tan(x) - 1) = 0 Split into two equations: tan(x) - 1 = 0 or tan(x) = 0 Add 1 to both sides: tan(x) = 1 or tan(x) = 0 Take the inverse tangent of both sides: x = π n_1 + π/4 for x element Z and n_1 element Z or tan(x) = 0 The roots x = π n_1 + π/4 violate x element Z for all n_1 element Z: tan(x) = 0 Take the inverse tangent of both sides: x = π n_2 for x element Z and n_2 element Z For π n_2 element Z, the assumptions on x are not violated. Assume these conditions: Answer: x = π n_2 for π n_2 element Z Related questions Why does sin x/cos x = tan x? What is the relation between sin x + cos x and tan x? Can you explain why sin (tan x) =cos(x)? Can you explain why tan(x) = sin(x) /cos(x) and not vice versa? Why is tan(x) sin(x) /cos(x) and not cos(x) /sin(x)? Can you explain why cos(-x) = -cos x? If tan x is positive, what can be said about the values of sin x and cos x? How do we prove that? cos(x) 1 ______ + tan(x) =_____________ sin(x) sin(x).cos(x) cos (x)/sin(x)+ tan(x) = 1/sin(x).cos(x) At what angle does tan(x) = sin(x) /cos(x)? Can we say that tan (sin x) = sec (1/cos x)? What is 1-sin(x) in terms of cos and tan? Why is cos x / sin x sometimes not equal to 1/tan x? What is the value of sin(x) cos(x) /tan(x)? Why does "arctan (tan x)" equal to "arctan (sin x/cos(x))"? If x is an acute angle, and tan x=3/4, what is the evaluation of cos x - sin x/cos x + sin x? 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10354	https://www.youtube.com/watch?v=VYH2FReL9Ac	Magnitude of a 2D Vector Daniel Kopsas 11400 subscribers 24 likes Description 1891 views Posted: 6 Nov 2019 1 comments Transcript: alright so in this video we want to talk about basic concept dealing with two-dimensional vectors namely we want to look at the magnitude the magnitude of a 2d two-dimensional vector okay and first of all what is magnitude it's just a fancy word for the length so magnitude is just the length of a vector alright so so nothing complex so if we want to find the magnitude of the vector we have a certain notation for that so let's let's consider consider let's consider the vector V we'll call it V and let's say that vector is written in component form as 2 negative 7 okay then if we want to describe the magnitude of this vector then the magnitude of V is written the magnitude of V is written as V within vertical bars okay so it looks like absolute value absolute value is what the vertical bars represent when you put a number inside so if you put a vector inside it means magnitude alright you could write it like that or if the vector does not have a name a label such as V you could just write the vector in component form within the vertical bars in either case that means magnitude alright so so let's let's actually see if we can determine let's determine the magnitude of this vector V alright so let's let's see if we can figure that out so consider this vector and determine vertical bars V the magnitude of V so the best thing you can do maybe especially at first is draw a picture of the vector and just see if you can figure out a way using mathematics to determine how long it is so vector to negative seven you pick some initial point for the vector to start at it's got a horizontal component of positive two that means it's going to move to the right two units and a vertical component of negative seven that means it's going to go down seven units so I need to pick some initial point on the board here that's going to give me enough room to move right to and down seven so I'll pick my initial point to be maybe you can like right here and that gives me room to move right and down so I'll move right two units and then I'll move down seven units so here's my initial point here's my terminal point and the vector in all of this is the vector that starts here and terminates down here so this purple guy is vector V and we would need to know how long this vector is what's its magnitude and the good news is for us is that vectors when they're given using the horizontal and vertical components horizontal is always perpendicular to vertical in in other words the angle here between the two and the negative seven that's always a 90-degree angle which means this is a simple right triangle so if you want to know how long this vector is or the magnitude of this vector you simply apply the Pythagorean theorem so in other words the magnitude of vector V or you could say the magnitude of two negative seven both of those are correct is simply going to be the square root of this leg squared so two squared plus this leg squared and notice when you consider it just as a right triangle the leg length is actually seven not negative seven however if you happen to use negative seven and put it here you'll still get the same thing as if you square positive seven so it's not really a big deal but we usually don't put the seven or the negatives were measuring lengths of triangle sides and so simplifying this you get four plus 49 that's 53 so the magnitude of this vector is the square root of 53 so that's our exact magnitude right we haven't rounded anything there so sometimes we'd like to have a better understanding of what that number is so I'll get out my calculator and type the square root of 53 and as a decimal that's approximately equal to let's say seven point three all right so this is an approximate magnitude and so in application type problems this may be a better sort of answer than this but if you if someone just gives you the approximate answer to a problem there's no way that you can go backwards and get the exact right but if someone gave you the square root of 53 is the answer you could always use that to approximate it yourself so they each have their pros and cons either way is fine lots of times you'll give the exact answer unless you're told to do otherwise okay so that's the magnitude of some 2-dimensional vector pretty simple let's do let's do one more example that we don't want over complicate by the way you might notice something there is a pattern here if these are horizontal and vertical then those are always the numbers you square here and your Pythagorean theorem alright so you can you can sort of take that shortcut but I recommend really understanding why this process works first all right now let's try a different vector that's even simpler but sometimes sometimes we over complicate the simple problems you can probably relate relate to that so let's say we want to determine determine the magnitude of the vector zero negative eight and so when you draw this vector zero negative eight there is no horizontal component you're not moving left or right you're just moving down so we'll choose our initial point up here to give us room to go down and then our terminal point is down here eight units away and the vector is the vector that starts here and terminates at the bottom so I need to draw my arrow in and now I can see the vector okay so the question is what is the magnitude of this vector well since there is no horizontal component there's no right triangle you literally just look at the vector and ask yourself how long is it well it is eight units long and that's it the magnitude of this vector is eight it's eight units long and notice the answer is not negative eight magnitudes will never give you negative answers because it's the length of a vector and so in that respect it's these vertical bars behave somewhat like absolute value right if you don't know absolute value measures a distance in a distance typically is not considered to be negative all right so that magnitude of a two-dimensional vector
10355	https://blog.csdn.net/wasane/article/details/122767380	求解等差数列公式_设一等差数列为(2, 4.6.8.),试求:(1) 公差d:(2) 通项a,:(3)前n项和s,.-CSDN博客博客下载学习社区 GitCode InsCodeAI 会议搜索 AI 搜索登录登录后您可以：复制代码和一键运行与博主大V深度互动解锁海量精选资源获取前沿技术资讯立即登录会员中心消息历史创作中心创作求解等差数列公式原创已于 2022-02-02 00:48:42 修改·6k 阅读 · 4 · 13· CC 4.0 BY-SA版权版权声明：本文为博主原创文章，遵循CC 4.0 BY-SA版权协议，转载请附上原文出处链接和本声明。文章标签： #等差数列于 2022-02-02 00:45:02 首次发布数据结构与算法专栏收录该内容 35 篇文章订阅专栏本文详细介绍了等差数列的定义及核心计算公式，包括公差、第n项值、前n项和及项数的求解。通过实例解析了如何从首项、末项和项数来确定等差数列的各项属性，帮助读者深入理解等差数列的基本性质。例如如下等差数列a: 2 4 6 8 10 。首项a 1 为2 , 末项a n 为10，项数n为5。公差d为2 。等差数列和S n 为30。 1、公差: d = (第n项值-第k项值) / (n-k) 。 d = (a n-a k)/(n-k) 如2 4 6 8等差数列中第二项减第一项求公差就是 d = (d 2-d 1) = 4-2=2 2、第n项值: 首项+(n-1)公差。 a n = a 1+(n-1)d 如求2 4 6 8等差数列中第二项的值: a 2 = a 1+d = 2+2 = 4 3、前n项和: S n = (第n项值+首项)n / 2 。S n=(a n+a 1)n/2 如求2 4 6 8前3项的和 : S 3 = (a 3+ a 1)3/2 = 12 4、项数: n = (第n项值-首项) / 公差 +1。 n = (a n-a 1)/d+1 如求 2 4 6 8 等差数列的项数： n = (a 4-a 1)/2+1 = 6/2+1 = 4 确定要放弃本次机会？福利倒计时 : : 立减 ¥ 普通VIP年卡可用立即使用 tanxinji 关注关注 4点赞踩 13 收藏觉得还不错? 一键收藏 0评论分享复制链接分享到 QQ 分享到新浪微博扫一扫打赏打赏打赏举报举报专栏目录 leetcode 1 50 2 . 判断是否形成等差数列 ( js ) 前端中的弟中之弟 03-15 1032 题目给你一个数字数组 arr 。如果一个数列中，任意相邻两项的差总等于同一个常数，那么这个数列就称为等差数列。如果可以重新排列数组形成等差数列，请返回 true ；否则，返回 false 。示例 1：输入：arr = [3,5,1] 输出：true 解释：对数组重新排序得到 [1,3,5] 或者 [5,3,1] ，任意相邻两项的差分别为 2 或 -2 ，可以形成等差数列。示例 2：输入：arr = [1,2,4] 输出：false 解释：无法通过重新排序得到等差数列。提示： 2&l 常用公差配合表图_机械制图基础知识，公差配合与技术测量技术，标准公差和基本偏差 . . . 热门推荐 weixin_39567013的博客 12-09 7万+ 一、标准公差 1．标准公差等级：确定尺寸精确程度的等级。国家标准设置了 2 0个公差等级。 2．公称尺寸分段：从理论上讲，同一公差等级的标准公差数值也应随公称尺寸的增大而增大。尺寸分段后，同一尺寸段内所有的公称尺寸，在相同公差等级的情况下，具有相同的公差值。二、基本偏差 1．基本偏差及其代号基本偏差——国家标准《极限与配合》中所规定的，用以确定公差带相对于零线位置的上偏差或下偏差。基本偏差的代 . . . 参与评论您还未登录，请先登录后发表或查看评论等差数列末项 _ 等差数列项数公式 weixin_32778881的博客 01-14 815 点击上方蓝字 · 关注珂珂的小园上次讲了等差数列末项公式▼等差数列末项 ( 第n 项 ) 公式这次讲项数，怎么求一个等差数列有多少个数。先思考，再对答案，不会做和没做对的，看视频讲解。题目 1 . 等差数列 3 ，5 ，7 ，9 ， … ，99．( 1 ) 其中 6 5是这个数列的第几个数？( 2 ) 这个数列共有多少项？2 . ( 1 ) 等差数列 1 ，4 ，7 ，1 0， … ，其中7 6 是这个数列的第多少项？ ( 2 ) 等差数 . . . 等差数列中两项为整数求公差 _Simplelife_新浪博客 jxzhzcf的博客 03-06 196 等差数列求和公式 wusj的博客 10-08 1万+ 等差数列是常见数列的一种，可以用AP表示，如果一个数列从第二项起，每一项与它的前一项的差等于同一个常数，这个数列就叫做等差数列，而这个常数叫做等差数列的公差，公差常用字母d表示。例如：1,3,5,7,9……（2 n-1 )。等差数列{an}的通项公式为：an=a 1+( n-1 ) d。前 n 项和公式为：Sn=na 1+n ( n-1 ) d/2 或Sn=n ( a 1+an )/2。注意：以上n均属于正整数。目录 1 . . . java 等差数列公式 _ 求解 java 等差数列问题！ weixin_42389522的博客 02-25 476 publicclassArithmeticSequence{publicstaticvoidmain ( Stringargs[]){Scannerinput=newScanner ( System . in );inta_ 1,d,N;doublea_n,S_n;System . out . print ("Entera_ 1 :");a=input . nextInt . . . public class ArithmeticSequ . . . 二级等差数列独得一段好程序 02-14 2413 二级等差数列可以用在EXECL上的公式等差数列 ( c++求解 ) zyang654321的博客 03-10 1185 【代码】等差数列 ( c++求解 ) python编写等差数列求和 python _等差数列求和公式前 1 00 项的和实例 weixin_39714307的博客 12-10 1522 最近跑去学了下python,一个很简单的题，结果发现数学公示忘了，在不用for循环的情况下居然有些懵，记录为下 . . 题：等差数列可以定义为每一项与它的前一项的差等于一个常数，可以用变量 x 1 表示等差数列的第一项，用 d 表示公差，请计算数列 1 4 7 1 0 1 3 1 6 1 9 …前 1 00 项的和。等差求和公示：和=( 首数+尾数 )项数/2;题的懵就是尾数忘了怎么求了，查了百度得到结果很简单。尾数公 . . . 高三数学等差数列前 N 项和公式 PPT课件 . pptx 10-10 本篇文章将依据“高三数学等差数列前 N 项和公式 PPT课件 . pptx”的内容，为读者详细解读等差数列的核心知识点和计算方法。首先，等差数列的定义是学习的起点。所谓等差数列，是指一个数列中，从第二项起，每一项与它 . . . 高中数学数列—等差数列求和公式 PPT课件 . pptx 10-10 这些性质使等差数列区别于其他类型的数列，并为求解等差数列相关问题提供了理论基础。在实际应用方面，等差数列的应用非常广泛，它不仅用于解决数学内部的问题，还广泛应用于经济、物理、工程等领域。例如，在计算 . . . 等差数列求和公式 . ppt 12-08 - 例子 2 求解等差数列-_1_ 0, -_6_, -_2_, _2_, _._ _._ _._的和为5 4 时的项数，解得n = 9。 - 例子 3 是一个实际问题，通过V形架的结构，我们可以将每层铅笔的数量视为等差数列的项，从而计算总铅笔数。 5 . 练习题目 - 练习 1 . . . 中职数学等差数列 . pptx 10-07 3 . 求解等差数列 -_1_, 5, _1_ _1_, _1_ 7,_._ _._ _._ 的第 50 项，同样利用通项公式，确定首项和公差后，可以求解。 4 . 当公差和项数已知，但首项未知时，可以反向使用通项公式求解首项。通过这些实例，学生可以熟练掌握等差数列 . . . 将中缀表达式转换成后缀表达式并求值（栈） tanxinji的博客 05-04 4315 中缀表达式转换成后缀表达式需要用到的是数据结构中的一种——栈。中缀表达式 : 生活中进行正常加减乘除计算的表达式。如 2+( 34-6/2 )。也就是算术表达式后缀表达式：也叫逆波兰表达式。只有运算符和数字没有括号，运算符的先后顺序存在优先级的关系。如 : 2 3 46 22+ 。（由于需要处理字符串的原因，目前只有个位数的算术运算）给出的算术表达式通过代码实现很难计算出来，而转换成了后缀表达式的就很容易实现出来了。当然，以下代码是很容易的出计算结果的。 public class Main Java 如何将数组所有元素转换成一个数 tanxinji的博客 02-08 4041 只适用于数组每一个下标存的是一位数，并且是有序的，如某个数组存的是百十个位，或者个十百位。一、将数组元素转换成一个数设 num是要转换的数 num = num1 0 + array[i] ; i = 0~n-1。乘以十是为了提高数的位数，每次加的数组元素是当前 num的个位。从在数组中所认为的数的个位开始加。代码示例 : public class Test { public static void main ( String[] args ) { int array[] = {5 Java 求数的全排列 ( dfs ) tanxinji的博客 02-07 3759 如何求 n个元素的全排列，如 1 2 3 的全排列为 1 2 3 ； 1 3 2 ； 2 1 3； 2 3 1； 3 1 2 ； 3 2 1；使用的是递归，暴力搜索所有可行的方案。可以用一个一维数组存储每次找到的一种方案。一、求 1~n的全排列代码示例 : // 输出一个n，输出 1~n的全排列 import java . util .; public class Main{ static int N = 1 0; static int n ; static int path[] = n 用邻接矩阵创建图 tanxinji的博客 06-06 3330 用一个字符数组vexs[]存储着图的所有顶点，在用一个二维数组arcs[][]存储边的关系，这就是邻接矩阵。另外v表示图的顶点数，e表示图的边数。w表示某条边的权值对于arcs[i][j]二维数组存储边的关系中，数组的i,j一般都是图的顶点数，即i = j = v；并且例如arcs[i][j] = w 就可以表示为，图存在顶点vexs[i] 到顶点vexs[j]的边，并且权值为w; 另外对于arcs[i][j]随着边的不同可以自行定义其规则。 1 . 当i = j的时候，arcs[i][j]的权值 2 Java实现单链表 tanxinji的博客 05-24 3076 实现一个可以用任意对象为结点的单链表。嘻嘻，埋伏笔了。因为我是用内部类来实现结点的操作（Java-内部类），和用上一篇文章的Student类来作为本链表的结点的测试案例，另外还有一个如何删除对象结点，哈哈，也是上一篇的，为了防止可能看不懂，记得去看看上一篇的文章哦（Java继承原来就是这样）。创建的任意类对象的实例可以成为单链表的结点，这里也用到的就是一点点泛型的知识。另外提一下，如何解决我的那篇多态的文章的向下转型带来的不安全性就是可以用泛型解决的啦（Java多态到底教了我干啥?）。一开始以为单链表只如何使用递推公式来求解等差数列和等比数列的通项和前 n 项和？请结合实例详细说明。最新发布 11-12 在解决等差数列和等比数列的求解问题时，递推公式是重要的工具之一。递推公式通过已知项来推导未知项，是求数列通项和前 n 项和的关键。首先，我们需要了解等差数列和等比数列的基本性质。参考资源链接：数列题型解析：从定义到求和方法全攻略等差数列的定义是每一项与前一项的差为常数，即d=a_n+1 - a_n。其递推公式可以表示为a_n+1 = a_n + d。对于求解等差数列的通项公式 an = a 1 + ( n - 1 ) d，我们可以通过已知的首项 a 1 和公差 d来使用公式法计算出任意项的值。对于求前 n 项和，等差数列的公式 Sn = n/2 ( a 1 + an ) 或Sn = n/2 [2 a 1 + ( n - 1 ) d]，我们只需将首项 a 1 和公差 d代入即可得出结果。在等比数列中，每项与其前一项的比为常数，即r=a_n+1 / a_n。其递推公式可以表示为a_n+1 = a_n r。等比数列的通项公式 an = a 1 r^( n - 1 ) 直接体现了这一关系。通过首项 a 1 和公比r的已知值，我们可以求出数列的任意一项。对于前 n 项和，等比数列的公式 Sn = a 1 ( 1 - r^n ) / ( 1 - r )，其中r≠1 时适用。当r=1 时，Sn=na 1。通过这个公式，我们可以快速计算出等比数列的前 n 项和。举个实例，对于等差数列 2, 5, 8, . . ., 我们知道首项 a 1=2，公差 d=3。求第 1 0 项 a 1 0 和前 1 0 项和 S 1 0。使用递推公式，我们得到a 1 0=2+( 1 0-1 )3=2 9。利用等差数列前 n 项和的公式，我们得到S 1 0=1 0/2( 2+2 9 )=1 55。在等比数列 2, 4, 8, . . . 中，首项 a 1=2，公比r=2。求第5 项 a5 和前 5 项和 S5。通过递推公式，我们得到a5=22^( 5-1 )=3 2。使用等比数列前 n 项和公式，我们得到S5=2( 1-2^5 )/( 1-2 )=-6 2（注意此处r=2>1，因此公式适用）。通过这些实例，我们能够更好地理解如何应用递推公式来求解等差数列和等比数列的相关问题。对于想要深入理解和掌握更多解题技巧的学习者，可以参考《数列题型解析：从定义到求和方法全攻略》这份资源。该文档不仅提供了等差数列和等比数列的递推公式解析，还包含了许多实用的练习题目，帮助学习者通过实践来巩固所学知识。参考资源链接：数列题型解析：从定义到求和方法全攻略关于我们招贤纳士商务合作寻求报道 400-660-0108 kefu@csdn.net 在线客服工作时间 8:30-22:00 公安备案号11010502030143 京ICP备19004658号京网文〔2020〕1039-165号经营性网站备案信息北京互联网违法和不良信息举报中心家长监护网络110报警服务中国互联网举报中心 Chrome商店下载账号管理规范版权与免责声明版权申诉出版物许可证营业执照 ©1999-2025北京创新乐知网络技术有限公司 tanxinji 博客等级码龄6年 299 原创1472 点赞 4218 收藏 591 粉丝关注私信热门文章高等数学-求曲线拐点 42892 Visio如何绘制数据流图 40689 Matlab中的循环 40181 Java中的随机数 34586 eclipse如何导入导出项目 34379 分类专栏 Java基础97篇 C#1篇数据结构与算法35篇数据库基础25篇 ASP.NET与C#5篇 ABAP2篇 Linux10篇大数据20篇 Spring24篇 C/C++5篇 Java常用API13篇工具和软件的使用15篇前端基础17篇 Vue.js13篇 pta13篇计算机基础8篇 Matlab5篇展开全部收起上一篇： Ajax入门下一篇： Java 求最大公约数最新评论 fatal: unable to access ‘ Encountered end of file 〆我們の明天❀:捣鼓了半天终于解决了补码移位规则 2401_83579988:### 答案你的描述在补码的移位规则中基本正确，但需注意以下几点细化： 1. 正数： - 正确：符号位不变，左移和右移时空出的位均补0。适用于所有编码（原码、反码、补码）。 2. 负数： - 补码（Two's Complement）： - 左移：符号位不变，数值左移，低位补0。但需确保符号位与最高有效位相同，否则溢出。 - 右移：符号位不变，数值右移，高位补符号位（即补1）。 - 反码（One's Complement）： - 左移：符号位不变，数值左移，低位补0。若最高数值位丢失0，可能导致错误。 - 右移：符号位不变，数值右移，高位补1（保持负数特性）。 3. 示例修正： - 原数 1.0001（假设为补码，十进制-15）： - 左移一位：1.0010（十进制-14）。正确。 - 右移一位：应保持位数固定，结果为 1.1000（高位补1，十进制-8）。若允许扩展位数，则 1.10001 可能表示扩展后的值，但需明确位数规则。 --- ### 详细对比表 \| 编码类型 \| 移位操作 \| 规则 \| 示例（假设5位存储） \| -----------------------------------------------------------\| \| 补码 \| 左移 \| 符号位不变，数值左移，低位补0。需符号位与最高有效位相同。 \| 1.0001（-15）→ 1.0010（-14） \| \| 补码 \| 右移 \| 符号位不变，数值右移，高位补符号位（1）。 \| 1.0001（-15）→ 1.1000（-8） \| \| 反码 \| 左移 \| 符号位不变，数值左移，低位补0。若最高数值位为0，可能导致错误。 \| 1.0101（-5）→ 1.1010（-2，错误） \| \| 反码 \| 右移 \| 符号位不变，数值右移，高位补1。 \| 1.0101（-5）→ 1.1010（-2） \| --- ### 关键总结 - 补码右移必须补符号位：确保负数右移后仍为负数。 - 反码右移需补1：保持反码的取反特性。 - 左移需检查溢出：补码左移时，若符号位与最高有效位不同，会溢出。你的规则在补码场景下正确，但需明确区分反码与补码的不同处理。例子中的位数扩展需谨慎，通常移位不改变位数。为何String类中的split方法字符.无法识别 dabidai:谢谢大佬的博客解除了疑惑 Spring Boot入门-快速搭建网页 LikeStars7:如果是要完成增删改查四种且与网页有交互的操作，后端和前端要如何完成互相的传值呢？ Visio如何绘制数据流图一条咸鱼酱:为啥子数据存储编辑不了文字佛了大家在看 MySQL面试（1）用OpenCV CSRT实现实时目标跟踪 3 2025年这5款三防手机各有各的优势，适用于危急特场景 359 Python 初学者最容易踩的 5 个坑及解决方法跨网络JVM调试新范式：一个方案实现远程代码诊断最新文章 SQLServer死锁监测方案:自定义死锁扩展事件生成 SQLServer死锁监测方案:如何使用XE.Core解析xel文件里包含死锁扩展事件的死锁xml ssms加载死锁信息扩展事件xel文件时，提示:值不能为 null。参数名: metadataFiles (Microsoft.SqlServer.XEvent.Linq) 2025年 4篇 2024年 2篇 2023年 59篇 2022年 144篇 2021年 91篇目录 1、公差: d = (第n项值-第k项值) / (n-k) 。 d = (an-ak)/(n-k) 2、第n项值: 首项+(n-1)公差。 an = a1+(n-1)d 3、前n项和: Sn = (第n项值+首项)n / 2 。Sn=(an+a1)n/2 4、项数: n = (第n项值-首项) / 公差 +1。 n = (an-a1)/d+1 展开全部收起相关专栏 Python \| 华为OD机试真题专栏 282 人学习华为机试实时速递，不定期更新最新机试题目，华为OD机试真题对应社招100/200分的分类下所有题目，备战华为机试的最佳题目华为od真题求解连续数列华为OD算法题详解专栏 9 人学习本专栏收录了上百篇华为od算法题及解法，代码主要为python代码。求解连续数列问题编程专栏专栏 17 人学习编程专栏是一个为程序员和编程爱好者提供信息和资源的平台。该专栏涵盖了各种编程语言、开发工具和技术的教程、实践经验和最新动态。读者可以在这里学习编程基础知识，掌握编程技巧，了解最新的编程趋势和技术发展。通过分享编程经验和解决问题的方法。 CSDN每日一练 \|『吃!吃!吃!』『交际圈』『等差数列』2023-08-13 目录 1、公差: d = (第n项值-第k项值) / (n-k) 。 d = (an-ak)/(n-k) 2、第n项值: 首项+(n-1)公差。 an = a1+(n-1)d 3、前n项和: Sn = (第n项值+首项)n / 2 。Sn=(an+a1)n/2 4、项数: n = (第n项值-首项) / 公差 +1。 n = (an-a1)/d+1 展开全部收起上一篇： Ajax入门下一篇： Java 求最大公约数分类专栏 Java基础97篇 C#1篇数据结构与算法35篇数据库基础25篇 ASP.NET与C#5篇 ABAP2篇 Linux10篇大数据20篇 Spring24篇 C/C++5篇 Java常用API13篇工具和软件的使用15篇前端基础17篇 Vue.js13篇 pta13篇计算机基础8篇 Matlab5篇展开全部收起登录后您可以享受以下权益：免费复制代码和博主大V互动下载海量资源发动态/写文章/加入社区 ×立即登录评论被折叠的条评论为什么被折叠?到【灌水乐园】发言查看更多评论添加红包祝福语请填写红包祝福语或标题红包数量个红包个数最小为10个红包总金额元红包金额最低5元余额支付当前余额 3.43 元前往充值 > 需支付：10.00 元取消确定成就一亿技术人! 领取后你会自动成为博主和红包主的粉丝规则 hope_wisdom 发出的红包打赏作者 tanxinji 你的鼓励将是我创作的最大动力 ¥1¥2¥4¥6¥10¥20 扫码支付：¥1 获取中扫码支付您的余额不足，请更换扫码支付或充值打赏作者实付元使用余额支付点击重新获取扫码支付钱包余额 0 抵扣说明： 1.余额是钱包充值的虚拟货币，按照1:1的比例进行支付金额的抵扣。 2.余额无法直接购买下载，可以购买VIP、付费专栏及课程。余额充值确定取消举报选择你想要举报的内容（必选）内容涉黄政治相关内容抄袭涉嫌广告内容侵权侮辱谩骂样式问题其他原文链接（必填）请选择具体原因（必选）包含不实信息涉及个人隐私请选择具体原因（必选）侮辱谩骂诽谤请选择具体原因（必选）搬家样式博文样式补充说明（选填）取消确定 AI助手 AI 搜索智能体 AI 编程 AI 作业助手下载APP 程序员都在用的中文IT技术交流社区公众号专业的中文 IT 技术社区，与千万技术人共成长视频号关注【CSDN】视频号，行业资讯、技术分享精彩不断，直播好礼送不停！客服返回顶部
10356	https://pmc.ncbi.nlm.nih.gov/articles/PMC3463765/	The Proline Rich Tetramerization Peptides in Equine Serum Butyrylcholinesterase - 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 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 FEBS J . Author manuscript; available in PMC: 2013 Oct 1. Published in final edited form as: FEBS J. 2012 Sep 7;279(20):3844–3858. doi: 10.1111/j.1742-4658.2012.08744.x Search in PMC Search in PubMed View in NLM Catalog Add to search The Proline Rich Tetramerization Peptides in Equine Serum Butyrylcholinesterase Kevser Biberoglu Kevser Biberoglu 1 Department of Biochemistry, School of Pharmacy, Hacettepe University 06100 Ankara, Turkey Find articles by Kevser Biberoglu 1, Lawrence M Schopfer Lawrence M Schopfer 2 University of Nebraska Medical Center, Eppley Institute, Omaha, NE. 68198-5950 Find articles by Lawrence M Schopfer 2, Ozden Tacal Ozden Tacal 1 Department of Biochemistry, School of Pharmacy, Hacettepe University 06100 Ankara, Turkey Find articles by Ozden Tacal 1, Oksana Lockridge Oksana Lockridge 2 University of Nebraska Medical Center, Eppley Institute, Omaha, NE. 68198-5950 Find articles by Oksana Lockridge 2 Author information Article notes Copyright and License information 1 Department of Biochemistry, School of Pharmacy, Hacettepe University 06100 Ankara, Turkey 2 University of Nebraska Medical Center, Eppley Institute, Omaha, NE. 68198-5950 ✉ Corresponding author: Lawrence M. Schopfer University of Nebraska Medical Center, Eppley Institute 985950 Nebraska Medical Center, Omaha, NE 68198-5950 phone (402)-559-6032 fax 402-559-4651 lmschopf@unmc.edu Issue date 2012 Oct. PMC Copyright notice PMCID: PMC3463765 NIHMSID: NIHMS400142 PMID: 22889087 The publisher's version of this article is available at FEBS J Summary Soluble, tetrameric, plasma butyrylcholinesterase from horse has previously been shown to include a non-covalently attached polyproline peptide in its structure. The polyproline peptide matched the polyproline rich region of human lamellipodin. Our goal was to examine the tetramer organizing peptides of horse butyrylcholinesterase in more detail. Horse butyrylcholinesterase was denatured by boiling, thus releasing a set of polyproline peptides ranging in mass from 1173 to 2098 Da. The peptide sequences were determined by fragmentation in the MALDI-TOF-TOF and LTQ-Orbitrap mass spectrometers. Twenty-seven polyproline peptides grouped into 13 families were identified. Peptides contained a minimum of 11 consecutive proline residues and as many as 21. Many of the peptides had a non-proline amino acid at the N-terminus. A search of the protein databanks matched peptides to 9 proteins, though not all peptides matched a known protein. It is concluded that polyproline peptides of various lengths and sequences are included in the tetramer structure of horse BChE. The function of these polyproline peptides is to serve as tetramer organizing peptides. Keywords: mass spectrometry, serum butyrylcholinesterase, polyproline peptide INTRODUCTION Butyrylcholinesterase (BChE) is a serine hydrolase which has attracted attention because it hydrolyzes cocaine, heroin, aspirin, bambuterol, and succinylcholine, and it scavenges organophosphorus pesticides and chemical warfare agents [1–6]. These properties endow BChE with therapeutic potential. The therapeutic effectiveness of BChE relies on its ability to stay in the circulation for a long time. The half-life of human plasma derived BChE is 12–14 days when injected into humans [7, 8]. Longevity in the circulation requires BChE to be glycosylated and to have a tetrameric structure [9–12]. Therefore it is of interest to understand the factors that maintain the tetrameric structure of BChE. In neuromuscular junctions and in the central nervous system, tetrameric forms of acetylcholinesterase (AChE) and BChE are anchored to the membrane. Attachment is via the proline-rich attachment domain of the collagen tail protein (ColQ) in the neuromuscular junction and via the proline-rich membrane anchor (PRiMA) in the central nervous system [13, 14]. Both ColQ and PRiMA use unique proline-rich attachment domains near their N-termini to interact with AChE and BChE C-Termini . Structurally, horse and human serum butyrylcholinesterase (BChE) are very similar. Alignment of their amino acid sequences (accession code Q9N1N9 for horse and P06276 for human BChE) shows 90% identity. Both have a 28 amino acid signal peptide followed by 574 amino acids per subunit. Horse BChE has 8 and human BChE has 9 Asn-linked glycans [16–18]. The glycans account for about 25% of the mass of the BChE protein. BChE has three intrachain and one interchain disulfide bonds per subunit . The BChE in plasma is a soluble, globular tetramer with a molecular weight of 340,000. The tetramer is composed of four identical subunits in a dimer of dimers structure where a dimer is formed by a disulfide bond at Cys 571. The catalytic triad residues are Ser 198, Glu 325 and His 438 . The 40 C-terminal residues of BChE constitute the tetramerization domain [21, 22]. It was found by Li et al. that a series of proline-rich peptides derived from lamellipodin function as tetramer-organizing peptides for both human and horse serum BChE . However, only the tetramer-organizing peptides from horse BChE with the most intense mass spectral signals were analyzed. Cholinesterases are the only proteins known to employ proline-rich peptides to stabilize their tetrameric structure. A novel addition to this tetramerization scheme was made by the demonstration that polyproline peptides, presumably derived from lamellipodin, were employed by serum BChE. The unique nature of this process begs for confirmation and extension from additional studies. In the present study, our goal was to identify additional polyproline peptides from tetrameric horse BChE by examining the smaller peptides that were noted, in passing, in the earlier study by Li et. al. We found a range of new polyproline peptides. Highly purified horse BChE was boiled to denature the protein and thus release non-covalently bound peptides. Peptides were separated from the bulk proteins by centrifugation through a YM10 cutoff filter or by HPLC. Released peptides were analyzed by MALDI-TOF-TOF and LTQ-Orbitrap mass spectrometry. We found 13 types of polyproline peptides ranging in mass from 1173 to 2098 Da, of which 9 could be matched to specific proteins. Only one of those peptides matched the human lamellipodin tetramer-organizing peptide reported by Li et al.. RESULTS MS analysis Peptides released by boiling were separated from BChE protein with a 10 000 Da cut-off centrifugal filter. The pass-through solution was collected and analyzed by MALDI-TOF mass spectrometry. Control BChE samples were treated identically except that they were not boiled. As shown in Figures 1 (A) and (B), the pass-through solution from unboiled, native horse BChE sample (upper panels) contained virtually no ions, whereas the pass-through solution from boiled human BChE (lower panels) gave a series of signals which turned out to be proline-rich peptides. These ranged in mass from about 1200 to 2700 Da. Masses marked in the figure correspond to peptides for which complete amino acid sequences could be extracted. Details for these peptides are in Table 1. Most of the remaining signals yielded partial sequence information indicating that they also were proline-rich peptides. Figure 1. Open in a new tab MALDI-TOF mass spectra of peptides released from horse BChE by boiling followed by filtration to separate peptides from bulk protein. The mass range has been divided into two sections (A and B) to better resolve the signals and to emphasize low intensity signals. Section A shows the masses from 1200 to 1500 Da and B shows the masses from 1500 to 2700 Da. The top panel in each section shows that no peptides were released when BChE was not boiled. The bottom panel in each section shows peaks for the released peptides. The right-hand y-axes indicate the absolute signal intensity in counts-per-second. Labeled masses correspond to peptides for which complete amino acid sequences were extracted. Table 1. Summary of the peptides released from horse BChE \| Sequencea \| # residues \| Series Typeb \| Instrument \| Mass, Dac \| :--- :--- \| Family 1 \| \| \| \| \| \| (PP)PPP PPPPP (PP) \| 12 \| y- & b-ion series \| Orbitrap \| 1183.6 \| \| PPPPP PPPPP PP potassium+ \| 12 \| y-ion series \| MALDI \| 1221.1 \| \| (PPP)PP PPPPP PPPPP (PP) \| 17 \| y- & b-ion series \| Orbitrap \| 1668.9 \| \| (PPPPP PPPPP) PPPPP PPPPP (P) \| 21 \| y-ion series \| Orbitrap \| 2057.1 \| \| Family 2 \| \| \| \| \| \| VPPPP PPPPP PP Na+ \| 12 \| y- & a-ion series \| MALDI \| 1207.1 \| \| VPPPP PPPPP PPP Na+ \| 13 \| y- & a-ion series \| MALDI \| 1304.8 \| \| Family 3 \| \| \| \| \| \| L/I/Hyp(HV)PP PPPPP (PPPPP PP) Na+ \| 17 \| y-ion series \| MALDI \| 1749.0 \| \| Family 4 \| \| \| \| \| \| (SP)PPP PPPPP (PP) \| 12 \| y- & b-ion series \| Orbitrap \| 1173.6 \| \| (SPP)PP PPPPP PPPP(P P) \| 16 \| y- & b-ion series \| Orbitrap \| 1561.8 \| \| Family 5 \| \| \| \| \| \| HPPPP PPPPP PP \| 12 \| y-ion series \| MALDI \| 1223.2 \| \| HPPPP PPPPP PP Na+ \| 12 \| y-ion series \| MALDI \| 1245.7 \| \| Family 6 \| \| \| \| \| \| L/I/HypHPPP PPPPP PPPP \| 14 \| y- & b-ion series \| MALDI & Orbitrap \| 1433.2 \| \| PL/I/HypHPP PPPPP PP(PP) Na+ \| 14 \| y-ion series \| MALDI \| 1455.9 \| \| PL/I/HypHPP PPP(PP PPPP) 2Na+ \| 14 \| y-ion series \| MALDI \| 1477.8 \| \| (PL/I/Hyp)HPP PPPPP PPPPP (PP) Na+ \| 17 \| y- & b-ion series \| MALDI \| 1747.2 \| \| Family 7 \| \| \| \| \| \| (L/I/HypP)PPP (PP)PPP (PP) \| 12 \| y- & b-ion series \| Orbitrap \| 1199.6 \| \| L/I/HypPPPP PP(PPP PP) Na+ \| 12 \| y-ion series \| MALDI \| 1221.7 \| \| (L/I/HypP)PPP PPPPP PPP \| 13 \| y- & b-ion series \| Orbitrap \| 1296.7 \| \| (L/I/HypP)PPP PP(PP)P PP(PP) \| 14 \| y- & b-ion series \| Orbitrap \| 1393.8 \| \| (L/I/HypPPPP P)PPPP PPPP(P P) \| 16 \| y-ion series \| Orbitrap \| 1587.9 \| \| (L/I/HypPPPP PP)PPP PPPPP (PP) \| 17 \| y-ion series \| Orbitrap \| 1684.9 \| \| (L/I/HypPPPP PPPP)P PPPPP PPP(PP P) \| 21 \| y-ion series \| Orbitrap \| 2073.1 \| \| Family 8 \| \| \| \| \| \| PGGGP PPPPP PPPPP P Na+ \| 16 \| y- & b-ion series \| MALDI \| 1473.8 \| \| Family 9 \| \| \| \| \| \| (EPPPP P)PPPP PPPP(P P) \| 16 \| y- & b-ion series \| Orbitrap \| 1603.9 \| \| Family 10 \| \| \| \| \| \| (APP)PP PPPPP P(PP) \| 13 \| y-ion series \| Orbitrap \| 1254.7 \| \| (APPP)P PPPPP PPPPP PP \| 17 \| y- & b-ion series \| Orbitrap \| 1642.9 \| \| (APPPP P)PPPP PPPPP P(PP) \| 18 \| y-ion series \| Orbitrap \| 1739.9 \| \| Family 11 \| \| \| \| \| \| PAPPP PPPPP PPP Na+ \| 13 \| y- & x-ion series \| MALDI \| 1276.7 \| \| Family 12 \| \| \| \| \| \| (AP)PPP PPPPP PPPPP L/I/HypPPPP M NH+4 \| 21 \| y-ion series \| MALDI \| 2098.3 \| \| Family 13 \| \| \| \| \| \| (TPPPP) PPPPP PPPPP L/I/HypT \| 17 \| x-ion series \| Orbitrap \| 1692.9 \| Open in a new tab a Peptide sequences are written with the N-terminus on the left and the C-terminus on the right. Residues enclosed in brackets indicate that separate signals for the masses of each residue were not resolved in the mass spectrum. However, masses of the bracketed fragments exactly fit the indicated sequence. Leucine, isoleucine, and hydroxyproline (L/I/Hyp) cannot be distinguished because they have the same nominal mass. b The “Series Type” column lists the major fragment series observed in the MSMS spectrum of each peptide. c “Mass” is the monoisotopic mass for the singly-charged peptide, i.e. [M + H]+ or [M + metal]+ . Observed masses from the Orbitrap were not singly-charged. Masses for such peptides have been transformed from the observed mass to their singly-charged equivalent for presentation in this column. We observed that freezing BChE in the absence of glycerol also caused release of proline-rich peptides. Separation of peptides from bulk protein was essential. If this step was omitted the protein suppressed ionization of peptides and no peptides were detected by MALDI-TOF mass spectrometry. Peptide masses similar to those in Figure 1 were found when HPLC was used to separate boiled BChE from peptides. HPLC purification reduced the complexity of the samples that were applied to the MALDI target thereby reducing signal suppression and improving the detection of peptides. Another advantage of HPLC purification was that it increased the concentration of purified peptides in the 0.5 μl aliquot analyzed by MALDI-TOF mass spectrometry. Better quality fragmentation spectra were obtained from the HPLC purified peptides. The peptides observed with the MALDI mass spectrometer were typically in complex with a single sodium atom (sodium replacing the added proton that makes the peptide positively charged). In one instance, a peptide was detected that was in complex with two sodium atoms. We also observed ammonium and potassium adducts (Table 1). Peptides released from horse BChE were also analyzed on the LTQ-Orbitrap mass spectrometer where peptides were separated by liquid chromatography before their masses were determined (Figure 2). Figure 2 shows the complete MS spectrum of a sample prepared by the filtration method. The annotated peaks indicate peptides for which complete sequence information could be obtained. Table 1 lists the sequences of peptides determined from collision induced dissociation in the Orbitrap mass spectrometer. Masses observed in the Orbitrap mass spectrometer were all protonated, no metalized peptides were observed. Peaks that are not annotated in Figure 2 yielded no useful MSMS data. Peptides with charge states up to +7 could be observed, but the MSMS spectra associated with charge states greater than +3 were typically not amenable to MSMS analysis. Occasionally, sequences of proline residues could be extracted from the peptides with higher charge states but complete sequences were not forthcoming. Figure 2. Open in a new tab LTQ-Orbitrap mass spectra of peptides released from denatured horse BChE by boiling. The mass range has been divided into two panels to better resolve the signals and to emphasize low intensity signals at high mass. Labeled peaks indicate peptides for which complete sequence information was obtained (numbers are mass-to-charge ratios for doubly and triply-charged ions). Intensity values are relative to the most intense peak in the spectrum. Peaks which are not marked gave either partial peptide sequence information or no sequence information. Twenty-seven peptides representing thirteen different families were extracted from the data in Figures 1A/B and 2. Assignment of a peptide to a family was based on the hetero atoms identified. Each family contains a unique set of hetero atoms, though in some instances one family might be considered to be a sub-set of another. For example, family 2 could be a sub-set of family 3; family 5 a sub-set of 6, family 10 a sub-set of 11, and family 11 a sub-set of 12 (but family 10 cannot be a sub-set of 12). These families of peptides are shown in Table 1. Family 1 is composed solely of proline residues (from 12 to 21 residues long). Family 2 has valine at the N-terminus of 11 or 12 prolines. Family 3 has leucine/isoleucine/hydroxyproline plus histidine and valine at the N-terminus of 14 prolines. Family 4 has serine at the N-terminus of 11 or 15 prolines. Family 5 has histidine at the N-terminus of 11 prolines. Family 6 has proline followed by leucine/isoleucine/hydroxyproline plus histidine or leucine/isoleucine/hydroxyproline plus histidine at the N-terminus. Family 7 has leucine/isoleucine/hydroxyproline at the N-terminus of 11 to 20 prolines. Family 8 has proline followed by three glycines at the N-terminus of 12 prolines. Family 9 has glutamate at the N-terminus of 15 prolines. Family 10 has alanine at the N-terminus of 12 to 17 prolines. Family 11 has proline and alanine at the N-terminus of 11 prolines. Family 12 has alanine at the N-terminus of 14 prolines followed by leucine/isoleucine/hydroxyproline, 4 more prolines and ending with methionine at the C-terminus. Family 13 has threonine at both the N-and C-termini, and leucine/isoleucine/hydroxyproline near the C-terminus. Of these 13 families, only the peptides from family 7 are consistent with the lamellipodin sequence that Li et. al. reported for the tetramer organizing peptides from human BChE . Examples of MSMS spectra from these peptides are given in Figures 3, 4 and 5. MALDI MSMS spectra generally yielded a-, b- and y-ion series. The intensity of the y-ion series was generally greater than that of the b-ion series. In addition, there were major fragments at high mass in MALDI MSMS spectra that were due to sodium promoted C-terminal fragmentation. Orbitrap MSMS spectra yielded b-ion and y-ion spectra and one x-ion series. Orbitrap MSMS spectra from triply-charged peptides were complicated by prominent signals due to fragments for +2 and +3 charge states. In spectra from both instruments, it was common to find a series of fragments associated with internal fragmentation initiated by proline, as well as stretches of minor peaks which corresponded to prolines to which a formal series could not be assigned. Figure 3. Open in a new tab MALDI MSMS spectrum of the 1433.2 Da peptide L/I/HypHPPP PPPPP PPPP that was isolated from boiled horse BChE by HPLC. Leucine, isoleucine and hydroxyproline cannot be distinguished because they have the same mass. Amino acid sequences are marked to show the y-ion series and b-ion series. The right-hand y-axis indicates the signal intensity in counts-per-second. Figure 4. Open in a new tab MALDI MSMS spectrum of the 1207.07 Da peptide VPPPP PPPPP PP Na+ that was isolated from boiled horse BChE by HPLC. Amino acid sequences are marked to show the y-ion and a-ion series. The sodium ion is associated with the N-terminal fragment in the a-ion series and with the C-terminal fragment in the y-ion series. The peaks marked with an asterisk () are a consequence of sodium ion promoted fragmentation. The presence of sodium is indicated by the peak at 23 Da. The right-hand y-axis indicates the absolute signal intensity in counts-per-second. Figure 5. Open in a new tab Orbitrap MSMS spectrum of peptide SPPPP PPPPP PP isolated from boiled horse BChE by HPLC. The parent ion mass was doubly charged m/z 587.3 [M+2H]2+ Amino acid sequences are annotated to show a singly-charged y-ion series, a singly-charge b-ion series and a doubly-charged b-ion series (marked with +2). The y-axis is labeled in abundance, relative to the most intense peak in the spectrum. MSMS fragmentation analysis Figure 3 shows the MSMS spectrum for the 1433.2 Da parent ion, taken with the MALDI mass spectrometer. The same peptide was detected and sequenced with the Orbitrap mass spectrometer. The parent ion mass is consistent with 12 prolines, a histidine and either leucine, isoleucine, or hydroxyproline. The spectrum shows both y-series and b-series ions for a peptide that contains 12 proline residues in a row with a Leu/Ile/Hyp His ion pair at the N-terminus. Starting at y11 there is an intense y-ion series wherein signals from alternating fragments i.e., y11, y9 and y7 are more intense than signals from intervening fragments i.e., y10, y8 and y6. This pattern is reminiscent of the polyproline fragmentation patterns described by Unnithan et. al. . In addition, there is a b-ion series. B-ions were not reported by Unnithan et. al. . They appear here in all likelihood because of the histidine residue at the b2 position. At low mass there are prominent signals for the immonium ion of histidine (110 Da), the immonium ion of proline (70 Da), and a characteristic proline fragment (126 Da) [listed on the Protein Prospector website; Additional prominent signals at low mass are due to internal polyproline fragments (not annotated). All major signals were annotated. Figure 4 shows MSMS data for the 1207.1 Da peptide, taken with the MALDI mass spectrometer. The parent ion mass is consistent with the presence of 11 prolines, a valine and a sodium atom. The presence of the sodium atom was confirmed by the mass at 23 Da. The spectrum revealed both a y-ion series and an a-ion series for a peptide containing 11 proline residues in a row with a Val at the N-terminus. The sequence interval between 1108.8 Da and the parent ion at 1207.8 Da in the y-series is consistent with a valine residue at the N-terminus. The intensities of the y-series do not follow the alternating pattern reported by Unnithan et. al. . The sodium atom remained associated with the N-terminal fragment in the a-ion series and with the C-terminal fragment in the y-ion series. Five residues in the low mass portion of the y-series gave no signals. However, the y5 mass (526.4 Da) is equal to the mass of a C-terminal proline (116 Da) plus the mass of 4 dehydro-prolines (4 × 97.05 Da), plus the mass of sodium (23 Da), and minus the mass of the proton displaced by the sodium (1 Da), results are consistent with the sequence PPPPP Na+ for the y5 ion. The a-ion series in Figure 4 shows masses in which each a-ion is heavier by 22 Da than the masses in the absence of sodium ion. The combined information from the y-ion and a-ion series gives complete confidence that the peptide contains 12 residues with the sequence VPPPP PPPPP PP 12. The most prominent peaks (marked with an asterisk) in Figure 4 are at high mass and are due to sodium ion promoted fragmentation. They include the most intense signal (at 1110.8 Da) that is consistent with the loss of a C-terminal proline; a mass that is 44 Da smaller than the parent ion (at 1163.5) which is consistent with the loss of CO 2; a mass at 1094.8 Da that is 113 Da smaller than the parent ion, and a mass at 1013.7 Da that is consistent with loss of a second proline residue from the 1110.8 Da fragment. Sodium ion promoted fragmentation is a unique form of mass spectral fragmentation that is promoted by complexes of peptides with metal ions such as sodium, lithium and potassium [25–29]. The principal fragmentation “involves the transfer of the hydroxyl group from the C-terminus of the peptide to the adjacent amino acid and subsequent loss of the residue mass of the C-terminal amino acid [97 Da for proline] leading ultimately to the production of a new alkali cationized peptide lacking the original C-terminal residue” . Additional fragmentation can occur for the newly formed alkali cationized peptide resulting in ions which are two and even three amino acids shorter than the original peptide. The signal intensity of the ion from the second fragmentation can be comparable to that from the first, but the signal intensity for the ion from third fragmentation is generally much smaller . Additional sodium promoted fragmentation can result in a prominent loss of 44 Da from the parent ion (probably due to loss of CO 2) . Finally, for polyproline peptides, loss of 113 Da from the parent ion (equal to 97 Da plus 16) is typically seen [L.M. Schopfer, unpublished observations]. All of these fragments were observed in the fragmentation spectrum for the 1207.7 Da peptide (Figure 4). Figure 5 shows MSMS data for a doubly-charged peptide ([M+2H]+2 = 587.32 Da) taken with an LTQ Orbitrap mass spectrometer. The parent ion mass is consistent with the presence of a serine and eleven prolines. Singly-charged peaks for both a y-ion series and a b-ion series are annotated. Doubly-charged b-ion masses dominate the spectrum and singly-charged b-ions yield more intense signals than singly-charged y-ions. The N-terminal b-ion pair is not resolved, but the mass interval is consistent with the presence of a serine and a proline. This pattern of signals is completely different from that reported by Unnithan et. al. demonstrating the strong influence of heteroatoms (even uncharged heteroatoms) on the fragmentation of strings of prolines. Polyacrylamide gel electrophoresis Nondenaturing gel electrophoresis was used to determine the migration of BChE from which the polyproline peptides had been released by boiling. The activity stained portion of Figure 6 (lanes 1–4) shows that human serum (lane 1) has fewer active forms of BChE than horse serum (lane 2). Human serum has 4 forms: a monomer, two types of dimer and a heavily staining tetramer. Horse serum has at least 9 forms, six of which correspond to the 4 bands in human serum: a monomer, a doublet at each dimer location, and a tetramer. Several additional forms appear between the positions of the tetramer and dimer. Purified horse BChE from Sigma contained only the tetrameric form (lane 3). Boiled horse BChE did not show any activity (lane 4). All forms of non-denatured horse BChE migrate more slowly than the corresponding forms of human BChE. We propose that this is due to the fact that horse BChE has 8 N-linked carbohydrates per subunit compared to 9 in human BChE [16–18, 31]. N-linked carbohydrates on BChE generally carry two sialic acids, therefore horse BChE would have 2 fewer negative charges per monomer than human BChE. The lower negative charge on horse BChE would explain a slower migration toward the positive electrode. Figure 6. Open in a new tab BChE resolved by nondenaturing polyacrylamide gel electrophoresis (PAGE). Lanes 1 to 4 were stained for BChE activity. Lanes 5 and 6 were stained with Coomassie Blue R-250. Lane 1, human serum 5 μl; lane 2, horse serum 5 μl; lane 3, not boiled purified horse BChE (0.015 units = 0.02 μg); lane 4, boiled purified horse BChE 0.02 μg; lane 5, not boiled purified horse BChE 2 μg; lane 6, boiled purified horse BChE 2 μg. The arrow indicates the direction of migration toward the positive electrode. Coomassie blue staining of purified horse BChE on a nondenaturing gel showed that the position of tetrameric horse BChE (not boiled, lane 5) was identical to the tetrameric BChE band in the gel stained for BChE activity (lane 3). Unexpectedly, the boiled form of horse BChE traveled at a position between tetramer and dimer, roughly at the position expected for a trimer (lane 6). Shape is an important determinant of migration on non-denaturing gels. We suggest that the boiled horse BChE dissociates into dimers and then rearranges to form aggregates such that normal dimeric migration of BChE is not seen. In support of this aggregation proposal, it can be seen that for the boiled sample in lane 6 new protein bands appear that migrate near the top of the gel, as expected for aggregates of BChE. Released polyproline peptides were not observed in the Coomassie stained gel. To further examine the quaternary structure of horse BChE we performed SDS-PAGE in the presence and absence of dithiothreitol. Figure 7 shows that in the absence of dithiothreitol denatured horse BChE migrates primarily as a dimer (lane 1). Boiling did not change this pattern (lane 2) indicating that boiling did not cause fragmentation of the subunits. Addition of dithiothreitol followed by boiling caused the majority of the BChE to be converted into monomers (lane 3). Thus, like human BChE, tetrameric horse BChE consists of a dimer of cysteine cross-linked dimers. It is known that the disulfide bond that crosslinks two subunits is located at Cys 571 in human BChE . We propose that the interchain disulfide bond in horse BChE is similarly located at Cys 571. Figure 7. Open in a new tab The SDS-PAGE analysis of horse BChE. The gradient gel was stained with Coomassie blue. Lane 1, horse BChE 2 μg; lane 2, boiled horse BChE 2 μg; lane 3, boiled horse BChE 2 μg with dithiothreitol. BLAST search: The origin of proline-rich peptides A total of 27 proline-rich peptides were released from horse BChE. These could be divided into 13 different families (Table 1). The longest representative from each family was submitted for a standard protein blastp search of the NCBI non-redundant protein sequence database using the Equus taxonomy (9789) or the mammalian taxonomy (40674). When the search was launched, it automatically switched to “short input sequence” status. Parameters for that status were: Expect Threshold = 200,000; Matrix = PAM30; Word Size = 2; GAP Costs = Existence 9, Extension 1; Compositional Adjustments = no adjustment; Filter = off; Masking = off; and 500 results requested. A perfect match was found for 9 families. Blast results are summarized in Table 2. Table 2. Blast search results on the peptides released from horse BChE \| Family \| Sequence \| Blast Resulta \| Speciesb \| GI numberc \| :--- :---: \| 1 \| PPPPPPPPPPPPPPPPPPPPP 21 \| Predicted: low quality protein: formin-like protein 2-like and UDP-N-acetylglucosamine transferase subunit ALG13 homolog \| Equus caballus \| 194222229 338729459 \| \| 2 \| VPPPPPPPPPPPP 13 \| ras-associated and pleckstrin homology domains-containing protein 1 (i.e. lamellipodin) \| Rattus norvegicus \| 341823648 \| \| 3 \| LHVPPPPPPPPPPPPPP 17 \| No perfect match \| \| \| \| 4 \| SPPPPPPPPPPPPPPP 16 \| Predicted: RIMS-binding protein 3A \| Equus caballus \| 194214017 \| \| 5 \| HPPPPPPPPPPP 12 \| R3H domain-containing protein 1 \| Homo sapiens \| 31543535 \| \| 6 \| PLHPPPPPPPPPPPPPP 17 \| no perfect match \| \| \| \| 7d \| LPPPPPPPPPPPPPPPPPPPP 21 \| Predicted: UDP-N-acetylglucosamine subunit ALG13 homolog and lamellipodin \| Equus caballus Homo sapiens \| 338729459 82581557 \| \| 8 \| PGGGPPPPPPPPPPPP 16 \| Predicted: low quality protein: hypothetical protein LOCI 00057058 \| Equus caballus \| 338729352 \| \| 9 \| EPPPPPPPPPPPPPPP 16 \| Predicted: low quality protein: formin-binding protein 4-like \| Equus caballus \| 194217903 \| \| 10 \| APPPPPPPPPPPPPPPPP 18 \| Leiomodin-2 \| Bos Taurus \| 157427900 \| \| 11 \| PAPPPPPPPPPPP 13 \| Predicted: sal-like protein 2 and leimodin-2 \| Equus caballus \| 149692720 338724213 \| \| 12 \| APPPPPPPPPPPPPPLPPPPM 21 \| no perfect match \| \| \| \| 13 \| TPPPPPPPPPPPPPPLT 17 \| no perfect match \| \| \| Open in a new tab a Data were BLAST searched using the Equus taxonomy (9789) or the Mammalian taxonomy (40674) in the NCBInr database. b If a matching sequence was not found in the equine proteome a sequence from another taxonomy was used. c The GI number is the protein accession number in the NCBI database in PubMed. d It should be emphasized that peptides from only family 7 are consistent with the sequence of the lamellipodin peptide reported by Li et. al. for the tetramer organizing peptide from human BChE. One peptide LPPPP PPPPP PPPPP PPPPP P 21 in family 7 was consistent with the tetramer-organizing peptide from lamellipodin that was reported by Li et al. for human butyrylcholinesterase PSPPL PPPPP PPPPP PPPPP PPPPP LP 27 . A second peptide VPPPP PPPPP PPP 13 in family 2 was consistent with a lamellipodin sequence from Rattus norvegicus. Four peptides were composed of only proline residues (family 1). A Blast search of the Equus caballus taxonomy (9789) using the 21-residue proline peptide from this family found two matching proteins - “Predicted: low quality protein: formin-like protein 2-like” and UDP-N-acetylglucosamine transferase subunit ALG13 homolog. Five other families were also matched with proteins from the equine proteome. When the L/I/Hyp PPPPP PPPPP PPPPP PPPPP 21 sequence in family 7 was searched using Ile or Hyp, no matches were obtained. However, when it was searched using Leu, it matched to “Predicted: UDP-N-acetylglucosamine subunit ALG13 homolog” in Equus caballus and lamellipodin in Homo sapiens. Consequently, we have assigned leucine as the N-terminal residue for this peptide. Of the remaining families, peptides from three were matched with proteins from other taxonomies: one from Rattus norvegicus, one from Homo sapiens and one from Bos Taurus. There was no perfect match from mammalian taxonomy for peptides in 4 families. DISCUSSION Function of polyproline peptides is to organize subunits into tetramers Native BChE in horse, mouse, and human plasma is 98% tetrameric. Previous studies have shown that full-length BChE expressed in culture medium is predominantly monomeric and dimeric, but that addition of poly(L-proline) to the culture medium or co-expression with the proline-rich N-terminus of ColQ increased the amount of tetrameric BChE to 70% [22, 23]. Similarly, live mice transfected with adenovirus expressing mouse BChE had predominantly dimeric BChE in plasma. Addition of 100 μM poly(L-proline) to mouse plasma converted nearly 100% of the BChE into tetramers . Conversely, native tetrameric BChE from horse and human plasma , as well as native tetrameric AChE from fetal bovine serum (Biberoglu, K, Schopfer, LM, Saxena, A, Tacal, O & Lockridge, O; submitted) have been shown to release polyproline peptides following protein denaturation. This means the polyproline peptides are part of the structure of soluble tetrameric BChE and AChE. We conclude that the function of polyproline peptides is to serve as tetramer organizing peptides. One polyproline peptide per tetramer There has been no X-ray crystal structure showing how full-length BChE subunits associate with the proline-rich peptides. In the following discussion we will use amino acid composition analysis to argue that there is one polyproline peptide per BChE tetramer. The amino acid sequences of human and horse plasma BChE have been determined. Both contain 574 amino acids per subunit [16–18]. There are 31 proline residues per subunit in the sequence of the horse BChE. However, amino acid composition analysis of tetrameric horse BChE showed that there are 36.6 proline residues per subunit (Table 3). The number of observed prolines per subunit is 5.6 greater than expected. This means that there are 22 extra proline residues per tetramer. The longest proline-rich peptide that we observed for horse BChE contained 21 proline residues. The similarity in these two values strongly argues that there is one proline-rich peptide per tetramer for horse BChE. Table 3. Number of polyproline peptides per BChE tetramer \| Number of residues \| Human BChE \| Horse BChE \| :---: \| accession number \| P06276 \| Q9N1N9 \| \| # residues per subunit \| 574a \| 574b \| \| # prolines from subunit sequence \| 30aa \| 31b \| \| # prolines from amino acid composition of tetramer \| 36a \| 36.6c \| \| # extra prolines per subunit \| 6 \| 5.6 \| \| # extra prolines per tetramer \| 24 \| 22 \| \| calculated polyproline length \| 24 \| 22 \| \| # polyproline peptides per BChE tetramer \| 1 \| 1 \| Open in a new tab a Taken from Lockridge et al., 1987b. b Taken from Wierdl et al.; 2000 and Moorad et al., 1999. c Taken from Teng et al.; 1979 The same argument can be drawn for human BChE, where the number of proline residues determined by amino acid composition analysis (36) is 6 greater than the number of proline residues in the sequence (30) . The extra proline residues can be accounted for by the presence of one polyproline peptide (containing approximately 24 prolines) per tetramer. Polyproline peptides of this size were isolated from human BChE . We conclude that soluble tetrameric BChE includes one polyproline peptide per tetramer. Many of the peptides that we identified are shorter than the 22-residues projected by the amino acid composition analysis (some as small as 12 residues, see Table 1). On this basis, one might suspect that more than one polyproline peptide could be found in the tetramerization domain. However, Dvir et. al. have shown that the 15-residue PRAD peptide is more than sufficient to occupy the entire tetramerization domain of cholinesterase (see Figure 8). Thus it is unlikely that two 12-residue peptides would be bound. Figure 8. Open in a new tab Model of the AChE tetramer in the presence of the full-length ColQ protein. Figure reproduced from . AChE monomers are displayed as globular units with their tryptophan-rich amphiphilic α-helices protruding out from the top of each unit. The ColQ protein, or alternatively, the tetramer organizing polyproline peptide (yellow) projects through the middle of the AChE tetramer complex, occupying the core of the tetramerization domain on top, and extending as a single chain on the bottom. Polyproline peptide interacts with the tetramerization domain Assembly of BChE subunits into tetramers requires the presence of the 40 amino acids at the C-terminus . Deletion of these residues yields monomeric BChE with full catalytic activity, but reduced stability. The 40 C-terminal residues of BChE constitute the tetramerization domain. The tetramerization domain adopts an amphiphilic alpha-helical structure with 7 conserved aromatic amino acids on one side of the helix. A similar tetramerization domain containing 7 conserved aromatic amino acids is present in human AChE. Dvir et al solved the crystal structure of a synthesized peptide representing the tetramerization domain of human AChE in complex with a synthesized proline-rich peptide. The complex contained 4 parallel alpha-helical tetramerization peptides wrapped around a single antiparallel 15-residue proline-rich peptide LLTPP PPPLF PPPFF. The conformation of the proline-rich peptide was that of a left-handed polyproline II helix. The global structure was described as a left-handed screw within a left-handed threaded hollow tube formed by the four tetramerization peptides . Dvir et al. produced a model of the AChE tetramer linked to the full length ColQ protein through the proline-rich region at the N-terminus of ColQ . Figure 8 shows that the ColQ protein traverses the entire AChE tetramer, interacting with the tetramerization domain at its C-terminal end, passing through the center of the AChE tetramer, and extending its N-terminal end out the other side. This configuration will accommodate a variety of sizes of polyproline peptides. The structures of AChE and BChE are very similar so it is reasonable to propose that the polyproline peptides obtained from horse BChE (including the 21-residue polyproline peptide) will form a similar structure. Thus, we propose that the relative size and location of the organizing peptides in fully formed BChE tetramers is represented by the structure in Figure 8. Multiple origins of polyproline peptides According to blast search results the tetramer-organizing, proline-rich peptides from horse BChE derive from at least 9 different proteins. The well-established tetramer-organizing proteins, ColQ and PRiMA are not in that list of proteins (Table 2) and the proline-rich attachment domains from ColQ, LLTPP PPPLF PPPFF 15, and PRiMA, PPPPL PPPPP PPPPP 15, do not appear in the list of observed peptides (Table 1). These observations indicate that soluble BChE in horse serum is not generated by cleavage of BChE from its well characterized membrane anchored sites. Furthermore, it suggests that serum BChE may be associated with a variety of proteins at some time in its synthesis. Because BChE appears to have associated with a variety of polyproline containing proteins it is of interest to explore the nature of mammalian proteins that contain extended sequences of prolines. We therefore performed a Blastp search of the NCBI non-redundant protein database for proteins containing at least 15 sequential prolines. A minimum of 15 sequential prolines was chosen because 10 of the tetramer promoting polyproline families in our study contained sequences of 16 to 21 prolines. We excluded matches for entries that were translated from DNA without identifying the associated protein (that is, entries from databases such as the Center for Genome Dynamics (CG), Kazusa cDNA sequencing project (KIAA), and the Japanese cDNA library (RIKEN) among others). We included isoforms of a given protein when they appeared. Eighty-one matches were found (8 contained 25 contiguous prolines or more, 25 contained 20–24 prolines, while 48 contained 15–19). Twenty-eight of the matches were to taxonomy Homo sapiens, two were to Equus caballus. A quarter of the matches were to proteins involved in transcription. Another quarter were to proteins that bind to actin and/or were involved in cytokinesis, cell polarity or neurite formation. Six bound to RNA or DNA. Three were involved in cell-cell adhesion and two were associated with ubiquitin. There was no obvious tendency for these proteins to be membrane bound. For most proteins the polyproline sequence was located 50 or more residues from the nearest end of the protein. To date soluble BChE and AChE tetramers are the only proteins reported in the literature that embed short polyproline peptides in their structure. The polyproline peptides fit no single precursor protein and therefore arise from multiple genes. More heterogeneity in short polyproline sequences Li et al. reported on a group of proline-rich, tetramer-organizing peptides from human and horse BChE that matched a 39-residue sequence in lamellipodin PSPPL PPPPP PPPPP PPPPP PPPPP LPSQS APSAG SAAP 39 . The peptides extracted from human BChE in that study ranged in mass from 2074 to 2878 Da, while those extracted from horse BChE ranged in mass from 2171 to 2878 Da. Additional peptides, at lower mass and of lower intensity, were visible in the mass spectra of both horse and human BChE extracts but they were not examined. In the current study, the peptides from horse BChE that we examined were in the mass range 1173 to 2098 Da. Larger peptides had much weaker intensities and did not yield productive sequence information. Only two of the peptides that we detected could be matched to lamellipodin. The obvious question is “What differences exist between the two experiments that might explain the differences in observation?” The biggest technical difference between the experiments is in the lasers on the mass spectrometers that were used. This is an important consideration because polyproline is quite sensitive to the laser intensity during desorption/ionization in the MALDI mass spectrometer. That is, increasing laser intensity causes proline-proline fragmentation in the source [LM Schopfer unpublished observation]. The laser used by Li et al. (AB Sciex Voyager DE Pro from Applied Biosystems) is arguably less intense than the one used in the current experiments (MALDI-TOF-TOF 4800 from Applied Biosystems). This conclusion is based on the following considerations: 1) The DE Pro uses a 337 nm nitrogen laser, while the 4800 uses a 355 nm Nd:YAG laser. The absorbance maximum for a thin film of dried α-CHCA matrix (which was used in both experiments) is at 371 nm . Therefore the α-CHCA/sample mixture will absorb more light per pulse in the 4800 source than in the DE Pro source because the laser excitation wavelength is closer to the α-CHCA absorbance maximum in the 4800. 2) The laser pulse rate in the 4800 is 10-fold higher (200 Hz) than that in the DE Pro (20 Hz). Therefore, light will be delivered to the sample faster in the 4800 than in the DE Pro increasing the sample temperature to a greater extent. Higher temperatures should promote greater fragmentation. 3) The laser spot size is smaller in the 4800 than in the DE Pro. Therefore the laser energy will be concentrated on a smaller area of the sample in the 4800 than in the DE Pro, thereby increasing the laser intensity experienced by the sample. The predicted consequences of these factors would be increased fragmentation of the polyproline peptides in the source of the 4800, leading to a decrease in the size of the ions entering the mass spectrometer. This in turn would provide us smaller peptides to study. Since the smaller peptides were not investigated by Li et al, they will not have seen the same results that we have. The ultimate conclusion is that our results differ from Li et al. because we studied a different set of peptides. It is noteworthy that the electrospray ionization source used on the Orbitrap mass spectrometer introduces the sample into the mass spectrometer more gently. The list of ions that we observed in the Orbitrap included four of the eight ions reported for horse BChE by Li et al (2171, 2566, 2663, and 2878 Da). However, fragmentation of the latter three was unproductive. Fragmentation of the 2171 Da ion showed a string of 11 prolines but the complete sequence could not be determined. Conclusion Soluble, tetrameric BChE in plasma is an assembly of four identical 574-residue BChE subunits in complex with a heterogeneous group of non-covalently bound, short polyproline peptides. EXPERIMENTAL Materials Highly purified BChE [EC 3.1.1.8] from equine serum (catalog number C1057), 5,5'-dithiobis-2-nitrobenzoic acid (catalog number D8130), sodium dodecylsulfate (SDS, electrophoresis grade, catalog number L3771), glycine (electrophoresis grade, G8898) and formic acid (puriss p.a. for mass spectrometry, catalog number 94318) were from Sigma (a member of the Sigma-Aldrich group St. Louis, MO, U.S.A.). α-Cyano-4-hydroxy cinnamic acid (α-CHCA, catalog number 70990) and butyrylthiocholine iodide (catalog number 20820) were from Fluka (a member of the Sigma-Aldrich group). Acetonitrile (DNA sequencing grade, catalog number BP-1170), dithiothreitol (electrophoresis grade, catalog number BP172), bromophenol blue (electrophoresis grade, catalog number BP114) and Coomassie Brilliant Blue R-250 (electrophoresis grade, catalog number BP101) were from Fisher Scientific (a member of the Thermo Fisher Scientific group, Waltham MA, U.S.A.). Trifluoroacetic acid was of sequencing grade (>99.9%) from Beckman (Brea, CA, U.S.A., catalog number 290204). All other chemicals were of biochemical grade. α-CHCA was prepared as a saturated solution (10 mg/ml) in 50% acetonitrile/water plus 0.3 % trifluoroacetic acid (v/v). BChE activity BChE activity was assayed by recording increase in absorbance at 412 nm of a 2 ml solution containing 1 mM butyrylthiocholine in 0.1 M potassium phosphate buffer pH 7.0 at 25 °C, in the presence of 0.5 mM 5,5'-dithiobis-(2-nitrobenzoic acid) in a Gilford spectrophotometer interfaced to MacLab 200 (ADInstruments Pty Ltd., Castle Hill, Australia). The molar extinction coefficient for the product is 13,600 M−1 cm−1 . Units of activity are defined as micromoles of butyrylthiocholine hydrolyzed per minute. The lyophilized horse BChE from Sigma contains approximately 30% protein and 70% buffer salts. A bottle of Sigma horse BChE (lot 084K7351) that contains 3325 mg solid and 1,000,000 Sigma units contains 326 mg BChE protein and 234,700 units as measured under our conditions. We estimated the Sigma horse BChE to be 81% pure based on our parameters for 100% pure BChE of 720 units/mg where we measured units of activity with 1 mM butyrylthiocholine in 0.1 M potassium phosphate buffer pH 7 at 25 °C, and we measured protein concentration by absorbance at 280 nm where a 1 mg/ml solution has an absorbance of 1.8. Filtration to separate tetramer organizing peptides from horse BChE Horse BChE was desalted by dialysis against 2 × 4 L of water. Twenty ml of desalted horse BChE (0.3 mg/ml in water) was concentrated to 1.5 ml in an Amicon stirred cell using a YM 30 membrane 30 kDa cutoff (Millipore, Billerica, MA, U.S.A.). The 1.5 ml of 4 mg/ml sample was divided into two fractions of equal volume. One fraction was boiled for 5 minutes to denature the protein. Both fractions were filtered through Micron ultracel YM 10 spin filters (Millipore) with a 10,000 Da cut-off to separate free peptides from residual protein. The solution that passed through the filter membrane was recovered, concentrated to 0.3 ml in a vacuum centrifuge (Speedvac model sc100 from Savant) and saved for mass spectrometry. In the end, the samples (boiled and not boiled) contained peptides equivalent to 10 mg BChE/ml (assuming no losses during handling) which is equivalent to 118 μM BChE monomer (MW = 84,551). Assuming one peptide per tetramer this would predict 30 μM peptide. HPLC purification of tetramer organizing peptides Lyophilized Sigma horse BChE, 29.8 mg containing 2.92 mg BChE protein, was dissolved in 2.92 ml water to make a 1 mg/ml BChE solution. Non-covalently bound peptides were released by heating the BChE in a boiling water bath for 5 minutes. After boiling, 2 ml of the solution (equivalent to 2 mg BChE or 24 nmoles) were filtered through a nylon 0.2 μm syringe filter to remove particulates, and injected onto a Zorbax 300 SB C-18 reverse-phase column (Agilent Technologies, Santa Clara, CA, U.S.A.) attached to a Waters 625 HPLC system (Milford, MA, U.S.A.). The HPLC was operated at room temperature (22 °C) at a flow rate of 0.5 ml/min. Buffer A was 0.1 % trifluoroacetic acid in water (v/v); buffer B was 0.09 % trifluoroacetic acid in acetonitrile (v/v). Peptides were eluted with a gradient of 0–60 % buffer B in 60 min. Absorbance was recorded at 210 nm. The HPLC eluent was collected in 1 min fractions (0.5 ml each). Each fraction was concentrated to 10–50 μl in a vacuum centrifuge (Savant SpeedVac) and saved at 4°C for mass spectrometry. In the end, the fractions contained peptides equivalent to 2 mg of BChE (assuming no losses) or 2340 to 468 μM BChE monomer (MW = 84,551). Assuming one peptide per tetramer this would predict 585 to 117 μM peptide. MALDI-TOF-TOF mass spectrometry Samples used for MALDI-TOF-TOF mass spectrometry were prepared by both the HPLC and the filtration methods. The filtered samples (boiled and not boiled) contained peptides equivalent to 10 mg BChE/ml (assuming no losses during handling) or 30 μM peptide. The HPLC fractions contained peptides equivalent to 2 mg of BChE (assuming no losses during handling) in 0.01 to 0.05 ml or 585 to 117 μM peptide. All MALDI-TOF-TOF experiments were performed on an Applied Biosystems MALDI TOF-TOF 4800 mass spectrometer equipped with a 355 nm Nd-YAD laser (Applied Biosystems, Framingham, MA, U.S.A.). The samples (0.5 μl) were spotted on a MALDI target plate (Opti-TOF 384 well Insert from Applied Biosystems), air-dried and overlaid with 0.5 μl of α-CHCA. Mass spectra were acquired in positive ion reflector mode, under delayed extraction conditions (500 ns), using an acceleration voltage of 20 kV, with laser intensity of 4000–6000 volt, a mass range of 1000–4000 Da, a detector voltage multiplier of 0.75, with low mass gate on and low mass gate offset equal to zero. Mass spectra shown are the average of 500 laser shots collected from randomly selected locations on the target spot (50 pulses per location). Selected ions were fragmented by collision induced dissociation (CID), using a 1 kV method, with air as the collision gas at 2×10−6 Torr, a precursor mass window of ± 1 Da, metastable suppression on, a detector voltage multiplier of 0.95, using factory calculated delayed extraction values (DE1 = 370 ns, DE2 = 37,271 ns), and with timed ion selector activated. The identity of the fragments in the MSMS spectra were assigned manually using the Data Explorer software (version 4.9 from Applied Biosystems) with the aid of the Proteomics Toolkit ( and the MS Product algorithm in Protein Prospector v 5.9.4 from the University of California, San Francisco mass spectrometry facility (prospector.ucsf.edu/prospector/mshome.htm). LTQ-Orbitrap Samples used for LTQ-Orbitrap mass spectrometry were prepared by the filtration method. Filtered samples (boiled and not boiled) contained peptides equivalent to 10 mg BChE/ml (assuming no losses during handling) or 30 μM peptide. A 3.5 μl aliquot of the filtered sample was diluted with 6.5 μl of 0.1 % formic acid to make a 10 μM peptide solution. LC/MSMS analysis was performed on an LTQ Orbitrap quadrupole mass spectrometer (Thermo Scientific a part of Thermo Fisher Scientific, Rockland, IL, U.S.A.) using electrospray ionization. Five microliters of sample were loaded onto a C18 reverse phase trap column (CapTrap Peptide from Michrom BioResources, Auburn, CA, U.S.A., catalog # TRI/25109/32) and washed with 2 % acetonitrile/water plus 0.1 % formic acid (v/v). Peptides were transferred to a C18 reverse phase separation column (Picofrit BioBasics C-18 from New Objective, Woburn, MA, U.S.A., catalog # PF 7515-100H052) and eluted with a 45 minute linear gradient starting with 98 % solvent A (2 % acetonitrile/water plus 0.1 % formic acid; v/v) and 2 % solvent B (98 % acetonitrile/water plus 0.1 % formic acid; v/v) and ending with 35 % solvent A and 65 % solvent B. The flow rate was 250 nl/min. The effluent was electrosprayed directly into the mass spectrometer. Data were collected in a data dependent manner with each cycle of data collection consisting of one high-resolution mass spectrum (over a 300 to 2000 Da mass range) taken with the Orbitrap and five MSMS fragmentation spectra taken with the LTQ ion trap. Collision induced dissociation was used for fragmentation with helium as the collision gas (at 1×10−3 Torr) at a normalized collision energy of 35 (or about 35 % of the maximum collision energy of 5 volts). The activation time was 30 ms, and the activation Q was 0.25. Parent ions of any charge state with a minimum signal intensity of 50,000 counts per second were accepted for fragmentation. After two data collections for a given parent ion mass were completed, that mass was excluded from further analysis for 60 s. Fragmentation spectra were analyzed manually using the Qual Browser feature of Xcalibur software (v 4.9 from Thermo Scientific) with the aid of the Proteomics Toolkit ( and the MS Product algorithm in Protein Prospector v 5.9.4 from the University of California, San Francisco mass spectrometry facility (prospector.ucsf.edu/prospector/mshome.htm). Non-denaturing gradient gel electrophoresis Polyacrylamide gradient gels (11×18 cm, 4–30%) with a 4% stacking gel, 0.75 mm thick, were prepared in a Hoefer SE6000 gel apparatus (Hoefer Scientific Instruments, San Francisco, CA, U.S.A.). The running gel was prepared in 375 mM TrisCl buffer pH 8.9. The stacking gel was prepared in 125 mM TrisCl buffer pH 6.8. The upper tank buffer was 20 mM Tris glycine pH 8.9, while the bottom tank buffer was 60 mM TrisCl pH 8.1. Electrophoresis was run at 208 V constant voltage for 24 h (5000 volt-hours) at 4 °C. BChE samples (in 20 mM TrisCl, 1 mM EDTA, pH 7.5) were diluted with distilled water and mixed (1 to 1) with 50 % (v/v) glycerol containing 0.004% (w/v) bromophenol blue, to give the final concentrations indicated in the following paragraphs. Some samples were boiled for 3 minutes (refer to Figure 5 legend). For total protein staining, the equivalent of 2 μg (1.44 units) of purified horse BChE was loaded per lane. After electrophoresis, the gel was stained with Coomassie Brilliant Blue R-250. Protein staining requires 100 times more BChE per lane than activity staining. For BChE activity staining, the equivalent of 5 μl of human plasma, 5 μl of horse serum or 5 μl of purified horse BChE (3 u/ml) were loaded per lane. After electrophoresis, the gel was stained for BChE activity by the method of Karnovsky and Roots . The staining solution contained 180 ml of 0.2 M sodium maleate pH 6.0, 15 ml of 0.1 M sodium citrate, 30 ml of 0.03 M cupric sulfate, 30 ml of 5 mM potassium ferricyanide, and 0.18 g butyrylthiocholine iodide in a total volume of 300 ml. The gel was incubated, with gentle shaking, at room temperature for 3 to 5 h until brown bands of activity appeared. The reaction was stopped by washing the gel with water. Denaturing gradient gel electrophoresis: SDS Polyacrylamide gradient gels (11×18 cm, 4–30%) with a 4% stacking gel, 0.75 mm thick, were prepared as described for non-denaturing gels. The tank buffers were prepared as described for non-denaturing gels except that the upper tank buffer was supplemented with 0.1% SDS (w/v). Electrophoresis was run at 125 V constant voltage for 24 h (3000 volt-hours) at 4°C. BChE samples were diluted (1 to 1) with SDS buffer (62.5 mM TrisCl, 10 (v/v) glycerol, 2% SDS (w/v), 0.012% bromophenol (w/v) blue, pH 7.2). Dithiothreitol (50 mM) was added to some samples; some samples were boiled for 3 minutes (refer to Figure 6 legend). The equivalent of 2 μg of horse BChE was loaded per lane. Gels were stained with Coomassie Brilliant Blue R-250. ACKNOWLEDGEMENTS Mass spectra were obtained with the support of the Mass Spectrometry and Proteomics core facility at the University of Nebraska Medical Center. This work was supported by a TÜBITAK grant from the Scientific and Technological Research council of Turkey to OT, a fellowship from Hacettepe University (to KB), and an NIH grant [P30CA36727 (to the Eppley Cancer Center)]. Abbreviations BChE butyrylcholinesterase α-CHCA α-cyano-4-hydroxycinnamic acid BLAST basic logical alignment search tool MSMS the process of mass spectral fragmentation LC/MSMS liquid chromatography/mass spectrometry including a peptide fragmentation stage LTQ linear ion trap quadrupole mass spectrometer MALDI-TOF matrix-assisted laser desorption ionization time of flight NCBI National Center for Biotechnology Information PRAD proline-rich attachment domain PRiMA proline-rich membrane anchor; ColQ, collagen tail protein Footnotes Conflict of interest: There are no conflicts of interest between the authors and the content of this paper. REFERENCES 1.Brimijoin S, Gao Y. Cocaine hydrolase gene therapy for cocaine abuse. 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[DOI] [PubMed] [Google Scholar] ACTIONS View on publisher site PDF (934.8 KB) Cite Collections Permalink PERMALINK Copy RESOURCES Similar articles Cited by other articles Links to NCBI Databases On this page Summary INTRODUCTION RESULTS DISCUSSION EXPERIMENTAL ACKNOWLEDGEMENTS Abbreviations Footnotes 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
10357	https://pmc.ncbi.nlm.nih.gov/articles/PMC5659657/	DNA topology in chromatin is defined by nucleosome spacing - 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 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 Sci Adv . 2017 Oct 27;3(10):e1700957. doi: 10.1126/sciadv.1700957 Search in PMC Search in PubMed View in NLM Catalog Add to search DNA topology in chromatin is defined by nucleosome spacing Tatiana Nikitina Tatiana Nikitina 1 Laboratory of Cell Biology, Center for Cancer Research, National Cancer Institute, National Institutes of Health, Bethesda, MD 20892, USA. Find articles by Tatiana Nikitina 1, Davood Norouzi Davood Norouzi 1 Laboratory of Cell Biology, Center for Cancer Research, National Cancer Institute, National Institutes of Health, Bethesda, MD 20892, USA. Find articles by Davood Norouzi 1, Sergei A Grigoryev Sergei A Grigoryev 2 Department of Biochemistry and Molecular Biology, Pennsylvania State University College of Medicine, Hershey, PA 17033, USA. Find articles by Sergei A Grigoryev 2,, Victor B Zhurkin Victor B Zhurkin 1 Laboratory of Cell Biology, Center for Cancer Research, National Cancer Institute, National Institutes of Health, Bethesda, MD 20892, USA. Find articles by Victor B Zhurkin 1, Author information Article notes Copyright and License information 1 Laboratory of Cell Biology, Center for Cancer Research, National Cancer Institute, National Institutes of Health, Bethesda, MD 20892, USA. 2 Department of Biochemistry and Molecular Biology, Pennsylvania State University College of Medicine, Hershey, PA 17033, USA. Corresponding author. Email: sag17@psu.edu (S.A.G.); zhurkin@nih.gov (V.B.Z.) Received 2017 Mar 27; Accepted 2017 Sep 27; Collection date 2017 Oct. Copyright © 2017 The Authors, some rights reserved; exclusive licensee American Association for the Advancement of Science. No claim to original U.S. Government Works. Distributed under a Creative Commons Attribution NonCommercial License 4.0 (CC BY-NC). This is an open-access article distributed under the terms of the Creative Commons Attribution-NonCommercial license, which permits use, distribution, and reproduction in any medium, so long as the resultant use is not for commercial advantage and provided the original work is properly cited. PMC Copyright notice PMCID: PMC5659657 PMID: 29098179 Nucleosome spacing variations cause topological polymorphism of chromatin that can affect DNA supercoiling and transcription. Abstract In eukaryotic nucleosomes, DNA makes ~1.7 superhelical turns around histone octamer. However, there is a long-standing discrepancy between the nucleosome core structure determined by x-ray crystallography and measurements of DNA topology in circular minichromosomes, indicating that there is only ~1.0 superhelical turn per nucleosome. Although several theoretical assumptions were put forward to explain this paradox by conformational variability of the nucleosome linker, none was tested experimentally. We analyzed topological properties of DNA in circular nucleosome arrays with precisely positioned nucleosomes. Using topological electrophoretic assays and electron microscopy, we demonstrate that the DNA linking number per nucleosome strongly depends on the nucleosome spacing and varies from −1.4 to −0.9. For the predominant {10 n + 5} class of nucleosome repeats found in native chromatin, our results are consistent with the DNA topology observed earlier. Thus, we reconcile the topological properties of nucleosome arrays with nucleosome core structure and provide a simple explanation for the DNA topology in native chromatin with variable DNA linker length. Topological polymorphism of the chromatin fibers described here may reflect a more general tendency of chromosomal domains containing active or repressed genes to acquire different nucleosome spacing to retain topologically distinct higher-order structures. INTRODUCTION In eukaryotic chromatin, the DNA double helix is repeatedly supercoiled in the nucleoprotein particles called nucleosomes. The core of a typical nucleosome contains 145 to 147 base pairs (bp) of DNA making 1.7 left-handed superhelical turns around a histone octamer (1–3). The nucleosomes play an essential role in regulating all DNA-dependent processes such as transcription, replication, recombination, and repair through local and dynamic unfolding of chromatin (4–6). In the cell nucleus, the nucleosome cores are connected by relatively extended DNA linkers forming beads-on-a-string nucleosome arrays. The nucleosome arrays fold into higher-order structures that mediate DNA packing and availability for the DNA recognizing machinery (7, 8). Transcriptionally active chromosomal domains accumulate unconstrained negative supercoiling (9, 10), indicating that active chromatin may have a different linear distribution of torsional stress compared to condensed heterochromatin. These findings necessitate experiments aimed at relating the intrinsic chromatin structural variations predicted in vitro and in silico to DNA topologies and associated functional interactions in vivo. Monitoring of DNA supercoiling in circular nucleosome arrays (minichromosomes) by topological gel assays has long been used to analyze topological properties of DNA in chromatin. These assays are based on the helical periodicity of DNA making one turn per ~10.5 bp under physiological conditions (11, 12), which corresponds to helical Twist ≈ 34.5°. When the ends of the DNA chains are constrained in covalently closed circular DNA (ccDNA), the changes in the DNA Twist (Tw) are compensated by the changes in the Writhe (Wr) according to the well-known equation Δ Lk = Δ Tw + Wr, where the linking number (Lk) is defined as the number of times one strand of the duplex turns around the other (13–16). Note that the three parameters describing the DNA topology (Δ Lk, Δ Tw, and Wr) are measured relative to the relaxed state of DNA. Therefore, Δ Wr = Wr because the Writhe of unconstrained DNA is zero. For a chain of nucleosomes, the x-ray crystal structure of the nucleosome core (1–3) predicts the generation of DNA supercoiling with Wr = Δ Lk = −1.7 per nucleosome if the DNA Twist is not changed (Δ Tw = 0). However, multiple measurements of Δ Lk in ccDNA from either native minichromosomes of SV40 virus (17, 18) or nucleosome arrays reconstituted from histones and circular DNA (19) persistently resulted in experimental values of Δ Lk = −1.0. This discrepancy is known as the linking number paradox (16, 20–22). When acetylated histones were used for reconstitution, the linking number changed to Δ Lk = −0.8 (23), which is consistent with contribution of the nucleosome core structure to the associated DNA topology, according to the surface linking theory (24). How could the nucleosome structure and topological properties of the nucleosome chain be reconciled? One explanation is that the DNA Twist in the nucleosome cores could differ from that of naked DNA in solution (20–22, 24–27). The other possibility is that a special path of linker DNA could alter Writhe (15, 28). Specific linker DNA geometries consistent with the observed DNA topology have been proposed (28–33). However, the actual linker DNA path within topologically constrained chromatin domains remained obscure because of the absence of adequate circular minichromosome models with precise nucleosome positions. During the past dozen years, there has been a significant progress in the structural studies of chromatin, largely due to a wide usage of strongly positioned “601” nucleosome sequence (34). In particular, the seminal studies of Richmond and colleagues showed the two-start zigzag fibers in solution (35) and in crystal form (36). Their results, together with the high-resolution cryo-electron microscopy (cryo-EM) data obtained by Song et al. (37), strongly support the two-start zigzag organization of chromatin fibers for relatively short nucleosome repeat lengths (NRLs) = 167, 177, and 187 bp. In addition, the EM images presented by Rhodes and co-workers (38) confirmed zigzag organization for NRL = 167 bp, whereas for NRL = 197 bp and longer, more tightly packed fibers were observed, especially in the presence of linker histones. All these structures were obtained for arrays of strongly positioned 601 nucleosomes (34), with NRL varying from 167 to 237 bp in increments of 10 bp. Provided that the nucleosome core is 147 bp (2), the linker length L varies from 20 to 90 bp, that is, L belongs to the {10 n} series. On the other hand, it is known that in vivo, the linker sizes are close to {10 n + 5} values (39–42). Thus, the structural data mentioned above correspond to linker lengths with occurrences in vivo that are relatively small. Recently, we have both experimentally and theoretically analyzed positioned nucleosomal arrays (NAs) with linker lengths belonging to the {10 n} and {10 n + 5} series. Using analytical ultracentrifugation and EM imaging, we demonstrated that fibers with L = 25 bp have less propensity to fold in a compact state than those with L = 20 or 30 bp (43). Furthermore, using computer modeling, we predicted the two families of energetically stable conformations of the two-start zigzag fibers, one with L = 10 n and the other with L = 10 n + 5 to be topologically different, the former imposing ~50% higher absolute \|Δ Lk\| values in the circular DNA (see Fig. 1) (44, 45). Here, we experimentally addressed these predictions for circular nucleosome arrays containing regularly spaced repeats of the 601 nucleosome positioning sequence (34). Fig. 1. Models of two-start chromatin fibers with NRLs = 167 and 172 bp. Open in a new tab (A and B) Energy-optimized regular fibers (43, 44) containing 12 nucleosomes, with NRL = 167 bp (A) and NRL = 172 bp (B). The DNA linking number per nucleosome, Δ Lk = −1.37 and −0.93, respectively. The DNA is shown in alternating blue and orange colors to emphasize the two stacks of nucleosomes; the DNA “entry” points are shown as red balls. The histone cores are shown in two colors—The entry sides are in yellow, and the “exit” sides are in white. In this manner, it is easier to distinguish the fiber configurations. In addition, the green arrows indicate different DNA folding pathways in the two topoisomers. (C and D) Representative configurations of the 167x11 and 172x11 circular NAs obtained during Monte Carlo simulations (see main text and Fig. 3 for details). Note that the circular topoisomers (C and D) are significantly distorted and extended compared to the regular conformers (A and B). The nucleosome-free DNA fragments are shown as white tubes. RESULTS Plasmid-based DNA circles We prepared circular DNA templates using two different methods. First, we constructed plasmid-based DNA circles containing 12 repeats, the 601 sequence with NRL = 167 or 172 bp inserted into pUC19 vector (see Materials and Methods). In these 4.7-kb-long circles, about half of DNA belongs to pUC19 vector, which is not a nucleosome positioning sequence. To ensure that the presence of vector DNA does not affect the resulting measurements of Δ Lk difference, we also prepared DNA minicircles consisting purely of 601 repeats (see below). The plasmid-based circular constructs (denoted p-167x12 and p-172x12) were prepared by standard methods of cell transformation, DNA extraction, and purification (see Materials and Methods). Plasmid DNA extracted from cells has a superhelical density σ ≈ −0.06 (46), which is expected to facilitate the formation of nucleosomes. Therefore, we used these plasmid-based circles for reconstitution of nucleosomes by a standard salt dialysis method. Examination of the resulting NAs by electrophoretic mobility retardation assay on native agarose gel (fig. S1) shows practically complete incorporation of DNA into the nucleosome bands at the 100% loading of the core histones. We also digested the NAs with restriction enzymes to mononucleosomes and verified the equal core histone loading by the nucleosome band shift assay (fig. S2A). Next, DNA in the reconstituted NAs was relaxed with topoisomerase I (Topo I), deproteinized, and run on agarose gels containing different concentrations of intercalator chloroquine (CQ). Comparison of the electrophoretic mobility of supercoiled circular DNA (scDNA) obtained with variable histone loading allows monitoring increase in the superhelical density of DNA that accompanies the formation of nucleosomes (Fig. 2). Our approach is illustrated in fig. S3. Fig. 2. Topological difference between NAs with 167- and 172-bp NRLs reconstituted on plasmid DNA. Open in a new tab Plasmid-based circular DNA templates p-167x12 (lanes 1, 4, 6, 8, and 10) and p-172x12 (lanes 2, 5, 7, 9, and 11) were reconstituted with 0, 25, 50, 75, and 100% core histones (top), treated with Topo I, and the circular DNA was isolated and separated on agarose gels run in the absence (A) and presence of CQ (1.5 μg/ml) (B) and CQ (4.0 μg/ml) (C) in the gel and TAE buffer. Lanes 3 and 12 show molecular weight markers. The strongest bands are indicated by blue circles for the 167-bp NRL and by red circles for the 172-bp NRL. The numbers of superhelical turns in the topoisomers corresponding to these bands are given in the same colors (B). In particular, for the bands denoted +8 and +9 (lanes 4 and 5), the shifts from the top of the gel (nicked circles, form II) are counted directly. For the bands denoted +14 (lanes 1 and 2), the shift from the top is evaluated from the relative shift of six supercoils between these bands and the band +8 (lane 4). Difference in DNA linking number between the 167- and 172-bp NRL constructs, or Δ(Δ Lk), equals 1 at 25%, 5 at 50%, 6 at 75%, and 7 at 100% loading of core histones. Note that the presence of intercalator CQ introduces additional positive supercoils in DNA and changes its electrophoretic mobility. For example, the DNA circles relaxed in the absence of histones (lanes 1 and 2) become strongly supercoiled and run at the bottom of the gel. By contrast, the strong negative superhelical density of DNA obtained at 100% loading of histones (lanes 10 and 11) is partially compensated by CQ, and thus, its mobility is decreased (see fig. S3 for details). As follows from Fig. 2A (lanes 1 and 2), both DNA circles, p-167x12 and p-172x12 (without core histones), are relaxed and run at the top of the gel. A gradual increase in the core histone loading increases the superhelical density and accordingly changes the electrophoretic mobility, entirely consistent with the scheme shown in fig. S3. Specifically, the DNA samples obtained with 25% loading of histones have a “modest” superhelical density (Fig. 2A, lanes 4 and 5), which is increased at 50 to 75% loading, as reflected in a relatively high gel mobility (lanes 6 to 9). Finally, the 100% loading of core histones further increases superhelical density of DNA, so that the tightly coiled DNA migrates to the very bottom of the gel and forms zones of poorly resolved bands (Fig. 2A, lanes 10 and 11). To resolve individual fast-migrating bands, we repeated the electrophoresis in the presence of various concentrations of CQ, which is known to induce additional positive supercoils in circular DNA (47). The DNA circles obtained with a small loading of histones (0 and 25%) become positively supercoiled and run at the bottom of the gel [Fig. 2, B (lanes 1 and 2) and C (lanes 1, 2, 4, and 5)]. The DNA purified from NAs with 50 to 75% loading has intermediate superhelical density; its gel mobility is reversed in the presence of CQ [Fig. 2, B (lanes 6 and 7) and C (lanes 6 to 9)], in agreement with published data (48). For DNA with the highest superhelical density (with 100% histone loading) in the presence of CQ (1.5 μg/ml), we observed a strong decrease in mobility so that individual bands can be resolved on the gel (Fig. 2B, lanes 10 and 11). Now, using gels shown in Fig. 2B (CQ = 1.5 μg/ml), we can accurately determine the change in the DNA linking number induced by the formation of nucleosomes. On the basis of the analysis presented above, we conclude that, for the low histone loading (0 and 25%), the p-167x12 and p-172x12 constructs have similar number of positive superhelical turns. The dominant topoisomers are either located at the same position, band +14 (lanes 1 and 2), or shifted by one band (lanes 4 and 5; bands +8 and +9). For the 100% loading, both DNA circles, p-167x12 and p-172x12, have negative superhelical turns and now show a marked difference between the two samples. Evaluation of the linking number difference between the p-167x12 and p-172x12 nucleosome arrays is straightforward and gives a change of ΔΔ Lk = 7 negative supercoils in 167-bp versus 172-bp arrays. In the cases of 50 and 75% loadings (lanes 6 and 9), the strongest bands are located too close to the top of the gels, and it is impossible to decide whether they correspond to the negative or positive supercoils from the one-dimensional (1D) gels. To resolve all topoisomers on a single gel, we run 2D electrophoreses (fig. S4). The CQ concentrations were selected so that the strongest bands for the p-172x12 construct would be located at the top of the gel, and for the p-167x12 construct, they would be on the left side of the gel. As a result, we find that the dominant topoisomers in the two constructs are shifted by 5 to 6 supercoils for 50% histone loading (fig. S4A) and by 6 to 7 supercoils for 75% loading (fig. S4B). Thus, we conclude that the topological difference between the p-167x12 and p-172x12 constructs, ΔΔ Lk, monotonically increases, reaching about seven supercoils when the loading of core histones increases to 100%. Provided that the nucleosome arrays have the same number of 601 nucleosomes and the vector-associated nucleosomes are not supposed to change between the two samples, our data are consistent with the model-predicted (Fig. 1) difference in ΔΔ Lk = 0.5 (per 601 nucleosome) between the p-167x12 and p-172x12 constructs. DNA minicircles To ensure that the Lk difference between the two types of chromatin fibers observed above is not affected by the presence of vector DNA, we analyzed DNA minicircles containing only the 601 repeats. This approach is similar to that previously used in construction of the minicircles based on sea urchin 5 S DNA nucleosome positioning sequences by Simpson et al. (19) and Norton et al. (23). The 2-kb-long supercoiled DNA circles were prepared by ligation and relaxation by Topo I in the presence of various concentrations of ethidium bromide (EtBr) (see Materials and Methods), which allowed to produce scDNA with defined superhelical density (49). By gradually increasing concentration of EtBr up to 2.0 μg/ml and comparing electrophoretic mobility of scDNA in the presence of variable amounts of CQ, we were able to modulate the number of supercoils in each sample (fig. S5). In particular, the scDNA obtained in the presence of EtBr (2.0 μg/ml) has 12 negative supercoils (fig. S5D), which corresponds to a superhelical density σ = −0.06 in the 2-kb-long minicircle [note that this superhelical density is comparable to that in native Escherichia coli plasmids (46)]. The 167x12 and 172x12 minicircles produced the same superhelical density (172x12 bands are shown as asterisks in fig. S5). We used 167x12 scDNA as a reference for measuring the number of DNA supercoils in scDNA extracted from the NAs. Reconstitution of nucleosomes on the scDNA formed in the presence of EtBr (2.0 μg/ml) gave a result very similar to that in the case of the plasmid-based circles, namely DNA in 167x12 NAs is more negatively supercoiled than that in 172x12 NAs (Fig. 3, A and B). However, the topological difference (ΔΔ Lk) between the two arrays is only three supercoils. In addition, the number of DNA supercoils in these arrays (11 and 8 supercoils, respectively) is less than expected for complete nucleosome saturation (Fig. 3, A and B). This result is consistent with the number of nucleosomes directly counted on the EM micrographs (Fig. 3, C and D, and the supplementary EM data set file (pdf). Instead of 12 nucleosomes expected for the 100% loading of core histones, only about nine nucleosomes, on average, are formed on both constructs, 167x12 and 172x12 (Fig. 3, C and D). Most likely, the limited number of nucleosomes formed on the 2-kb-long circles is due to the decreased conformational flexibility of DNA during nucleosome reconstitution in the minicircles (compared to the 4.7-kb-long plasmid-based circles). Fig. 3. Topological difference between 167x12 and 172x12 NAs reconstituted on minicircular DNA. Open in a new tab Nucleosomes were reconstituted on the scDNA templates prepared in the presence of EtBr (2 μg/ml) (A to D) and EtBr (4 μg/ml) (E to H). (A and E) Scans of gels in (B) and (F). The strongest bands are marked with “+”. (B and F) Electrophoretic mobility of the DNA topoisomers extracted from NAs was compared with the mobility of scDNA templates obtained at [EtBr] = 2 μg/ml (lanes 1 and 2). Gels were run in the presence of CQ (8 μg/ml) (B) and CQ (32 μg/ml) (F). Lane 5 shows molecular weight markers. Note that the scDNAs with 167-bp (lane 1) and 172-bp (lane 2) NRLs have similar distributions of topoisomers, with the strongest bands corresponding to Δ Lk = −12 (fig. S5D). NAs reconstituted on these scDNAs (obtained at [EtBr] = 2 μg/ml) have Δ Lk = −11 and −8 (lanes 3 and 4) (B). Arrays reconstituted on scDNA obtained at [EtBr] = 4 μg/ml have Δ Lk = −15 and −10 (lanes 3 and 4 ) (F). (C and G) Graphs showing the average number of nucleosomes (10.9) on 167x12 and 172x12 arrays and distributions of the number of nucleosomes per one minicircle (calculated from EM_images_interactive PDF file). Normalization of the values Δ Lk = −15 and −10 [shown in (F)] gives the DNA linking number per nucleosome, Δ Lk = −1.38 and −0.92 for the 167x12 and 172x12 arrays, respectively. (D and H) Representative transmission EM images of relaxed minicircular 167x12 and 172x12 arrays reconstituted on scDNA obtained at [EtBr] = 2 μg/ml (D) and [EtBr] = 4 μg/ml (H). Scale bars, 50 nm. To facilitate the formation of nucleosomes, we increased the number of negative supercoils in minicircular DNA using scDNA prepared in the presence of EtBr (4.0 μg/ml) and maintaining equal distribution of superhelical topoisomers between the two constructs (fig. S6). The efficient saturation of the DNA minicircles by histones was confirmed by Nla III restriction digestion assay excising equal ratios of occupied nucleosomes from the 167 and 172 minicircles at equal histone loading (fig. S2B). As a result, the number of nucleosomes detectable on the EM micrographs increased to ~11, on average, the distribution of the corresponding numbers narrowed, and the minicircles containing the full amount of 12 nucleosomes became rather abundant (Fig. 3, G and H). Accordingly, the number of supercoils increased up to 15 and 10 for NRLs = 167 and 172 bp, respectively (Fig. 3, E and F). For illustration, the representative conformations of 167x11 and 172x11 NAs obtained in the course of energy minimization are shown in Fig. 1 (C and D). Note that the average numbers of nucleosomes in the 167-bp NRL arrays (10.86) and in the 172-bp NRL arrays (10.94) are virtually equal (Fig. 3G); therefore, we can exclude the possibility that the stronger superhelical density observed for NRL = 167 bp is the result of the higher histone loading in this case. Thus, the topological difference between the uniform {10 n} and {10 n + 5} NAs increased to five supercoils, or 45% of the total number of nucleosomes, in excellent agreement with our model (Fig. 1) predicting the dependence of the nucleosome array folding and topology on the linker DNA length. Previously, in similar experiments with minicircular reconstitutes of 207-bp NRL sea urchin 5 S DNA nucleosomes, it had been shown that the reconstitution persistently resulted in experimental values of Δ Lk = −1.01 (±0.08) (19). Consistent with this result, we found here that, in minicircular DNA containing 12 repeats of 601 nucleosome positioning sequence with NRL = 207 bp, the Δ Lk change was about −1.1 (figs. S7 and S8). We also observed that reconstitution of nucleosomes on another minicircular DNA construct with NRL = 188 bp brought changes of Δ Lk ≈ −1.3. Thus, we see that, in the three NAs belonging to the {10 n} series, 12x167, 12x188, and 12x207, the absolute value of DNA linking number, \|Δ Lk\|, gradually decreases from 1.4 to 1.1 with increase in the linker length, in a good agreement with our predictions based on energy minimization of the fiber topoisomers (fig. S9) (45). DISCUSSION Our results reconcile the nucleosome core crystal structure with topology of the nucleosome array. For the first time, we show that the level of DNA supercoiling in the NAs varies by as much as ~50% depending on the DNA linker length. The DNA linking number per nucleosome, Δ Lk, changes from −1.4 for L = 20 bp to −0.9 for L = 25 bp (in DNA minicircles consisting purely of 601 repeats). Here, like in the previous nucleosome minicircle reconstitution experiments (19, 23), the negative supercoiling of template DNA was used to promote histone octamer assembly efficiency. Our results show that the number of assembled nucleosomes is equal for minicircles with equal number of nucleosome templates and initial supercoiling, although the resulting numbers of unconstrained supercoils are different. Thus, the observed Δ Lk changes are due to the changes in the nucleosome linker length rather than initial supercoiling density or the number of nucleosomes. Because the same sequence 601 was used in both cases, one can be certain that the observed change in DNA supercoiling is due to the changes in the global DNA pathway (Writhe) rather than to the alteration of DNA twisting in nucleosomes. Note that the nucleosome positioning signals embedded in this sequence are so strong that they secure the same positioning of the 601 nucleosomes both in array and in a single nucleosome with a single-nucleotide precision (50, 51). Furthermore, changing the DNA linker length in the 601 repeats is sufficient to significantly alter the overall DNA topology in hybrid plasmids containing a mixture of positioned and nonpositioned nucleosomes and thus mimicking native NAs. By showing that altered linker spacing is sufficient to change the DNA topology in circular minichromosomes, our results provide a decisive proof to the notion that the chromatin higher-order structure is defined by nucleosome rotational settings (43, 44). Previous crystallographic and EM studies of Richmond and colleagues (35, 36), together with the high-resolution cryo-EM data (35), revealed the two-start zigzag organization of chromatin fibers for the linkers belonging to the {10 n} series (L = 20, 30, and 40 bp). Overall, the DNA folding in these fibers remains the same, the main difference being gradual monotonic changes in dimensions of fibers due to increase in linker length. By contrast, we observe here an abrupt change in DNA topology when comparing the fibers with L = 20 and 25 bp, consistent with the folding trajectories of DNA linkers being markedly different in the two cases (Fig. 1). The topological polymorphism of a chromatin fiber depending on linker DNA trajectory, first noted by Crick (15), is reflected in the early studies by Worcel et al. (30), Woodcock et al. (31), and Williams et al. (32) who presented space-filling models of the chromatin fiber with the DNA linking number Δ Lk varying from −1 to −2 per nucleosome, depending on the DNA trajectory. The increased fiber “plasticity” observed for the {10 n + 5} linkers (43–45) is biologically relevant because the {10 n + 5} values are frequently found in vivo (39–43). On the other hand, the more tightly folded {10 n} structures appear to facilitate nucleosome disk stacking by interactions between histone H4 N-terminal domain and histone H2A/H2B acidic patch at the nucleosomal interface (35) and make the structure to be especially sensitive to the effects of histone H4 acetylation (52). The experimentally observed topological polymorphism and an extrapolation of our model to include a wide range of variations of linker lengths provide a simple explanation for the observed DNA linking number Δ Lk ≈ −1.0 in native SV40 minichromosomes (17, 18). These minichromosomes do not have a strong nucleosome positioning and thus may give rise to Δ Lk values widely distributed between −0.8 and −1.5 per nucleosome (fig. S9). For nucleosome linker lengths close to 60 bp (NRL = 207 bp), which is the most common value for eukaryotic chromatin, the mean value of the Δ Lk is about −1.05 per nucleosome (fig. S9). Earlier analysis of the distribution of Δ Lk for a nucleosome array with randomized nucleosome orientation (29) also showed the Δ Lk values close to −1 per nucleosome. In living cells, the DNA-dependent processes, such as transcription, constantly generate positive and negative local DNA supercoiling in accord with the twin-domain model of transcription (53). Because Topo I and Topo II do not completely remove this dynamically induced supercoiling, transcriptionally active chromosomal domains accumulate regions with the higher and lower superhelical densities, which can be detected by a combination of psoralen cross-linking with genome-wide sequencing (9, 54, 55). Recent genome-wide mapping with psoralen revealed extended (on average, 100 kb) overtwisted domains correlated with transcriptional activity and chromatin decondensation (10). In contrast to the transcriptionally active chromatin, transcriptionally silent domains in yeast acquire negative supercoiling (56). Our data suggest that the level of DNA supercoiling is closely tied with the nucleosome spacing, so that in the repressed state, the supercoiling could become more negative (Δ Lk ≈ −1.5), and the overall chromatin structure could become more compact (Fig. 1A) because of a larger fraction of nucleosomes acquiring the {10 n} class of repeats (for example, NRL = 167 bp). This is because, in the nucleosome arrays belonging to the {10 n} series, the nucleosome stacking is enhanced by histone H4 N-tail/histone H2A/H2B acidic patch interactions (1, 44). We believe that our in vitro findings reflect important aspects of chromatin structure adapted to rendering the repressed chromatin more compact by acquiring special nucleosome spacing. More specifically, we propose that, by altering nucleosome spacing, transcription may impose long-lasting topological changes in the nucleosome arrays and thus generate topologically and structurally distinct chromosomal domains of {10 n} class of nucleosome repeats that will maintain their more compact and repressed states. Future studies, especially detailed 3D analysis by advanced imaging techniques such as high-resolution cryo-EM tomography, should reveal the actual DNA configurations in active and repressed chromatin. MATERIALS AND METHODS Preparation of circular DNA constructs The plasmid-based DNA circles have 12 repeats of 601 nucleosome positioning sequence with repeat length of either 167 or 172 bp inserted into pUC19 vector (43). For the pUC19-167x12 plasmid, additional 60 bp were added to make it the same DNA length as the pUC19-172x12 plasmid. Two self-complementary single-stranded DNA fragments, purchased from Integrated DNA Technologies, were annealed so that the resulting double-stranded DNA (dsDNA) had single-stranded overhangs with Xba I site on one end and Eco RI site on the other. The pUC19-167x12 construct was digested with Eco RI and Xba I restriction enzymes and ligated with dsDNA with overhangs using standard procedures. Ligation products were used to transform Stbl2 competent cells; the transformants were grown, and plasmid DNA was prepared using a plasmid purification kit (Qiagen) according to the manufacturer’s instructions. DNA minicircles containing only 12-mer 601 nucleosomes with repeat lengths of 167, 172, and 207 bp were prepared by ligations of linear DNA fragments at low concentration to ensure circle formation rather than ligation of multiple linear fragments. The (167x12)+60-, 172x12-, and 207x12-bp linear DNA fragments were prepared by cutting off DNA fragments between Xba I and Spe I restriction sites from pUC19 plasmids with corresponding inserts. The DNA fragments were run on 1% agarose gel, bands with inserts were isolated, and DNA was extracted with Promega Wizard SV Gel and PCR Clean-Up System, cleaned with phenol/chloroform, and precipitated. Linear DNA fragments were suspended in 10 mM tris-HCl (pH 7.5). For ligation reaction, DNA was diluted to 1 ng/μl in a total volume of 800 μl in multiple tubes. A total of 3200 units of T4 ligase (New England Biolabs) were added to each tube. Reaction was kept at 16°C for about 48 hours. Every 2 to 3 hours, additional 800 ng of DNA was added to each reaction mixture. Ligation mixtures were concentrated and washed in 10 mM tris-HCl (pH 7.5) using Amicon Ultra-4 centrifugal filters units with a molecular mass cutoff of 100 K. To prepare scDNA with defined superhelical densities (49), 3 μg of DNA samples were treated with various concentrations of EtBr in the presence of 10 units of human Topo I (cat. no. TG2005H, TopoGEN) in Topo I buffer [10 mM tris-HCl (pH 7.9), 1 mM EDTA, 150 mM NaCl, 0.1% bovine serum albumin (BSA), 0.1 mM spermidine, and 5% glycerol]. Mixtures were kept at 37°C for 2 hours. The reaction was stopped by the addition of Na acetate to 0.33 M, and samples were cleaned with phenol/chloroform. After precipitation, scDNA was suspended in 10 mM tris-HCl (pH 7.5). Preparation of circular NAs NAs were reconstituted by mixing purified chicken erythrocyte core histones (57, 58) with circular DNA. The reconstitution was performed as described (43) by a salt dialysis from 2.0 to 0.5 M NaCl, followed by dialysis to 10 mM NaCl, 10 mM tris-HCl (pH 8.0), and 0.25 mM EDTA. For Topo I treatment, 1 μg of NAs was mixed with 10 units of Topo I in Topo I buffer in 50-μl reaction volume for 30 min at 37°C. The core histones were then removed by SDS/proteinase K treatment, and DNA was cleaned with Promega Wizard SV Gel and PCR Clean-Up System. DNA samples were run on 1% agarose gels containing different concentrations of CQ in tris-acetate-EDTA (TAE) buffer. Restriction enzyme digestion assay The plasmid-based circular NAs were diluted to 40 μg/ml in a 40-μl reaction buffer [50 K acetate, 20 mM tris-acetate (pH 7.9), 1 mM Mg acetate, and BSA (100 μg/ml)] and digested with the mixture of restriction enzymes (Xba I, Spe I, and Nla III), 20 units each, overnight at 37°C. Reaction was stopped by the addition of 10 mM EDTA. The minicircular arrays were digested in the same buffer with restriction enzyme Nla III only. Samples were run on a type IV agarose gel for 2 hours 20 min at 80 V/cm in a horizontal gel electrophoresis apparatus with constant buffer recirculation. 2D electrophoresis Gels were run in 1% agarose at 3 V/cm for 18 hours in TAE buffer with CQ concentration as indicated. The gels were removed, soaked for 3 hours in TAE buffer with appropriate CQ concentration, placed into e/f apparatus perpendicular to the first direction, and run again for 18 hours at 3 V/cm. Gels were stained in GelRed stain (cat. no. 41002, Biotium) and visualized using Bio-Rad Image Lab imager. Transmission EM of NAs For EM, the reconstituted and Topo I–treated nucleosome arrays were dialyzed in 10 mM Hepes, 10 mM NaCl, and 0.1 mM EDTA buffer for 4 hours and then fixed with 0.1% glutaraldehyde overnight at +4°C. Glutaraldehyde was then removed by dialysis in the same buffer (without glutaraldehyde). Samples were diluted with 50 mM NaCl to ~1 μg/ml, applied to carbon-coated and glow-discharged EM grids (T1000-Cu, Electron Microscopy Science), and stained with 0.04% aqueous uranyl acetate, as previously described (43). Dark-field images were obtained and digitally recorded as described using JEM-1400 electron microscope (JEOL USA) at 120 kV with SC1000 Orius 11-megapixel charge-coupled device camera (Gatan Inc.). The number of nucleosomes formed in each array was counted as described (50). Electron micrographs of individual particles and the overlaying counting masks are included in the layered EM data set file (pdf). The counting masks can be easily removed from the view using free Adobe Acrobat Reader DC software ( or any other Adobe Acrobat version that supports layered PDF documents. Supplementary Material supp_3_10_e1700957__index.html (2.3KB, html) Acknowledgments We are grateful to D. Clark and F. Kouzine for valuable discussions and to H. Chen for technical assistance with EM conducted at the Penn State Hershey Imaging Facility. Funding: This work was supported by Intramural Research Program of the NIH, National Cancer Institute (to V.B.Z.), and NSF grant 1516999 (to S.A.G.). Author contributions: S.A.G. and V.B.Z. designed the experiments. T.N. and S.A.G. performed the experiments. D.N. performed the structural computations. T.N., S.A.G., and V.B.Z. wrote the manuscript. All authors discussed the results and commented on the manuscript. Competing interests: The authors declare that they have no competing interests. Data and materials availability: All data needed to evaluate the conclusions in the paper are present in the paper and/or the Supplementary Materials. Additional data related to this paper may be requested from the authors. SUPPLEMENTARY MATERIALS Supplementary material for this article is available at fig. S1. Analysis of circular nucleosome array reconstitution by electrophoretic mobility shift assay. fig. S2. Analysis of circular nucleosome array reconstitution by restriction enzyme digestion and electrophoretic DNA band-shift assay. fig. S3. Binding of intercalator CQ to supercoiled DNA changes its electrophoretic mobility. fig. 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[DOI] [PubMed] [Google Scholar] Associated Data This section collects any data citations, data availability statements, or supplementary materials included in this article. Supplementary Materials supp_3_10_e1700957__index.html (2.3KB, html) 1700957_SM.pdf (1.3MB, pdf) 1700957_Dataset_file.pdf (9.3MB, pdf) Supplementary material for this article is available at fig. S1. Analysis of circular nucleosome array reconstitution by electrophoretic mobility shift assay. fig. S2. Analysis of circular nucleosome array reconstitution by restriction enzyme digestion and electrophoretic DNA band-shift assay. fig. S3. Binding of intercalator CQ to supercoiled DNA changes its electrophoretic mobility. fig. S4. 2D gel electrophoresis of DNA extracted from NAs with 167- and 172-bp NRLs. fig. S5. Modulating negative supercoiling in DNA minicircles in the presence of various concentrations of EtBr. fig. S6. Preparation of supercoiled DNA constructs 167x12 and 172x12 relaxed by Topo I in the presence of EtBr (4.0 μg/ml). fig. S7. Negative supercoiling of DNA minicircles with 207-bp NRL in the presence of EtBr. fig. S8. Negative supercoiling of circular NAs with 207-bp NRL. fig. S9. Linking number per nucleosome, Δ Lk, in regular fibers with various linker lengths. 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10358	https://www.quora.com/What-is-the-complete-solution-of-quadratic-inequality-ax%C2%B2-bx-c-0	Something went wrong. Wait a moment and try again. Quadratic Inequality Math Problem Solution Inequalities (general) Algebra Class Solving Quadratic Equatio... Quadratic Polynomials Basic Algebra 5 What is the complete solution of quadratic inequality ax²+bx+c>0? J Deurwaarder Former TVET Consultant at Various, Consultant (2004–2018) · Author has 1.6K answers and 684.1K answer views · 3y Quadratic inequalities are solved by first solving the quadratic equality ax²+bx+c = 0 Cases to consider i. a > 0 and D < 0, D = 0, D > 0 ii. a < 0 and D < 0, D = 0, D > 0 i. Case a> 0 and D <0 [no real solutions for x] ax²+bx+c > 0 for all real x. The graph of y = ax²+bx+c is completely above the x-axis. a > 0 and D = 0 [the graph of y = ax²+bx+c touches the x-axis, say for x = p, hence ax²+bx+c > 0 for all real x, except x = p a > 0 and D > 0 The quadratic equation has 2 different real solutions p and q [assume p < q] ax²+bx+c > 0 for x < p or x > q Case ii Similar to case i Case a< 0 and D <0 [no r Quadratic inequalities are solved by first solving the quadratic equality ax²+bx+c = 0 Cases to consider i. a > 0 and D < 0, D = 0, D > 0 ii. a < 0 and D < 0, D = 0, D > 0 i. Case a> 0 and D <0 [no real solutions for x] ax²+bx+c > 0 for all real x. The graph of y = ax²+bx+c is completely above the x-axis. a > 0 and D = 0 [the graph of y = ax²+bx+c touches the x-axis, say for x = p, hence ax²+bx+c > 0 for all real x, except x = p a > 0 and D > 0 The quadratic equation has 2 different real solutions p and q [assume p < q] ax²+bx+c > 0 for x < p or x > q Case ii Similar to case i Case a< 0 and D <0 [no real solutions for x of the quadratic equation] ax²+bx+c > 0 for NO real x. The graph of y = ax²+bx+c is completely below the x-axis. a >< 0 and D = 0 [the graph of y = ax²+bx+c touches the x-axis, say at x = p, but lays for the rest below the x axis, hence ax²+bx+c > 0 for NO real x a < 0 and D > 0 The quadratic equation has 2 different real solutions p and q [assume p < q] ax²+bx+c > 0 for p <x < q Related questions What is the quadratic expression, ax^2+bx+c, such that the quadratic equation ax^2+bx+c=0 will have(1) -5 and 1 as solution AND 'a' will be smaller than 'c'? How can I know if the given points are solution of the quadratic inequality y≥x²-5x+1? What are the solutions for quadratic inequality in the general case? What is the solution set of the quadratic inequalities then graph x²+9x+14>0? How can we write a quadratic inequality with a solution: -2/7 < x < 6? Would it just be Y= (7x+2) (x-6)? John Fryer Author has 890 answers and 885.5K answer views · 3y Difficult to get a complete solution to an equation with three unknowns. BUT Consider the constants are all positive with a = 1 or a = -1 Dividing throughout by a will not significantly change the inequality and we get x² + bx/a + c/a > 0 Where the results for other values of x will be similar CASE 1 a is positive If a is positive and equal to 1 then large areas will agree with the inequality except around the origin. For example take a, b and c to be positive and only the region -b to 0 will not agree with the inequality. The constant having a small effect close to x = 0 and x = -b restricting the par Difficult to get a complete solution to an equation with three unknowns. BUT Consider the constants are all positive with a = 1 or a = -1 Dividing throughout by a will not significantly change the inequality and we get x² + bx/a + c/a > 0 Where the results for other values of x will be similar CASE 1 a is positive If a is positive and equal to 1 then large areas will agree with the inequality except around the origin. For example take a, b and c to be positive and only the region -b to 0 will not agree with the inequality. The constant having a small effect close to x = 0 and x = -b restricting the part that is in disagreement with the inequality. If x = 2 then the area in disagreement is halved CASE 2 a is negative If a is negative and equal to -1 then the reverse applies Now only a narrow band of values of x are in accord with the inequality from x = 0 to x = b Again the constant has a small effect extending these restricted values of x both near x = 0 and x = b If x = - 2 then the area in agreement is reduced by a half. Philip Lloyd Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist. · Author has 6.8K answers and 52.8M answer views · 4y Related Let x1 and x2 be the roots of the quadratic equation ax2+bx+c=0 By using quadratic formula, find the roots of ax2+bx+c=0 Find x1+ x2 and x1 x2. What relationships do you observe? The use of x1 and x2 will make this question must too messy so I am going to let the solutions (or roots) be α and β The above is a very awkward w... The use of x1 and x2 will make this question must too messy so I am going to let the solutions (or roots) be α and β The above is a very awkward w... Sponsored by Grammarly Is your writing working as hard as your ideas? Grammarly’s AI brings research, clarity, and structure—so your writing gets sharper with every step. Ananwaya Shankharupa Studied at Ahmedabad International School · 3y Using the quadratic equation x=[-b +- sqrt(b^-4ac)]/2a, when ax^2+bx+c=0. Now to make this above inequality true, x will simply have to be either: x>[-b +- sqrt(b^-4ac)]/2a or: x<[-b +- sqrt(b^-4ac)]/2a Gordon M. Brown Math Tutor at San Diego City College (2018-Present) · Upvoted by BowTangey , PhD Mathematics, Iowa State University (1988) · Author has 6.2K answers and 4.3M answer views · 2y Related How do you solve the quadratic inequality in the form x²+bx+c<0 or >0 with solution set {x:-3<x<2}? (x + 3) (x - 2) = x^2 + x - 6 b = 1, c = -6 Now I urge you to study carefully the two graphs below, and determine the connections between them. (x + 3) (x - 2) = x^2 + x - 6 b = 1, c = -6 Now I urge you to study carefully the two graphs below, and determine the connections between them. Promoted by The Penny Hoarder Lisa Dawson Finance Writer at The Penny Hoarder · Updated Sep 16 What's some brutally honest advice that everyone should know? Here’s the thing: I wish I had known these money secrets sooner. They’ve helped so many people save hundreds, secure their family’s future, and grow their bank accounts—myself included. And honestly? Putting them to use was way easier than I expected. I bet you can knock out at least three or four of these right now—yes, even from your phone. Don’t wait like I did. Cancel Your Car Insurance You might not even realize it, but your car insurance company is probably overcharging you. In fact, they’re kind of counting on you not noticing. 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What is the best definition for showing that x^2 > 8x – 9 is a quadratic inequality? What is the zeros of quadratic polynomial ax2+ bx +c? Philip Lloyd Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist. · Author has 6.8K answers and 52.8M answer views · 2y Related How do you find the solution set of this quadratic inequality and how do you graph it x^2-3x-18 ≥0? Instead of messing about with test points, I prefer to do a quick sketch graph as follows… Consider the graph y = x^2 – 3x – 18 = (x + 3)(x – 6) So y values are positive ABOVE the x axis so the sets of x values are… x ≤ – 3 or x ≥ 6 —————————————————————————————————————- EDIT This idea can easily be applied to cubics and other functions and the sketch graphs are easy to produce. See this example too… Instead of messing about with test points, I prefer to do a quick sketch graph as follows… Consider the graph y = x^2 – 3x – 18 = (x + 3)(x – 6) So y values are positive ABOVE the x axis so the sets of x values are… x ≤ – 3 or x ≥ 6 —————————————————————————————————————- EDIT This idea can easily be applied to cubics and other functions and the sketch graphs are easy to produce. See this example too… Jörg Straube M.Sc. in Computer Science, ETH Zurich (Graduated 1987) · Upvoted by BowTangey , PhD Mathematics, Iowa State University (1988) · Author has 6.2K answers and 1.7M answer views · 2y Related How do you solve the quadratic inequality in the form x²+bx+c<0 or >0 with solution set {x:-3<x<2}? x^2 + bx + c < 0 means: “What x-values will result in negative y-values”. Or “In what x-interval the parabola is below the x-axis” As the solution set is {x: -3 < x < 2} it means that at x = -3 and x = 2 the parabola crosses the x-axis: (x + 3)(x - 2) = x^2 + x - 6 → b = 1, c = -6 For all x between -3 .. 2, the parabola is below the x-axis If the the quadratic > 0 for the solution set {x: -3 < x < 2}, then the same parabola is open downwards: As you see the sign of the x^2 < 0. So, this can not be a solution to the original question. x^2 + bx + c < 0 means: “What x-values will result in negative y-values”. Or “In what x-interval the parabola is below the x-axis” As the solution set is {x: -3 < x < 2} it means that at x = -3 and x = 2 the parabola crosses the x-axis: (x + 3)(x - 2) = x^2 + x - 6 → b = 1, c = -6 For all x between -3 .. 2, the parabola is below the x-axis If the the quadratic > 0 for the solution set {x: -3 < x < 2}, then the same parabola is open downwards: As you see the sign of the x^2 < 0. So, this can not be a solution to the original question. Robert Paxson BSME in Mechanical Engineering, Lehigh University (Graduated 1983) · Author has 3.9K answers and 4M answer views · 3y Related How do you find the solution set of this quadratic inequality and how do you graph it 6y^2-5y≥6? 6y2−5y≥6 6y2−5y−6≥0 (2y−3)(3y+2)≥0 The critical lines can be found by setting each factor equal to zero: 2y−3=0 y=32 and 3y+2=0 y=−23 These two horizontal lines partition the x-y plane into three regions. Since the given inequality includes zero, both of these lines are in the solution set. For the inequality to be true, both factors must have the same sign such that their product is positive. Therefore, the solution set is: y≥32 y≤−23 A plot of this solution set looks like this: 6y2−5y≥6 6y2−5y−6≥0 (2y−3)(3y+2)≥0 The critical lines can be found by setting each factor equal to zero: 2y−3=0 y=32 and 3y+2=0 y=−23 These two horizontal lines partition the x-y plane into three regions. Since the given inequality includes zero, both of these lines are in the solution set. For the inequality to be true, both factors must have the same sign such that their product is positive. Therefore, the solution set is: y≥32 y≤−23 A plot of this solution set looks like this: Philip Lloyd Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist. · Upvoted by BowTangey , PhD Mathematics, Iowa State University (1988) · Author has 6.8K answers and 52.8M answer views · 5y Related What is the following quadratic inequality in vertex form, 2(x-2) ^2+3<1? This question “seems” to be a little ambiguous. Do you actually mean “Solve this quadratic inequality which is in vertex form”? I would answer like this… There are no real number solutions but there certainly are complex solutions. To draw this graph I need an extra axis for the imaginary parts of the x values. Here is the fabulous result! This question “seems” to be a little ambiguous. Do you actually mean “Solve this quadratic inequality which is in vertex form”? I would answer like this… There are no real number solutions but there certainly are complex solutions. To draw this graph I need an extra axis for the imaginary parts of the x values. Here is the fabulous result! Herb Weil Author has 68 answers and 48.1K answer views · 4y Related How do I solve the quadratic inequality x^2+11x+10>_ O? If you just graph the function, you will see a parabola, with x-intercepts at (-10,0) and (-1,0). The parabola divides the real plane into two sets of points, those inside the parabola, those with x-coordinates between -10 and -1, and those outside the parabola, all the remaining x-coordinates. The question now becomes which set of points will solve the inequality. The easiest way is to chose a point in one of the two areas, if that point solves the inequality then so will all the others in that area. So lets chose the point where x=0. That would give us 0^2+11(0)+10 which is 10. Is 10 greater If you just graph the function, you will see a parabola, with x-intercepts at (-10,0) and (-1,0). The parabola divides the real plane into two sets of points, those inside the parabola, those with x-coordinates between -10 and -1, and those outside the parabola, all the remaining x-coordinates. The question now becomes which set of points will solve the inequality. The easiest way is to chose a point in one of the two areas, if that point solves the inequality then so will all the others in that area. So lets chose the point where x=0. That would give us 0^2+11(0)+10 which is 10. Is 10 greater than or equal to zero? Yes it is, therefore, all the points on the outside of the parabola would be solutions to the inequality. Since the inequality is greater than or equal to, then all the points on the parabola would also be solutions. We can verify that by checking the point where x=-1; (-1)^2+11(-1)+10 is equal to zero, which satisfies the inequality. I have included a copy of my graph. Hope this helps. MargeryW Trying to score A+ in math... · Author has 706 answers and 1.3M answer views · 3y Related What is the graph of this quadratic inequality: x^2+4x+3<0. How do you explain it? First draw the graph of the function y=x2+4x+3. Find the x-intercept: x2+4x+3=0, (x+1)(x+3)=0, x=−1 or x=−3. The inequality x2+4x+3<0 means that only the part below the y-axis of this function is kept, that is, the part in the interval (-3, -1). Then we can sketch the graph of x2+4x+3<0: First draw the graph of the function y=x2+4x+3. Find the x-intercept: x2+4x+3=0, (x+1)(x+3)=0, x=−1 or x=−3. The inequality x2+4x+3<0 means that only the part below the y-axis of this function is kept, that is, the part in the interval (-3, -1). Then we can sketch the graph of x2+4x+3<0: Gordon M. Brown Math Tutor at San Diego City College (2018-Present) · Author has 6.2K answers and 4.3M answer views · 4y Related What is the solution to the quadratic inequality, 2x^2-11x-6<=0? This inequality is about as straightforward as they get. Let’s see first whether we can factor the expression on the left: 2x^2 – 11x - 6 (2x + 1) (x - 6) By setting each factor equal to zero, we can establish critical points with which to test the intervals: 2x + 1 = 0 x = -1/2 x - 6 = 0 x = 6 Thus our critical points will be -1/2 and 6. Next, we’ll select three test points between and around the critical points. The test points -1, 1, and 7 will do nicely here: (2(-1) + 1) (-1 - 6) = (-1) (-7) = 7; false; not ≤ 0 (2(1) + 1) (1 - 6) = (3) (-5) = -15; true; ≤ 0 (2(7) + 1) (7 - 6) = (15) (1) = 15; fa This inequality is about as straightforward as they get. Let’s see first whether we can factor the expression on the left: 2x^2 – 11x - 6 (2x + 1) (x - 6) By setting each factor equal to zero, we can establish critical points with which to test the intervals: 2x + 1 = 0 x = -1/2 x - 6 = 0 x = 6 Thus our critical points will be -1/2 and 6. Next, we’ll select three test points between and around the critical points. The test points -1, 1, and 7 will do nicely here: (2(-1) + 1) (-1 - 6) = (-1) (-7) = 7; false; not ≤ 0 (2(1) + 1) (1 - 6) = (3) (-5) = -15; true; ≤ 0 (2(7) + 1) (7 - 6) = (15) (1) = 15; false; not ≤ 0 The test points reveal that the only region in the domain that fulfills our inequality is [-1, 6]. The graph below corroborates what we found algebraically. Note that the shaded region is closed on both sides, consistent with the inequality being non-strict—i.e., including equality to 0. Mohammad Afzaal Butt B.Sc in Mathematics & Physics, Islamia College Gujranwala (Graduated 1977) · Author has 24.6K answers and 22.9M answer views · 6y Related What is the root of equation ax^2 +bx+c = 0? ax2+bx+c=0⟹x2+bax+ca=0 ⟹x2+bax+b24a2−b24a2+ca=0 ⟹(x+b2a)2−b2−4ac4a2=0 ⟹(x+b2a)2−(√b2−4ac2a)2=0 ⟹(x+b2a−√b2−4ac2a)(x+b2a+√b2−4ac2a)=0 ⟹(x+b−√b2−4ac2a)(x+b+√b2−4ac2a)=0 ⟹x ax2+bx+c=0⟹x2+bax+ca=0 ⟹x2+bax+b24a2−b24a2+ca=0 ⟹(x+b2a)2−b2−4ac4a2=0 ⟹(x+b2a)2−(√b2−4ac2a)2=0 ⟹(x+b2a−√b2−4ac2a)(x+b2a+√b2−4ac2a)=0 ⟹(x+b−√b2−4ac2a)(x+b+√b2−4ac2a)=0 ⟹x=−b±√b2−4ac2a Related questions What is the quadratic expression, ax^2+bx+c, such that the quadratic equation ax^2+bx+c=0 will have(1) -5 and 1 as solution AND 'a' will be smaller than 'c'? How can I know if the given points are solution of the quadratic inequality y≥x²-5x+1? What are the solutions for quadratic inequality in the general case? What is the solution set of the quadratic inequalities then graph x²+9x+14>0? How can we write a quadratic inequality with a solution: -2/7 < x < 6? Would it just be Y= (7x+2) (x-6)? If ɑ and β are the zeroes of the quadratic polynomial ax²+bx+c such that x does not lie between ɑ and β, then (A) a>0 and ax²+bx+c<0 (B) ax²+bx+c<0 and a<0 (C) a>0 and ax²+bx+c >0 (D) Both B and C. Which is the correct answer with solution? How do you prove a quadratic inequality? What is a quadratic inequality with the solution -2/7 < x < 6? What is the best definition for showing that x^2 > 8x – 9 is a quadratic inequality? What is the zeros of quadratic polynomial ax2+ bx +c? Is x2-x-6 a quadratic inequality? What is the sum of the roots of quadratic equations ax2+bx+cis -6/a? Using the method of completing squares, how do you solve the general quadratic equation ax²+bx+c=0 to develop the quadratics formula? How do you fill in each blank so that the resulting statement is true the solutions of a quadratic equation in the general form ax^2+bx+c=0, a<>0, are given by the quadratics formula x=? What are the roots at quadratic equations ax²+bx+c=0 (a is not =0)? About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
10359	https://en.wikipedia.org/wiki/List_of_viscosities	Published Time: 2018-11-18T15:59:01Z List of viscosities - Wikipedia Jump to content [x] Main menu Main menu move to sidebar hide Navigation Main page Contents Current events Random article About Wikipedia Contact us Contribute Help Learn to edit Community portal Recent changes Upload file Special pages Search Search [x] Appearance Appearance move to sidebar hide Text Small Standard Large This page always uses small font size Width Standard Wide The content is as wide as possible for your browser window. Color (beta) Automatic Light Dark This page is always in light mode. Donate Create account Log in [x] Personal tools Donate Create account Log in Pages for logged out editors learn more Contributions Talk [x] Toggle the table of contents Contents move to sidebar hide (Top) 1 Units and conversion factors 2 Viscosities at or near standard conditionsToggle Viscosities at or near standard conditions subsection 2.1 Gases 2.1.1 Noble gases 2.1.2 Diatomic elements 2.1.3 Hydrocarbons 2.1.4 Organohalides 2.1.5 Other gases 2.2 Liquids 2.2.1 n-Alkanes 2.2.2 1-Chloroalkanes 2.2.3 Other halocarbons 2.2.4 Alkenes 2.2.5 Other liquids 2.3 Aqueous solutions 2.4 Substances of variable composition 3 Viscosities under nonstandard conditionsToggle Viscosities under nonstandard conditions subsection 3.1 Gases 3.2 Liquids (including liquid metals) 3.3 Solids 4 References List of viscosities [x] Add languages Add links Article Talk [x] English Read Edit View history [x] Tools Tools move to sidebar hide 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 Add interlanguage links Print/export Download as PDF Printable version In other projects Wikidata item From Wikipedia, the free encyclopedia Dynamic viscosity is a material property which describes the resistance of a fluid to shearing flows. It corresponds roughly to the intuitive notion of a fluid's 'thickness'. For instance, honey has a much higher viscosity than water. Viscosity is measured using a viscometer. Measured values span several orders of magnitude. Of all fluids, gases have the lowest viscosities, and thick liquids have the highest. The values listed in this article are representative estimates only, as they do not account for measurement uncertainties, variability in material definitions, or non-Newtonian behavior. Kinematic viscosity is dynamic viscosity divided by fluid density. This page lists only dynamic viscosity. Units and conversion factors [edit] For dynamic viscosity, the SI unit is Pascal-second. In engineering, the unit is usually Poise or centiPoise, with 1 Poise = 0.1 Pascal-second, and 1 centiPoise = 0.01 Poise. For kinematic viscosity, the SI unit is m^2/s. In engineering, the unit is usually Stoke or centiStoke, with 1 Stoke = 0.0001 m^2/s, and 1 centiStoke = 0.01 Stoke. For liquid, the dynamic viscosity is usually in the range of 0.001 to 1 Pascal-second, or 1 to 1000 centiPoise. The density is usually on the order of 1000 kg/m^3, i.e. that of water. Consequently, if a liquid has dynamic viscosity of n centiPoise, and its density is not too different from that of water, then its kinematic viscosity is around n centiStokes. For gas, the dynamic viscosity is usually in the range of 10 to 20 microPascal-seconds, or 0.01 to 0.02 centiPoise. The density is usually on the order of 0.5 to 5 kg/m^3. Consequently, its kinematic viscosity is around 2 to 40 centiStokes. Viscosities at or near standard conditions [edit] Here "standard conditions" refers to temperatures of 25°C and pressures of 1 atmosphere. Where data points are unavailable for 25°C or 1 atmosphere, values are given at a nearby temperature/pressure. The temperatures corresponding to each data point are stated explicitly. By contrast, pressure is omitted since gaseous viscosity depends only weakly on it. Gases [edit] Noble gases [edit] The simple structure of noble gas molecules makes them amenable to accurate theoretical treatment. For this reason, measured viscosities of the noble gases serve as important tests of the kinetic-molecular theory of transport processes in gases (see Chapman–Enskog theory). One of the key predictions of the theory is the following relationship between viscosity μ{\displaystyle \mu }, thermal conductivityk{\displaystyle k}, and specific heat c v{\displaystyle c_{v}}: k=f μ c v{\displaystyle k=f\mu c_{v}} where f{\displaystyle f} is a constant which in general depends on the details of intermolecular interactions, but for spherically symmetric molecules is very close to 2.5{\displaystyle 2.5}. This prediction is reasonably well-verified by experiments, as the following table shows. Indeed, the relation provides a viable means for obtaining thermal conductivities of gases since these are more difficult to measure directly than viscosity. \| Substance \| Molecular formula \| Viscosity (μPa·s) \| Thermal conductivity (W m−1 K−1) \| Specific heat (J K−1 kg−1) \| f≡k/(μ c v){\displaystyle f\equiv k/(\mu c_{v})} \| Notes \| Refs. \| --- --- --- --- \| \| Helium \| He \| 19.85 \| 0.153 \| 3116 \| 2.47 \| \| \| \| Neon \| Ne \| 31.75 \| 0.0492 \| 618 \| 2.51 \| \| \| \| Argon \| Ar \| 22.61 \| 0.0178 \| 313 \| 2.52 \| \| \| \| Krypton \| Kr \| 25.38 \| 0.0094 \| 149 \| 2.49 \| \| \| \| Xenon \| Xe \| 23.08 \| 0.0056 \| 95.0 \| 2.55 \| \| \| \| Radon \| Rn \| ≈26 \| ≈0.00364 \| 56.2 \| \| T = 26.85°C; k{\displaystyle k} calculated theoretically; μ{\displaystyle \mu } estimated assuming f=2.5{\displaystyle f=2.5} \| \| Diatomic elements [edit] \| Substance \| Molecular formula \| Viscosity (μPa·s) \| Notes \| Ref. \| --- --- \| Hydrogen \| H 2 \| 8.90 \| \| \| \| Nitrogen \| N 2 \| 17.76 \| \| \| \| Oxygen \| O 2 \| 20.64 \| \| \| \| Fluorine \| F 2 \| 23.16 \| \| \| \| Chlorine \| Cl 2 \| 13.40 \| \| \| Hydrocarbons [edit] \| Substance \| Molecular formula \| Viscosity (μPa·s) \| Notes \| Ref. \| --- --- \| Methane \| CH 4 \| 11.13 \| T = 20°C \| \| \| Acetylene \| C 2 H 2 \| 10.2 \| T = 20°C \| \| \| Ethylene \| C 2 H 4 \| 10.28 \| T = 20°C \| \| \| Ethane \| C 2 H 6 \| 9.27 \| T = 20°C \| \| \| Propyne \| C 3 H 4 \| 8.67 \| T = 20°C \| \| \| Propene \| C 3 H 6 \| 8.39 \| T = 20°C \| \| \| Propane \| C 3 H 8 \| 8.18 \| T = 20°C \| \| \| Butane \| C 4 H 10 \| 7.49 \| T = 20°C \| \| Organohalides [edit] \| Substance \| Molecular formula \| Viscosity (μPa·s) \| Notes \| Ref. \| --- --- \| Carbon tetrafluoride \| CF 4 \| 17.32 \| \| \| \| Fluoromethane \| CH 3 F \| 11.79 \| \| \| \| Difluoromethane \| CH 2 F 2 \| 12.36 \| \| \| \| Fluoroform \| CHF 3 \| 14.62 \| \| \| \| Pentafluoroethane \| C 2 HF 5 \| 12.94 \| \| \| \| Hexafluoroethane \| C 2 F 6 \| 14.00 \| \| \| \| Octafluoropropane \| C 3 F 8 \| 12.44 \| \| \| Other gases [edit] \| Substance \| Molecular formula \| Viscosity (μPa·s) \| Notes \| Ref. \| --- --- \| Air \| \| 18.46 \| \| \| \| Ammonia \| NH 3 \| 10.07 \| \| \| \| Nitrogen trifluoride \| NF 3 \| 17.11 \| T = 26.85°C \| \| \| Boron trichloride \| BCl 3 \| 12.3 \| Theoretical estimate at T = 26.85°C; estimated uncertainty of 10% \| \| \| Carbon dioxide \| CO 2 \| 14.90 \| \| \| \| Carbon monoxide \| CO \| 17.79 \| \| \| \| Hydrogen sulfide \| H 2 S \| 12.34 \| \| \| \| Nitric oxide \| NO \| 18.90 \| \| \| \| Nitrous oxide \| N 2 O \| 14.90 \| \| \| \| Sulfur dioxide \| SO 2 \| 12.82 \| \| \| \| Sulfur hexafluoride \| SF 6 \| 15.23 \| \| \| \| Molybdenum hexafluoride \| MoF 6 \| 14.5 \| Theoretical estimates at T = 26.85°C \| \| \| Tungsten hexafluoride \| WF 6 \| 17.1 \| \| Uranium hexafluoride \| UF 6 \| 17.4 \| Liquids [edit] n-Alkanes [edit] Substances composed of longer molecules tend to have larger viscosities due to the increased contact of molecules across layers of flow. This effect can be observed for the n-alkanes and 1-chloroalkanes tabulated below. More dramatically, a long-chain hydrocarbon like squalene (C 30 H 62) has a viscosity an order of magnitude larger than the shorter n-alkanes (roughly 31 mPa·s at 25°C). This is also the reason oils tend to be highly viscous, since they are usually composed of long-chain hydrocarbons. \| Substance \| Molecular formula \| Viscosity (mPa·s) \| Notes \| Ref. \| --- --- \| Pentane \| C 5 H 12 \| 0.224 \| \| \| \| Hexane \| C 6 H 14 \| 0.295 \| \| \| \| Heptane \| C 7 H 16 \| 0.389 \| \| \| \| Octane \| C 8 H 18 \| 0.509 \| \| \| \| Nonane \| C 9 H 20 \| 0.665 \| \| \| \| Decane \| C 10 H 22 \| 0.850 \| \| \| \| Undecane \| C 11 H 24 \| 1.098 \| \| \| \| Dodecane \| C 12 H 26 \| 1.359 \| \| \| \| Tridecane \| C 13 H 28 \| 1.724 \| \| \| \| Tetradecane \| C 14 H 30 \| 2.078 \| \| \| \| Pentadecane \| C 15 H 32 \| 2.82 \| T = 20°C \| \| \| Hexadecane \| C 16 H 34 \| 3.03 \| \| \| \| Heptadecane \| C 17 H 36 \| 4.21 \| T = 20°C \| \| 1-Chloroalkanes [edit] \| Substance \| Molecular formula \| Viscosity (mPa·s) \| Notes \| Ref. \| --- --- \| Chlorobutane \| C 4 H 9 Cl \| 0.4261 \| \| \| \| Chlorohexane \| C 6 H 11 Cl \| 0.6945 \| \| \| Chlorooctane \| C 8 H 17 Cl \| 1.128 \| \| \| Chlorodecane \| C 10 H 21 Cl \| 1.772 \| \| \| Chlorododecane \| C 12 H 25 Cl \| 2.668 \| \| \| Chlorotetradecane \| C 14 H 29 Cl \| 3.875 \| \| \| Chlorohexadecane \| C 16 H 33 Cl \| 5.421 \| \| \| Chlorooctadecane \| C 18 H 37 Cl \| 7.385 \| Supercooled liquid \| Other halocarbons [edit] \| Substance \| Molecular formula \| Viscosity (mPa·s) \| Notes \| Ref. \| --- --- \| Dichloromethane \| CH 2 Cl 2 \| 0.401 \| \| \| \| Trichloromethane (chloroform) \| CHCl 3 \| 0.52 \| \| \| \| Tribromomethane (bromoform) \| CHBr 3 \| 1.89 \| \| \| \| Carbon tetrachloride \| CCl 4 \| 0.86 \| \| \| \| Trichloroethylene \| C 2 HCl 3 \| 0.532 \| \| \| \| Tetrachloroethylene \| C 2 Cl 4 \| 0.798 \| T = 30°C \| \| \| Chlorobenzene \| C 6 H 5 Cl \| 0.773 \| \| \| \| Bromobenzene \| C 6 H 5 Br \| 1.080 \| \| \| \| 1-Bromodecane \| C 10 H 21 Br \| 3.373 \| \| \| Alkenes [edit] \| Substance \| Molecular formula \| Viscosity (mPa·s) \| Notes \| Ref. \| --- --- \| 2-Pentene \| C 5 H 10 \| 0.201 \| \| \| \| 1-Hexene \| C 6 H 12 \| 0.271 \| \| \| \| 1-Heptene \| C 7 H 14 \| 0.362 \| \| \| \| 1-Octene \| C 8 H 16 \| 0.506 \| T = 20°C \| \| \| 2-Octene \| C 8 H 16 \| 0.506 \| T = 20°C \| \| \| n-Decene \| C 10 H 20 \| 0.828 \| T = 20°C \| \| Other liquids [edit] \| Substance \| Molecular formula \| Viscosity (mPa·s) \| Notes \| Ref. \| --- --- \| Acetic acid \| C 2 H 4 O 2 \| 1.056 \| \| \| \| Acetone \| C 3 H 6 O \| 0.302 \| \| \| \| Benzene \| C 6 H 6 \| 0.604 \| \| \| \| Bromine \| Br 2 \| 0.944 \| \| \| \| Ethanol \| C 2 H 6 O \| 1.074 \| \| \| \| Glycerol \| C 3 H 8 O 3 \| 1412 \| \| \| \| Hydrazine \| H 4 N 2 \| 0.876 \| \| \| \| Iodine pentafluoride \| IF 5 \| 2.111 \| \| \| \| Mercury \| Hg \| 1.526 \| \| \| \| Methanol \| CH 4 O \| 0.553 \| \| \| \| 1-Propanol (propyl alcohol) \| C 3 H 8 O \| 1.945 \| \| \| \| 2-Propanol (isopropyl alcohol) \| C 3 H 8 O \| 2.052 \| \| \| \| Squalane \| C 30 H 62 \| 31.123 \| \| \| \| Water \| H 2 O \| 1.0016 \| T = 20°C, standard pressure \| \| Aqueous solutions [edit] The viscosity of an aqueous solution can either increase or decrease with concentration depending on the solute and the range of concentration. For instance, the table below shows that viscosity increases monotonically with concentration for sodium chloride and calcium chloride, but decreases for potassium iodide and cesium chloride (the latter up to 30% mass percentage, after which viscosity increases). The increase in viscosity for sucrose solutions is particularly dramatic, and explains in part the common experience of sugar water being "sticky". \| Table: Viscosities (in mPa·s) of aqueous solutions at T = 20°C for various solutes and mass percentages \| \| Solute \| mass percentage = 1% \| 2% \| 3% \| 4% \| 5% \| 10% \| 15% \| 20% \| 30% \| 40% \| 50% \| 60% \| 70% \| \| Sodium chloride (NaCl) \| 1.020 \| 1.036 \| 1.052 \| 1.068 \| 1.085 \| 1.193 \| 1.352 \| 1.557 \| \| \| \| \| \| \| Calcium chloride (CaCl 2) \| 1.028 \| 1.050 \| 1.078 \| 1.110 \| 1.143 \| 1.319 \| 1.564 \| 1.930 \| 3.467 \| 8.997 \| \| \| \| \| Potassium iodide (KI) \| 0.997 \| 0.991 \| 0.986 \| 0.981 \| 0.976 \| 0.946 \| 0.925 \| 0.910 \| 0.892 \| 0.897 \| \| \| \| \| Cesium chloride (CsCl) \| 0.997 \| 0.992 \| 0.988 \| 0.984 \| 0.980 \| 0.966 \| 0.953 \| 0.939 \| 0.922 \| 0.934 \| 0.981 \| 1.120 \| \| \| Sucrose (C 12 H 22 O 11) \| 1.028 \| 1.055 \| 1.084 \| 1.114 \| 1.146 \| 1.336 \| 1.592 \| 1.945 \| 3.187 \| 6.162 \| 15.431 \| 58.487 \| 481.561 \| Substances of variable composition [edit] \| Substance \| Viscosity (mPa·s) \| Temperature (°C) \| Reference \| --- --- \| \| Whole milk \| 2.12 \| 20 \| \| \| Blood \| 2 - 9 \| 37 \| \| \| Olive oil \| 56.2 \| 26 \| \| \| Canola oil \| 46.2 \| 30 \| \| \| Sunflower oil \| 48.8 \| 26 \| \| \| Honey \| ≈{\displaystyle \approx } 2000-10,000 \| 20 \| \| \| Ketchup[a] \| ≈{\displaystyle \approx } 5000-20,000 \| 25 \| \| \| Peanut butter[a] \| ≈{\displaystyle \approx } 10 4-10 6 \| \| \| \| Pitch \| 2.3×10 11 \| 10-30 (variable) \| \| ^ Jump up to: abThese materials are highly non-Newtonian. Viscosities under nonstandard conditions [edit] Gases [edit] Pressure dependence of the viscosity of dry air at 300, 400 and 500 kelvins All values are given at 1 bar (approximately equal to atmospheric pressure). \| Substance \| Chemical formula \| Temperature (K) \| Viscosity (μPa·s) \| --- --- \| \| Air \| \| 100 \| 7.1 \| \| 200 \| 13.3 \| \| 300 \| 18.5 \| \| 400 \| 23.1 \| \| 500 \| 27.1 \| \| 600 \| 30.8 \| \| Ammonia \| NH 3 \| 300 \| 10.2 \| \| 400 \| 14.0 \| \| 500 \| 17.9 \| \| 600 \| 21.7 \| \| Carbon dioxide \| CO 2 \| 200 \| 10.1 \| \| 300 \| 15.0 \| \| 400 \| 19.7 \| \| 500 \| 24.0 \| \| 600 \| 28.0 \| \| Helium \| He \| 100 \| 9.6 \| \| 200 \| 15.1 \| \| 300 \| 19.9 \| \| 400 \| 24.3 \| \| 500 \| 28.3 \| \| 600 \| 32.2 \| \| Water vapor \| H 2 O \| 380 \| 12.498 \| \| 400 \| 13.278 \| \| 450 \| 15.267 \| \| 500 \| 17.299 \| \| 550 \| 19.356 \| \| 600 \| 21.425 \| \| 650 \| 23.496 \| \| 700 \| 25.562 \| \| 750 \| 27.617 \| \| 800 \| 29.657 \| \| 900 \| 33.680 \| \| 1000 \| 37.615 \| \| 1100 \| 41.453 \| \| 1200 \| 45.192 \| Liquids (including liquid metals) [edit] Viscosity of water as a function of temperature \| Substance \| Chemical formula \| Temperature (°C) \| Viscosity (mPa·s) \| --- --- \| \| Mercury \| Hg \| -30 \| 1.958 \| \| -20 \| 1.856 \| \| -10 \| 1.766 \| \| 0 \| 1.686 \| \| 10 \| 1.615 \| \| 20 \| 1.552 \| \| 25 \| 1.526 \| \| 30 \| 1.495 \| \| 50 \| 1.402 \| \| 75 \| 1.312 \| \| 100 \| 1.245 \| \| 126.85 \| 1.187 \| \| 226.85 \| 1.020 \| \| 326.85 \| 0.921 \| \| Ethanol \| C 2 H 6 O \| -25 \| 3.26 \| \| 0 \| 1.786 \| \| 25 \| 1.074 \| \| 50 \| 0.694 \| \| 75 \| 0.476 \| \| Bromine \| Br 2 \| 0 \| 1.252 \| \| 25 \| 0.944 \| \| 50 \| 0.746 \| \| Water \| H 2 O \| 0.01 \| 1.7911 \| \| 10 \| 1.3059 \| \| 20 \| 1.0016 \| \| 25 \| 0.89002 \| \| 30 \| 0.79722 \| \| 40 \| 0.65273 \| \| 50 \| 0.54652 \| \| 60 \| 0.46603 \| \| 70 \| 0.40355 \| \| 80 \| 0.35405 \| \| 90 \| 0.31417 \| \| 99.606 \| 0.28275 \| \| Glycerol \| C 3 H 8 O 3 \| 25 \| 934 \| \| 50 \| 152 \| \| 75 \| 39.8 \| \| 100 \| 14.76 \| \| Aluminum \| Al \| 700 \| 1.24 \| \| 800 \| 1.04 \| \| 900 \| 0.90 \| \| Gold \| Au \| 1100 \| 5.130 \| \| 1200 \| 4.640 \| \| 1300 \| 4.240 \| \| Copper \| Cu \| 1100 \| 3.92 \| \| 1200 \| 3.34 \| \| 1300 \| 2.91 \| \| 1400 \| 2.58 \| \| 1500 \| 2.31 \| \| 1600 \| 2.10 \| \| 1700 \| 1.92 \| \| Silver \| Ag \| 1300 \| 3.75 \| \| 1400 \| 3.27 \| \| 1500 \| 2.91 \| \| Iron \| Fe \| 1600 \| 5.22 \| \| 1700 \| 4.41 \| \| 1800 \| 3.79 \| \| 1900 \| 3.31 \| \| 2000 \| 2.92 \| \| 2100 \| 2.60 \| In the following table, the temperature is given in kelvins. \| Substance \| Chemical formula \| Temperature (K) \| Viscosity (mPa·s) \| --- --- \| \| Gallium \| Ga \| 400 \| 1.158 \| \| 500 \| 0.915 \| \| 600 \| 0.783 \| \| 700 \| 0.700 \| \| 800 \| 0.643 \| \| Zinc \| Zn \| 700 \| 3.737 \| \| 800 \| 2.883 \| \| 900 \| 2.356 \| \| 1000 \| 2.005 \| \| 1100 \| 1.756 \| \| Cadmium \| Cd \| 600 \| 2.708 \| \| 700 \| 2.043 \| \| 800 \| 1.654 \| \| 900 \| 1.403 \| Solids [edit] \| Substance \| Viscosity (Pa·s) \| Temperature (°C) \| --- \| granite \| 3×10 19 - 6×10 19 \| 25 \| \| asthenosphere \| 7.0×10 19 \| 900 \| \| upper mantle \| 7×10 20 – 1×10 21 \| 1300–3000 \| \| lower mantle[citation needed] \| 1×10 21 – 2×10 21 \| 3000–4000 \| References [edit] ^ Jump up to: abChapman, Sydney; Cowling, T.G. 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ISSN0021-9568. ^Lal, Krishan; Tripathi, Neelima; Dubey, Gyan P. (2000). "Densities, Viscosities, and Refractive Indices of Binary Liquid Mixtures of Hexane, Decane, Hexadecane, and Squalane with Benzene at 298.15 K". Journal of Chemical & Engineering Data. 45 (5): 961–964. doi:10.1021/je000103x. ISSN0021-9568. ^ Jump up to: abcdFellows, P.J. (2009), Food Processing Technology: Principles and Practice (3rd ed.), Woodhead Publishing, ISBN978-1845692162 ^Pries, A. R.; Neuhaus, D.; Gaehtgens, P. (1992-12-01). "Blood viscosity in tube flow: dependence on diameter and hematocrit". American Journal of Physiology. Heart and Circulatory Physiology. 263 (6): H1770 –H1778. doi:10.1152/ajpheart.1992.263.6.H1770. ISSN0363-6135. ^Yanniotis, S.; Skaltsi, S.; Karaburnioti, S. (February 2006). "Effect of moisture content on the viscosity of honey at different temperatures". Journal of Food Engineering. 72 (4): 372–377. doi:10.1016/j.jfoodeng.2004.12.017. ^Koocheki, Arash; Ghandi, Amir; Razavi, Seyed M. A.; Mortazavi, Seyed Ali; Vasiljevic, Todor (2009), "The rheological properties of ketchup as a function of different hydrocolloids and temperature", International Journal of Food Science & Technology, 44 (3): 596–602, doi:10.1111/j.1365-2621.2008.01868.x ^Citerne, Guillaume P.; Carreau, Pierre J.; Moan, Michel (2001), "Rheological properties of peanut butter", Rheologica Acta, 40 (1): 86–96, doi:10.1007/s003970000120, S2CID94555820 ^Edgeworth, R; Dalton, B J; Parnell, T (1984), "The pitch drop experiment", European Journal of Physics, 5 (4): 198–200, Bibcode:1984EJPh....5..198E, doi:10.1088/0143-0807/5/4/003, S2CID250769509 ^Suhrmann, Von R.; Winter, E.-O. (1955), "Dichte- und Viskositätsmessungen an Quecksilber und hochverdünnten Kalium- und Cäsiumamalgamen vom Erstarrungspunkt bis + 30 C", Zeitschrift für Naturforschung, 10a (12): 985, doi:10.1515/zna-1955-1211, S2CID97692836, archived from the original on 2020-02-15, retrieved 2021-10-17 ^ Jump up to: abcdAssael, Marc J.; Armyra, Ivi J.; Brillo, Juergen; Stankus, Sergei V.; Wu, Jiangtao; Wakeham, William A. (2012), "Reference Data for the Density and Viscosity of Liquid Cadmium, Cobalt, Gallium, Indium, Mercury, Silicon, Thallium, and Zinc"(PDF), Journal of Physical and Chemical Reference Data, 41 (3): 033101, doi:10.1063/1.4729873, archived(PDF) from the original on 2021-10-17, retrieved 2019-12-12 ^Kumagai, Naoichi; Sasajima, Sadao; Ito, Hidebumi (15 February 1978). "Long-term Creep of Rocks: Results with Large Specimens Obtained in about 20 Years and Those with Small Specimens in about 3 Years". Journal of the Society of Materials Science (Japan). 27 (293): 157–161. Archived from the original on 2011-05-21. Retrieved 2008-06-16. ^ Jump up to: abFjeldskaar, W. (1994). "Viscosity and thickness of the asthenosphere detected from the Fennoscandian uplift". Earth and Planetary Science Letters. 126 (4): 399–410. Bibcode:1994E&PSL.126..399F. doi:10.1016/0012-821X(94)90120-1. Retrieved from " Categories: Viscosity Chemistry-related lists Hidden categories: Articles with short description Short description is different from Wikidata All articles with unsourced statements Articles with unsourced statements from November 2018 This page was last edited on 11 November 2024, at 01:31(UTC). Text is available under the Creative Commons Attribution-ShareAlike 4.0 License; additional terms may apply. By using this site, you agree to the Terms of Use and Privacy Policy. Wikipedia® is a registered trademark of the Wikimedia Foundation, Inc., a non-profit organization. Privacy policy About Wikipedia Disclaimers Contact Wikipedia Code of Conduct Developers Statistics Cookie statement Mobile view Edit preview settings Search Search [x] Toggle the table of contents List of viscosities Add languagesAdd topic
10360	https://www.tiger-algebra.com/en/solution/power-of-i/i%5E101/	Copyright Ⓒ 2013-2025 tiger-algebra.com This site is best viewed with Javascript. If you are unable to turn on Javascript, please click here. Solution - Powers of i Other Ways to Solve Step-by-step explanation 1. Find the highest multiple of 4 that is less than or equal to the exponent of i When i is raised to increasing powers, its values will begin repeating themselves every four terms indefinitely: i0=1,i1=i,i2=−1,i3=−i, i4=1,i5=i,i6=−1,i7=−i, i8=1 and so on. The results start repeating after i4, which is a pattern that continues every four terms forever. We can use this pattern to figure out i raised to any power. Divide the power of the i (101) by 4: 1014=25.25 Multiply 4 by 25: 4⋅25=100 100 is the highest multiple of 4 that is less than or equal to 101. 2. Calculate the power of i Expand the power using the rule: x(a+b)=xa·xb i101=i100⋅i1 Rewrite 100 as a multiple of 4: i100⋅i1=i4⋅25⋅i1 Expand the power using the rule: xab=(xa)b i4⋅25⋅i1=(i4)25⋅i1 Because i4=1: (i4)25⋅i1=125⋅i1 Because 1 raised to any power equals 1: 125⋅i1=1⋅i1 Simplify according to the pattern of the powers of i: i0=1, i1=i, i2=-1, i3=-i 1⋅i1=1⋅(i)=i The power of i101 equals i i101=i How did we do? Why learn this Despite their misleading name, imaginary numbers - almost always written as i - are not exactly "imaginary". They were originally described as "imaginary" as an insult because they represent an abstract concept that, when first discovered, did not seem to be particularly useful. They became more widely used and accepted over time, but by that point it was too late! The name stuck. Today, imaginary numbers are frequently used in scientific contexts, such as understanding the behavior of soundwaves, concepts in quantum mechanics, and relativity. Because imaginary numbers represent the solutions to the square roots of negative numbers, we can use them to solve quadratic equations that have no real roots (meaning they do not intercept the x-axis when graphed). Terms and topics Related links Latest Related Drills Solved Copyright Ⓒ 2013-2025 tiger-algebra.com
10361	https://math.libretexts.org/Courses/Prince_Georges_Community_College/MAT_1130_Mathematical_Ideas_Mirtova_Jones_(PGCC%3A_Fall_2022)/01%3A_Sets/1.02%3A_Operations_with_Sets	1.2: Operations with Sets - Mathematics LibreTexts Skip to main content Table of Contents menu search Search build_circle Toolbar fact_check Homework cancel Exit Reader Mode school Campus Bookshelves menu_book Bookshelves perm_media Learning Objects login Login how_to_reg Request Instructor Account hub Instructor Commons Search Search this book Submit Search x Text Color Reset Bright Blues Gray Inverted Text Size Reset +- Margin Size Reset +- Font Type Enable Dyslexic Font - [x] Downloads expand_more Download Page (PDF) Download Full Book (PDF) Resources expand_more Periodic Table Physics Constants Scientific Calculator Reference expand_more Reference & Cite Tools expand_more Help expand_more Get Help Feedback Readability x selected template will load here Error This action is not available. chrome_reader_mode Enter Reader Mode 1: Sets MAT 1130: Mathematical Ideas { } { "1.01:_Basics_of_Sets" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "1.02:_Operations_with_Sets" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "1.03:_Applications_of_Sets" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "1.04:_Review_Exercises" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } { "00:_Front_Matter" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "01:_Sets" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "02:_Logic" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "03:_Probability" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "04:_Statistics" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "05:_Personal_Finance" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "06:_Voting_and_Apportionment" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "zz:_Back_Matter" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } Thu, 12 Dec 2024 17:18:15 GMT 1.2: Operations with Sets 92947 92947 Andy Jones { } Anonymous Anonymous 2 false false [ "article:topic", "license:ccbysa", "showtoc:no", "authorname:lippman" ] [ "article:topic", "license:ccbysa", "showtoc:no", "authorname:lippman" ] Search site Search Search Go back to previous article Sign in Username Password Sign in Sign in Sign in Forgot password Expand/collapse global hierarchy 1. Home 2. Campus Bookshelves 3. Prince George's Community College 4. MAT 1130: Mathematical Ideas 5. 1: Sets 6. 1.2: Operations with Sets Expand/collapse global location 1.2: Operations with Sets Last updated Dec 12, 2024 Save as PDF 1.1: Basics of Sets 1.3: Applications of Sets Page ID 92947 David Lippman Pierce College via The OpenTextBookStore ( \newcommand{\kernel}{\mathrm{null}\,}) Table of contents 1. Venn Diagram 2. Universal Set 3. Example 1 4. Complement 5. Example 2 6. Union and Intersection 7. Example 3 1. Solution Example 4 Solution Example 5 Solution Example 6 Solution Example 7 Example 8 Solution Example 9 Solution Sets can interact. For example, you and a new roommate decide to have a house party, and you both invite your circle of friends. At this party, two sets are being combined, though it might turn out that there are some friends that were in both sets. To visualize the interactions and operations with sets, John Venn in 1880 thought to use overlapping circles, building on a similar idea used by Leonhard Euler in the 18 th century. These illustrations now called Venn Diagrams. Venn Diagram A Venn diagram represents each set by a circle, usually drawn inside of a containing box representing the universal set. Overlapping areas of circles indicate elements common to both sets. Non-overlapping areas of circles indicate that the sets have no elements in common. Universal Set The universal set is a set that contains all elements of interest and is usually denoted, U. The universal set is defined by the context of the problem. Example 1 If we were searching for books, the universal set might be all the books in the library. If we were grouping your Facebook friends, the universal set would be all your Facebook friends. If you were working with sets of numbers, the universal set might be all whole numbers, all integers, or all real numbers. There are three common operations that can be performed on sets: complement, union, and intersection. Each of these operations is defined below and is illustrated using a shaded area of a Venn diagram. The universal set is necessary to find the complement of set. Complement The complement of a set contains everything that is not in the set but still within the universal set. The complement of set A is notated A′. More formally, we write x∈A′ if x∈U and x∉A. The shaded region of the Venn diagram to the right shows A′. Note: Occasionally you may see other notation such as A c and A― used to represent the complement of A. Example 2 Suppose the universal set is U= all whole numbers from 1 to 9. If A={1, 2, 4}, then A′={3, 5, 6, 7, 8, 9}. The union and intersection are operations that work on two sets. Union and Intersection The union of two sets contains all the elements contained in either (or both) sets. The union of sets A and B is notated A∪B. More formally, we write x∈A∪B if x∈A or x∈B (or both). The shaded region of the Venn diagram to the right shows A∪B. The intersectionof two sets contains only the elements that are in both sets. The intersection of sets A and B is notated A∩B. More formally, we write x∈A∩B if x∈A and x∈B. The shaded region of the Venn diagram to the right shows A∩B. Example 3 Suppose the universal set is the letters in the word elastic: U={e, l, a, s, t, i, c}. Consider these sets: A={s, c, a, l, e}B={c, a, t} Find A∪B. Find A∩B. Find A′. Solution The union contains all the elements in either set: A∪B={s, c, a, l, e, t}. Notice we only list c and a once. The intersection contains all the elements in both sets: A∩B={c, a}. Here look for all the elements that are not in set A but still in U: A′={t, i}. Even though letters like f and g are not in set A, they cannot be in A′ because f and g are not in the universal set. Example 4 Consider the sets: A={red, green, blue}B={red, yellow, orange}C={red, orange, yellow, green, blue, purple} Find A∪B. Find A∩B. Find A′∩C. Solution The union contains all the elements in either set: A∪B={red, green, blue, yellow, orange}. The intersection contains all the elements in both sets: A∩B={red}. Here we're looking for all the elements that are not in set A but are in set C: A′∩C={orange, yellow, purple}. Try it Now 1 Using the sets from the previous example, find A∪C and B′∩A Answer A∪C={red, orange, yellow, green, blue purple} B′∩A={green, blue} As we saw earlier with the expression A′∩C, set operations can be grouped together. Grouping symbols can be used with sets like they are with arithmetic - to force an order of operations. When there are multiple set operations to perform, they are performed in the following order: First, perform any operation within parentheses, ( ). Then, find the complement. Next, perform the union ∪ and the intersection ∩ in order from left to right. Example 5 Suppose U={1, 2, 3, 4, 5, 6, 7, 8}A={2, 4, 7}B={1, 2, 3, 8} Find (A∪B)′. Find A′∪B′. Solution We start with the grouping symbols and find the union of set A and set B: A∪B={1, 2, 3, 4, 7, 8}. Now find the complement of that result with reference to the universal set: (A∪B)′={5, 6}. We start by finding the complements of sets A and B: A′={1, 3, 5, 6, 8} and B′={4, 5, 6, 7}. Now, union the two results: A′∪B′={1, 3, 4, 5, 6, 7, 8}. Example 6 Suppose H={cat, dog, rabbit, mouse}F={dog, cow, duck, pig, rabbit}W={duck, rabbit, deer, frog, mouse} Find (H∩F)∪W Find H∩(F∪W) Find (H∩F)′∩W Solution We start with the intersection: H∩F={dog, rabbit}. Now we union that result with W:(H∩F)∪W={dog, rabbit, duck, deer, frog, mouse}. We start with the union: F∪W={dog, cow, duck, pig, rabbit, deer, frog, mouse}. Now we intersect that result with H:H∩(F∪W)={dog, rabbit, mouse}. We start with the intersection: H∩F={dog,rabbit}. Now we want to find the elements of W that are not in H∩F: (H∩F)′∩W={duck, deer, frog, mouse}. Sometimes it is useful to represent set operations using Venn diagrams when the elements of the sets are unknown or the number of elements in the sets is too large. Basic Venn diagrams can illustrate the interaction among two or three sets. Example 7 Use Venn diagrams to illustrate A∪B,A∩B, and A′∩B. A∪B contains all elements in either set (or both.) A∩B contains only those elements in both sets - in the overlap of the circles. A′∩B contains those elements that are not in set A but are in set B. Example 8 Use a Venn diagram to illustrate (H∩F)′∩W. Solution We'll start by identifying everything in the set H∩F. Now, (H∩F)′∩W will contain everything not in the region shown above but that is in set W. Example 9 Write an expression to represent the outlined part of the Venn diagram shown. Solution The elements in the outlined set are in sets H and F, but are not in set W. So we could represent this set as (H∩F)∩W′. Try it Now 2 Write an expression to represent the outlined portion of the Venn diagram shown: Answer (A∪B)∩C′ This page titled 1.2: Operations with Sets is shared under a CC BY-SA license and was authored, remixed, and/or curated by David Lippman (The OpenTextBookStore) . Back to top 1.1: Basics of Sets 1.3: Applications of Sets Was this article helpful? Yes No Recommended articles 1.1: Basics of Sets 1.3: Applications of Sets 1.4: Review Exercises 5: Personal Finance 6: Voting and Apportionment Article typeSection or PageAuthorDavid LippmanLicenseCC BY-SAShow Page TOCno Tags This page has no tags. © Copyright 2025 Mathematics LibreTexts Powered by CXone Expert ® ? 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10362	https://math.stackexchange.com/questions/103280/asymptotics-for-a-partial-sum-of-binomial-coefficients	Skip to main content Asymptotics for a partial sum of binomial coefficients Ask Question Asked Modified 5 years, 3 months ago Viewed 5k times This question shows research effort; it is useful and clear 11 Save this question. Show activity on this post. Good afternoon, I would like to ask, if anyone knows how to evaluate a sum ∑k=0λn(nk) for fixed λ<1/2 with absolute error O(n−1), or better. In Concrete Mathematics (Graham, Knuth, Patashnik), it is shown how to evaluate this sum with absolute error O(1), but it is not clear to me, how to obtain better absolute error in a straightforward manner. Thank you in advance. discrete-mathematics binomial-coefficients asymptotics Share CC BY-SA 3.0 Follow this question to receive notifications edited Jan 28, 2012 at 18:20 Srivatsan 26.8k77 gold badges9595 silver badges148148 bronze badges asked Jan 28, 2012 at 15:04 042042 1,0751010 silver badges2222 bronze badges Add a comment \| 2 Answers 2 Reset to default This answer is useful 10 Save this answer. Show activity on this post. This is more of an extended comment than an answer, but you may find it useful. In Exercise 9.42 of Concrete Mathematics (page 492 in the second edition), the authors establish the asymptotic formula ∑k=0λn(nk)=2nH(λ)−lg(n)/2+O(1) where 0<λ<1/2, H(λ)=λlg(1λ)+(1−λ)lg(11−λ), and lg is the binary logarithm. The sum on the left is a small fraction of the full sum 2n. Note that this is a multiplicative approximation, the ratio of the sum and the approximation remains bounded as n→∞, not the difference. Their result has an interpretation using probability. Write ∑k=0λn(nk)=2nP(Xn/n≤λ) where Xn is a binomial(n,1/2) random variable. The theory of large deviations suggests an approximation P(Xn/n≤λ)≈exp(−nI(λ)) where I is the rate function I(x)=xlog(x)+(1−x)log(1−x)+log(2). This gives the leading factor in their approximation; they also divide by n−−√ for further accuracy. If you are willing to use the standard normal distribution function Φ(z)=P(Z≤z), then by the central limit theorem, we have P(Xn≤λn)=P(Xn−n/2n/4−−−√≤λn−n/2n/4−−−√)≈P(Z≤n−−√(2λ−1)). In other words, we have the approximation ∑k=0λn(nk)=2nΦ(n−−√(2λ−1)). This seems to be at least as accurate as the Concrete Mathematics approximation, and you can get more accuracy by using the "continuity correction". More detailed asymptotic results for binomial tails can be found in this paper by Andrew Carter and David Pollard. In particular, see Theorem 1. I hope you find what you want there; happy hunting! Share CC BY-SA 3.0 Follow this answer to receive notifications answered Jan 28, 2012 at 18:13 user940user940 4 4 Let me mention that CLT cannot yield large deviations estimates. (I remember having explained that to Qiaochu on the site hence this part of your answer indirectly confirms the surprising extent of this confusion.) Did – Did 11/26/2012 08:48:44 Commented Nov 26, 2012 at 8:48 1 @Did Do you mean that CLT based approximations can only be sharp for λ close to 1/2, and not in the tail? Thomas Ahle – Thomas Ahle 09/17/2015 22:44:33 Commented Sep 17, 2015 at 22:44 @Byron Would it be possible to expand the O(1) from concrete mathematics further? Thomas Ahle – Thomas Ahle 09/17/2015 22:54:45 Commented Sep 17, 2015 at 22:54 4 @ThomasAhle What I mean is that, in the present context, CLT provides the asymptotics of P(Xn≤n/2+λn−−√/2) when n→∞ for every fixed λ, but that CLT says nothing about the asymptotics of P(Xn≤n/2+λnn−−√/2) when n→∞ if λn varies with n, especially if λn→∞ or if λn→0. There are tools to deal with the latter, but CLT is not one of them. Did – Did 09/18/2015 08:42:49 Commented Sep 18, 2015 at 8:42 Add a comment \| This answer is useful 2 Save this answer. Show activity on this post. Asked for is an approximation to the sum ∑j=0cn(nj) for some fixed constant c. We provide more than this, an approximation to the finite sum over binomial coefficients ∑j=ab(nj) for integers 0≤a≤b≤n The asymptotic approximation (which proves to be useful numerically, even for modest values of n) can be obtained by combining two ideas. First, the sum of a function g over the integers can be written as an integral over g plus an asymptotic series of the odd derivatives of g by using the midpoint form of the Euler-Maclaurin sum formula, ∑j=abg(j)∼∫a+1/2a−1/2g(x)dx+∑j=1⋯−(1−21−2j)B2j(2j)!(g(2j−1)(b+1/2)−g(2j−1)(a−1/2)) Here B2j is a Bernoulli number. Second, the binomial coefficient itself is the real function g(x)=(nx)=Γ(n+1)Γ(n−x+1)Γ(x+1) ,x>−1 evaluated at the integers; and an asymptotic series for that real function can be developed using the asymptotic expansion of for the Γ-function (Stirling's approximation) lnΓ(x)∼(x−1/2)lnx−x+12ln(2π)+∑m=0⋯B2m(2m)(2m−1)x2m−1 Combining the two series gives an asymptotic approximation to the desired sum where all the terms are known, the various integrations required being tractable. The lead approximation, as other contributors have noted, involves the error function erf(x)=1π−−√∫x0e−t2dt and we have S≡12n∑j=ab(nj)∼12erf(yb)−12erf(ya)+O(1/n) The advantage to working with S and not the original sum is that S is O(1) as n→∞ (and S=1 for a=0 and b=n). We have made the definitions ξaya=a−n/2−1/2=ξa/n/2−−−√ξbyb=b−n/2+1/2=ξb/n/2−−−√ Using MapleTM to do the necessary integrations exactly and to organize the necessary algebra, we can find the higher terms as S≈f(yb)−f(ya)+2–√6π−−√1n3/2(h(ξb)−h(ξa)) Here f(y)=12erf(y)+1π−−√∑j=1⋯e−y2Ljnj is a sum whose coefficients Lj are all O(1) for large n, even if both a and b are O(n). We also find h(y)=e−2ξ2/n∑j=0⋯ξMj(ξ)nj which is a more complicated sum; the coefficients ξMj(ξ) can be large since we can have ξ=O(n), but when ξ is large the contribution to h is heavily suppressed by the lead factor of e−2ξ2/n, so that h functions numerically as a series whose terms diminish successively, though I do not know if they fall as O(1/n) or not. In practice a keeping few terms works well if n is not too small. The first few corrections are L1L2L3L4=112(2y2−3)y=−11440(40y6−292y4+410y2−45)y=1362880(1120y10−20048y8+103248y6−159768y4+33390y2+14175)y=−187091200(22400y14−745920y12+8413728y10−38540496y8+65383848y6−19336212y4−15416730y2+893025)y and M0(ξ)M1(ξ)M2(ξ)M3(ξ)=1=−910+4315λ−43λ2=+25224−1153252λ+3296315λ2−26845λ3+89λ4=+4964−69160λ−12329600λ2+625061575λ3−1126364725λ4+728135λ5−3281λ6 where ξ and λ=ξ2/n can both be O(n) if both a and b are O(n). With the terms up to L4 and M3 included, the first term missing from the expansion has an error of O(1/n5). Asymptotic series have the characteristic numerical defect that if one adds too many terms, the partial sum of the series departs from the function being approximated. To check that, I have evaluated the series with including the terms out to L7 and M6, when the first term missing from the series is O(1/n8); and for n as modest as 100 then the absolute error in S improves steadily with an increasing number of terms to <10−14 for a=0 and all b with 0≤b≤n. So if the series begins to depart from the true sum it will have to do so at higher order than that. Evaluation is fast because the evaluation to approximate a sum involves but two calls to erf (which are fast); two calls to exp; and one call to sqrt, outside of the evaluation of a few polynomials. Share CC BY-SA 4.0 Follow this answer to receive notifications answered May 24, 2020 at 22:35 user438111user438111 14911 silver badge22 bronze badges 2 @CharliesMungerJr: OP is asking for the asymptotic behaviour of a sum, where the number of summands increase with increasing n. Markus Scheuer – Markus Scheuer 05/25/2020 13:53:58 Commented May 25, 2020 at 13:53 Would you mind to have a look at math.stackexchange.com/questions/3912163/… ? I think that you could bring a lot for the case a=0, b=m. I would enjoy to improve my continuous solution. Thanks and cheers :-) Claude Leibovici – Claude Leibovici 11/28/2020 10:09:59 Commented Nov 28, 2020 at 10:09 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 discrete-mathematics binomial-coefficients asymptotics See similar questions with these tags. 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10363	https://www.quora.com/How-do-you-find-the-domain-of-function-f-x-9-x%C2%B2	How to find the domain of function f(x) = √ (9-x²) - Quora Something went wrong. Wait a moment and try again. Try again Skip to content Skip to search Sign In Mathematics Function Domain Square Root Index Basic Functions Quadratic Function Algebra Class Domain (mathematics) Solving Square Root Equat... Functions (general) 5 How do you find the domain of function f(x) = √ (9-x²)? All related (42) Sort Recommended Assistant Bot · Sep 7 We require the expression inside the square root to be defined and nonnegative. For f(x) = √(9 − x²) this gives: Solve 9 − x² ≥ 0 → x² ≤ 9. Equivalent inequality: −3 ≤ x ≤ 3. Therefore the domain is the closed interval [−3, 3]. Additional notes: On this domain f(x) is real-valued; at the endpoints x = ±3 the value is 0. If considering a complex-valued function, all real x are allowed but the principal square root becomes imaginary when \|x\| > 3. Upvote · Related questions More answers below What is the domain and range of f(x) =√9-x²? How do you find the domain of function given by f(x) = - √ (9-x²)? What is the domain of the function f(x) = √ (4x-x²)? How do I find the domain F(x) = √ (x + 2)? What is the domain of f(x) = -2-√9-x²? Gurjaspreet Singh 6y Domain: 9-x² is greater than or equal to 0 (3-x)(3+x) is greater than or equal to 0 Representing on number line we get[-3,3] Range:Maximum when x=0 , i.e., √9=3 Minimum when x=3 or -3 , i.e., √(9–9)=0 [0,3] Upvote · 9 5 Sponsored by Grammarly Stuck on the blinking cursor? Move your great ideas to polished drafts without the guesswork. Try Grammarly today! Download 99 34 Suvadip Mazoomder I know enough mathematics to work as an Engineer. ·6y y = f(x) The domain of a function f(x) is the set of all values for which the function is defined. Given function y= f(x) = √(9-x²) We have a function with a variable inside a radical sign. We set (9-x²) ≥0 ,considering f(x) to be a real valued function.. 9-x² ≥0 if and only if -3≤x≤3 {x : -3≤x≤3 , x ∈ R} ⇒ Domain of the function. Hope this helps.. Upvote · 9 5 9 1 Rukmangadareddy Former Student at Sri Venkateswara (2017–2021) ·6y The given function is f(x)=√(9-x²) The function in the square root can allow the values which gives a positive number or zero as output. The function in the square root is (9-x²) which should be always greater than or equal to zero. 9-x²≥0 x²-9≤0 (x+3)(x-3)≤0 -3≤x≤3 The values of x should be between -3 and 3 including those two values. x€[-3,3]. Upvote · 9 1 Related questions More answers below What is the domain and range for the function f(x) =√ (4-x^2)? What's the domain of f(x) =2x-√4x-x²? How do I find the domain and range of the function f(x) = 5-(x+3) ²? What is the domain and range for the function f(x) = -√-5-6x-x²? What is the domain and range of following function f(x) = [√ (x²-4)] / [5-√ (36-x²)]? Jayanthi Ragu 6y Domain of any number should be real.To get real number, first I should know condition to get real number: square root is real if and only if 9-x^2>=0.or else we will get non real complex number called imaginary number. If x=+3,-3,then 9-x^2=0.0^1/2 =0 is real . x>3 & x <-3 gives 9-x^2 <0 (9-4^2=-7). x <3 &x>-3 gives 9-x^2 >0 (9-2^2=5). So,x should lie between -3 & 3. the domain of f (x)=(9-x^2)^1/2 is [-3,3]. Upvote · S.R Maths Faculty for JEE-love to do Maths problems (2012–present) · Author has 637 answers and 546.9K answer views ·4y Originally Answered: What is the domain and range of f(x) =√9-x²? · f(x)=√9−x 2 f(x)=9−x 2 9−x 2≥0 9−x 2≥0 x 2–9≤0 x 2–9≤0 (x−3)(x+3)≤0(x−3)(x+3)≤0 So domain[−3,3]So domain[−3,3] For range replace f(x) by y, f(x) always be positve number. y 2=9−x 2 y 2=9−x 2 x 2=9−y 2 x 2=9−y 2 x=√9−y 2 x=9−y 2 So range[0,3][0,3] Upvote · 9 9 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 999 620 Dale Brandt M.S. in Physics&Mathematics, Oakland University (Graduated 1987) · Author has 301 answers and 131.5K answer views ·6y In math, on an x-y graph, x is the independent variable … the domain … the horizontal axis (the abscissa). y is the dependent variable … the range … the vertical axis (the ordinate). The range is dependent on the domain: y = f(x). You may want to check out this link. Domain and Range of a Function Upvote · Sivakumar Manickam YouTuber \| SoI JEE, NEET Physics & Maths [IITB] · Author has 447 answers and 537.5K answer views ·7y Originally Answered: What is the domain and range of f(x) =√9-x²? · values of x will be restricted such that ( 9 - x^2 ) > 0 9 - x^2 = ( 3 + x ) ( 3 - x ) this is parabola of inverted U shape, whose value is +ve in between roots. so ( 9 - x^2 ) > 0 for x in between -3 & 3 so domain is x = [ -3 , 3 ] range = ( 0 , max of f(x) ) f(x) is maximum at mid point of roots, i.e. f max = f(0) = 3 so range = ( 0 , 3 ) Upvote · 99 10 9 1 9 1 Sponsored by RedHat Customize AI for your needs, with simpler model alignment tools. Your AI needs context, not common knowledge. Learn More 9 7 John Tsang Manager ·6y Originally Answered: What is the domain and range of f(x) =√9-x²? · Actully, I think we can ^2 same time on lefe and right between the equal, thus it will be Y^2+X^2=9. it is a circle with diameter of 3. but when you called it as a function, so different value of x will produce different ONE y, so the answer is -3<=X<=3 and 0<=Y<=3. is right? discussing please. Upvote · 9 2 Antonio P Garza Jr MST 1970 in Mathematics&Education, Boston College · Author has 571 answers and 168.2K answer views ·3y Originally Answered: What is the domain of f(X) =square root of 9-x^2? · Solution: If f(x) = square root (9.- x^2), then find its domain. 9 - x^2 >= 0 x^2 - 9 <= 0 (x + 3)•(x - 3) <= 0 Case I: x+3 >= 0 and x-3 <= 0 x >= -3 and x <= 3 -3 <= x <= 3 is the solution to Case I. OR Case Ii: x+3 <= 0 and x-3 >= 0 x <= -3 and x >= 3 The empty set is the solution to Case Ii. The union of the two solutions is the domain of f. Therefore, the domain of f is {x\|-3<=x<=3}. Upvote · Sponsored by Book Geists If you're a Kiwi, this could be the best day of your life! Available to Kiwis only. Read today. Learn More 1.3K 1.3K Michael Michaelsen Knows Danish · Author has 60 answers and 10.2K answer views ·3y Originally Answered: What is the domain of f(X) =square root of 9-x^2? · The domain is found by finding the values that make the function inside of the square root negative. You find the values that x can’t take on by solving the inequality 9-x^2<0. You can find the values that x can take on by solving the inequality 9-x^2>0. Upvote · Adwait Patwardhan Studied at Resonance Eduventures ·7y Originally Answered: What is the domain and range of f(x) =√9-x²? · First we should define the function in interval. Since there is square root, 9−x 2>=0 9−x 2>=0 (3+x)(3−x)>=0(3+x)(3−x)>=0 (3+x)(x−3)<=0(3+x)(x−3)<=0 Therefore x is defined over [-3,3] This is the domain of function. For range, Y 2=9−x 2 Y 2=9−x 2 X 2=9−y 2 X 2=9−y 2 Since x is only real Discreminant of 9−y 2 i s>=0 9−y 2 i s>=0 Now do the math and get the range. Upvote · 99 12 9 1 Akash S (Ak19) Studied at Kendriya Vidyalaya · Author has 107 answers and 578.1K answer views ·5y Related What is the domain of f(x) =arcsin (√x²-1)? Hope you... Continue Reading Hope you... Upvote · 99 22 Related questions What is the domain and range of f(x) =√9-x²? How do you find the domain of function given by f(x) = - √ (9-x²)? What is the domain of the function f(x) = √ (4x-x²)? How do I find the domain F(x) = √ (x + 2)? What is the domain of f(x) = -2-√9-x²? What is the domain and range for the function f(x) =√ (4-x^2)? What's the domain of f(x) =2x-√4x-x²? How do I find the domain and range of the function f(x) = 5-(x+3) ²? What is the domain and range for the function f(x) = -√-5-6x-x²? What is the domain and range of following function f(x) = [√ (x²-4)] / [5-√ (36-x²)]? How do I find the domain and range of the function f(x) = (x+1) / (x²-1)? If f(x) = √ (x - 1) / (x2 - 9), then how do I find the domain of f given by the interval? What is the domain of the following function: f(x) = (√(x-1)) (3-x)? What is domain of this function f(x) =2\√t²-16? If f(x) = √ (x - 1) / (x² - 9), then what is the domain of f given by the interval? Related questions What is the domain and range of f(x) =√9-x²? How do you find the domain of function given by f(x) = - √ (9-x²)? What is the domain of the function f(x) = √ (4x-x²)? How do I find the domain F(x) = √ (x + 2)? What is the domain of f(x) = -2-√9-x²? What is the domain and range for the function f(x) =√ (4-x^2)? Advertisement About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
10364	https://zhuanlan.zhihu.com/p/690968807	直答 4.3 函数凹凸性概念、性质、判别、Jensen不等式、拐点首发于微积分-数学分析切换模式 4.3 函数凹凸性概念、性质、判别、Jensen不等式、拐点收录于 · 微积分-数学分析 21 人赞同了该文章凸(凹)函数 1.1 凸函数概念 1.2 凸函数的几何意义 1.3 凸函数的例子凸函数的简单命题 2.1 数乘 2.2 加法 2.3 复合 2.4 反函数 2.5 最值 2.6 弧和弦关系 2.7 补充说明函数凸性的判别 3.1 凸性的另一表述 3.2 凸性的令一几何描述 3.3 凸函数的充要条件 3.5 凸函数的充要条件3 3.6 凸函数的更多例 3.7 两个凸函数乘积为凹函数的例 Jensen不等式及应用 4.1 Jensen不等式 4.2 幂函数形式的Jensen不等式 4.3 对数函数形式 4.4 xlnx形式拐点 5.1 基本概念 5.2 利用导数说明 5.3 拐点的寻找和判别 5.4 高阶导在拐点的应用 5.5 函数与拐点相关的性质 5.6 附加说明 1. 凸(凹)函数前面我们研究过函数的单调性，接下来继续了解一个比较重要的函数性质凹凸性。 1.1 凸函数概念注意: 凹凸函数定义本身不要求函数的连续性，但是凹凸性能得到连续性，但是比较麻烦。因此为了省去不必要的麻烦，我们这里加上了连续的条件。这种做法叫做关注重点，忽略细节。凸函数(向下凸): 设 f(x) \in C[\mathcal{X}] , 若 \forall\ x_1,x_2\in\mathcal{X} 有不等式: \bbox[8pt,border:1pt]{\begin{aligned}f(q_1x_1+q_2x_2) \leq q_1f(x_1) + q_2f(x_2)\end{aligned}}\tag{1}其中 q_1 + q_2 = 1, q_1, q_2 > 0 , 则函数 f(x) 是凸函数。凹函数: 设 f(x) \in C[\mathcal{X}] , 若 \forall\ x_1,x_2\in\mathcal{X} 有不等式: \bbox[8pt,border:1pt]{\begin{aligned}f(q_1x_1+q_2x_2) \geq q_1f(x_1) + q_2f(x_2)\end{aligned}}\tag{1a}其中 q_1 + q_2 = 1, q_1, q_2 > 0 , 则函数 f(x) 是凹函数。说明: 若 f(x) 为凸函数，则 -f(x) 为凹函数，反之亦然。这个说明就允许我们后面仅讨论凸函数就够了。凹凸函数的概念是Jensen引入的，但是他使用的是 q_1=q_2 = 1/2 。 1.2 凸函数的几何意义 \bbox[8pt,border:1pt]{\begin{aligned}x=q_1x_1+q_2x_2(x_1 < x_2)\end{aligned}}\tag{2} 则上述 x\in(x_1, x_2) ; 反过来， \forall\ x\in(x_1, x_2), \exists\ q_1,q_2 >0, q_1+q_2=1: x = q_1x_1 + q_2x_2\ 其中 \bbox[8pt,border:1pt]{\begin{aligned}q_1 = \frac{x_2-x}{x_2-x_1}, q_1 = \frac{x-x_1}{x_2-x_1}\end{aligned}}\tag{2a} 上图函数 y = f(x), A_1(x_1, y_1), A_2(x_2, y_2), y_1 = f(x_1), y_2 = f(x_2) , 弧 A_1A_2 上的横坐标为 x 的点 A 的纵坐标为 f(x) = f(q_1x_1 + q_2x_2) , 其中系数为(2a)，而A上面的弦 A_1A_2 上的点 B 的纵坐标为 q_1f(x_1) + q_2f(x_2) ，即: \bbox[8pt,border:1pt]{\begin{aligned}y= \frac{x_2-x}{x_2-x_1}y_1+ \frac{x-x_1}{x_2-x_1}y_2\end{aligned}}\tag{3} 于是凸函数可以描述为: 函数图像(弧)上所有的点都在相应的弦下面，或位于弦本身上(取等的时候)。同样凹函数描述为: 函数图像上所有的点都在相应的弦的上面或弦本身上。 1.3 凸函数的例子 1) 线性函数 f(x) = ax + b 是特殊的既凸又凹的函数，因为函数图像上的点也是对应弦本身的点。 2) 函数 f(x) = x^2 是凸函数，这可以直接用定义来验证。 2. 凸函数的简单命题 2.1 数乘 1) 正的数乘凸函数得到的仍是凸函数。 2.2 加法 2) 两个凸函数的和仍是凸函数。 2.3 复合 3) 若 \varphi(u) 是常增的凸函数， u = f(x) 也是凸函数，则复合函数 \varphi(f(x)) 也是凸函数。 2.4 反函数 4) 单值互反函数 y = f(x), x=g(y) ，具有下面性质: 2.5 最值 5) 非常数凸函数，不可能在区间内部达到最大值。 2.6 弧和弦关系 6) 若区间 [x_1, x_2], x_1 < x_2 包含在使函数 f(x) 为凸的区间 \mathcal{X} 内，那么关系式(1)要么总是取等，要么总是不带等号。换句话说，要么弧和弦重叠，要么弧整个落在弦下面。 2.7 补充说明两个凸函数的乘积不一定是凸函数，这个后面举例说明。凹凸函数也有类似的狭义凹凸和广义凹凸。 3. 函数凸性的判别 3.1 凸性的另一表述利用下面的公式 \bbox[8pt,border:1pt]{\begin{aligned}f(q_1x_1+q_2x_2) \leq q_1f(x_1) + q_2f(x_2)\end{aligned}}\tag{1}\bbox[8pt,border:1pt]{\begin{aligned}x=q_1x_1+q_2x_2(x_1 < x_2)\end{aligned}}\tag{2}\bbox[8pt,border:1pt]{\begin{aligned}q_1 = \frac{x_2-x}{x_2-x_1}, q_1 = \frac{x-x_1}{x_2-x_1}\end{aligned}}\tag{2a} 可以将不等式(1)改写为: \begin{aligned}f(x) \leq \frac{x_2-x}{x_2-x_1}f(x_1) + \frac{x-x_1}{x_2-x_1}f(x_2)\end{aligned}\ 或者写得更加对称一些: \bbox[8pt,border:1pt]{\begin{aligned}(x_2-x)f(x_1) + (x_1-x_2)f(x) + (x-x_1)f(x_2) \geq 0\end{aligned}}\tag{4} 最后这个式子改为行列式写成: \bbox[8pt,border:1pt]{\begin{aligned}\left\| \begin{array}{ccc} 1 & x_{1} & f(x_1) \ 1 & x & f(x) \ 1 & x_2 & f(x_2) \end{array} \right\|\geq 0\end{aligned}}\tag{5} 上述所有情况 x_1 < x < x_2 。 3.2 凸性的令一几何描述将函数凸性条件改写为(5)的形式后，就得到关于凸性的另一个几何解释。 (5)式中的行列式表示的是 \Delta A_1AA_2 面积的两倍，它取正值的充要条件是三角形是正定向的，换句话说这个三角形的周界 A_1-A-A_2 是依逆时针方向划过的。特别地，如果指的狭义凸性，那么在这些条件中将等号除去。 3.3 凸函数的充要条件 3.4 凸函数的充要条件2 设 f(x), f^\prime(x)\in C[\mathcal{X}] , 且在区间内有有限导数 f^{\prime\prime}(x) , 则 f(x) 在区间内是凸函数的充要条件是 f^{\prime\prime}(x)\geq 0 。设 f(x)\in C[\mathcal{X}] , 且在区间内有有限导数 f^\prime(x) , 则 f(x) 在区间内是凸函数的充要条件是 f^\prime(x) 是增函数(广义增)。必要性: f(x) 凸函数 \Rightarrow f^\prime(x) \uparrow 将 \bbox[8pt,border:1pt]{\begin{aligned}(x_2-x)f(x_1) + (x_1-x_2)f(x) + (x-x_1)f(x_2) \geq 0\end{aligned}}\tag{4}\begin{aligned}(x_2-x)f(x_1) - ((x_2 - x) - (x_1-x))f(x) + (x-x_1)f(x_2) \geq 0\end{aligned}\\begin{aligned}(x_2-x)[f(x_1)-f(x)] + (x-x_1)[f(x_2)-f(x)] \geq 0\end{aligned}\改写为 \bbox[8pt,border:1pt]{\begin{aligned}\frac{f(x) - f(x_1)}{x-x_1} \leq \frac{f(x_2) - f(x)}{x_2-x}\end{aligned}}\tag{6} 令 x\to x_1或x\to x_2 并求极限，则得到: \bbox[8pt,border:1pt]{\begin{aligned}f^\prime(x_1) \leq \frac{f(x_2) - f(x_1)}{x_2-x_1}\end{aligned}}\tag{7a} 和 \bbox[8pt,border:1pt]{\begin{aligned}f^\prime(x_2) \geq \frac{f(x_2) - f(x_1)}{x_2-x_1}\end{aligned}}\tag{7b} 由此就得到 f^\prime(x_1) \leq f^\prime(x_2) , 就得到导数是增函数的结论。充分性: f^\prime(x)\uparrow \Rightarrow f(x) 凸函数我们只需要证明满足不等式(6)即可，对(6)利用有限增量公式: \begin{aligned}\frac{f(x) - f(x_1)}{x-x_1} = f^\prime(\xi_1), \frac{f(x_2) - f(x)}{x_2-x} = f^\prime(\xi_2)\end{aligned}\ 其中 x_1 < \xi_1 < x < \xi_2 < x_2 ，而导函数为增函数，则有 f^\prime(\xi_1) \leq f^\prime(\xi_2) , 也就证明了不等式(6)成立。 3.5 凸函数的充要条件3 凸函数还有一个重要的几何特征。定理3: 设 f(x)\in C[\mathcal{X}] , 则 f(x) 在区间内是凸函数的充要条件是: 它的图像上的任意点落在它的任意切线上面(或切线上)。必要性: 凸性 \Rightarrow 曲线 y=f(x) 在点 A_0(x_0,f(x_0)) 处有斜率 f^\prime(x_0) , 则切线方程为: \begin{aligned}y = f(x_0) + f^\prime(x_0)(x-x_0)\end{aligned}\ 下面我们需要证明，由函数的凸性，对于 \forall\ x_0, x\in\mathcal{X} , 成立不等式: \bbox[8pt,border:1pt]{\begin{aligned}f(x)\geq f(x_0) + f^\prime(x_0)(x-x_0)\end{aligned}}\tag{10} 它等价于两个不等式: \bbox[8pt,border:1pt]{\begin{aligned}f^\prime(x_0)\leq \frac{f(x)-f(x_0)}{x-x_0}\ (x>x_0)\end{aligned}}\tag{11a}和 \bbox[8pt,border:1pt]{\begin{aligned}f^\prime(x_0)\geq \frac{f(x)-f(x_0)}{x-x_0}\ (x<x_0)\end{aligned}}\tag{11b} 这两个不等式在定理1的证明中已经得到, 就是在凸函数的条件假设下得到的, 满足(7a), (7b)不等式。充分性: \Rightarrow 凸性反过来，假定不等式(10)也就是(11a), (11b)成立，那么，由它们同样可以确立(7a), (7b)，由此可得 f^\prime(x_1) \leq f^\prime(x_2) , 也就是证得了 f^\prime(x) 是增函数，于是就证得了凸函数。 3.6 凸函数的更多例 1) y = a^x(a>0,a\neq 1), x\in(-\infty,+\infty) 内是凸的。因为 y^{\prime\prime} = a^x\cdot(\ln a)^2 > 0 2) y = \ln x, x\in(-\infty, +\infty) 是凹的，可以考虑二阶导小于零，也可以利用反函数的凹凸性关系。 3) y = x\cdot\ln x(x\in (0, +\infty)) , 二阶导数为 \begin{aligned}y^{\prime\prime} = \frac{1}{x} > 0\end{aligned} ，因而为凸。 4) y = x^r(x\in (0, +\infty)) , 二阶导 y^{\prime\prime} = r(r-1)x^{r-2} , 因此 r\in (-\infty, 0)\cup(1,+\infty) 函数是凸的，而 r\in(0,1) 函数是凹的。 3.7 两个凸函数乘积为凹函数的例说明: \begin{aligned}y = -x^{\frac{1}{3}}\end{aligned} 是凸函数，它的平方 \begin{aligned}z = x^{\frac{2}{3}}\end{aligned} 是凹函数。这个例子就表明两个凸函数的乘积，未必是凸函数。 4. Jensen不等式及应用 4.1 Jensen不等式依照凸函数定义: \bbox[8pt,border:1pt]{\begin{aligned}f(q_1x_1 + q_2x_2) \leq q_1f(x_1) + q_2f(x_2)\ (q_1,q_2 > 0; q_1 + q_2 = 1)\end{aligned}}\tag{def-1} 推广至: \bbox[8pt,border:1pt]{\begin{aligned}f(q_1x_1 + q_2x_2+\cdots +q_nx_n) \leq q_1f(x_1) + q_2f(x_2) + \cdots + q_nf(x_n)\ (q_i > 0; \sum q_i = 1)\end{aligned}}\tag{J-01} 说明: 上述 x_i 为基本区间 \chi 内的任意数值。 n=2 时就是我们的凸函数定义。对于 n\geq 2 的，我们只需要利用数学归纳法即可证得。这里不赘述。进一步推广: 通常我们并不使用总和为1的诸因子 q_i , 而引用一些任意的正数 p_i , 在不等式(J-01)中令: q_i = \frac{p_i}{p_1 + \cdots + p_n} 则有: \sum\limits_{i=1}^n q_i = 1 , 于是就有: \bbox[8pt,border:1pt]{\begin{aligned}f(\frac{\sum p_ix_i}{\sum p_i}) \leq \frac{\sum p_i\cdot f(x_i)}{\sum p_i}\end{aligned}}\tag{J-02} 注: 这里是凸函数情形，而对应的凹函数情形不等号反过来。下面根据函数 f(x) 的具体不同，我们得到关于Jensen不等式得不同形式: 4.2 幂函数形式的Jensen不等式 f(x) = x^k (x>0, k>1) 凸函数 \begin{aligned}\bigg(\frac{\sum p_ix_i}{\sum p_i}\bigg)^k \leq \frac{\sum p_i\cdot x_i^k}{\sum p_i}\end{aligned}\ 即 \bbox[8pt,border:1pt]{\begin{aligned}(\sum p_ix_i)^k \leq (\sum p_i)^{k-1}\sum p_i\cdot x_i^k\ (x_i>0, k>1)\end{aligned}}\tag{J-03} Cauchy-Holder不等式将上面的符号进行代换 p_i = {b_i}^\frac{k}{k-1}, x_i = \frac{a_i}{{b_i}^\frac{1}{k-1}} , 则 p_ix_i ={b_i}^\frac{k}{k-1}\cdot \frac{a_i}{{b_i}^\frac{1}{k-1}} = b_i^ka_i , p_ix_i^k ={b_i}^\frac{k}{k-1}\cdot (\frac{a_i}{{b_i}^\frac{1}{k-1}})^k = a_i^k 得到Cauchy-Holder不等式: \bbox[8pt,border:1pt]{\begin{aligned}\sum a_ib_i \leq (\sum a_i^k)^{\frac{1}{k}}(\sum b_i^{\frac{k}{k-1}} x_i^k)^{\frac{k-1}{k}}\end{aligned}}\tag{J-04} 备注: 我没有推出来，左侧我得到的是 \sum a_ib_i^k , 暂时没找到哪里算错了。 4.3 对数函数形式令 f(x) = \ln x, x>0 凹函数 \bbox[8pt,border:1pt]{\begin{aligned}f(\frac{\sum p_ix_i}{\sum p_i}) \geq \frac{\sum p_i\cdot f(x_i)}{\sum p_i}\end{aligned}}\tag{J-02a} 即: \bbox[8pt,border:1pt]{\begin{aligned}\ln\frac{\sum p_ix_i}{\sum p_i} \geq \frac{\sum p_i\cdot \ln x_i}{\sum p_i}\end{aligned}}\tag{J-02a} 去掉对数: \bbox[8pt,border:1pt]{\begin{aligned}\frac{\sum p_ix_i}{\sum p_i} \geq \bigg[\prod x_i^{p_i}\bigg]^{\frac{1}{\sum p_i}}\end{aligned}}\tag{J-02a} 4.4 xlnx形式最后，取 f(x) = x\ln x ( x>0 ) 凸函数 \bbox[8pt,border:1pt]{\begin{aligned}\frac{\sum p_ix_i}{\sum p_i} \ln\frac{\sum p_ix_i}{\sum p_i} \leq \frac{\sum p_i\cdot x_i\ln x_i}{\sum p_i}\end{aligned}}\tag{J-02a} \bbox[8pt,border:1pt]{\begin{aligned}\frac{\sum p_ix_i}{\sum p_i} \leq \bigg[\prod x_i^{p_ix_i}\bigg]^{\frac{1}{\sum p_ix_i}}\end{aligned}}\tag{J-02b}特例，令 p_i = \frac{1}{x_i} \bbox[8pt,border:1pt]{\begin{aligned}\frac{n}{\sum\frac{1}{x_i}} \leq \bigg[\prod x_i\bigg]^{\frac{1}{n}}\end{aligned}}\tag{J-02b} 这个就是: \sqrt[n]{a_1a_2\cdots a_n} \geq \frac{n}{\frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_n}} 几何中值大于等于调和中值。详细参照 5. 拐点接下来介绍函数图像绘制中一个重要的概念，就是关于函数图像的拐点的。 5.1 基本概念拐点: 简单来说就是函数 f(x) 图像分割开凸和凹部分的点称为拐点。拐点: 无非就有由凹到凸的衔接点；由凸到凹的衔接点。 5.2 利用导数说明假设函数 f(x) 在考察区间内有有限导数 f^\prime(x) ，那么我们就可以利用前面定理来对拐点进行说明。 a) 凹凸的衔接点: 导函数 f^\prime(x) 先减后增，它会经过一个极小值点 x_0 ，导函数的这个极小值点 x_0 就是该情形下的拐点。 b) 凸凹的衔接点: 导函数f^\prime(x) 先增后减，它会经过一个极大值点 x_0 ，导函数的这个极大值点 x_0 就是该情形下的拐点。如果假设函数在该区域还有二阶导数 f^{\prime\prime}(x) , 那么拐点 x_0 处一定有 f^{\prime\prime}(x_0) = 0 。请注意，这个条件只是必要的，而非充分的。比如说 y = x^4 , y^{\prime\prime} = 12x^2 \geq 0 , 这个函数在整个区间内是凸的，虽然我们有 f^{\prime\prime}(0) = 0 。这点在图像上也可以看出来: 5.3 拐点的寻找和判别从上面的说明，其实也给出我们一个寻找拐点的思路: 若函数存在二阶导，那么我们就可以找出二阶导的根，然后再在二阶导的根中去找拐点，也就是利用拐点的定义来判断根是否为拐点，就是观察这些根左右邻域中二阶导的符号是否相反，如果是，则是拐点；否则就不是拐点。拐点的判别方法: 若二阶导 f^{\prime\prime}(x) 在所考察的区间内部处处存在，那么二阶导的根 x_0 , 若二阶导 f^{\prime\prime}(x) 经过这个点是保持定号，则该点不是拐点；若经过该点二阶导变号，则是拐点。拐点将函数曲线分成两部分，一部分是狭义凸曲线，一部分是狭义凹曲线。例如 y = \sin x, y^{\prime\prime} = -\sin x , 二阶导在 x = k\pi 处为零，同时变更符号。因此正弦函数落在 x 轴上的一切点都是拐点。从函数图像上也很容易看出这一点。 5.4 高阶导在拐点的应用这部分可以和极值部分结合起来考虑。若在点 x_0 处第一个不为零的高于二阶的导数是奇数阶导数，那么有拐点；如果是偶数阶导数，那么没有拐点。 5.5 函数与拐点相关的性质曲线 y=f(x) 有一个值得注意的性质，这个性质与拐点处的切线有关(前提是切线存在): 曲线在拐点处，曲线会从切线的一侧通过拐点穿过到另一侧，或者换句话来说，曲线与切线相交。说明: 1) 如果切线铅锤的: 这个情况显然。 2) 斜切线情况: 这个时候假定存在有限导数 f^\prime(x_0) 。为了好说明问题，我们假设拐点左侧凸，右侧凹，也就是: x_0-\delta \leq x < x_0 时，曲线为凸的；x_0 \leq x < x_0 + \delta 时，曲线为凹的。我们现在就是要确定: x < x_0 时，曲线落在切线以上或切线本身上； x > x_0 时，曲线落在切线以下或切线本身上。即: f(x) \geq f(x_0) + f^\prime(x_0)(x-x_0) (x < x_0)\和 f(x) \leq f(x_0) + f^\prime(x_0)(x-x_0) (x > x_0)\ 这两个不等式就是我们前面的(10)在凸凹函数下的表述。 5.6 附加说明在关于极值的讨论中，虽然我们给出了三种判别法，但是这些判别法并非万能的，还是有些情况无法处理的。同样对于拐点，判别法也非万能的，有些函数它们也是无能为力的。另外，在前面我们提到的一个关于拐点的性质，有些时候我们也用它来简单的作为拐点的定义，这可以称作拐点的第二个定义: 拐点处的切线与曲线相交。但是这第二个定义和我们前面的第一个定义绝不等价。原因如下: 可能曲线在拐点处没有切线，那么这个时候第二个定义无从谈起。可能某点处切线与曲线相交，但该点不是曲线的凸凹分界点，第二个定义也无法应用。参考上面的图第三个和第四个。还有一种情况，比较有趣，我们用例子来说明比如曲线 \begin{aligned}f(x) = \left{\begin{array}{l}x^5\cdot\sin^2\frac{1}{x}, x\neq 0\ 0, x=0\end{array}\right.\end{aligned}\ 这条曲线在原点处和 x 轴相切并和它相交；这个函数甚至存在连续的二阶导数，但是曲线在点 x=0 的附近，不管是左边还是右边，都无限次的变更符号。编辑于 2024-11-27 04:33・广东凸优化非凸优化凸分析写下你的评论... 3 条评论默认最新天空是蔚蓝色国内凹凸性和国外正好反过来了 02-21 · 山东余着推导c-h不等式时，pixi算错了，应为aibi 2024-11-07 · 安徽学数相伴作者 2024-11-07 · 湖北关于作者学数相伴提升数学思维，拓宽数学眼界，高维俯瞰，相伴而学，学数有我。教师资格证持证人回答 1,698 文章 329 8,313 推荐阅读 # 一个看上去匪夷所思的题(凹凸函数的性质) 题目：设 f(x) 是定义在 R 上的三阶连续可导函数，证明: \exists a\in R s.t. f(a)f'(a)f''(a)f'''(a)\geq0 证明:这个题第一感觉就是通过反证法来证明，因为真正的… # 关于函数凹凸性与二阶导数的证明 # 严格凹凸函数的等价定义证明及其判定定理注意1：除在定理5、6中，我们所说的凹函数都是严格凹函数。注意2：我们规定：“对区间上（或曲线上）任意两点”这句话蕴含着这两点不重合。这样的规定旨在不想多写x1≠x2 注意3：我们规定… # 【数学分析新讲笔记】8.3函数的凹凸与拐点 8.3.1前言上一节: 8.2泰勒公式下一节: 8.4不等式的证明数学分析新讲笔记整理在：数学分析新讲笔记目录本文阐述两个部分: 凸函数利用导数判别凹凸与拐点 8.3.2凸函数凸(凹)函数引言图像凹… 想来知乎工作？请发送邮件到 jobs@zhihu.com 未注册手机验证后自动登录，注册即代表同意《知乎协议》《隐私保护指引》扫码下载知乎 App
10365	https://people.tamu.edu/~yvorobets//MATH433-2023A/Lect1-09web.pdf	MATH 433 Applied Algebra Lecture 9: Chinese Remainder Theorem. Linear congruences Linear congruence is a congruence of the form ax ≡b mod n, where x is an integer variable. We can regard it as a linear equation in Zn: [a]nX = [b]n, where X = [x]n. Theorem 1 If the congruence class [a]n is invertible, then the equation [a]nX = [b]n has a unique solution in Zn, which is X = [a]−1 n [b]n. Theorem 2 The linear congruence ax ≡b mod n has a solution if and only if d = gcd(a, n) divides b. If this is the case, then the solution set consists of d congruence classes modulo n. Chinese Remainder Theorem Theorem Let n, m ≥2 be relatively prime integers and a, b be any integers. Then the system x ≡a mod n, x ≡b mod m of congruences has a solution. Moreover, this solution is unique modulo nm. Proof: Since gcd(n, m) = 1, we have sn + tm = 1 for some integers s, t. Let c = bsn + atm. Then c = bsn + a(1 −sn) = a + (b −a)sn ≡a (mod n), c = b(1 −tm) + atm = b + (a −b)tm ≡b (mod m). Therefore c is a solution. Also, any element of [c]nm is a solution. Conversely, if x is a solution, then n\|(x −c) and m\|(x −c), which implies that nm\|(x −c), i.e., x ∈[c]nm. Problem. Solve simultaneous congruences x ≡3 mod 12, x ≡2 mod 29. The moduli 12 and 29 are coprime. First we use the Euclidean algorithm to represent 1 as an integral linear combination of 12 and 29: 1 0 12 0 1 29 → 1 0 12 −2 1 5 → 5 −2 2 −2 1 5 → 5 −2 2 −12 5 1 → 29 −12 0 −12 5 1 . Hence (−12) · 12 + 5 · 29 = 1. Let x1 = 5 · 29 = 145, x2 = (−12) · 12 = −144. Then x1 ≡1 mod 12, x1 ≡0 mod 29. x2 ≡0 mod 12, x2 ≡1 mod 29. It follows that one solution is x = 3x1 + 2x2 = 147. The other solutions form the congruence class of 147 modulo 12 · 29 = 348. Problem. Solve a system of congruences x ≡3 mod 12, x ≡2 mod 10. The system has no solutions. Indeed, any solution of the ﬁrst congruence must be an odd number while any solution of the second congruence must be an even number. Problem. Solve a system of congruences x ≡6 mod 12, x ≡2 mod 10. The general solution of the ﬁrst congruence is x = 6 + 12y, where y is an arbitrary integer. Substituting this into the second congruence, we obtain 6 + 12y ≡2 mod10 ⇐ ⇒ 12y ≡−4 mod 10 ⇐ ⇒6y ≡−2 mod 5 ⇐ ⇒y ≡3 mod 5. Thus y = 3 + 5k, where k is an arbitrary integer. Then x = 6 + 12y = 6 + 12(3 + 5k) = 42 + 60k or, equivalently, x ≡42 mod 60. Note that the solution is unique modulo 60, which is the least common multiple of 12 and 10. Chinese Remainder Theorem (generalized) Theorem Let n1, n2, . . . , nk ≥2 be pairwise coprime integers and a1, a2, . . . , ak be any integers. Then the system of congruences        x ≡a1 mod n1, x ≡a2 mod n2, . . . . . . . . . x ≡ak mod nk has a solution which is unique modulo n1n2 . . . nk. Idea of the proof: The theorem is proved by induction on k. The base case k = 1 is trivial. The induction step uses the usual Chinese Remainder Theorem. Problem. Solve simultaneous congruences    x ≡1 mod 3, x ≡2 mod 4, x ≡3 mod 5. First we solve the ﬁrst two congruences. Let x1 = 4, x2 = −3. Then x1 ≡1 mod 3, x1 ≡0 mod 4 and x2 ≡0 mod 3, x2 ≡1 mod 4. It follows that x1 + 2x2 = −2 is a solution. The general solution is x ≡−2 mod 12. Now it remains to solve the system x ≡−2 mod 12, x ≡3 mod 5. We need to represent 1 as an integral linear combination of 12 and 5: 1 = (−2) · 12 + 5 · 5. Then a particular solution is x = 3 · (−2) · 12 + (−2) · 5 · 5 = −122. The general solution is x ≡−122 mod 60, which is the same as x ≡−2 mod 60. Problem. Solve a system of congruences 2x ≡3 mod 15, x ≡2 mod 31. We begin with solving the ﬁrst linear congruence. Since gcd(2, 15) = 1, all solutions form a single congruence class modulo 15. Namely, x is a solution if [x]15 = −1 15 15. We ﬁnd that −1 15 = 15. Hence [x]15 = 1515 = 15 = 15. Equivalently, x ≡9 mod 15. Now the original system is reduced to x ≡9 mod 15, x ≡2 mod 31. Next we represent 1 as an integral linear combination of 15 and 31: 1 = (−2) · 15 + 31. It follows that one solution to the system is x = 2 · (−2) · 15 + 9 · 31 = 219. All solutions form the congruence class of 219 modulo 15 · 31 = 465.
10366	https://www.reddit.com/r/AskPhysics/comments/he0nz6/why_is_the_speed_of_light_exactly_299792458/	Why is the speed of light exactly 299792458 meters per second? Why is it constant? And most importantly, why is that speed the limit and why can’t it be greater? Taking physics course online and would appreciate assistance, thanks. : r/AskPhysics Skip to main content Why is the speed of light exactly 299792458 meters per second? Why is it constant? And most importantly, why is that speed the limit and why can’t it be greater? Taking physics course online and would appreciate assistance, thanks. : r/AskPhysics Open menu Open navigation Go to Reddit Home r/AskPhysics A chip A close button Get App Get the Reddit app Log In Log in to Reddit Expand user menu Open settings menu Go to AskPhysics r/AskPhysics r/AskPhysics 1.5M Members Online • 5 yr. ago [deleted] Why is the speed of light exactly 299792458 meters per second? Why is it constant? And most importantly, why is that speed the limit and why can’t it be greater? Taking physics course online and would appreciate assistance, thanks. So I have a test tomorrow and I was warned by someone that a question very similar to what I said in the title will appear. The main part of the question which I have issues looking up and finding in the textbook is why can’t the speed of light be faster. Why is 299792458 m/s and not 299792500 m/s for example? Read more adobe • Official • Promoted Frame it how you like it. With Photoshop's enhanced Frame tool, you can now frame an image in a custom shape or triangle. Plus, generate assets inside a shape. Learn More adobe.com Sort by: Best Open comment sort options Best Top New Controversial Old Q&A Pakketeretet • 5y ago The speed of light is exactly 299792458 meters per second because this is how the SI units for meter and second are implicitly defined. I am not 100% sure, but this probably came about something like this: Way back when the meter was invented, it was just defined as some fraction of some reference distance. Similarly, a second was probably defined as some fraction of a year or something. These units were based on some canonical reference, so somewhere in a vault in Paris, there is a stick made of some material that is very sturdy and is exactly one meter. Something similar must have been set up for a reference second. However, while refining the SI units, people at some point realized it is easier to measure the speed of light than it is to measure one second. Therefore, instead of measuring the time it takes for light to traverse one meter in one second, they said "you know what, let's just define the speed of light to be 299792458 m/s (this was close to earlier measurements of the speed of light) and instead define the second as the time it takes for light to traverse exactly 299792458 meters. Reply reply PaukAnansi • 5y ago You are almost spot on! The modern day definition of the second is the frequency of a hyperfine transition in a Cs-133 atom. In other words, when an electron in the Cs-133 atom undergoes a specific decay (falls from a specific higher energy level to a lower one), it releases a photon. The frequency of the photon is defined to be 9,192,631,770 1/sec. So that's the definition of a second. The speed of light is also a definition. Then the meter is derived from those two definitions as 1/299,792,458 of the distance that light travels in a second. Historically the meter and second where the fundamental units. The second was defined as 1/86400 of a day, and a meter was defined as 1/10,000,000 of the distance between the equator and North Pole along a great circle (a great great circle is the path you would take along a sphere if you don't turn, or in this case, a meridian (or line of constant longitude). Small side note, parallels, or lines of constant latitude are not great circles). Since then the SI unit system has been revised whenever we were able to make a more precise measurent than allowed by the inherent uncertainty in the definition of the units. Reply reply 5 more replies 5 more replies More replies dmitrden • 5y ago • Edited 5y ago You're almost right. The SI second is actually defined using the frequency of ground state hyperfine transition of Cs 133 atom. And it's the SI meter that's defined through the fixed speed of light: it's the distance light travels in 1/299792458 of a second Reply reply trivialgroup • 5y ago Yes, initially (when the SI system was first developed under Napoleon) the meter was defined as 1/10,000,000th the distance from a pole to the equator of Earth. Later it was defined as the length of a reference meter bar, and finally in 1983 it was defined as 1/299,792,458th of the distance light travels in a second in vacuum. The reason for the changes is that a length can only be measured as precisely as the meter standard, so as measurements became more precise, more precise standards were needed. It’s possible to measure time more precisely than distance, and the speed of light is constant, so that’s why the meter standard is based on time and the speed of light. The kilogram also was defined as the mass of a reference platinum-iridium cylinder in a vault near Paris, but is now based on the Planck constant. That change was just made in May of last year. Reply reply 6 more replies 6 more replies More replies Taylor7500 • 5y ago There isn't a "why" to those questions in physics, it's just true by every metric we can measure. Relativity is the study which largely covers this but the gist of it is that as you approach the speed of light, the energy required to get a little faster increases exponentially, with it requiring infinite energy to reach the speed of light. Since we know that there isn't infinite energy in the universe, we know that it is impossible to reach the speed of light, and so exceed it. As for why it's not a rounder number, why should a fundamental universal constant fit our (relatively arbtrary) measurement system? Reply reply 5 more replies 5 more replies More replies gautampk • 5y ago The numerical value (299792458 m/s) is what it is because the metre is defined such that this is case. You can choose units where the speed of light is 1 [distance]/s or even units where distance=time and then the speed of light is just 1 (with no units). Basically, why is the speed of light what it is and not faster? The way we explain why quantities have the values they do is by producing a model that predicts those quantities: this is reductionism. The thing about fundamental constants is that they're fundamental, so there isn't any (known) model that predicts their value. The answer is ultimately 'because 299792458 m/s is the value that makes the correct predictions'. N.B.: due to the way the metre and second are defined, the fundamental constant is actually not the speed of light, but the frequency of a certain Caesium hyperfine transition. The point is the same, however. Reply reply rtroshynski • 5y ago I have always thought that because photons have no mass they cannot go slower in a vacuum because of the fabric of spacetime. Does this mean that spacetime itself expands at the speed of light? No - spacetime can expand faster than the measured speed of light. Mind-blowing Reply reply officialsquarespace • Promoted PSA, Redditors: You don't need a business to have a website. All you need is yourself. And with Squarespace, you can easily create a website that reflects your personal brand, individuality, and identity–all by using its intuitive design, AI, and expressibility tools. Sign Up squarespace.com [deleted] • 5y ago It is the speed limit because it is an observed fact that regardless of an observers speed, light travels c faster than the observer in their reference frame. (c = speed of light) This may be super counter-intuitive, but as I said, this is an observed fact that we must reconcile with everything else. Think of this as being the premise for which everything else comes out of. You can't go faster than light because light will always be going c faster than you. As for your first question, who knows. Maybe the universe was just tuned this way. Reply reply 3 more replies 3 more replies More replies chaomera888 • 5y ago As others have stated there's no real reason for why the values are exactly as they are, these are just facts of experiments and measurement. However, I want to address the question of "why aren't faster speeds possible" by asking a different question. The vast majority of people's confusion about going faster than the speed of light I think stems from the naming, and associating that speed with a physical thing when, in our every day experience, we can almost always get one thing to travel faster than another. But the speed of light is not just the speed that light travels; all massless particles travel at that speed. In fact, the speed of light has nothing to do with light at all! It's just the first thing we noticed that travels at that speed. So what's it about if not about light? The hint is in the theories of motion where that speed is held as a default assumtion: the special and general theories of relativity. These theories are all about comparing motion across perspectives, and the different phenomena that get observed as a result of looking at the "same" events from different perspectives. It becomes obvious with a little thought that for th ed universe to be consistent, there has to be an exact rate at which interactions happen: I need a definite way of saying "this happened before that which happened before that which happened before that...", because otherwise we cant even agree on the existence of objects, let alone how they move. To enforce this rate of interaction, there has to be a single, finite and fixed speed at which information from one event can get to the next one. This "speed of causality" is the speed at which all massless interactions, including light, occurs. Going faster than light would mean going "faster than causality", literally faster than the information of your existence can be received by the rest of the universe, so hopefully it's clear that being able to do that would kind of mess up the universe rather quickly! Theres no need for it to be a specific number, but it has to be finite, because if there was no limit then all interactions would happen infinitely fast no matter how far things were from each other and the universe would run its entire course in an infinitesimally small amount of time. And it has to be the same speed for everyone because otherwise no one can agree on anything that happens ever, which means you dont really have a consistent universe to begin with. This video from PBS Space Time has more detail and is where I got the idea to explain it this way to my students. Reply reply restwonderfame • 5y ago There are several constants in the universe. Pi being 3.14 and not 3.16, for example. Things sometimes fit together in an elegant and simple way (e.g. E=MC2) other times they are a bit more messy. Reply reply New to Reddit?Create your account and connect with a world of communities. 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10367	https://www.physicsclassroom.com/concept-builder/newtons-laws/solve-it-newtons-secton-law	Physics Classroom is making strides to make our site accessible to everyone. Our site contains 6 navigation areas. The Primary, Secondary, and Page Level navigations have a screen reader version of their nav structure that allows using the left and right keys to navigate sibling navigation items, and up or down keys to navigate parent or child navigation items. The others can be navigated using tabs. The Primary Navigation handles the first 2 levels of site pages. The Secondary (which is not always available) handles the 3rd and 4th level of structure. The Page level navigation allows you to navigate the current page's headings quickly. The Header Navigation contains the Light/Dark Mode toggle, Search, Notifications and account login. The Breadcrumb Navigation contains the breadcrumb of the current page. If the current page has a breadcrumb, you can get to it by skipping to the content and tabbing in reverse (shift plus tab). The Footer Navigation contains links such as Privacy, Contact, about and terms. Some resources contain an Audio Player that can be activated by holding down the T key for 3 seconds, and then using K to pause and resume. While not every area of Physics Classroom is usable purely from keyboard and screen reader, we are committed to continue work on making this possible. If you have questions or need additional help, please use this link to contact us. Solve It! with Newton's Second Law The Solve It! (with Newton's Second Law) Concept Builder provides learners plenty of practice using the Fnet = m•a equation to analyze situations involving unbalanced forces and accelerations. Much more than the usual Concept Builder, this activity demands that learners solve numerical problems. A word story problem with numerical information and a free-body diagram are provided. Learners will have to determine the values of as many as eight quantities. There are a total of eight problem types spread across three difficulty levels. The difficulty levels are different in terms of the complexity of the problem. Learners can have as many tries at the problem as needed, but each miss results in a deduction from the health rating. Question-specific help is provided for each of the eight problems. Use of this Concept Builder with our Task Tracker system allows teachers to track student progress. Users are encouraged to open the Concept Builder and begin. There is no need for an activity sheet for this Concept Builder. Learners and Instructors may also be interested in viewing the accompanying Notes page. Technical information, teaching suggestions, and related resources that complement this Interactive are provided on the View Notes. We now offer a video tailored to this Concept Builder.
10368	https://www.youtube.com/watch?v=C64jIbvU5sQ	Why is sqrt (-1) = i ? Functions & Calculus by Professor Calculish 2080 subscribers 31 likes Description 1846 views Posted: 29 Aug 2024 So why is the sqrt of -1 = to an i? Because -1 represents a complex number and the square root operation brings out the imaginary component of this complex number. Each complex number has a real component "a" and an imaginary component "bi". The purpose of the channel is to learn, familiarize, and review the necessary precalculus and trigonometry/geometry topics that form a basis for calculus (with a focus on functions). Calculus topics include limits, derivatives, integrals, formula derivations, derivations/proofs, area/volumes of curves and calculus applications. Utilize simple and easy techniques to get the material across to you. Disclaimer: I am not a professor by profession. Thank you! complexnumbers #imaginarynumbers #trigonometric #polarform #precalculus #professorcalculish #squareroot 7 comments Transcript: I hope everyone's doing well I am Str thank you for joining me for this video where we are examining this why is the square < TK of minus1 = 2 and I you learned this you know it we're asking the question why when we are looking at something like this sare root of minus 5 we know we have a IUN 5 the minus comes out as an I if I have here a square root of let's say - 4 it'll come out as a 2 I this is coming out as a i this is coming out as a two we have a 2 I but why is this negative coming out as a i let's examine it we have a certain complex number which is represented by a plus and minus bi this is a complex number representation real part of your complex number and the imaginary part when I'm looking here at this minus representation what is my complex number my complex number when I'm looking at this is specifically 3us one but if I write it out toly I'm looking at -1 plus and - 0 I when I graph this out on a complex plane I'm looking at a minus1 Z it's a point which lands right over here a complex plane representation we know that the distance from the origin to here which is your R your modulus is equal to a 1 we know the angle from zero up to this right here the argument is equal to a pi when you bring this into the polar or the trigonometric form of a complex number we have this Z is equal to R and then you know you have a cosine Theta + I sin Theta this right here is your Polar form the trigonometric form of a complex number this form converted into that well you know what you have over here with regards to minus one which is my complex number it is = to 1 cosine piun and + I sin of piun that's what it is that's the Polar form of this number -1 now what we are looking here at is root of -1 which you can write as -1 ^ of 1 /2 now we're suddenly looking at a power of a complex number and we have a formula for that if you think about the D moous theorem the powers of a complex number we have a formula the power of a complex number is equal to the modulus raised to that value n time the argument which has cine n theta plus I sin n Theta the power has come here as the exponent of your modulus and as values affecting the the angle with which the trigonometric functions are applied to the argument has the N value and the modulus has the n as a exponent and what is n over here n here is equal to 1 2 bring this representation with regards to what you see over here we're looking here Z the^ of n but here particularly we have Z ^ of 1 2 which we know is min-1 plus and minus or we can just say plus 0 i^ 1/2 when you bring this de Mo theorem with regards to what we have over here we can bring everything into the template I have 1^ 1/2 cosine n which is 1 2 < + I sin 1 / 2 < and then we close it out when we solve it we know we are looking here at our complex number 0^ 102 which you know is all of this right over here which is same as this right here is equal to 1^ 1 2 which is 1 cine < / 2 < 1 2 is < 2+ I sin < / 2 cosine of < / 2 is cosine of 90° at s of < 2 is s of 90 which is a 1 this is equal to what it multiplies with this I which multiplies with this one over here so my complex number which was really what we were looking at as -1 ^ 1 or2 which was root of -1 is equal to this I that 1 which is equal to I and it's done and it's shown and now we know yunk -1 is = to I we have utilized here the D Mo theorem the powers of a complex number I've converted a square root into an exponential form so I can utilize this formula and that right there is the end thank you
10369	https://math.stackexchange.com/questions/2005509/conjecture-if-a-k-k-pia-k-0-then-a-k1-a-k-to-e?rq=1	number theory - Conjecture: If $a_k-k \pi(a_k)=0$ then $a_{k+1}/a_{k} \to e$. - 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 Conjecture: If a k−k π(a k)=0 a k−k π(a k)=0 then a k+1/a k→e a k+1/a k→e. Ask Question Asked 8 years, 10 months ago Modified8 years, 10 months ago Viewed 69 times This question shows research effort; it is useful and clear 2 Save this question. Show activity on this post. A natural question to ask is for what n n does π(n)∣n π(n)∣n where π(n)π(n) is the prime counting function. It can be shown with a discrete variant of the intermediate value theorem that such an n n exists for any k=n/π(n)k=n/π(n). (Intuition: f(n)=n−k π(k)f(n)=n−k π(k) can only increase by 1 1 and decrease by k−1 k−1. Since f(2)<0 f(2)<0 for k>3 k>3 and f(n)>0 f(n)>0 for large n n, there must be a zero). Computations suggest that all solutions to n−k π(n)=0 n−k π(n)=0 are grouped at some n k n k in an interval of around n k−−√n k width. But more surprisingly to me, for any pair of solutions a k a k and a k+1 a k+1, with the former being in the k k group and the latter in the k+1 k+1 group, the ratio a k+1/a k a k+1/a k seems to be about e e. My conjecture: If a k−k π(a k)=0 a k−k π(a k)=0 then a k+1/a k→e a k+1/a k→e. In other words, the solutions of f(n)=n−k π(k)f(n)=n−k π(k) grow exponentially as e k e k. I am at a loss why this is the case. My guess is the gaps between primes [n!,n!+n][n!,n!+n] comes into play, giving us Stiring's formula and hence our e e factor, but I really have no clue. Can someone either (1) provide a proof or disproof of this conjecture or (2) give an extremely plausible method of attack? I would also appreciate suggestions in the comments for ideas/theorems/techniques to investigate to help me make my own crack at the problem. number-theory distribution-of-primes Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Nov 8, 2016 at 18:33 abnryabnry asked Nov 8, 2016 at 18:26 abnryabnry 15k 2 2 gold badges 38 38 silver badges 79 79 bronze badges 1 2 oeis.org/A057809 is relevant.Barry Cipra –Barry Cipra 2016-11-08 19:29:31 +00:00 Commented Nov 8, 2016 at 19:29 Add a comment\| 1 Answer 1 Sorted by: Reset to default This answer is useful 2 Save this answer. Show activity on this post. This appears to be due to the distribution of the primes. Here's a heuristic. By the Prime Number Theorem, π(x)≈x log x π(x)≈x log⁡x, so that x x log x≈log x x x log⁡x≈log⁡x and so to increase the value of x x log x x x log⁡x by one, we need to increase x x by a factor of e e. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Nov 8, 2016 at 19:16 Matthew ConroyMatthew Conroy 11.4k 4 4 gold badges 34 34 silver badges 36 36 bronze badges 2 Ah, this should have been obvious. So the presence of e e really goes back to the mystery of the prime number theorem.abnry –abnry 2016-11-09 01:41:27 +00:00 Commented Nov 9, 2016 at 1:41 You're welcome! Cheers!Matthew Conroy –Matthew Conroy 2016-11-09 02:12:21 +00:00 Commented Nov 9, 2016 at 2:12 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 number-theory distribution-of-primes 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 0Legendre's Conjecture 0Is this equivalent to the Goldbach conjecture (modified slightly)? 0Goldbach’s conjecture 2Generalization of Opperman's Conjecture 14Hard system in integers related to natural number representations 1Show that if ∥a p n∥>0‖a p n‖>0 for primes p p, then ∥a n∥≥0‖a n‖≥0 3is there any possibility to write and calculate this sum in pari gp, which is very related to hardy littlewood first conjecture? 1Primality Formula Conjecture Hot Network Questions в ответе meaning in context Why are LDS temple garments secret? The rule of necessitation seems utterly unreasonable Are there any world leaders who are/were good at chess? 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10370	https://www.chronicliverdisease.org/disease_focus/downloads/Clinicians%20and%20Patients%20Confront%20Practical%20Issues%20in%20Wilson%20Disease.pdf	Clinicians and Patients Confront Practical Issues in Wilson Disease Sammy Saab, MD, MPH1; Sherona Bau, NP2; Jeff M. Bronstein, MD, PhD3; Rhonda Rowland4; Regino P . Gonzalez-Peralta, MD5 1David Geffen School of Medicine at UCLA, Department of Hepatology, Los Angeles, CA 2UCLA Health, Department of Hepatology, Los Angeles, CA 3David Geffen School of Medicine at UCLA, Department of Neurology, Los Angeles, CA 4Wilson Disease Association, New York, NY 5AdventHealth for Children, AdventHealth Transplant Institute, Orlando, FL Introduction Wilson disease (WD), first identified in 1911,1 is a genetic disease of copper metabolism caused by mutations in the ATP7B gene.2 WD is considered a rare disorder that can affect males and females in equal numbers and all races, ages, and ethnic groups. Since the major route of copper excretion (95%) is through the liver, excess copper first accumulates there and over time leads to progressive liver damage (fibrosis and subsequently cirrhosis). Copper accumulation eventually spills into the blood and impacts numerous organ systems, especially the liver and central nervous system, including the eyes (Figure 1). Although only about 2,000 to 3,000 cases have been diagnosed in the United States, other affected individuals may be misdiagnosed with different liver, neurological, or psychiatric disorders.2 WD affects nearly 1 in every 30,000 people worldwide3 and, with the US population at about 340 million, this indicates that approximately 10,000 patients in the United States have WD. Figure 1. Several Organ Systems Impacted by Wilson Disease4 1 Clinicians and Patients Confront Practical Issues in Wilson Disease 2 This disease is progressive and, if left unidentified and untreated, may cause chronic liver disease leading to cirrhosis, neuropsychiatric dysfunction, and death. Due to the rarity of WD, multisystemic involvement and clinical heterogeneity, it is difficult for a single practitioner to develop sufficient expertise to diagnose and treat WD through clinical experience alone.5 WD is often difficult to detect as its signs and symptoms are frequently mistaken for other conditions or diseases.6 Early diagnosis of WD is crucial to ensure that adequate treatment can be started, which can lead to a full and normal life. Although most patients respond to early institution of appropriate therapy,7 in some patients, WD is associated with considerable disability, the need for liver transplantation, and death.1, 8 This indicates that there are unmet needs associated with this disease. “With over 100 years of having this disease identified and 4 available treatments, patients should not be slipping through the cracks” Rhonda Rowland, Patient with WD, Patient Advocate and President of the Wilson Disease Association. Goals of the Whitepaper and Author Perspectives In 2022, the American Association for the Study of Liver Diseases (AASLD) set forth practice guidance on the diagnosis and treatment of WD, which replaced the previous 2008 guidelines.9, 10 In 2023, the Chronic Liver Disease Foundation (CLDF) published a summary of the updated AASLD Practice Guidance,4 in which the authors reviewed and commented on the latest recommendations and provided evidence-based and practical tools for clinicians who manage WD. This whitepaper seeks to further build upon the success of these publications and is geared toward the practical issues facing providers caring for patients with WD, including: • Strategies to improve the diagnosis of WD • Establishing safe, effective, and accessible treatment strategies • Identifying and overcoming adherence and compliance challenges • Understanding individual patient needs and building rapport As with the previous two publications, the goal of this whitepaper is to ultimately improve the long-term outcomes for patients with this disease. As depicted in Figure 1, the presentation of WD is clinically heterogeneous and can involve many organ systems. Therefore, in most instances, the way the disease presents in a patient dictates who ultimately diagnoses WD. For example, a patient with neurological symptoms will be directed to a neurologist, whereas an asymptomatic patient with abnormal liver function tests will be referred to a hepatologist. Furthermore, although the age at WD presentation is considered mainly between 3 and 55 years old, it is actually “both younger and older than generally appreciated.”4 Patients with WD can therefore be diagnosed and managed by clinicians of different specialties and with expertise in patients of various ages. As such, clinicians across multiple disciplines have collaborated to write this whitepaper (See Box 1). In addition, a patient with WD, who is also a patient advocate and the president of the Wilson Disease Association, has provided a patient’s perspective on her experience with the disease. Clinicians and Patients Confront Practical Issues in Wilson Disease 3 Box 1. About the Authors of and Contributors to this Whitepaper Regino P. Gonzalez-Peralta, MD, is a specialist in pediatric hepatology who treats WD in children younger than 21 and adults and serves on the WD Association Advisory Committee. Children with undiagnosed WD most often visit his practice with abnormal liver tests and he usually identifies WD after detecting low ceruloplasmin. His adult patients are typically established patients with WD who are seeking a disease expert to provide maintenance care. Dr. Gonzalez-Peralta builds an “incredible rapport” with his patients having cared for them through their journey from young children accompanied by their parents, to teenagers facing adolescent issues compounded by a chronic liver disease diagnosis, to young adults seeking college degrees and ultimately starting their adult lives. In his office, Dr. Gonzalez-Peralta works closely with Jordan Perno, BSN, RN, who, in addition to providing thorough patient care, assists patients in other essential areas, such as WD medication access. Sammy Saab, MD, is a hepatologist who diagnoses and treats WD in adult patients. In his practice, Dr. Saab works closely with Sherona Bau, NP, an advanced practice provider who specializes in hepatology. Both are independent providers, and their patients with WD have hepatic manifestations on presentation but are otherwise asymptomatic, unless they are at the point of decompensation. In their clinic, Dr. Saab and Ms. Bau diagnose and treat WD in new patients and provide maintenance care for established patients with WD. Jeff Bronstein, MD, PhD, is a neurologist who sees a mix of patients with WD in terms of symptoms and ages and directs the WD Association Center of Excellence. He also serves on the WD Association Medical Advisory Committee. In his experience, Dr. Bronstein reports that approximately 50% of patients with WD will have neurological and/or psychiatric symptoms on presentation. They may suffer from a long list of neurological symptoms, with the most common being dysarthria (change in voice), dystonia (abnormal contractions of movements), and tremors. Other symptoms can include ataxia, difficultly walking, gait disorders, Parkinsonism, and seizures. Depression, anxiety, and cognitive impairment are also common in patients with WD. Brain MRI results are usually abnormal in patients with WD who have neurological symptoms. Abnormalities of the brain usually involve the basal ganglia. Rhonda Rowland is a patient with established WD. In 1983, as a senior in college, Ms. Rowland went into liver failure and, because she presented with a low-end-of-normal ceruloplasmin, was initially misdiagnosed with chronic active hepatitis. After her liver biopsy showed high levels of copper, and an ophthalmologist referral led to the detection of Kayser-Fleischer rings, the diagnosis of WD was made. Despite living with WD for more than 40 years, Ms. Rowland has had a successful career at CNN as a medical correspondent and in freelance medical communications and, in recent years, has focused her attention to WD. She has interviewed Dr. John Walshe, discoverer of D-penicillamine, is working on a book about her experiences and in 2020 joined the board of the Wilson Disease Association, where she currently serves as president. Ms. Rowland is well-controlled on WD medication and considers herself one of the “lucky ones”, but has observed WD patients receive liver transplants and experience severe liver, neurologic and psychiatric complications and become emotionally impacted as a result of this disease. Ms. Rowland is doing what she can to raise awareness so other WD patients can avoid these types of outcomes. Clinicians and Patients Confront Practical Issues in Wilson Disease 4 Strategies to Improve the Diagnosis of WD If WD is diagnosed and treated early, patients can achieve normal life expectancy.7 According to the AASLD guidelines, WD should be considered when a patient has one or more of the following on presentation: liver abnormalities of uncertain cause, regardless of age; unexplained liver disease associated with neurological or psychiatric disorder(s); acute liver failure with hemolytic anemia; or recurrent, self-limited, autoimmune hepatitis. A gold-standard test to confirm WD does not exist. Therefore, after a thorough history and physical examination, a combination of the following is recommended: liver biochemistries; complete blood count and international normalized ratio; serum ceruloplasmin; basal 24-hour urinary copper excretion; slit-lamp or optical tomography examination for Kayser-Fleischer rings; a neurological evaluation; and a molecular genetic investigation of ATP7B.4, 9, 10 In selected cases, a liver biopsy will be performed to confirm the diagnosis of WD. Despite these recommendations, patients with WD may experience inaccurate or late diagnoses.11, 12 One study noted that 72% (129/179) of patients with WD were misdiagnosed with hepatitis, cirrhosis, splenomegaly, hepatomegaly, encephalitis, encephalopathy, peripheral neuropathy, psychosis, osteoarthrosis, nephrosis and anemia.12 The authors reinforce that these data are seen in their clinical practice, where WD diagnoses are often missed or “brushed off,” and added that they have seen WD misdiagnosed as metabolic dysfunction—associated steatotic liver disease, autoimmune hepatitis, chronic hepatitis, stroke, Parkinson’s disease, multiple sclerosis, and even dental disorders due to neurological mouth manifestations. One major problem is a misdiagnosis of depression, particularly in teenagers. These patients are not correctly diagnosed until they eventually experience severe WD symptoms. Increased awareness of WD is imperative for making an early diagnosis and for treatment. This is, however, a simple concept that is difficult to achieve. Keeping up with medical knowledge is demanding, and rare diseases rarely attract attention. Hepatology guidelines/guidances exist for WD, but there are no WD guidelines for neurologists. Case presentations of WD can make a profound impact on clinicians and, as such, a case study is included here (Box 2). In addition, Figure 2 illustrates some steps for improving WD diagnoses. Box 2. Wilson Disease Case Study Mary is an example of a patient with WD whose condition was misdiagnosed. Mary was 18 years old when she moved away from home to begin her college studies. After 2 months, Mary became very depressed and returned home for winter break. Her primary care physician (PCP) attributed the depression to the stress from her current circumstances and prescribed an antidepressant. Although Mary’s depression improved, she developed a tremor. Blood tests revealed mild liver enzyme abnormalities. Her PCP attributed these findings to the antidepressant and made no changes. Nine months after her first complaint of depression, Mary started falling and ended up in the emergency room. There, they noted an abnormal brain MRI and detected compromised liver function. Upon admission, common causes of liver disease were ruled out, and Mary received a diagnosis of viral syndrome. She was discharged from the hospital to a rehabilitation facility where a neurologist reviewed the MRI and considered the diagnosis of WD, which was quickly confirmed with blood and urine tests. After 6 months of chelation therapy and rehabilitation, Mary recovered almost completely and was able to return to college. Clinicians and Patients Confront Practical Issues in Wilson Disease 5 Figure 2. Methods to Improve the Diagnoses of Wilson Disease Establishing Safe, Effective, and Accessible Treatment Strategies All patients with a newly established diagnosis of WD should be initiated on lifelong medical therapy. Pharmacotherapy of WD involves lowering copper load, which is achieved either by blocking intestinal copper absorption with zinc or facilitating urinary excretion of copper with copper chelators trientine or D-penicillamine.4, 10 Zinc was traditionally reserved for the maintenance treatment of WD, but is now recommended by the AASLD as a first-line therapy for asymptomatic patients,4, 10 including children identified during family screening. This is because zinc eliminates copper at a much slower rate than chelators, making it more suitable for patients who are not affected by WD symptoms. Zinc, available in prescription and over-the-counter formulations, is more accessible and affordable than chelating agents. However, patients may complain of gastrointestinal adverse effects (nausea, in particular) and should be counseled to take zinc on an empty stomach. Patients who take over-the-counter zinc should be advised that for WD, a daily dose of 150 mg of elemental zinc is required. There are many formulations available, which can cause confusion. For example, the 220 mg formulation of zinc contains only 50 mg of elemental zinc, so patients with WD taking this formulation require 1 tablet, 3 times daily. Chelation therapy with both d-penicillamine and trientine increases free copper and can cause early worsening of neurological symptoms in a small subset of patients, with the risk of patients deteriorating before they improve.13 Chelating slowly appears to reduce the risk, but this has not been well tested and therefore not completely established as a mechanism. Both D-penicillamine and trientine are effective copper chelators but since trientine has fewer systemic side effects and is thought to have a lower risk of neurological worsening, it is considered the chelator of choice by most WD experts. When administering trientine, begin at 20 mg/kg/d (to 2000 mg/d daily maximum) in 2 to 3 divided doses and incrementally increase the dose over 2 to 3 weeks. D-penicillamine should be started at 250 to 500 mg and increased by 250 mg increments every 4 to 7 days to about 1,000 to 1,500 mg/d (15-20 mg/kg/d to a maximum of 2,000 mg/d) in 2 to 4 divided doses.4 Clinicians and Patients Confront Practical Issues in Wilson Disease 6 Some patients experience intolerance to D- penicillamine, such as worsening neurological symptoms, skin and kidney manifestations. and bone marrow suppression. Ms. Rowland chronicled that, despite 21 years of successful disease control with D-penicillamine treatment, she discontinued the medication due to skin changes in her neck, which personally affected her given her profession as an on-air medical correspondent. Ms. Rowland switched to zinc gluconate, since D-penicillamine can be linked to dermatological issues14, 15, and has remained well-controlled for the past 20 years without gastrointestinal adverse effects. Other patients have experienced safe and effective control of their WD on long-term D-penicillamine and should remain on that treatment. Trientine hydrochloride, the original formulation, must be refrigerated at all times or the drug potency is lost.16 Patients often point out that this is particularly stressful when navigating through airport security. Dr. Saab shared a story about a patient with WD taking trientine hydrochloride. While traveling to India, the patient had to request ice from the flight attendants to maintain the correct temperature to store his medication. Patients with WD also report that trientine hydrochloride is sometimes not shipped properly and arrives in a pool of melted ice. A new formulation, trientine tetrahydrochloride, was approved by the Food and Drug Administration in 2022 as a maintenance therapy for adults with stable WD who tolerate D-penicillamine.4, 17 Trientine tetrahydrochloride offers the advantage of twice daily dosing and refrigeration is not required,18 allowing patients the convenience of carrying a chronic medication in their purse or pocket rather than a cooler. Although this new formulation has been described as a “game changer” and a specialty product, prescribers of trientine tetrahydrochloride may need to take extra steps in order to access the medication. (For additional details, visit: patient-support/). If insurance authorization is denied for any WD medication, it is resourceful to have a medical necessity letter template on-hand to facilitate the appeal to the insurance company. Appendix A provides a table with a list of details that should be included in the letter. This table can be printed and used to collect information for each patient appeal. Identifying and Overcoming Adherence and Compliance Challenges There is considerable patient variability in regard to WD medication compliance. This is particularly common with asymptomatic patients and teenagers, but can also occur with patients who have experienced decades of disease control with chronic medication compliance. Noncompliance from WD treatment is different from that of other diseases. In conditions like diabetes or depression, a missed dose or two can result in the return of severe symptoms, whereas in WD, lack of compliance does not translate to an immediate symptom recurrence. Progression comes on so slowly that patients are unaware of how sick they are becoming. Clinical experience has demonstrated that the more convenient the medication regimen, the more likely the patient will be to adhere to it. For example, Ms. Rowland said that in her experience, because zinc typically needs to be taken three times daily either one hour before or two hours after a meal, the mid-day dose is most frequently (and unintentionally) missed. This was especially true when traveling for work or staying at home with her children, where her routine may change each day. Furthermore, patients, especially those who are frequently on the go or working out of the home, are more likely to be compliant with medications that are easiest to transport, such as those stored at room temperature. Appendix B provides a patient checklist to simplify adherence and compliance which can be printed and distributed to patients. As summarized in Appendix B, recommending visual and auditory cues to remind patients to take their medication can also be very helpful. For example, rather than storing medications in the cupboard or medicine cabinet, patients can create visual cues by placing the bottle next to places in the home that are frequently visited during the daily routine (eg, by their toothbrush or next to the stove). Smartphone or watch alarms can serve as useful auditory cues throughout the day. Basal 24-hour urinary copper excretion is an important aspect in the diagnosis of WD and requires 6-to-12 month, or even more frequent, monitoring while on treatment.4, 10 Adherence to this laboratory test is a big issue in WD. Collecting a full day of urine is challenging, particularly in younger patients, but this is currently the only method to estimate treatment Clinicians and Patients Confront Practical Issues in Wilson Disease 7 response and compliance. Novel assays are being studied to replace the 24-hour collection, but none are commercially available.17, 19 In the meantime, clinicians can provide clear instructions and helpful tips to make this easier for patients (Appendix B). Urine collection is difficult to store at work, so patients can be advised to choose the easiest day of the week to perform the test; most patients choose Sundays. All urine produced that day needs to be collected. If this is done incorrectly, they will need to repeat the test. Once patients need to repeat the test, they are usually more compliant and mindful of potential errors. WD researchers are pursuing blood-based copper tests, which are currently being studied,17, 19 to replace this standard yet cumbersome approach. Understanding Individual Patient Needs and Building Rapport Adherence and compliance challenges with medications and lab tests can ultimately lead to patients lost to follow-up. Improvement requires heightened clinician awareness, an understanding of patients’ individual situations and needs, and effective strategies. This can be hard to navigate in a busy clinical practice: It is comparable to opening Pandora’s box, which by definition means this may bring up great troubles and emotions (family problems, divorces, abuse), but also holds hope. This “box” needs to be opened if you want to treat the patient wholistically. Factors that affect adherence and compliance change as patients transition through life. Patients who receive a WD diagnosis at a very young age tend to have families who are very involved in their care. Parents or caregivers act as the primary advocate and decision-makers; therefore, adherence to the necessary steps in a patient’s care is more likely. Furthermore, siblings of patients with WD are typically screened for the disease as well, so caregivers may be experienced in dealing with WD in more than one child. The transition to adolescence can be a tumultuous time for patients, with cognitive and physical changes, peer pressure, and yearning for conventionality. Chronic medication requirements for any disease are particularly difficult in adolescents and college-aged patients. Although some executive functions aren’t fully formed until after 22 years of age20, young patients with WD are faced with adult responsibilities and consequences. Adherence may become a point of contention between the parent/caregiver and the patient. Parents report concerns with the transition to college, since they are not there to help ensure their children with WD continue to comply with treatment. Improvements in adherence are expected as patients transition to adulthood, but this isn’t always the case. Young adults may bypass regular physical examinations and laboratory assessments. Also, young adults are sometimes uninsured (as they lose parental coverage into their late 20’s) and, as previously discussed, are not immediately feeling the benefits of treatment or the consequences of noncompliance. The more rapport the clinician has with the patient, the more successful the results. Some clinicians see patients more frequently to build that relationship. At that time, laboratory samples can be drawn and patients can be shown the physical consequences of treatment noncompliance (eg, elevated liver function tests). Understanding the patient’s unique needs can help with rapport. For example, communication with teenagers may be more effective one-on-one, with the caregiver stepping out of the room. Battling with adolescents tends to be unsuccessful, whereas demonstrating empathy for their situation and providing realistic advice is helpful. They can be reminded that their friends don’t need to know their diagnosis and medications can be passed off as a “vitamin” or taken privately in the early morning and late at night. In some cases, patients of any age may experience anxiety, depression, and coping issues that accompany this type of diagnosis and a psychiatry referral may be appropriate. Figure 3 summarizes some of the points mentioned above and provides additional strategies to build patient rapport and promote adherence and compliance. Patient motivation to comply with treatment often comes from connecting with other patients with WD via the WD Association. Some patients are willing to share their stories with fellow patients who have WD. Hearing about how others have coped with WD treatment successes and Clinicians and Patients Confront Practical Issues in Wilson Disease 8 failures can also promote compliance. These types of stories and additional resources can be found on the WD Association YouTube channel ( Ms. Rowland said that her early encounters with patients who had severe WD and their families, where the consequences of uncontrolled disease were evident, was a “wakeup call” that made her even more dedicated to keeping her own disease under control. At present, she complies with zinc treatment and her WD is monitored twice a year by a general practitioner and annually by a hepatologist. Imaging studies, including transient elastography, ultrasound, and magnetic resonance imaging, demonstrate no evidence of cirrhosis. Keep in mind that this monitoring is specific for Ms. Rowland and the approach may vary in other patients, depending on how their disease presents (eg, liver vs neuropsychiatric disease), and the stage of the disease. Regardless, providers should commend patients like Ms. Rowland who comply with treatment by recognizing their efforts and successes. Figure 3. Strategies to Build Patient Rapport and Promote Adherence and Compliance Conclusions Prior to Ms. Rowland receiving a diagnosis of WD, she experienced cirrhosis and liver failure. Originally her disease was misdiagnosed as chronic active hepatitis, and she was told that she was very sick and her life expectancy was uncertain. When she received the correct WD diagnosis and was told this disease could be controlled with life-long treatment, the “hope of normalcy” motivated her. “I felt normal and because of that, I felt that I could go on living my life.” Early access to diagnosis and treatment from a skilled professional can provide this type of hope to all patients with WD. In the future, more accessible genetic testing, simpler follow-up assessments, and novel, convenient treatments can push us further toward this goal. Acknowledgments: This whitepaper was funded by Orphalan. The selection of the authors and the creation of this paper were done independently, and Orphalan did not play a role. Jordan Perno, BSN, RN, provided clinical input; Rachel E. Bejarano, PharmD provided medical writing assistance. References 9 1. Wilson SAK. Progressive lenticular degeneration: a familial nervous disease associated with cirrhosis of the liver. Brain. 1912;34(4):295-507. 2. National Organization for Rare Disorders. Wilson’s Disease. Updated 11/12/2024. Accessed February 8, 2024. https:// rarediseases.org/rare-diseases/wilson-disease/. 3. Immergluck J, Anilkumar AC. Wilson Disease. [Updated 2023 Aug 7]. In: StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan-. Available from: 4. Alkhouri N, Gonzalez-Peralta RP, Medici V. Wilson disease: a summary of the updated AASLD Practice Guidance. Hepatol Commun. 2023;7(6):e0150. doi: 10.1097/HC9.0000000000000150. 5. Aggarwal A, Bhatt M. Advances in Treatment of Wilson Disease. Tremor Other Hyperkinet Mov (N Y). 2018;8:525. 6. Roberts EA, Schilsky ML. Diagnosis and treatment of Wilson disease: an update. Hepatology. 2008;47(6):2089-111. 7. Ferenci P, Ott P. Wilson’s disease: Fatal when overlooked, curable when diagnosed. J Hepatol. 2019;71(1):222-24. 8. Nemeth D, Folhoffer A, Laszlo SB, et al. Liver transplantation in Wilson’s disease patients, 1996-2017. Orv Hetil. 2019;160(51):2021-25. 9. Schilsky ML, Roberts EA, Bronstein JM, et al. A multidisciplinary approach to the diagnosis and management of Wilson disease: Executive summary of the 2022 Practice Guidance on Wilson disease from the American Association for the Study of Liver Diseases. Hepatology. 2023;77(4):1428-55. 10. Schilsky ML, Roberts EA, Bronstein JM, et al. A multidisciplinary approach to the diagnosis and management of Wilson disease: 2022 Practice Guidance on Wilson disease from the American Association for the Study of Liver Diseases. Hepatology. Dec 7 2022. doi: 10.1002/hep.32801. 11. Ferenci P. Diagnosis of Wilson disease. Handb Clin Neurol. 2017;142:171-180. 12. Yu M, Ren L, Zheng M, et al. Delayed diagnosis of Wilson’s disease report from 179 newly diagnosed cases in China. Front Neurol. 2022;13:884840. 13. Antos A, Członkowska A, Smolinski L, et al. Early neurological deterioration in Wilson’s disease: a systematic literature review and meta-analysis. Neurol Sci. 2023;44(10):3443-55. 14. Levy RS, Fisher M, Alter JN. Penicillamine: review and cutaneous manifestations. J Am Acad Dermatol. 1983;8(4):548-58. 15. Yin JL, Salisbury J, Ala A. Skin changes in long-term Wilson’s disease. Lancet Gastroenterol Hepatol. 2024;9(1):92. 16. Will RG. Cold comfort pharm. BMJ Case Rep. 2009 Feb 2;2009:bcr0920080915. 17. Schilsky ML, Czlonkowska A, Zuin M, et al. Trientine tetrahydrochloride versus penicillamine for maintenance therapy in Wilson disease (CHELATE): a randomised, open-label, non-inferiority, phase 3 trial. Lancet Gastroenterol Hepatol. 2022;7(12):1092-1102. 18. Cuvrior. Prescribing information. Orphalan; 2024. Accessed February 15, 2025. 19. Del Castillo Busto ME, Cuello-Nunez S, Ward-Deitrich C, et al. A fit-for-purpose copper speciation method for the determination of exchangeable copper relevant to Wilson’s disease. Anal Bioanal Chem. 2022;414(1):561-73. 20. Baum GL, Ciric R, Roalf DR, et al. Modular Segregation of Structural Brain Networks Supports the Development of Executive Function in Youth. Curr Biol. 2017;27(11):1561-72 Appendices 10 Background Details Date Patient name Patient date of birth Insurance authorization number Drug requested, dose, number of tablets, days’ supply requested Proof-of-Diagnosis Age at diagnosis Initial presenting symptom(s) How the WD diagnosis was established (include at least 3 of the following: low ceruloplasmin, elevated 24-hour urine copper, elevated liver tissue copper, Kayser-Fleischer rings, positive genetic test for ATP7B) Patient Clinical Background Use and failure of other WD agents, if applicable Treatment failure is indicated by >500 µg/24-hour urine copper, >25 µg/dL of non-ceruloplasmin-bound copper, elevated AST and ALT)10 Patient Clinical Notes and Summary of Laboratory Results Relevant clinical notes Lab results Genetic tests Appendix A: WD Medication Insurance Appeal Letter Details Appendices 11 Visual Medication Reminders Auditory Medication Reminders 24-Hour Urine Collection Appendix B: A Patient Checklist to Simplify Adherence and Compliance I have placed my medication in a place that I frequent throughout the day (next to my toothbrush, near my oven) I have set an alarm on a device that is with me throughout the day (my smartphone, my watch) I have chosen the easiest day of the week for me to perform my test All the urine that I produced that day needs to be collected
10371	https://artofproblemsolving.com/wiki/index.php/Functional_equation?srsltid=AfmBOoppV6Q-e42kh6oyuumiH9NXR89djkwTpjv58zsEh_rlsLRjEmMQ	Art of Problem Solving Functional equation - 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 Functional equation Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Functional equation A functional equation, roughly speaking, is an equation in which some of the unknowns to be solved for are functions. For example, the following are functional equations: Contents 1 Introductory Topics 1.1 The Inverse of a Function 2 Intermediate Topics 2.1 Cyclic Functions 2.2 Problem Examples 3 Advanced Topics 3.1 Functions and Relations 3.2 Injectivity and Surjectivity 4 See Also Introductory Topics The Inverse of a Function The inverse of a function is a function that "undoes" a function. For an example, consider the function: . The function has the property that . In this case, is called the (right) inverse function. (Similarly, a function so that is called the left inverse function. Typically the right and left inverses coincide on a suitable domain, and in this case we simply call the right and left inverse function the inverse function.) Often the inverse of a function is denoted by . Intermediate Topics Cyclic Functions A cyclic function is a function that has the property that: A classic example of such a function is because . Cyclic functions can significantly help in solving functional identities. Consider this problem: Find such that . Let and in this functional equation. This yields two new equations: Now, if we multiply the first equation by 3 and the second equation by 4, and add the two equations, we have: So, clearly, Problem Examples 2006 AMC 12A Problem 18 2007 AIME II Problem 14 Advanced Topics Functions and Relations Given a set and , the Cartesian Product of these sets (denoted ) gives all ordered pairs with and . Symbolically, A relation is a subset of . A function is a special time of relation where for every in the ordered pair , there exists a unique . Injectivity and Surjectivity Consider a function be a function from the set to the set , i.e., is the domain of and is the codomain of . The function is injective (or one-to-one) if for all in the domain , if and only if . Symbolically, f(x)is injective⟺(∀a,b∈X,f(a)=f(b)⟹a=b). The function is surjective (or onto) if for all in the codomain there exists a in the domain such that . Symbolically, f(x)is surjective⟺∀a∈Y,∃b∈X:f(b)=a. The function is bijective (or one-to-one and onto) if it is both injective and subjective. Symbolically, f(x)is bijective⟺∀a∈Y,∃!b∈X:f(b)=a. The function has an inverse function , where , if and only if it is a bijective function. See Also Functions Cauchy Functional Equation Retrieved from " Categories: Algebra Definition 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.
10372	https://study.com/skill/learn/interpreting-a-bar-graph-explanation.html	Interpreting a Bar Graph \| Algebra \| 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 Interpreting a Bar Graph Algebra 1 Skills Practice Click for sound 4:04 You must c C reate an account to continue watching Register to access this and thousands of other videos Are you a student or a teacher? I am a student I am a teacher Try Study.com, risk-free As a member, you'll also get unlimited access to over 88,000 lessons in math, English, science, history, and more. Plus, get practice tests, quizzes, and personalized coaching to help you succeed. Get unlimited access to over 88,000 lessons. Try it risk-free It only takes a few minutes to setup and you can cancel any time. It only takes a few minutes. Cancel any time. Already registered? Log in here for access Back What teachers are saying about Study.com Try it risk-free for 30 days Already registered? Log in here for access 0:04 Interpreting a bar… 2:24 Interpreting a bar… Jump to a specific example Speed Normal 0.5x Normal 1.25x 1.5x 1.75x 2x Speed Nidhi Agarwal, AMY MAYERS Instructors Nidhi Agarwal Nidhi holds a Bachelor's degree in Secondary Education with a teaching major Biological Sciences and Master's degree in education from Lucknow University and has taught middle and high school math. I have over 15 years of experience working with students of different ages including two years of teaching experience internationally. View bio AMY MAYERS Amy has taught middle school math and algebra for over seven years. She has a Bachelor's degree in Mathematical Sciences from the University of Houston and a Master's degree in Curriculum and Instruction from The University of St. Thomas. She is a Texas certified teacher for grades 4-12 in Mathematics. View bio Example SolutionsPractice Questions Interpreting a Bar Graph Step 1: Take a look at the graph and interpret what the x-axis and the y-axis are representing. Step 2: Read the given word problem and see the bar that pertains to the question. Read the value of the bar to answer the question. Use these tips and tricks to interpret a bar graph. If there is a question about one category, read the value of the bar graph that pertains to that category and answer the question. If there is a question about the "most" or "greatest" of something, look at the bars and choose the category with the longest bar as the answer. Similarly, if there is a question for "least", look at the bars and choose the category with the smallest bar as the answer. If there is a comparison of two categories, read the value of the bar graph that pertains to both categories and answer the question. If the word problem is asking about the percentage, then identify the part that is being asked for and the whole and then substitute the values in this formula part whole×100 Interpreting a Bar Graph Vocabulary Bar Graph: A bar graph is a simple, neat visual way of organizing data. Rectangular bars of the same width are used to display and compare data of different categories. The height of the bar for each category in the graph tells us the value of that category. Bar graphs can use can be horizontal or vertical bars to display and compare data. One axis in the bar graph gives us the name of the categories we are comparing and the other axis gives us the scale to measure those categories. We will take a look at two bar graphs and answer questions about them to get a better understanding of the concept of interpreting a bar graph. Interpreting a Bar Graph Example 1 The bar graph below shows a distribution of students in each grade at Star High School. Read the bar graph and answer the following questions. a) Which grade has the most number of students? b) How many more students are enrolled in 11th grade than in 10th grade? c) What percentage of total students in the school are 9th graders? Step 1: Take a look at the graph and interpret what the X-axis and the Y-axis are representing. The x-axis shows the category or the grades in the high school and the y-axis represents the scale or the number of students in each grade. Step 2: Read the given word problem and see the bar that pertains to the question. a) To find out the grade that has the most number of students take a look at the graph and identify the longest bar. The longest bar in the graph is for 12th-grade students. Thus, 12th grade has the most number of students. b) To answer how many more students are enrolled in 11th grade than in 10th grade find the number of students in 10th and 11th grade. Read the value that pertains to the bar showing 10th and 11th-grade students. By taking a close look at the bar graph, we can see that 10th grade has 240 students and 11th grade has 260 students. 260-240 = 20 Thus, there are 20 more students enrolled in 11th grade than in 10th grade. c) To find the percentage of total students in the school that are 9th graders we must first find the total number of students in the school. By seeing the value of each bar we can interpret that there are 300 students in 9th grade 240 students in 10th grade 260 students in 11th grade 320 students in 12th grade Thus the total number of students are 300 + 240 + 260 + 320 = 1,120 Percentage of 9th graders =part whole×100=300 1120×100=26.8 Percentage of 9th graders in the school is 26.8%. Interpreting a Bar Graph Example 2 The bar graph below shows the result of a survey done on 90 second-graders in Morse's elementary school for their favorite lunch choices. Read the bar graph and answer the following questions. a) Which lunch choice is the least favorite? b) How many more students like a sandwich than hotdogs as their favorite lunch choice? c) What percentage of total students have pizza as their favorite lunch choice? Step 1: Take a look at the graph and interpret what the X-axis and the Y-axis are representing. The x-axis shows the favorite lunch choice categories and the y-axis represents the scale or the number of students that have each of the lunch choices as their favorite. Step 2: Read the given word problem and see the bar that pertains to the question. a) To find out the lunch choice is the least favorite take a look at the graph and identify the smallest bar. Thus, rice is the least favorite lunch choice. b) To answer how many more students like a sandwich than hotdogs as their favorite lunch choice we find the number of students who like a sandwich and a hotdog as their lunch choices. By taking a close look at the bar graph we can see that 25 students like a sandwich and 20 students like hotdogs. 25 - 20 = 5 Thus, 5 more students like a sandwich than hotdogs as their lunch choice. c) To find the percentage of total students who have pizza as their favorite lunch choice, we analyze the data as follows. The total number of students surveyed = 90 Percentage of students who like pizza =30 90×100=33.3 Percentage of students who have pizza as their favorite lunch choice is 33.3%. Get access to thousands of practice questions and explanations! Create an account Table of Contents Interpreting a Bar Graph Vocabulary Example 1 Example 2 Test your current knowledge Practice Interpreting a Bar Graph Related Courses Study.com ACT® Test Prep: Help and Review NY Regents Exam - Geometry: Help and Review Introduction to Statistics: Help and Review High School Geometry: Help and Review PSAT Prep: Tutoring Solution Related Lessons Pie Chart Definition & Examples \| What is a Pie Chart? Tally Chart Definition, History & Example Line Graph Definition, Uses & Examples Venn Diagram \| Definition, Components & Examples Line Plot Definition & Examples Recently updated on Study.com Videos Courses Lessons Articles Quizzes Concepts Teacher Resources Attraction: Types, Cultural Differences & Interpersonal... 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10373	https://ieeexplore.ieee.org/document/781584/	Opens in a new window Opens an external website Opens an external website in a new window This website utilizes technologies such as cookies to enable essential site functionality, as well as for analytics, personalization, and targeted advertising. To learn more, view the following link: Privacy Policy Loading [MathJax]/extensions/MathZoom.js A smooth rational spline for visualizing monotone data \| IEEE Conference Publication \| IEEE Xplore Skip to Main Content IEEE.org IEEE Xplore IEEE SA IEEE Spectrum More Sites Subscribe Donate Cart Create Account Personal Sign In "Sign In") Browse My Settings Help Institutional Sign In Institutional Sign In ADVANCED SEARCH Conferences>1999 IEEE International Confe... A smooth rational spline for visualizing monotone data Publisher: IEEE Cite This PDF M. Sarfraz All Authors Sign In or Purchase 1 Cites in Paper 53 Full Text Views AlertsAlerts Manage Content Alerts Add to Citation Alerts Abstract Authors References Citations Keywords Metrics More Like This Download PDF Download References Request Permissions Save to Alerts Abstract: A C/sup 2/ curve interpolation scheme for monotonic data has been developed. This scheme uses piecewise rational cubic functions. The two families of parameters, in the d...Show More Metadata Abstract: A C/sup 2/ curve interpolation scheme for monotonic data has been developed. This scheme uses piecewise rational cubic functions. The two families of parameters, in the description of the rational interpolant, have been constrained to preserve the shape of the data. The monotone rational cubic spline scheme has a unique representation. Published in:1999 IEEE International Conference on Information Visualization (Cat. No. PR00210) Date of Conference: 14-16 July 1999 Date Added to IEEE Xplore: 06 August 2002 Print ISBN:0-7695-0210-5 Print ISSN: 1093-9547 DOI:10.1109/IV.1999.781584 Publisher: IEEE Conference Location: London, UK Authors References Citations Keywords Metrics More Like This Rational spline interpolation preserving the shape of the monotonic data Proceedings Computer Graphics International Published: 1997 Data Visualization Using Shape Preserving C2 Rational Spline 2011 15th International Conference on Information Visualisation Published: 2011 Show More References References is not available for this document. IEEE Personal Account Change username/password Purchase Details Payment Options View Purchased Documents Profile Information Communications Preferences Profession and Education Technical interests Need Help? US & Canada: +1 800 678 4333 Worldwide: +1 732 981 0060 Contact & Support Follow About IEEE Xplore \| Contact Us \| Help \| Accessibility \| Terms of Use \| Nondiscrimination Policy \| IEEE Ethics Reporting \| Sitemap \| IEEE Privacy Policy A public charity, IEEE is the world's largest technical professional organization dedicated to advancing technology for the benefit of humanity. © Copyright 2025 IEEE - All rights reserved, including rights for text and data mining and training of artificial intelligence and similar technologies. IEEE Account Change Username/Password Update Address Purchase Details Payment Options Order History View Purchased Documents Profile Information Communications Preferences Profession and Education Technical Interests Need Help? US & Canada: +1 800 678 4333 Worldwide: +1 732 981 0060 Contact & Support About IEEE Xplore Contact Us Help Accessibility Terms of Use Nondiscrimination Policy Sitemap Privacy & Opting Out of Cookies A not-for-profit organization, IEEE is the world's largest technical professional organization dedicated to advancing technology for the benefit of humanity. © Copyright 2025 IEEE - All rights reserved. Use of this web site signifies your agreement to the terms and conditions.
10374	https://www.dentalschooldeclassified.com/blog/cube-counting-hacks-for-the-pat-how-to-visualize-like-a-pro	Cube Counting Hacks for the PAT: How to Visualize Like a Pro — Dental School Declassified Skip to Content Open Menu Close Menu Dental School Declassified Newsletter Digital Downloads About Blog Services Contact (0) Cart (0) Open Menu Close Menu (0) Cart (0) Dental School Declassified Newsletter Digital Downloads About Blog Services Contact Newsletter Digital Downloads About Blog Services Contact Cube Counting Hacks for the PAT: How to Visualize Like a Pro Jun 8 Written By DSD . Master Cube Counting with Speed, Accuracy, and Strategy for a Higher DAT Score Struggling with the cube counting section of the PAT? Learn step-by-step visualization strategies, cube counting hacks, and test-day tips to improve speed and accuracy on the DAT. The Perceptual Ability Test (PAT) on the Dental Admission Test (DAT) includes six challenging subsections, and Cube Counting is one of the most score-boosting—if you know how to approach it. While it may seem straightforward at first, Cube Counting questions are designed to test your 3D visualization, attention to detail, and pattern recognition—all under timed pressure. In this guide, you’ll learn: What Cube Counting is and how it works on the PAT The most common mistakes students make Proven visualization strategies that actually work Cube counting hacks to improve accuracy and speed Let’s dive in and help you approach Cube Counting like a pro. 🧠 What Is Cube Counting on the PAT? In this section, you're shown a stack of cubes, some of which may be hidden behind others. Your job is to determine how many cubes have: 0, 1, 2, 3, 4, or 5 visible faces You’ll then be asked questions like: "How many cubes have exactly three visible sides?" There are 15 cube counting questions, and they rely on your ability to: Visualize hidden parts of the structure Keep track of how many faces are visible Stay organized under pressure ❌ Common Mistakes in Cube Counting (And How to Fix Them) 1. Miscounting Hidden Cubes Students often overlook cubes that are partially or completely blocked in back rows or under other cubes. ✅ Fix: Train yourself to count the total number of cubes layer by layer, then calculate visible sides per cube. 2. Rushing Without a System Jumping into cube counting without a structure leads to skipping cubes, double-counting, or misjudging exposed sides. ✅ Fix: Use a systematic method: count each cube from front to back and left to right, recording visibility for each one. 3. Guessing the Number of Visible Sides Some students rely on intuition instead of logic, especially for cubes not directly visible in the front view. ✅ Fix: Visualize or sketch the cube layout and analyze how many faces each cube can possibly show. ✅ Cube Counting Hacks That Actually Work 1. Use a Tally Chart for Each Visibility Category Set up a quick tally like this before each question set: Each time you count a cube, place a mark in the correct row. This prevents double-counting and keeps you organized. 2. Work From Top to Bottom, Front to Back Start at the top layer, and then move row by row—counting left to right in each row. This top-down, front-to-back strategy helps ensure that you: Don’t miss hidden cubes Don’t double-count Catch all visible faces systematically 3. Remember the Max Number of Visible Sides Is 5 A cube can show a maximum of 5 faces—not 6—because one face must always be hidden where it touches the ground or another cube. Here’s a quick reference: 1 cube on top with nothing around it = 5 visible faces Corner cube surrounded on 2 sides = 3 faces Center cube surrounded on all sides = 1 or 2 faces 4. Use the DOT Method for Sketching If you're practicing on paper, lightly draw dots on the exposed faces of each cube and count them: Front face = ● Top face = ● Side face = ● Use this method to confirm how many sides are visible before tallying. 5. Don’t Let Shadows Trick You Test designers often shade areas to create visual complexity—but remember: shadows don’t hide cubes. Every cube is drawn purposefully, so look carefully at each level and corner. 🕒 Timing Tips for the Cube Counting Section Each PAT subsection is timed, so efficiency matters. Aim to spend under 1 minute per question, giving you time to double-check answers. Time-saving tip: Group cubes with the same number of visible faces and count them together. 🔁 Weekly Practice Plan for Cube Counting Success 🧰 Best Tools to Practice Cube Counting DATBooster – Offers detailed cube counting explanations and visual tutorials. Bootcamp PAT Generator – Interactive 3D questions with real-time cube views. PATCrusher – Custom analytics on your weakest cube types. Printable cube blocks or 3D modeling apps – Practice physical or virtual builds to strengthen spatial awareness. 🦷 Why Cube Counting Matters for Future Dentists As a dentist, spatial reasoning is critical. From reading X-rays to placing restorations, you'll need to mentally visualize: Tooth positions Surface angles Layered structures in 3D space Cube Counting helps build the spatial skills you’ll use every day in clinical dentistry. 🚀 Final Takeaway: Accuracy Beats Guesswork With practice, Cube Counting becomes a fast and reliable PAT section for scoring high. Use structured methods like tally charts, top-down analysis, and layer-by-layer visualization to boost your accuracy and reduce careless errors. Want to Master the PAT with Expert Help? At Dental School Declassified, our tutors scored in the 98th percentile or higher on the DAT. We offer: School supplies 1-on-1 PAT tutoring (including cube counting) Personalized DAT study schedules Strategy-based coaching with real-time feedback Book your PAT session now Build your Custom Study Plan Today #DATPrep#CubeCounting#PATStrategy#PerceptualAbilityTest#DATSuccess#DentalAdmissions#PreDentalTips DSD . Previous Previous Top 10 Strategies to Improve Your PAT Score on the DAT ------------------------------------------------------Next Next Hole Punching Tips That Actually Work: Avoid Common Mistakes ------------------------------------------------------------ EXPLORE Shop About Contact LET’S BE FRIENDS Drop your email address to receive news and updates. 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10375	https://emedicine.medscape.com/article/1217083-clinical	For You News & Perspective Tools & Reference Edition English Medscape Editions About You Professional Information Newsletters & Alerts Your Watch List Formulary Plan Manager Log Out Register Log In EN Medscape Editions English Deutsch EspaÃ±ol FranÃ§ais PortuguÃªs UK X Univadis from Medscape About You Professional Information Newsletters & Alerts Your Watch List Formulary Plan Manager Log Out Register Log In close Please confirm that you would like to log out of Medscape. If you log out, you will be required to enter your username and password the next time you visit. Log out Cancel processing.... Tools & Reference>Ophthalmology Adult Optic Neuritis Clinical Presentation Updated: Aug 11, 2022 Author: Andrew A Dahl, MD, FACS; Chief Editor: Edsel B Ing, MD, PhD, MBA, MEd, MPH, MA, FRCSC more...;) Share Print Feedback Close Facebook Twitter LinkedIn WhatsApp Email Sections Adult Optic Neuritis Sections Adult Optic Neuritis Overview Practice Essentials Background Etiology Epidemiology Prognosis Show All Presentation History Physical Examination Show All DDx Workup Approach Considerations Magnetic Resonance Imaging Visual Evoked Potentials Show All Treatment Approach Considerations Steroid Therapy Medical Care Show All Medication Medication Summary Monoclonal Antibodies Corticosteroids Show All References;) Presentation History A history of preceding viral illness and pain on eye movement may be present. Typically, patients with first-time acute optic neuritis (ON) are otherwise healthy young adults. Patients with ON experience rapidly developing impairment of vision in 1 eye or, far less commonly, both eyes during an acute attack. Symptoms of dyschromatopsia (change in color perception) in the affected eye may occasionally be more prominent than the decreased vision. In nearly all patients with ON, the visual changes are associated with retro-orbital or ocular pain, usually exacerbated by eye movement. The pain may precede the visual loss. Patients may report vision loss exacerbated by heat or exercise (Uhthoff phenomenon). Objects moving in a straight line may appear to have a curved trajectory (Pulfrich effect) when viewed bilaterally, presumably as a result of asymmetric conduction between the optic nerves. Patients with MS may have recurrent attacks of ON ; therefore, a history of episodes of decreased vision in the same or the fellow eye may be elicited. A history of neurologic problems, such as transient episodes of extremity or facial numbness, weakness, or balance difficulties, suggests a diagnosis of MS; a family history of MS may exist. Neuromyelitis optica (NMO) or neuromyelitis optica spectrum disorder (NMOSD), a rare autoimmune disease until recently was thought to be a type of multiple sclerosis (MS). Neuromyelitis optica is characterized by ON and myelitis in a close temporal relationship [6, 7, 8, 9] ; however, ON can occasionally precede the myelopathy. Some patients with neuromyelitis optica (NMO) have relapses limited to the optic nerves and spinal cord. NMOSD has recently been identified as a distinct autoimmune illness with more severe symptoms than MS and with a separate set of causes. The demyelinating process in NMOSD is only a product of a newly identified and more fundamental disease process: astrocytopathy, or destruction of astrocytes. These specialized and star-shaped central nervous system cells perform a range of functions, including delivery of nutrients to nervous tissue, regulation of blood flow in the brain, and repair processes after injury or infection. This astrocyte degeneration comes in 4 main types that the researchers have named astrocyte lysis, progenitor, protoplasmic gliosis, and fibrous astrogliosis, each with their own set of characteristic markers identifiable from astrocyte lesions, or damage to the astrocytes. Astrocyte lysis, or extensive loss or complete destruction of astrocytes, a characteristic of the most acute type of such damage (meaning sudden onset and short duration), is a feature highly specific to NMOSD. The other 3 types describe subacute or chronic forms of the disease (meaning slow onset that can worsen over time). In male patients with bilateral, sequential optic neuropathy with little recovery of vision, Leber hereditary optic neuropathy, rather than demyelinating optic neuritis, should be considered as the diagnosis. Patients with Leber hereditary optic neuropathy may have a history of vision loss in maternal uncles. Chronic relapsing inflammatory optic neuropathy (CRION) is a form of inflammatory optic neuropathy that is frequently bilateral and often painful. It is characterized by relapses and remissions. Magnetic resonance imaging (MRI) findings in the brains of patients with CRION are normal, and MRI of the optic nerves often, but not always, shows high-signal abnormalities that enhance. Corticosteroid treatment is effective in reducing symptoms of CRION, although longterm immunosuppression is often necessary. The syndrome behaves in a way that is typical of granulomatous optic neuropathy, but during long-term follow-up, no systemic sarcoidosis has been implicated. Next: Physical Examination In a patient with a typical initial acute case of optic neuritis (ON), findings on a general physical examination are normal. Pupillary light reaction is decreased in the affected eye, and a relative afferent pupillary defect (RAPD) or Marcus Gunn pupil is commonly found. In bilateral cases, the RAPD may not be apparent. Measurement of visual acuity reveals variation in reduction from no reduction to complete visual loss. Patients with ON will all have at least mild subjective decreases in color vision and perception of brightness in the affected eye. All patients with decreased visual acuity also have abnormal contrast sensitivity and color vision, as revealed by examination using a Pelli-Robson chart and Ishihara color plates, respectively. A central scotoma is most commonly seen in patients with ON; however, results of the Optic Neuritis Treatment Trial (ONTT) suggest that altitudinal field defects, arcuate defects, and nasal steps are more common than central scotomas and centrocecal scotomas. Visual field examination may show a central scotoma, often with peripheral extension in any direction. Generalized depression of the entire visual field in the affected eye may be encountered. In two thirds of cases of acute optic neuritis, the optic nerve appears normal because of only retrobulbar involvement of the nerve. One third of patients with ON have a swollen disc (papillitis). With time, the optic nerve may become pale. The disc edema of ON, when present, is often diffuse. The presence of segmental changes, altitudinal swelling, pallor, arterial attenuation, and splinter hemorrhages suggest other diagnoses (eg, anterior ischemic optic neuropathy). Findings on retinal examination usually are normal. If a dilated fundus examination is not performed, other conditions such as central serous retinopathy and retinal detachment may be mistaken for ON. Patients with NMO often develop a severe, bilateral form of ON and myelitis. Bitemporal or junctional visual field defects, indicating chiasm involvement, may be present. Myelitis may be associated with localized back or radicular pain and the Lhermitte sign (spine or limb paresthesias elicited by neck flexion) early in the course of the disease. Severe neurologic deficits, including paraplegia, are typical. Symptoms such as respiratory failure or hiccups may occur when the cervical spinal cord lesions extend into the medulla. Previous Differential Diagnoses References Beck RW, Smith CH, Gal RL, et al. Neurologic impairment 10 years after optic neuritis. Arch Neurol. 2004 Sep. 61(9):1386-9. [QxMD MEDLINE Link]. Beck RW, Gal RL, Bhatti MT, et al. 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Overlapping clinical profiles. Arch Ophthalmol. 1991 Dec. 109(12):1668-72. [QxMD MEDLINE Link]. Lennon VA, Kryzer TJ, Pittock SJ, Verkman AS, Hinson SR. IgG marker of optic-spinal multiple sclerosis binds to the aquaporin-4 water channel. J Exp Med. 2005 Aug 15. 202(4):473-7. [QxMD MEDLINE Link]. Lennon VA, Wingerchuk DM, Kryzer TJ, et al. A serum autoantibody marker of neuromyelitis optica: distinction from multiple sclerosis. Lancet. 2004 Dec 11-17. 364(9451):2106-12. [QxMD MEDLINE Link]. Beck RW, Trobe JD, Moke PS, et al. High- and low-risk profiles for the development of multiple sclerosis within 10 years after optic neuritis: experience of the optic neuritis treatment trial. Arch Ophthalmol. 2003 Jul. 121(7):944-9. [QxMD MEDLINE Link]. Trobe JD, Sieving PC, Guire KE, Fendrick AM. The impact of the optic neuritis treatment trial on the practices of ophthalmologists and neurologists. Ophthalmology. 1999 Nov. 106(11):2047-53. [QxMD MEDLINE Link]. 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Yamamura T, Kleiter I, Fujihara K, Palace J, Greenberg B, Zakrzewska-Pniewska B, et al. Trial of Satralizumab in Neuromyelitis Optica Spectrum Disorder. N Engl J Med. 2019 Nov 28. 381 (22):2114-2124. [QxMD MEDLINE Link].[Full Text]. Traboulsee A, Greenberg BM, Bennett JL, Szczechowski L, Fox E, Shkrobot S, et al. Safety and efficacy of satralizumab monotherapy in neuromyelitis optica spectrum disorder: a randomised, double-blind, multicentre, placebo-controlled phase 3 trial. Lancet Neurol. 2020 May. 19 (5):402-412. [QxMD MEDLINE Link]. Diem R, Molnar F, Beisse F, Gross N, DrÃ¼schler K, Heinrich SP, et al. Treatment of optic neuritis with erythropoietin (TONE): a randomised, double-blind, placebo-controlled trial-study protocol. BMJ Open. 2016 Mar 1. 6 (3):e010956. [QxMD MEDLINE Link]. Collongues N, de Seze J. An update on the evidence for the efficacy and safety of rituximab in the management of neuromyelitis optica. Ther Adv Neurol Disord. 2016 May. 9 (3):180-8. [QxMD MEDLINE Link]. Kim SH, Kim W, Park MS, Sohn EH, Li XF, Kim HJ. Efficacy and safety of mitoxantrone in patients with highly relapsing neuromyelitis optica. Arch Neurol. 2011 Apr. 68(4):473-9. [QxMD MEDLINE Link]. Deschamps R, Gueguen A, Parquet N, Saheb S, Driss F, Mesnil M, et al. Plasma exchange response in 34 patients with severe optic neuritis. J Neurol. 2016 May. 263 (5):883-887. [QxMD MEDLINE Link]. Morrow MJ, Ko MW. Should Oral Corticosteroids Be Used to Treat Demyelinating Optic Neuritis?. J Neuroophthalmol. 2017 Dec. 37 (4):444-450. [QxMD MEDLINE Link]. Naumovska M, Sheikh R, Bengtsson B, MalmsjÃ¶ M, Hammar B. Visual outcome is similar in optic neuritis patients treated with oral and i.v. high-dose methylprednisolone: a retrospective study on 56 patients. BMC Neurol. 2018 Sep 29. 18 (1):160. [QxMD MEDLINE Link]. Watanabe S, Nakashima I, Misu T, Miyazawa I, Shiga Y, Fujihara K. Therapeutic efficacy of plasma exchange in NMO-IgG-positive patients with neuromyelitis optica. Mult Scler. 2007 Jan. 13(1):128-32. [QxMD MEDLINE Link]. Ruprecht K, Klinker E, Dintelmann T, Rieckmann P, Gold R. Plasma exchange for severe optic neuritis: treatment of 10 patients. Neurology. 2004 Sep 28. 63(6):1081-3. [QxMD MEDLINE Link]. McKinney AM, Lohman BD, Sarikaya B, Benson M, Benson MT, Lee MS. Accuracy of routine fat-suppressed FLAIR and diffusion-weighted images in detecting clinically evident optic neuritis. Acta Radiol. 2013 Feb 5. [QxMD MEDLINE Link]. Wilhelm H, Schabet M. The Diagnosis and Treatment of Optic Neuritis. Dtsch Arztebl Int. 2015 Sep 11. 112 (37):616-26. [QxMD MEDLINE Link]. Media Gallery A case of acute optic neuritis. A. 1.5 Tesla, contrast-enhanced spin echo T1-weighted, fat-suppressed coronal MRI through the orbits shows enlargement and contrast enhancement of the left optic nerve in the retrobulbar portion (arrow). B. Coronal spin echo T1-weighted, fat-suppressed MRI of the same patient shows enlargement and contrast enhancement of the nerve in a parasagittal oblique section (arrow). of 1 Tables Back to List Contributor Information and Disclosures Author Andrew A Dahl, MD, FACS Assistant Professor of Surgery (Ophthalmology), New York College of Medicine (NYCOM); Director of Residency Ophthalmology Training, The Institute for Family Health and Mid-Hudson Family Practice Residency Program; Staff Ophthalmologist, Telluride Medical CenterAndrew A Dahl, MD, FACS is a member of the following medical societies: American Academy of Ophthalmology, American College of Surgeons, American Intraocular Lens Society, American Medical Association, American Society of Cataract and Refractive Surgery, Contact Lens Association of Ophthalmologists, Medical Society of the State of New York, New York State Ophthalmological Society, Outpatient Ophthalmic Surgery SocietyDisclosure: Nothing to disclose. Chief Editor Edsel B Ing, MD, PhD, MBA, MEd, MPH, MA, FRCSC Professor, Department of Ophthalmology and Vision Sciences, Sunnybrook Hospital, University of Toronto Faculty of Medicine; Incoming Chair of Ophthalmology, University of Alberta Faculty of Medicine and Dentistry, CanadaEdsel B Ing, MD, PhD, MBA, MEd, MPH, MA, FRCSC is a member of the following medical societies: American Academy of Ophthalmology, American Association for Pediatric Ophthalmology and Strabismus, American Society of Ophthalmic Plastic and Reconstructive Surgery, Canadian Medical Association, Canadian Ophthalmological Society, Canadian Society of Oculoplastic Surgery, Chinese Canadian Medical Society, European Society of Ophthalmic Plastic and Reconstructive Surgery, North American Neuro-Ophthalmology Society, Ontario Medical Association, Royal College of Physicians and Surgeons of Canada, Statistical Society of CanadaDisclosure: Nothing to disclose. Additional Contributors Erhan Ergene, MD Clinical Assistant Professor, Department of Neurology, University of Illinois College of Medicine at Peoria; Medical Director, Comprehensive Epilepsy Program and Clinical Neurophysiology, Illinois Neurological Institute at OSF Saint Francis Medical CenterErhan Ergene, MD is a member of the following medical societies: American Academy of NeurologyDisclosure: Nothing to disclose. Nancy A Machens, APN, CNP Professor of Nursing, Bradley University; Advanced Practice Nurse, Nurse Practitioner, Department of Neurology, Illinois Neurological Institute at OSF Saint Francis Medical CenterDisclosure: Nothing to disclose. Reuben M Valenzuela, MD Clinical Assistant Professor of Neurology, Section of Neuro-ophthalmology, Section of Multiple Sclerosis, Illinois Neurological Institute, University of Illinois College of Medicine Peoria; Neurophthalmologist, OSF HealthCare Illinois Neurological InstituteDisclosure: Nothing to disclose. Acknowledgements Edsel Ing, MD, FRCSC Associate Professor, Department of Ophthalmology and Vision Sciences, University of Toronto Faculty of Medicine; Consulting Staff, Toronto East General Hospital, Canada Edsel Ing, MD, FRCSC is a member of the following medical societies: American Academy of Ophthalmology, American Association for Pediatric Ophthalmology and Strabismus, American Society of Ophthalmic Plastic and Reconstructive Surgery, Canadian Ophthalmological Society, North American Neuro-Ophthalmology Society, and Royal College of Physicians and Surgeons of Canada Disclosure: Nothing to disclose. Simon K Law, MD, PharmD Associate Professor of Ophthalmology, Jules Stein Eye Institute, University of California, Los Angeles, David Geffen School of Medicine Simon K Law, MD, PharmD is a member of the following medical societies: American Academy of Ophthalmology, American Glaucoma Society, and Association for Research in Vision and Ophthalmology Disclosure: Nothing to disclose. Brian R Younge, MD Professor of Ophthalmology, Mayo Clinic School of Medicine Brian R Younge, MD is a member of the following medical societies: American Medical Association, American Ophthalmological Society, and North American Neuro-Ophthalmology Society Disclosure: Nothing to disclose. Acknowledgments The authors and editors of Medscape Reference gratefully acknowledge the assistance of Ryan I Huffman, MD, with the literature review and referencing for this article. Close;) What would you like to print? What would you like to print? Print this section Print the entire contents of Print the entire contents of article Sections Adult Optic Neuritis Overview Practice Essentials Background Etiology Epidemiology Prognosis Show All Presentation History Physical Examination Show All DDx Workup Approach Considerations Magnetic Resonance Imaging Visual Evoked Potentials Show All Treatment Approach Considerations Steroid Therapy Medical Care Show All Medication Medication Summary Monoclonal Antibodies Corticosteroids Show All References;) encoded search term (Adult Optic Neuritis) and Adult Optic Neuritis What to Read Next on Medscape Related Conditions and Diseases Multiple Sclerosis Brain Imaging in Multiple Sclerosis Multiple Sclerosis Spine Imaging The Patient's Journey Through Multiple Sclerosis Biomarker Testing in Multiple Sclerosis Trending Clinical Topic: Multiple Sclerosis Promising New Biomarkers in Multiple Sclerosis News & Perspective Migraine a Forerunner of Multiple Sclerosis? 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10376	https://www.youtube.com/watch?v=6JHcNdHI-Fk	Khan Academy Tutorial: Justify triangle congruence West Explains Best 4960 subscribers 10 likes Description 320 views Posted: 13 Sep 2024 maths #khanacademy #geometry #proofs #congruenceoftriangles Please follow me on Instagram! @westexplainsbest Link: Or my Facebook Page! Link: I hope you enjoyed the video! Please leave a comment if you'd like to see a topic covered or have any mathematics related question. Be sure to search for any other concept you need and check out some of my non-math videos in the special features playlist. 6 comments Transcript: hi this is Mr Weston today we're doing a con Academy tutorial on justifi triangle congruence this was requested by Leb 5624 thanks for requesting this video and if you have your own video request make sure to leave a comment below this is actually really tough Con Academy there's just a lot of language around it and it's it's hard to sip through the information so I understand why this was requested as a video I'm going to try to break it down as simply as possible and class over a lot of stuff that you don't need to know so we're given this triangle here uh two triangles and we're told that they're going to be congruent right and it's kind of laying out the proof for it if I were doing this I would just see that we have a side and an angle and a side okay so you see there we have an angle in between these this is side angle side congruence and I would call it good but we're going to do a proof here and we're going to analyze how they did the proof and justify why they able to make some of these claims so the first thing it does is it kind of transforms one on top of the other so it takes uh I can't remember which one it moves it it does a transformation so that a gets moved over to D and they call that a prime okay so we're taking the ABC triangle we're moving it directly on top of the DF triangle okay so there it is both of them on top if you were to do two different colors it would look like this okay so then you can see the Green from the overlap of the two triangles okay that's not too super important um that's the first thing it's saying one of the things it's saying though if you're looking at the drawing why it has question marks is if you're not if you're not told any information about the sides and you just know that angle a and angle D are are the same then we don't know exactly where Point C is going to overlap with f and how where point B is going to overlap with uh e so part of this is it tells us that it does overlap looking at this second drawing step too that a and sorry with b and e overlap because of this because that side is the same so it's not here it's not here they have the same side length so they have to match up right there and then step C sorry step three is talking about C and F and it's the same deal so um before it was a question mark look up here it's a question mark because because if we don't know the side length then we don't know exactly where C is but we do know the side length so we go to this step and I'm just looking at the drawing here and you can see that it's given that the side length is the same and if it's on that same angle and the side length is the same then they must overlap okay so now we just need to justify it for step three that final step with CNF sorry C Prime and F on top of each other why are they on top of each other well it's because they're the same distance distance away from d and they're on the same Ray when it says along the same Ray I found that confusing at first but basically it's just saying it's on the it has the same angle okay so it's on the same directional path another way to say okay there's an angle that it shares between them and it has that same distance therefore it's the same point so we're going to choose a for this one um C and F uh C Prime and F lying on this intersection of circles when it's talking about intersection of circles uh let's see it with um Center at D so in order for that one to work I gota remember here I'm going to try to get this lined up more exactly almost got it there okay so if it's talking about being centered at the intersection of two circles and I think it's like one and then we have another one I think it's talking about from centered at d and e okay so there's a a circle centered at D and yeah we can see that the radius that's a really bad Circle let me try that again okay so here's the red circle okay so it's actually a pretty big big circle centered at D so we see yeah we know the the length of the radius for that one this is for justification of B and why it doesn't work okay so this part of it works but if we were to draw the second Circle centered at e like it suggests and it's like okay if we draw a circle like this okay it's the intersection of those two circles okay but the problem is we don't know this radius right here okay so we can't use that justification because that's an unknown radius so instead we're going to say okay it's option A because we know that it's along the same Ray and it's the same distance it's a more simple explanation and then C doesn't work um because it's not the intersection of two rays because we don't know anything about let me erase all this we don't know we don't know anything about the second Ray okay we don't know anything about this second angle that angle unknown we don't know about it so we can't say it's the intersection of those two rays cuz we don't know anything about that one right there the one I just erased okay so that's going to be the correct answer and we got our next question okay so now we have two more and you can tell if I were to do this I would say okay all three sides are the same whoops not side angle it would just be side side sides they have all three sides side side side okay so we're going to see how they justify it here they're going to map it they're going to put one on top of the other rigid Transformations and then they're going to say that um a so they move a on top of D so they call that a prime and d and then basically what it's saying is B Prime and E are going to be the same because of this shared side okay that shared side is exactly the same so we have those two points locked in and it's saying where's C I don't know okay cuz technically you could have a triangle that looks like this or like that okay so that's why it's like where's C we don't know and so we need a further step to justify where exactly it is to make sure it's congruent that it lines up with f because that's the whole point we want to make sure that c is the same distance as F so it says C Prime and F are on the same side of De then C Prime equals F okay and the reason why C Prime equals f is you can see that let me erase this is because it has the same distance away okay so if if it's on this left side okay left side then we know it's equal because it's got two different sides that are like rigid it's like not going to move and it has to match up with f because it shares those sides okay but if we go over here okay let's say ended up on the wrong side that was like the first like that drawing I showed you is like the question mark which side is it on if it is on this side then you can do a transformation it says you can reflect it across de de is this line and it's going to end up in the same spot okay so if it ends up on this side you can just do one more transformation that's all it's saying and it's going to be fine so then you'd call it cble Prime and then it's going to match up based on the same reason for step two okay so we need to say oh oh step one we didn't even need to look step two or three okay so what fact can we use to justify step one step one um is this one and it says that we knew where B Prime was okay and how do we know that well it's because a b and d are the same length so these two sides so that's ab and de those are the both the same side because that little line right there so let's go ahead and look for that answer so ab and de in segments with the same length are congruent that's the reason why okay we don't really need to know anything about AC or these other lengths because Step One is talking about how do we know where B Prime and E why they overlap okay so that's that's the reason why for that one step one we're only concerned about that step next one okay if I were looking at this one I would just say this is a side there's an angle and a side and we see that's the same on both of those side angle side congruence okay so I like having that as a frame of mind as I begin and if you need more information on side angle side I do other triangle congruence videos if you want to check those out just do a search on my main page uh so here's a rough uh outline of the the proof we can map ABC using a sequence of rigid Transformations that a prime equals D okay so we're matching up the angles B Prime and E are on the same Ray okay so it's t this the same Ray stuff is probably confusing for many people so first off we're going to map these over okay and then if it's talking about it's on the same Ray it just means it's in the same like Direction because of the angle okay so if it's like oh it's on the same Ray same Ray that thing it's just saying because we mapped a prime onto D and that this angle is the same then there's two rays here there's one I'll make the other one blue here's the other one we have two rays and and those points are going to be on the same Ray because that angle is the same right there now again it's like where's B where's C we need side information for that to know exactly where they are but that's just step one of the proof we can easily do it with side angle side um but they're going to do it a different way so as a result okay you can see here it says this is just like our first one B is equal or sorry B Prime is equal to E because of that shared side and then I'm guessing C is going to say same thing because that shared side how do we show the triangles who are congruent well we ma one figure onto the other using any kind of transformation well we didn't use any kind of transformation um we used a sequence of rigid Transformations not any we didn't use a dilation okay dilation would be a type of transformation we ma one figure out to the other using rigid transform yes we use rigid Transformations let me just check see we showed that all corresponding sides had equal lengths and all corresponding angles had equal measures we didn't do that we only used three measures a side an angle and a side we didn't prove all the sides and all the angles okay I just scrolled up just to take a look at that first picture again let me do that kind of probably giving you vertigo or making you dizzy sorry about that so yeah we didn't find all the other sides we only did it in three steps okay so we just used rigid for Transformations we didn't use dilations which is a type of transformation but it's not rigid we want to make sure the Integrity of the triangles are kept intact what triangles do we show are congruent triangles where one pair of corresponding sides have the same length and one pair of corresponding angles have the same measure no okay we did two uh corresponding sides and one corresponding angle triangles were two pairs of corresponding sides have the same length and including corresponding angles have the same measure B okay all triangles no we didn't do that because that's you know that's not what we did we only did uh two sides and the angle in between all right last question here we go ABCDE EF okay looking at this oh this one's interesting we have an angle side angle okay angle with a side shared in between so let's take a look here um we can map using a sequence of rig Transformations a prime is equal to D and B Prime equals e now the reason why I can say that b Prime is equal to E I change positions if my voice sound is different so a prime equal to D okay that's one we just put on top of the other one and then we know that this one's equal because of that shared side and we can put at angle any angle we want okay so that side's only the important one and we know it's the same because it's locked in it's that same side length but again it's like Hey where's C we don't know the angle in between so which side is it on so we're going to take a look so then we know that if it's on the same side Okay C and F are on the same side of D then C Prime equals F now here's the thing about this it's not like the other one where it's like oh we have the same side information this time it's just the angle so that's going to change our answer for the last part let's see which question it wants us to justify step three okay so step three is oh I should probably pull up the picture for step three I would show the drawing for every single one that's going to be really critical for you so step three is if it's on the other side that's there we can reflect it okay so it's like oh it's a question mark which side it's on this one and this one but if it is on this side we can just reflect it over and then it's going to match up and then we can say it's congruent we'll call it couble Prime because we did two Transformations so what is the justification that c Prime equals f c Prime and F are the same distance from E along the same Ray that is wrong because I mean after the after you establish that they're congruent then you can say that but initially we don't don't know the information about the sides how long is that side no clue okay so we can't say it's the same distance way because we don't know what the distance is so that one is wrong a is going to be wrong there because we don't know anything about the distance okay I'm going to jump down to C and then I'll go to B at the end C Prime and F are the at the intersection of the same pair of rays okay actually I want to go to B so B both C Prime and F lie on the intersection points of circles centered at d and e with right DF and EF respectively there are two such possible points one on each side of De okay this one will work if we know that the the lengths of those so we said centered at d and e so here's my D and like Okay so this it's saying if we had a circle like that with this radius but we don't know that radius and then another circle like this with this radius but we don't know that radius notice how there's two spots where it intersects obviously my circles are terrible okay but if you have two circles it could intersect at two possible locations with those radi but the problem is we don't know the lengths of those radi you have to know that okay so we can't use that one um because we don't know the lengths of the radi the the radius of each circle so instead we're going to choose option C they're at the intersection of the same pair of rays how do I know that well if you're looking at the ray the the ray means it's going in the same direction often times when you say with this exercise if it's saying the Ray it's talking about the angle measure is the angle measured the same so here's the direction of the ray okay now we don't know anything about the distance so we're like is CP Prime here is CP Prime over here okay we don't know the the point is if we know that we have a second Ray that is going in the same direction of it well that was way too big here let me do it like this there we go okay and again if you only had the green you would be like is uh C Prime over here is it over in this area okay you don't know but the thing is if there if you have two known rays that are going that direction even though you don't know the distance there's only one possible spot they meet up and that's right there and then you can be like ah now I know where it is okay so that's what it's talking about for this one when it's talking about the Rays it's talking about the direction it's talking about angles so we can check this answer and be pretty confident that is correct so we made it through I hope that made it a little bit more clear this is a tough one just because the language used in it uh thanks again for requesting this video uh make sure to request your own video if you need it and I look forward to seeing you next time right here on West explains best
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10378	https://www.youtube.com/watch?v=imxAo-yWwsE	Number of ways of forming groups \| Permutations and Combinations \| Grade 11 \| Math \| Khan Academy Khan Academy India - English 540000 subscribers 3 likes Description 287 views Posted: 18 Apr 2025 In this video, we solve two problems of forming groups with given constraints on selections. We realise that we need to consider all possible cases which are satisfied by the constraint and add the number of ways for each of them. For the first one, we form teams of boys and girls for three different scenarios - the team has no girls, has at least one boy and one girl, and has at least three girls. For the second one, the constraints are - exactly 3 girls, at least 3 girls, and at most 3 girls. Courses on Khan Academy are always 100% free. Start practicing—and saving your progress—now! ( Timestamps: 0:00 Problem 1 - Forming the team with boys and girls 0:40 Constraint - the team has no girls 1:40 Constraint - At least one boy and one girl 4:20 Constraint - At least three girls 4:47 Problem 2 - Forming the committe with boys and girls 8:30 Constraint - exactly 3 girls 8:35 Constraint - at least 3 girls 8:42 Constraint - at most 3 girls Khan Academy India is a nonprofit organization with the mission of providing a free, world-class education for anyone, anywhere. We have videos and exercises that have been translated into multiple Indian languages, and 15 million people around the globe learn on Khan Academy every month. Support Us: Created by Ashish Gupta Transcript: Problem 1 - Forming the team with boys and girls A group consists of seven boys and four girls. In how many ways can a team of five members be formed if the team has no girl, at least one boy and one girl, at least three girls? There are three parts to this problem. Pause the video. Try to solve this. Okay, let's do this together. Before we begin, let's ask ourselves, is this a permutation problem or a combination problem? Well, we're forming teams where the order does not matter. It doesn't matter which one we pick first as long as we have a team of five members. So this is a combination problem. All right, let's take this case first. How can we form a team of five Constraint - the team has no girls members when we have no girl in the team? Well, that can only happen if all five of them are boys. So this means we can do this if we select five boys and we select zero girls. So how do we do this? We select five boys from seven. That's 7 C5 and 4 C 0. Selecting zero girls out of four. This is 4 C 0. Now 7 C5 is same as 7 C2. We can reduce this number 5 to 2. We can do this because ncr is same as n c n minus r. So 7 c5 is same as 7 c 7 - 5 that's 2. So this is 7 c2 4 c 0. Anything c 0 is 1. So that's 1. Now 7 C2 is 7 6 by 2 1 this is 7 3 this is equal to 21. So there are 21 ways to form a team of five members where all of them are boys or we can say Constraint - At least one boy and one girl that none of them are girls. Let's take the next case. At least one boy and one girl. How many scenarios can we form here? Pause the video. Think about it. Okay. So at least one boy and one girl. We can do this by having four boys in the team and one girl. In this way, we can have at least one boy and at least one girl. But there are more scenarios. We can have this scenario or we can have three boys and two girls. Reduce the number of boys by one, increase the number of girls by one. Well, there are more scenarios. This or two boys and three girls. And we can keep going. or one boy and four girls. After this, we'll have zero boys and five girls. That's not possible. We need at least one boy and at least one girl. So, these are the four scenarios. We'll figure out the number of ways for each scenario and we'll add all of them because we have or here. So, let's write them down. Four boys out of seven, that's 7 C4. and which means multiply one girl out of four that's 4 C1 plus three boys out of seven that's 7 C3 times two girls out of four that's 4 C2 plus two boys out of seven that's 7 C2 and three girls out of four that's 4 C3 plus one boy out of seven that's 7 C1 and four girls out of four that's 4 C4 let's solve each of them 7 C4 is same as 7 C3 that's 7 into 6 into 5 by 3 into 2 into 1 and 4 C1 is 4. So let's multiply this by 4. 3 into 2 is 6. So this cancels 7 20 that's 140. So there are 140 ways to form a team of five members where we have four boys and one girl. Let's solve the next three. 7 C3 that's 765 by 3 2 1 4 C2 that's 4 into 3 by 2 into 1 6 cancels out 2 cancels out here this is 7 into 5 into 6 that's 7 into 30 that's 210 let's solve the next one 7 C2 that's 7 into 6 by 2 into 1 4 C3 that's the same as 4 C1 so that's 4 2 cancels out we have 7 into 3 4 that's 12. 7 12 is 84. Then we have 7 C1 that's 7. 4 C4 is 1. 7 1 is 7. Now adding all of them up. This is 140 + 210 + 84 + 7 that's a total of 441 ways. Now let's solve this Constraint - At least three girls case at least three girls. How can we have a group where we have at least three girls? This means we can have either three girls or four girls. We can't have more than four girls because we have a total of four. So we'll take this scenario and this scenario and add them. So that's 84 + 7 that's equal to 91. So for this third case, the answer is 91. Now let's solve one more problem. Problem 2 - Forming the committe with boys and girls A committee of seven members has to be formed from a group of seven boys and four girls. In how many ways can this be done when the committee consists of exactly three girls, at least three girls, at most three girls? Again, we have three parts. Pause the video. Try this out. All right. So, let's write down the scenarios. Exactly three girls. This means if three of them are girls, the remaining four are boys. So, we have four boys and three girls. That's the scenario for exactly three girls. There's no other way in which we can do this. What about at least three girls? Now, if we have at least three girls, we can either have three or four girls. So, we have two scenarios. For three girls, we have four boys because we need a total of seven members. So, this is four boys and three girls. This same scenario or three boys and four girls. So, we'll add these two scenarios to get at least three girls. What about at most three girls? This says that the maximum number of girls is three. So let's start with three. When we have three girls, we'll have four boys. This is four boys and three girls. Or we can reduce the number of girls by one. We can have five boys and two girls. Or we can have six boys and one girl. We want at most three girls. The minimum number of girls could be zero as well. So there's one more scenario. Or seven boys. We pick all seven as boys and zero girls. Now let's find the values of each scenario. We'll start with zero girls and then one girl, 2, 3, and four. These are our five scenarios. When we have zero girls, all seven are boys. This means we have 77 ways of picking boys and four C0 ways of picking girls. When there's one girl, the remaining six are boys. So we have 7 C6 4 C1 ways. When there's two girls, we have five boys. So 7 C5 4 C2. For three girls, we have four boys. So 7 C4 4 C3. And for four girls, we have three boys. That's 7 C3 4 C4. Let's solve each of them. 7 C7 that's 1. 4 C 0 that's also 1. So this is 1 1. 7 C6 is same as 7 C1 because NCR is same as NCUS R. So 7 C1 that's equal to 7. 4 C1 that's equal to 4. This is 7 4. 7C5 is the same as 7 C2. So that's 7 into 6 by 2 into 1. And 4 C2 is 4 into 3 by 2 into 1. Next one is 7C4. That's the same as 7 C3. So that's 7 6 5. 3 terms up by 3 2 1 3 terms down. So 765 by 32 1. 4 C3 is same as 4 C1 that's equal to 4. So multiplying this by 4. And lastly 7 C3 4 C4 7 C3 is 765 by 3 to 1 4 C4 that's equal to 1. Now let's evaluate each of them. 1 1 is 1. 7 4 is 28. Here we have 2 and 2 both of them cancel out with this 4. We have 7 into 6 3 that's 7 18 that's equal to 126. Then we have 76 5 4 / 6. So 6 cancel out. 7 20 that's 140. Here we have 3 2 6. 6 cancels out. 7 5 that's 35. So we have all five scenarios and number of ways for each. Now let's figure out these cases. Exactly three girls that's this Constraint - exactly 3 girls case. three girls, that's 140 ways. At least three girls, that's three girls or Constraint - at least 3 girls four girls. So 140 + 35, that's 175. At most three girls, that's 0, 1, Constraint - at most 3 girls 2, or 3. These four scenarios. So that's 1 + 28 + 126 + 140. Let's add them. 100 + 100 is 200. Then we have 40 + 26, that's 66. 66 + 20 is 86. + 8 is 94 + 1 is 95. So that's 295 phase.
10379	https://leanprover-community.github.io/mathlib4_docs/Mathlib/MeasureTheory/MeasurableSpace/CountablyGenerated.html	Mathlib.MeasureTheory.MeasurableSpace.CountablyGenerated - [x] Documentation Mathlib.MeasureTheory.MeasurableSpace.CountablyGenerated Search return to top source Imports Init Mathlib.Data.Set.MemPartition Mathlib.MeasureTheory.MeasurableSpace.Embedding Mathlib.Order.Filter.CountableSeparatingOn Imported by MeasurableSpace.CountablyGenerated MeasurableSpace.countableGeneratingSet MeasurableSpace.countable_countableGeneratingSet MeasurableSpace.generateFrom_countableGeneratingSet MeasurableSpace.empty_mem_countableGeneratingSet MeasurableSpace.nonempty_countableGeneratingSet MeasurableSpace.measurableSet_countableGeneratingSet MeasurableSpace.natGeneratingSequence MeasurableSpace.generateFrom_natGeneratingSequence MeasurableSpace.measurableSet_natGeneratingSequence MeasurableSpace.CountablyGenerated.comap MeasurableSpace.CountablyGenerated.sup MeasurableSpace.instCountablyGeneratedOfCountable MeasurableSpace.instCountablyGeneratedSubtype MeasurableSpace.instCountablyGeneratedProd MeasurableSpace.SeparatesPoints MeasurableSpace.separatesPoints_def MeasurableSpace.exists_measurableSet_of_ne MeasurableSpace.separatesPoints_iff MeasurableSpace.separating_of_generateFrom MeasurableSpace.SeparatesPoints.mono MeasurableSpace.CountablySeparated MeasurableSpace.countablySeparated_of_hasCountableSeparatingOn MeasurableSpace.hasCountableSeparatingOn_of_countablySeparated MeasurableSpace.countablySeparated_def MeasurableSpace.CountablySeparated.mono MeasurableSpace.CountablySeparated.subtype_iff MeasurableSpace.Subtype.separatesPoints MeasurableSpace.Subtype.countablySeparated MeasurableSpace.separatesPoints_of_measurableSingletonClass MeasurableSpace.MeasurableSingletonClass.of_separatesPoints MeasurableSpace.hasCountableSeparatingOn_of_countablySeparated_subtype MeasurableSpace.countablySeparated_subtype_of_hasCountableSeparatingOn MeasurableSpace.countablySeparated_of_separatesPoints MeasurableSpace.exists_countablyGenerated_le_of_countablySeparated MeasurableSpace.mapNatBool MeasurableSpace.measurable_mapNatBool MeasurableSpace.injective_mapNatBool MeasurableSpace.measurableEquiv_nat_bool_of_countablyGenerated MeasurableSpace.measurable_injection_nat_bool_of_countablySeparated MeasurableSpace.measurableSingletonClass_of_countablySeparated MeasurableSpace.measurableSet_succ_memPartition MeasurableSpace.generateFrom_memPartition_le_succ MeasurableSpace.measurableSet_generateFrom_memPartition_iff MeasurableSpace.measurableSet_generateFrom_memPartition MeasurableSpace.generateFrom_iUnion_memPartition MeasurableSpace.generateFrom_memPartition_le_range MeasurableSpace.generateFrom_iUnion_memPartition_le MeasurableSpace.generateFrom_memPartition_le MeasurableSpace.measurableSet_memPartition MeasurableSpace.measurableSet_memPartitionSet MeasurableSpace.countablePartition MeasurableSpace.measurableSet_enumerateCountable_countableGeneratingSet MeasurableSpace.finite_countablePartition MeasurableSpace.instFinite_countablePartition MeasurableSpace.disjoint_countablePartition MeasurableSpace.sUnion_countablePartition MeasurableSpace.measurableSet_generateFrom_countablePartition_iff MeasurableSpace.measurableSet_succ_countablePartition MeasurableSpace.generateFrom_countablePartition_le_succ MeasurableSpace.generateFrom_iUnion_countablePartition MeasurableSpace.generateFrom_countablePartition_le MeasurableSpace.measurableSet_countablePartition MeasurableSpace.countablePartitionSet MeasurableSpace.countablePartitionSet_mem MeasurableSpace.mem_countablePartitionSet MeasurableSpace.countablePartitionSet_eq_iff MeasurableSpace.countablePartitionSet_of_mem MeasurableSpace.measurableSet_countablePartitionSet MeasurableSpace.CountableOrCountablyGenerated MeasurableSpace.instCountableOrCountablyGeneratedOfCountable MeasurableSpace.instCountableOrCountablyGeneratedOfCountablyGenerated MeasurableSpace.instCountableOrCountablyGeneratedProd MeasurableSpace.countableOrCountablyGenerated_left_of_prod_left_of_nonempty MeasurableSpace.countableOrCountablyGenerated_right_of_prod_left_of_nonempty MeasurableSpace.countableOrCountablyGenerated_prod_left_swap Countably generated measurable spaces # We say a measurable space is countably generated if it can be generated by a countable set of sets. In such a space, we can also build a sequence of finer and finer finite measurable partitions of the space such that the measurable space is generated by the union of all partitions. Main definitions # MeasurableSpace.CountablyGenerated: class stating that a measurable space is countably generated. MeasurableSpace.countableGeneratingSet: a countable set of sets that generates the σ-algebra. MeasurableSpace.countablePartition: sequences of finer and finer partitions of a countably generated space, defined by taking the memPartition of an enumeration of the sets in countableGeneratingSet. MeasurableSpace.SeparatesPoints : class stating that a measurable space separates points. Main statements # MeasurableSpace.measurableEquiv_nat_bool_of_countablyGenerated: if a measurable space is countably generated and separates points, it is measure equivalent to a subset of the Cantor Space ℕ → Bool (equipped with the product sigma algebra). MeasurableSpace.measurable_injection_nat_bool_of_countablySeparated: If a measurable space admits a countable sequence of measurable sets separating points, it admits a measurable injection into the Cantor space ℕ → Bool``ℕ → Bool (equipped with the product sigma algebra). The file also contains measurability results about memPartition, from which the properties of countablePartition are deduced. source classMeasurableSpace.CountablyGenerated(α : Type u_3)[m : MeasurableSpaceα] : Prop We say a measurable space is countably generated if it can be generated by a countable set of sets. isCountablyGenerated : ∃ (b : Set(Setα)), b.Countable∧m=generateFromb Instances source defMeasurableSpace.countableGeneratingSet(α : Type u_3)[MeasurableSpaceα][h : CountablyGeneratedα] : Set(Setα) A countable set of sets that generate the measurable space. We insert ∅ to ensure it is nonempty. Equations MeasurableSpace.countableGeneratingSetα=insert∅⋯.choose Instances For source theoremMeasurableSpace.countable_countableGeneratingSet{α : Type u_1}[MeasurableSpaceα][h : CountablyGeneratedα] : (countableGeneratingSetα).Countable source theoremMeasurableSpace.generateFrom_countableGeneratingSet{α : Type u_1}[m : MeasurableSpaceα][h : CountablyGeneratedα] : generateFrom(countableGeneratingSetα)=m source theoremMeasurableSpace.empty_mem_countableGeneratingSet{α : Type u_1}[MeasurableSpaceα][CountablyGeneratedα] : ∅∈countableGeneratingSetα source theoremMeasurableSpace.nonempty_countableGeneratingSet{α : Type u_1}[MeasurableSpaceα][CountablyGeneratedα] : (countableGeneratingSetα).Nonempty source theoremMeasurableSpace.measurableSet_countableGeneratingSet{α : Type u_1}[MeasurableSpaceα][CountablyGeneratedα]{s : Setα}(hs : s∈countableGeneratingSetα) : MeasurableSets source defMeasurableSpace.natGeneratingSequence(α : Type u_3)[MeasurableSpaceα][CountablyGeneratedα] : ℕ → Setα A countable sequence of sets generating the measurable space. Equations MeasurableSpace.natGeneratingSequenceα=Set.enumerateCountable⋯∅ Instances For source theoremMeasurableSpace.generateFrom_natGeneratingSequence(α : Type u_3)[m : MeasurableSpaceα][CountablyGeneratedα] : generateFrom(Set.range(natGeneratingSequenceα))=m source theoremMeasurableSpace.measurableSet_natGeneratingSequence{α : Type u_1}[MeasurableSpaceα]CountablyGeneratedα : MeasurableSet(natGeneratingSequenceα n) source theoremMeasurableSpace.CountablyGenerated.comap{α : Type u_1}{β : Type u_2}[m : MeasurableSpaceβ]h : CountablyGeneratedβ : CountablyGeneratedα source theoremMeasurableSpace.CountablyGenerated.sup{β : Type u_2}{m₁ m₂ : MeasurableSpaceβ}(h₁ : CountablyGeneratedβ)(h₂ : CountablyGeneratedβ) : CountablyGeneratedβ source @[instance 100] instanceMeasurableSpace.instCountablyGeneratedOfCountable{α : Type u_1}[MeasurableSpaceα][Countableα] : CountablyGeneratedα Any measurable space structure on a countable space is countably generated. source instanceMeasurableSpace.instCountablyGeneratedSubtype{α : Type u_1}[MeasurableSpaceα][CountablyGeneratedα]{p : α → Prop} : CountablyGenerated{x:α//p x} source instanceMeasurableSpace.instCountablyGeneratedProd{α : Type u_1}{β : Type u_2}[MeasurableSpaceα][CountablyGeneratedα][MeasurableSpaceβ][CountablyGeneratedβ] : CountablyGenerated (α×β) source classMeasurableSpace.SeparatesPoints(α : Type u_3)[m : MeasurableSpaceα] : Prop We say that a measurable space separates points if for any two distinct points, there is a measurable set containing one but not the other. separates(x y : α) : (∀ (s : Setα), MeasurableSets → x∈s → y∈s) → x=y Instances source theoremMeasurableSpace.separatesPoints_def{α : Type u_1}[MeasurableSpaceα][hs : SeparatesPointsα]{x y : α}(h : ∀ (s : Setα), MeasurableSets → x∈s → y∈s) : x=y source theoremMeasurableSpace.exists_measurableSet_of_ne{α : Type u_1}[MeasurableSpaceα][SeparatesPointsα]{x y : α}(h : x≠y) : ∃ (s : Setα), MeasurableSets∧x∈s∧y ∉ s source theoremMeasurableSpace.separatesPoints_iff{α : Type u_1}[MeasurableSpaceα] : SeparatesPointsα↔∀ (x y : α), (∀ (s : Setα), MeasurableSets → (x∈s↔y∈s)) → x=y source theoremMeasurableSpace.separating_of_generateFrom{α : Type u_1}(S : Set(Setα))h : SeparatesPointsα : (∀ s ∈ S, x∈s↔y∈s) → x=y If the measurable space generated by S separates points, then this is witnessed by sets in S. source theoremMeasurableSpace.SeparatesPoints.mono{α : Type u_1}{m m' : MeasurableSpaceα}hsep : SeparatesPointsα : SeparatesPointsα source classMeasurableSpace.CountablySeparated(α : Type u_3)[MeasurableSpaceα] : Prop We say that a measurable space is countably separated if there is a countable sequence of measurable sets separating points. countably_separated : HasCountableSeparatingOnαMeasurableSetSet.univ Instances source instanceMeasurableSpace.countablySeparated_of_hasCountableSeparatingOn{α : Type u_1}[MeasurableSpaceα][h : HasCountableSeparatingOnαMeasurableSetSet.univ] : CountablySeparatedα source instanceMeasurableSpace.hasCountableSeparatingOn_of_countablySeparated{α : Type u_1}[MeasurableSpaceα][h : CountablySeparatedα] : HasCountableSeparatingOnαMeasurableSetSet.univ source theoremMeasurableSpace.countablySeparated_def{α : Type u_1}[MeasurableSpaceα] : CountablySeparatedα↔HasCountableSeparatingOnαMeasurableSetSet.univ source theoremMeasurableSpace.CountablySeparated.mono{α : Type u_1}{m m' : MeasurableSpaceα}hsep : CountablySeparatedα : CountablySeparatedα source theoremMeasurableSpace.CountablySeparated.subtype_iff{α : Type u_1}[MeasurableSpaceα]{s : Setα} : CountablySeparated↑s↔HasCountableSeparatingOnαMeasurableSets source @[instance 100] instanceMeasurableSpace.Subtype.separatesPoints{α : Type u_1}[MeasurableSpaceα][h : SeparatesPointsα]{s : Setα} : SeparatesPoints↑s source @[instance 100] instanceMeasurableSpace.Subtype.countablySeparated{α : Type u_1}[MeasurableSpaceα][h : CountablySeparatedα]{s : Setα} : CountablySeparated↑s source @[instance 100] instanceMeasurableSpace.separatesPoints_of_measurableSingletonClass{α : Type u_1}[MeasurableSpaceα][MeasurableSingletonClassα] : SeparatesPointsα source @[instance 50] instanceMeasurableSpace.MeasurableSingletonClass.of_separatesPoints{α : Type u_1}[MeasurableSpaceα][Countableα][SeparatesPointsα] : MeasurableSingletonClassα source instanceMeasurableSpace.hasCountableSeparatingOn_of_countablySeparated_subtype{α : Type u_1}[MeasurableSpaceα]{s : Setα}[h : CountablySeparated↑s] : HasCountableSeparatingOnαMeasurableSets source instanceMeasurableSpace.countablySeparated_subtype_of_hasCountableSeparatingOn{α : Type u_1}[MeasurableSpaceα]{s : Setα}[h : HasCountableSeparatingOnαMeasurableSets] : CountablySeparated↑s source instanceMeasurableSpace.countablySeparated_of_separatesPoints{α : Type u_1}[MeasurableSpaceα][h : CountablyGeneratedα][SeparatesPointsα] : CountablySeparatedα source theoremMeasurableSpace.exists_countablyGenerated_le_of_countablySeparated(α : Type u_1)[m : MeasurableSpaceα][h : CountablySeparatedα] : ∃ (m' : MeasurableSpaceα), CountablyGeneratedα∧SeparatesPointsα∧m'≤m If a measurable space admits a countable separating family of measurable sets, there is a countably generated coarser space which still separates points. source noncomputable defMeasurableSpace.mapNatBool(α : Type u_1)[MeasurableSpaceα]CountablyGeneratedα(n : ℕ) : Bool A map from a measurable space to the Cantor space ℕ → Bool induced by a countable sequence of sets generating the measurable space. Equations MeasurableSpace.mapNatBoolα x n=decide (x∈MeasurableSpace.natGeneratingSequenceα n) Instances For source theoremMeasurableSpace.measurable_mapNatBool(α : Type u_1)[MeasurableSpaceα][CountablyGeneratedα] : Measurable(mapNatBoolα) source theoremMeasurableSpace.injective_mapNatBool(α : Type u_1)[MeasurableSpaceα][CountablyGeneratedα][SeparatesPointsα] : Function.Injective(mapNatBoolα) source theoremMeasurableSpace.measurableEquiv_nat_bool_of_countablyGenerated(α : Type u_1)[MeasurableSpaceα][CountablyGeneratedα][SeparatesPointsα] : ∃ (s : Set(ℕ → Bool)), Nonempty (α≃ᵐ↑s) If a measurable space is countably generated and separates points, it is measure equivalent to some subset of the Cantor space ℕ → Bool (equipped with the product sigma algebra). Note: s need not be measurable, so this map need not be a MeasurableEmbedding to the Cantor Space. source theoremMeasurableSpace.measurable_injection_nat_bool_of_countablySeparated(α : Type u_1)[MeasurableSpaceα][CountablySeparatedα] : ∃ (f : α → ℕ → Bool), Measurablef∧Function.Injectivef If a measurable space admits a countable sequence of measurable sets separating points, it admits a measurable injection into the Cantor space ℕ → Bool (equipped with the product sigma algebra). source theoremMeasurableSpace.measurableSingletonClass_of_countablySeparated{α : Type u_1}[MeasurableSpaceα][CountablySeparatedα] : MeasurableSingletonClassα source theoremMeasurableSpace.measurableSet_succ_memPartition{α : Type u_1}(t : ℕ → Setα)(n : ℕ){s : Setα}(hs : s∈memPartitiont n) : MeasurableSets source theoremMeasurableSpace.generateFrom_memPartition_le_succ{α : Type u_1}(t : ℕ → Setα)(n : ℕ) : generateFrom(memPartitiont n)≤generateFrom(memPartitiont (n+1)) source theoremMeasurableSpace.measurableSet_generateFrom_memPartition_iff{α : Type u_1}(t : ℕ → Setα)(n : ℕ)(s : Setα) : MeasurableSets↔∃ (S : Finset(Setα)), ↑S⊆memPartitiont n∧s=⋃₀↑S source theoremMeasurableSpace.measurableSet_generateFrom_memPartition{α : Type u_1}(t : ℕ → Setα)(n : ℕ) : MeasurableSet(t n) source theoremMeasurableSpace.generateFrom_iUnion_memPartition{α : Type u_1}(t : ℕ → Setα) : generateFrom(⋃ (n : ℕ), memPartitiont n)=generateFrom(Set.ranget) source theoremMeasurableSpace.generateFrom_memPartition_le_range{α : Type u_1}(t : ℕ → Setα)(n : ℕ) : generateFrom(memPartitiont n)≤generateFrom(Set.ranget) source theoremMeasurableSpace.generateFrom_iUnion_memPartition_le{α : Type u_1}[m : MeasurableSpaceα]{t : ℕ → Setα}(ht : ∀ (n : ℕ), MeasurableSet(t n)) : generateFrom(⋃ (n : ℕ), memPartitiont n)≤m source theoremMeasurableSpace.generateFrom_memPartition_le{α : Type u_1}[m : MeasurableSpaceα]{t : ℕ → Setα}(ht : ∀ (n : ℕ), MeasurableSet(t n))(n : ℕ) : generateFrom(memPartitiont n)≤m source theoremMeasurableSpace.measurableSet_memPartition{α : Type u_1}[MeasurableSpaceα]{t : ℕ → Setα}(ht : ∀ (n : ℕ), MeasurableSet(t n))(n : ℕ){s : Setα}(hs : s∈memPartitiont n) : MeasurableSets source theoremMeasurableSpace.measurableSet_memPartitionSet{α : Type u_1}[MeasurableSpaceα]{t : ℕ → Setα}(ht : ∀ (n : ℕ), MeasurableSet(t n))(n : ℕ)(a : α) : MeasurableSet(memPartitionSett n a) source defMeasurableSpace.countablePartition(α : Type u_3)[MeasurableSpaceα][CountablyGeneratedα] : ℕ → Set(Setα) For each n : ℕ, countablePartition α n is a partition of the space in at most 2^n sets. Each partition is finer than the preceding one. The measurable space generated by the union of all those partitions is the measurable space on α. Equations MeasurableSpace.countablePartitionα=memPartition(Set.enumerateCountable⋯∅) Instances For source theoremMeasurableSpace.measurableSet_enumerateCountable_countableGeneratingSet(α : Type u_3)[MeasurableSpaceα]CountablyGeneratedα : MeasurableSet(Set.enumerateCountable⋯∅n) source theoremMeasurableSpace.finite_countablePartition(α : Type u_3)[MeasurableSpaceα]CountablyGeneratedα : (countablePartitionα n).Finite source instanceMeasurableSpace.instFinite_countablePartition{α : Type u_1}[m : MeasurableSpaceα]h : CountablyGeneratedα : Finite↑(countablePartitionα n) source theoremMeasurableSpace.disjoint_countablePartition{α : Type u_1}[m : MeasurableSpaceα][h : CountablyGeneratedα]{n : ℕ}{s t : Setα}(hs : s∈countablePartitionα n)(ht : t∈countablePartitionα n)(hst : s≠t) : Disjoints t source theoremMeasurableSpace.sUnion_countablePartition(α : Type u_3)[MeasurableSpaceα]CountablyGeneratedα : ⋃₀countablePartitionα n=Set.univ source theoremMeasurableSpace.measurableSet_generateFrom_countablePartition_iff{α : Type u_1}[m : MeasurableSpaceα]h : CountablyGeneratedα(s : Setα) : MeasurableSets↔∃ (S : Finset(Setα)), ↑S⊆countablePartitionα n∧s=⋃₀↑S source theoremMeasurableSpace.measurableSet_succ_countablePartition{α : Type u_1}[m : MeasurableSpaceα]h : CountablyGeneratedα{s : Setα}(hs : s∈countablePartitionα n) : MeasurableSets source theoremMeasurableSpace.generateFrom_countablePartition_le_succ(α : Type u_3)[MeasurableSpaceα]CountablyGeneratedα : generateFrom(countablePartitionα n)≤generateFrom(countablePartitionα (n+1)) source theoremMeasurableSpace.generateFrom_iUnion_countablePartition(α : Type u_3)[m : MeasurableSpaceα][CountablyGeneratedα] : generateFrom(⋃ (n : ℕ), countablePartitionα n)=m source theoremMeasurableSpace.generateFrom_countablePartition_le(α : Type u_3)[m : MeasurableSpaceα]CountablyGeneratedα : generateFrom(countablePartitionα n)≤m source theoremMeasurableSpace.measurableSet_countablePartition{α : Type u_1}[m : MeasurableSpaceα]h : CountablyGeneratedα{s : Setα}(hs : s∈countablePartitionα n) : MeasurableSets source defMeasurableSpace.countablePartitionSet{α : Type u_1}[m : MeasurableSpaceα]h : CountablyGeneratedα(a : α) : Setα The set in countablePartition α n to which a : α belongs. Equations MeasurableSpace.countablePartitionSetn a=memPartitionSet(Set.enumerateCountable⋯∅)n a Instances For source theoremMeasurableSpace.countablePartitionSet_mem{α : Type u_1}[m : MeasurableSpaceα]h : CountablyGeneratedα(a : α) : countablePartitionSetn a∈countablePartitionα n source theoremMeasurableSpace.mem_countablePartitionSet{α : Type u_1}[m : MeasurableSpaceα]h : CountablyGeneratedα(a : α) : a∈countablePartitionSetn a source theoremMeasurableSpace.countablePartitionSet_eq_iff{α : Type u_1}[m : MeasurableSpaceα][h : CountablyGeneratedα]{n : ℕ}(a : α){s : Setα}(hs : s∈countablePartitionα n) : countablePartitionSetn a=s↔a∈s source theoremMeasurableSpace.countablePartitionSet_of_mem{α : Type u_1}[m : MeasurableSpaceα][h : CountablyGeneratedα]{n : ℕ}{a : α}{s : Setα}(hs : s∈countablePartitionα n)(ha : a∈s) : countablePartitionSetn a=s source theoremMeasurableSpace.measurableSet_countablePartitionSet{α : Type u_1}[m : MeasurableSpaceα]h : CountablyGeneratedα(a : α) : MeasurableSet(countablePartitionSetn a) source classMeasurableSpace.CountableOrCountablyGenerated(α : Type u_5)(β : Type u_6)[MeasurableSpaceβ] : Prop A class registering that either α is countable or β is a countably generated measurable space. countableOrCountablyGenerated : Countableα∨CountablyGeneratedβ Instances source instanceMeasurableSpace.instCountableOrCountablyGeneratedOfCountable{β : Type u_2}{α : Type u_3}[h1 : Countableα][MeasurableSpaceβ] : CountableOrCountablyGeneratedα β source instanceMeasurableSpace.instCountableOrCountablyGeneratedOfCountablyGenerated{β : Type u_2}{α : Type u_3}[MeasurableSpaceβ][h : CountablyGeneratedβ] : CountableOrCountablyGeneratedα β source instanceMeasurableSpace.instCountableOrCountablyGeneratedProd{β : Type u_2}{α : Type u_3}{γ : Type u_4}[MeasurableSpaceγ][hα : CountableOrCountablyGeneratedα γ][hβ : CountableOrCountablyGeneratedβ γ] : CountableOrCountablyGenerated (α×β) γ source theoremMeasurableSpace.countableOrCountablyGenerated_left_of_prod_left_of_nonempty{β : Type u_2}{α : Type u_3}{γ : Type u_4}[MeasurableSpaceγ][Nonemptyβ][h : CountableOrCountablyGenerated (α×β) γ] : CountableOrCountablyGeneratedα γ source theoremMeasurableSpace.countableOrCountablyGenerated_right_of_prod_left_of_nonempty{β : Type u_2}{α : Type u_3}{γ : Type u_4}[MeasurableSpaceγ][Nonemptyα][h : CountableOrCountablyGenerated (α×β) γ] : CountableOrCountablyGeneratedβ γ source theoremMeasurableSpace.countableOrCountablyGenerated_prod_left_swap{β : Type u_2}{α : Type u_3}{γ : Type u_4}[MeasurableSpaceγ][h : CountableOrCountablyGenerated (α×β) γ] : CountableOrCountablyGenerated (β×α) γ
10380	https://uen.pressbooks.pub/uvumqr/chapter/7-1-rules-of-exponents/	7-1: Exponential Expressions – Numeracy Skip to content Menu Primary Navigation Home Read Sign in Search in book: Search Book Contents Navigation Contents Cover Copyright Introduction Authors What is Numeracy? Innumeracy–Why it Matters Why it matters if we become innumerate The Importance of Numeracy Accessibility Funding Acknowledgement CHAPTER 1: NUMBER SENSE Chapter 1: NUMBER SENSE Chapter 1: NUMBER SENSE 1-0: Numbers in the Real World 1-1: Integers What are Integers? 1-2: The Power of 10 Decimals 1-3: Scientific Notation 1-4: Number Sense and Relative Size Number Sense and Relative Size 1-5 The Order of Operations The Order of Operations 1-6: Number Sense with Fractions Fractions 1-7 The Relationship Between Fractions and Decimals Fractions and Decimals Chapter 1 Project: Reading Level Reading Level CHAPTER 2: MEASUREMENT Chapter 2: MEASUREMENT Chapter 2: MEASUREMENT 2-1: U.S. Customary System of Measurement: Length and Area 2-2: US Customary System: Volume and Capacity Volume and Capacity 2-3: U.S. Customary System: Weight Weight 2-4: International System of Measurement: Length and Area 2-5: SI Measurement: Volume and Capacity 2-6: SI Measurement: Weight Mass versus Weight 2-7: Dimensions and Dimensional Analysis Dimensional Analysis Chapter 2 Project: Designing a juice carton Designing a Juice Carton CHAPTER 3: RELATIVE MEASUREMENT Chapter 3: RELATIVE MEASUREMENT Chapter 3: RELATIVE MEASUREMENT 3-1: Ratios and Rates Absolute and Relative Measures 3-2: Unit Rates What is a Unit Rate? 3-3: Percentages Percent or Percentage? 3-4: Percentages in Life Part of a Whole 3-5: Absolute and Relative Change “The Only Constant in Life is Change.”- Heraclitus Chapter 3 Project: The Use and Abuse of Percentages The Use and Abuse of Percentages CHAPTER 4: DATA MEASUREMENT CHAPTER 4: DATA MEASUREMENT Chapter 4: DATA MEASUREMENT 4-1: Averages Data 4-2: Weighted Means 4-3: Types of Data and Appropriate Representations 4-4: Interpreting Graphs and Charts Chapter 4 Project CHAPTER 5: LINEAR EQUATIONS IN ONE VARIABLE CHAPTER 5: LINEAR EQUATIONS IN ONE VARIABLE Chapter 5: Linear Equations in One Variable 5-1: Variables and Algebraic Expressions 5-2: Equations and Solutions 5-3: Properties of Equality 5-4: Solving Linear Equations in One Variable Chapter 5 Project CHAPTER 6: LINEAR GROWTH CHAPTER 6: LINEAR GROWTH Chapter 6: Linear Growth 6-1: Linear Patterns 6-2: Linear Reasoning - Rate and Initial Value 6-3: Linear Reasoning - Slope and Intercept 6-4: Linear Modeling - Linear Equations in Two Variables 6-5: Solving Linear Equations in Two Variables Chapter 6 Project CHAPTER 7: EXPONENTIAL AND LOGARITHMIC GROWTH CHAPTER 7: EXPONENTIAL AND LOGARITHMIC GROWTH Chapter 7: Exponential Growth and Logarithms 7-1: Exponential Expressions 7-2: Exponential Patterns 7-3: Exponential Equations and Graphs 7-4: Exponential Growth and Decay 7-5: Solving Exponential Equations using Logarithms Chapter 7 Project: Octaves in Music Appendix Numeracy CHAPTER 7: EXPONENTIAL AND LOGARITHMIC GROWTH 7-1: Exponential Expressions Exponents We use multiplication to represent repeated addition. For example, is shortcut notation for . Likewise, we use exponents to represent repeated multiplication. For example, can be written . Thebase 5 is multiplied by itself 7 times. 7 is called the exponent. Writing multiple additions or multiple multiplications is tedious and is susceptible to human error, so using multiplication to abbreviate multiple additions, and exponents to abbreviate multiple multiplications makes sense. Exponents are used regularly to represent area in square units like ft 2 (e.g., a hotel room is 200 square feet) or m 2 (e.g., a bedroom is 22 square meters) and volume in cubic units like cm 3 (e.g., a box has a volume of 3600 cubic centimeters) or yd 3 (e.g., 2 cubic yards of topsoil). So, measuring in two or more dimensions requires the use of exponents on the measurement units. Any number raised to a power is known as an exponential expression. In section 1-3, we considered powers of 10. We discovered some basic rules that we will see can apply to all numbers. Exponential Expressions Explore 1 – Exponential expressions Suppose a single bacterium is placed in a petri dish. The number of bacteria in the petri dish doubles every hour. The table shows this growth of bacteria. Time (hours)0 1 2 3 4 5 Bacteria 1 1×2=2 (2=2 1)(1×2) x2=4 (4=2 2)(1x2x2) x2=8 (8=2 3)(1x2x2x2) x2=16 (16=2 4)(1x2x2x2x2) x2=32 (32=2 5) How do we determine the number of bacteria from one hour to the next? Solution We multiply the number of bacteria in the first hour by 2. How many bacteria are there after 6 hours? Solution After six hours, the number of bacteria is (i.e., ) times the initial number of bacterium, 1. How many bacteria are there after 1 day? Solution Since 1 day = 24 hours, there are bacteria. Is it possible to write the number of bacteria at any time as , where n is the number of hours? Solution Yes. The pattern for the number of bacteria is for days 0, 1, 2, 3, 4, … The only day that doesn’t appear to be a power of 2 is day 1 when there is 1 bacterium. But, , so the number of bacteria at any time is , where n is the number of hours. To evaluate an exponential expression using a calculator we use either the carat key or the exponent key .To evaluate 2 24, the input will be Alternatively, the input is Either way, the result is 16,777,216. Negative Exponents Explore 2 – Negative exponents Knowing that 8 = 2 3, 4 = 2 2, 2 = 2 1, and 1 = 2 0, follow the pattern to determine what 2-1 and 2-2 equal. Solution The pattern 8, 4, 2, 1, … divides each number by 2. Consequently it continues to . At the same time the pattern 2 3, 2 2, 2 1, 2 0, continues by reducing the exponent each time by 1 to 2-1, 2-2, 2-3, … This means that and . Use the pattern to write as an expression with a positive exponent. Solution Use the pattern to write as an expression with a positive exponent. Solution From the pattern, . So, Expand your expression to write for any non-zero value , as an expression with a positive exponent. Solution The Product and Quotient Rules Explore 3 – The product rule Is it true that 2 × 2 × 2 × 2 × 2 is the same as (2 × 2) × (2 × 2 × 2)? Explain your reasoning. Solution Yes, they are the same. and There are two sets of parentheses that change the grouping of the multiplication but do not change the value of the multiplication of the five 2s. This is an example of the associative property of multiplication. Considering your answer to #1, is it true that . Explain your reasoning. Solution Yes. #1 shows us that . Since , , and , then . Since , is there an easy rule to multiply exponential expressions with the same base? Solution Yes. Since 2 + 3 = 5, we can keep the common base and add the exponents. This is called the product rule of exponents: Use the product rule to simplify Solution Since 2 8 and 2 12 have the same base 2, we keep the common base and add the exponents: Use the product rule to simplify Solution Since 5 5, 5 4, and 5 8 all have the same base, 5, we keep the common base and add the exponents: Explore 4 – The quotient rule Simplify by simplifying the fraction. Solution . Use multiplication and the fact that to show that Solution Since and , is there an easy rule to simplify the division? Solution For exponential expressions with the same base, we can divide the expressions by keeping the common base and subtracting the exponents. Use your rule to simplify: Solution With a common base of 5, we subtract the exponents: Explain the product rule and the quotient rule . In particular, explain why it is and respectively in the rules. Show/Hide Answer It is in the product rule because is multiplied times and then multiplied times. Therefore, is multiplied times. It is in the quotient rule because is multiplied times, and then divided times, which means that is multiplied times. The Product/Quotient to a Power Rule Explore 5 – The product to a power rule Evaluate the expression following the order of operations. Solution The order of operation tells us to first complete the multiplication inside the parentheses: Then we evaluate the exponent: Evaluate the expression following the order of operations. Solution First we evaluate the exponents: Then we multiply: What do the solutions to 1. and 2. tell us? Solution Since the answers are the same (i.e., 7776), . Write a rule in English that describes how to simplify , then use it to complete the simplification. Solution A product to a power, like , can be simplified by applying the exponent to both factors, then multiplying: Complete the algebraic rule: Solution Explore 6 – The quotient to a power rule Write the expression as a product of fractions, then simplify. Solution Write the expression as a product of fractions, then simplify. Solution Write the expression as a product of fractions, then simplify. Solution Complete the algebraic rule: Solution Does your rule work for all values of and ? Solution No. Since we cannot divide by zero, Explain the product to a power rule and the quotient to a power rule . In particular, explain why the exponent n is applied to to a and b. Show/Hide Answer For the product to a power rule, the reason may be applied to and is that the product is multiplied together times. The associative (i.e., ) and commutative (i.e., ) properties of multiplication can be applied to get a result that is equal to . In other words, we rearrange the s so that all the s are together and all the s are together. For the quotient to a power rule, the reason may be applied to and is that the fraction is multiplied together times. We multiply all the numerators and put the result in the numerator and multiply all the denominators and put the result in the denominator. There are terms of ‘s multiplied together in the numerator and terms of in the denominator multiplied together. Therefore, the result is . Power to a Power Rule Explore 7 – Power to a power rule To understand the exponential expression , Nina starts with the exponent 4, and comes up with the equivalent expression . Explain why there are four terms. Solution The meaning of the exponent 4 means there are four of the base that are multiplied together. Therefore, . How many 2’s are there in total in the multiplication? Solution Based on the product rule of exponents, . Therefore, there are 24 twos multiplied together. Nina explains that the total number of 2’s is 6 x 4 = 24. Why can Nina multiply the two powers? Solution It is because there are four multiplied together. The total number of 2 that are multiplied together is . Write a rule that simplifies a power raised to a power, then use it to simplify . Solution Explore 8 – Power to a power rule A population of bacteria doubles six times in one hour. Jonas says the population after 4 hours will be times the initial population. Is he correct? Explain your reasoning. Solution Doubling six times in one hour means the population doubles every 10 minutes. If the initial population is then after the first 10 minutes there will be ; then after the second 10 minutes there are ; then after the third 10 minutes there are ; …so after the sixth 10 minutes (i.e., one hour) there will be bacteria. Consequently, after 4 hours there will be . Simplify without using any exponents. Solution Simplify using the product rule of exponents. Solution . Simplify using the power to a power rule. Solution . Explain the power to a power rule . In particular, explain why it is . Show/Hide Answer Since there are of multiplied together, it shows . Since there are of , the sum of the addition of ‘s is . Therefore, . Rules of Exponents Exponent of 1:Any number to the power 1 equals the number. Zero Exponent:Any non-zero number to the power zero equals 1. Product Rule Two or more exponential expressions with the same base can be multiplied by keeping the common base and adding the exponents. Quotient Rule Two exponential expressions with the same base can be divided by keeping the common base and subtracting the exponents. Negative Exponents When an exponential expression has a negative exponent the number is a fraction of a whole. Product to a Power Rule A product can be raised to a power by multiplying each factor raised to the power. Quotient to a Power Rule A quotient raised to a power can be split into the numerator raised to the power divided by the denominator raised to the power. Power to a Power Rule To raise a power to a power, keep the base and multiply the exponents. Example The number of people waiting for a train that is running late doubles every 15 minutes. a) Complete the table to show how the number of people has grown over time. b) Write an expression that shows the growth after an hour. c) The next train is almost due, but the late train has not yet arrived. Consequently, the number of people waiting for the train triples every 15 minutes after the first hour. Complete the table for the second hour. c) Simplify the exponential form after 2 hours. d) If there were originally three people waiting for the train, how many people are waiting on the train after 2 hours? e) If each train carriage holds 630 passengers, how many carriages will be needed to accommodate the waiting passengers after 2 hours? Time (hours)0 0.25 0.5 0.75 1 1.25 1.5 1.75 2 People Exponential form Show/Hide Answer 1. Time (hours)0 0.25 0.5 0.75 1 1.25 1.5 1.75 2 People Exponential form c) d) e) 3888/630 = 6.2; 7 carriages will be required. Practice Exercises Use the rules of exponents to simplify. Leave your answers with only positive exponents. Nathan says is the same as . Is he correct? Explain your reasoning. Show/Hide Answer 1. 2. 3. 4. 5. 6. 7. No. , while . In this section, we will take what we have learned and apply the concepts to new situations. Perspectives Jean is moving apartments and doesn’t want to take all of her junk with her when she moves. She has decided to throw away 1 item the first day, then double that number of items she discards each day over the next week. a) Create a table to show how many items will be discarded each day. b) How many items will she discard on day ? A population of rabbits doubles every 10 days. a) Create a table to show how many rabbits there are in the population over time if there were initially rabbits in the population. b) Write an algebraic expression showing how many rabbits there are after days. c) If left unchecked, how many rabbits will there be after 360 days? Adriel deposits $5000 into his bank account. The account balance increases to 1.015 times the balance every year. What is the account balance after 10 years? Show your work using exponents. The population of a species doubles every month from January to April, and then reduces by half every month for the following two months. a) Write the population increase-decrease pattern in a mathematical sentence without using exponents.b) Write the population increase-decrease pattern in a mathematical sentence using exponents.c) Simplify the expression using the rules of exponents. A video clip goes viral. The number of people who view the clip doubles every 15 minutes for the first six hours, and then triples every 12 hours for the next 4 days. Write the population growth as a mathematical sentence using exponents. Show/Hide Answer a) Day 1 2 3 4 5 6 7 Number of items 1 2 4 8 16 32 64 b) a) Number of days 0 10 20 30 40 50 Number of rabbits b) c) = $5802.70 a) b) c) In this section, we will use what we have learned so far to practice skill problems. Skill Exercises Evaluate: 3 4 0.2 5 4 0 1.2 4 2-3 1.5-2 Simplify: 3 4· 3 5 14 8 · 14 1.5 7 · 1.5 3 6 7 ÷ 6 4 1.03 12 ÷ 1.03 9 5-2÷ 5 7 (2 · 3)4 Show/Hide Answer 1. 81 2. 0.00032 3. 1 4. 3.8416 5. 6. 7. 3 9 8. 14 9 9. 1.5 10 10. 6 3 11. 1.03 3 12. 17-4 = 13. 14. 6 4 15. 16. 17. 18. 19. 1 20. 4 15 21. 22. 0.03 8 definition the number of times a base number is multiplied by itself ×Close definition the amount of space inside a two-dimensional object measured in square units ×Close definition a measure of the amount of 3-dimensional space an object takes up ×Close definition Previous/next navigation Previous: CHAPTER 7: EXPONENTIAL AND LOGARITHMIC GROWTH Next: 7-2: Exponential Patterns Back to top License Numeracy Copyright © 2023 by Utah Valley University is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted. 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10381	https://math.stackexchange.com/questions/1767045/calculus-critical-points-inner-boundary-multivariable	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 Calculus Critical points (inner & boundary) multivariable Ask Question Asked Modified 9 years, 4 months ago Viewed 3k times 2 $\begingroup$ I have this assignment I need help with, it follows: We have the function: $$f(x,y)=\frac{(1+3xy)}{(1+x^2+y^2)}$$ for all $(x,y)$ in the closed disk $C$ with radius 2 and its center in $(x,y)=(0,0) \in \Bbb{R}^2$ Find and decide all critical points inside $f$. Do a parametrization of $C$ to find all extreme points on the boundary. Use Lagrange method to find all candidates for extreme points on the boundary of $C$. Use the result of the previous points to find all extreme points for $f$. So I've solved the first, I think. I first derivated the function for $x$ and then for $y$, and I got that the only critical point inside $f$ is $(0,0)$. But on point 2) I get stuck, I'm not sure if I'm on the right track, but I believe what I have to do is since the disk has radius 2, I have to write it as $x^2+y^2=4$, and then parameterize it as $2\cos t+2\sin t=4$? But after that then what? Thanks in advance. calculus Share edited May 1, 2016 at 18:27 EulersfunctionEulersfunction asked May 1, 2016 at 17:56 EulersfunctionEulersfunction 4711 silver badge99 bronze badges $\endgroup$ 7 $\begingroup$ I don't think using sines and cosines will help. Let $g(x,y)=x^2+y^2-4$ and then use Lagrange multipliers: $\nabla f=\lambda \nabla g$. $\endgroup$ Rick Sanchez – Rick Sanchez 2016-05-01 18:01:48 +00:00 Commented May 1, 2016 at 18:01 $\begingroup$ Alright, I'll try that, thanks :) $\endgroup$ Eulersfunction – Eulersfunction 2016-05-01 18:03:51 +00:00 Commented May 1, 2016 at 18:03 $\begingroup$ Just a quesiton, you have the gradient sign infront of f and g, does it mean that I have you take their derivatives? I thought the function alone was enough? $\endgroup$ Eulersfunction – Eulersfunction 2016-05-01 18:05:18 +00:00 Commented May 1, 2016 at 18:05 $\begingroup$ Yes, both functions. en.wikipedia.org/wiki/Lagrange_multiplier $\endgroup$ Rick Sanchez – Rick Sanchez 2016-05-01 18:06:44 +00:00 Commented May 1, 2016 at 18:06 $\begingroup$ EDIT: check b again, I have to do a parameterization.. $\endgroup$ Eulersfunction – Eulersfunction 2016-05-01 18:06:50 +00:00 Commented May 1, 2016 at 18:06 \| Show 2 more comments 1 Answer 1 Reset to default 1 $\begingroup$ Another way to solve 2., which uses a parametrization, is as follows: parametrize the circle as $x=2\cos t$, $y=2\sin t$, $0\leq t\leq 2\pi$. Then your function $f$ becomes a function of one variable, call it $h$: $$h(t)=f(x(t),y(t))=\frac{1+12\cos t\sin t}{5}, \; 0\leq t\leq 2\pi.$$ You can now find the extrema of this functions using one-variable calculus. The critipal points you'll find might be different than the ones obtained with the Lagrange multiplier method, but the max/min should be exactly the same. Share edited May 1, 2016 at 18:32 answered May 1, 2016 at 18:14 Îµ-Î´Îµ-Î´ 26811 silver badge55 bronze badges $\endgroup$ 20 $\begingroup$ Thanks, but I think you mean 1+12costsint/5 and not 25?^ And the next step is to derivate the function, put equal to 0, solve for t, and then find the value for x and y? $\endgroup$ Eulersfunction – Eulersfunction 2016-05-01 18:20:59 +00:00 Commented May 1, 2016 at 18:20 $\begingroup$ denominator is (1+x^2 + y^2)^2 = (1+4)^2 = 25. $\endgroup$ user317176 – user317176 2016-05-01 18:23:07 +00:00 Commented May 1, 2016 at 18:23 $\begingroup$ You're right, I missed the last squared sign lol.. But I then got t=pi/4 and x=y=square root of 2, is it correct? $\endgroup$ Eulersfunction – Eulersfunction 2016-05-01 18:25:17 +00:00 Commented May 1, 2016 at 18:25 $\begingroup$ @Eulersfunction I fixed the typo. Yes, that's how you'd proceed. $\endgroup$ ε-δ – ε-δ 2016-05-01 18:28:11 +00:00 Commented May 1, 2016 at 18:28 1 $\begingroup$ Sounds right. Remember you can wolfram to check your answers. $\endgroup$ ε-δ – ε-δ 2016-05-01 21:03:39 +00:00 Commented May 1, 2016 at 21:03 \| Show 15 more comments 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 See similar questions with these tags. 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10383	https://www.sciencedirect.com/science/article/pii/S0753332220307435	Immune defence to invasive fungal infections: A comprehensive review - ScienceDirect JavaScript is disabled on your browser. Please enable JavaScript to use all the features on this page. Skip to main contentSkip to article Journals & Books ViewPDF Download full issue Search ScienceDirect Outline Highlights Abstract Abbreviations Keywords 1. Introduction 2. Pathogenesis of medically important fungi 10. Conclusions and prospectives Declaration of Competing Interest Ethical approval Acknowledgments References Show full outline Cited by (171) Figures (3) Biomedicine & Pharmacotherapy Volume 130, October 2020, 110550 Review Immune defence to invasive fungal infections: A comprehensive review Author links open overlay panel Balaji Pathakumari a, Guanzhao Liang a, Weida Liu a b Show more Outline Add to Mendeley Share Cite rights and content Under a Creative Commons license Open access Highlights •New Insights into immune mechanisms is key for developing anti-fungal vaccines and drugs. •Discussed pathogenesis of fungal pathogens & their interaction with host immune system. •Discussed the critical role of innate and adaptive immune cells that respond to IFI. •Emphasized the role of memory cells in long-term protection in subsequent infections. •Summarize the current vaccine strategies, co-infections studies and drug resistance. Abstract The fungal infections are relatively common in humans that can range from common, mild superficial infections to life-threatening invasive infections. Most of the pathogenic fungi are opportunistic that cause disease under immunocompromised conditions such as HIV infection, cancer, chemotherapy, transplantation and immune suppressive drug users. Efficient detection and treatment of high-risk population remain the highest priority to avert most of the deaths. Majority of invasive infections are caused by Candida, Aspergillus and Cryptococcus species. Lack of effective vaccines, standardised diagnostic tools, efficient antifungal drugs and the emergence of drug-resistant species/strains pose a global threat to control Invasive fungal infections (IFI). A better understanding of the host immune response is one of the major approaches to developing new or improved antifungal strategies to control the IFIs. In this review, we have discussed pathogenesis of medically important fungi, fungal interaction with the host through pattern recognition receptors (PRRs) and the interplay of innate and adaptive immune cells in shaping host immunity to IFI. Further, we emphasized the role of memory cells by offering long-term protection in secondary or subsequent infections. Moreover, we depicted the role of unconventional innate-like immune cells in anti-fungal immunity. We also summarize the available information on the current vaccine strategies, genetic susceptibility to fungal infections, recent co-infections studies and the emergence of drug-resistance, a growing trend throughout the world. Finally, we emphasized the steps to be taken for the control of IFIs. Previous article in issue Next article in issue Abbreviations IFI invasive fungal infections HIV human immunodeficieny virus MDSC myeloid-derived suppressor cells iNKT Invariant natural killer T cells MAIT mucosal-associated invariant T cells SFI superficial infections IC invasive Candidiasis HWP hyphal wall protein SOD superoxide dismutase GT gliotoxin PAMPs pathogen-associated molecular patterns PRR pathogen recognition receptors PMN polymorphonuclear leukocytes TLR toll-like receptors NOD nucleotide-binding oligomerization domain NLR NOD like receptors CARD caspase recruiting domain-containing protein adaptor molecule AMP antimicrobial peptides APC antigen presenting cells ROI reactive oxygen intermediates RNI reactive nitrogen intermediates NOS2 nitric oxide synthase-2 DC dendritic cells mDCs myeloid dendritic cells pDCs plasmacytoid dendritic cells PMNs polymorphonuclear neutrophils NET or NETosis neutrophils extracellular traps PBMC peripheral blood mononuclear cells NK cells natural killer cells CMCC chronic mucocutaneous candidiasis MHC major histocompatibility complex TCR T cell receptors VVC vulvovaginal candidiasis RVVC recurrent vulvovaginal Candidiasis Tregs Regulatory T-cells NKT cells natural killer T cells DTH delayed-type hypersensitivity ADCC antibody-dependent cellular cytotoxicity mab monoclonal antibodies HPV hepatitis papilloma vaccine HBV hepatitis B vaccine Sap2 secreted aspartic protease Hsp90 stress associated proteins Hyr1 hyphal associated protein ABC ATP-binding cassette transporters Keywords Invasive fungal infections Immunocompromised patients Fungal-host interactions Immune response Candida Aspergillus Cryptococcus 1. Introduction Though there are enormous fungal species across the world, more than 500 species are predicted to be infectious to human. Most of the fungi are present in the environment and also part of the normal flora in humans and animals. The fungal infections in humans can range from common, mild, superficial infections to life-threatening invasive infections, especially in immunocompromised patients. The majority of these infections contributing to increased morbidity and mortality in healthy and immunocompromised patients respectively. It is a well-known fact that immune compromised patients, premature infants, cancer patients are more prone to fungal infections than immune competent individuals . Approximately, billions of world population is believed to be infected with pathogenic fungi, of which 1.5 million deaths are recorded each year [2,3]. The outcome of the infection depends on the pathogenicity of fungus, host immune response and the site of infection. Broadly, fungal infections are classified as superficial, cutaneous/mucosal, subcutaneous and Invasive. Superficial infections (SFI) are commonly caused by Malassezia species, Tinea nigra, that affects the surface of the body, like the skin, nails, and hair. There are no notable pathological changes and cellular immune response in SFI. Cutaneous mycosis is also a kind of superficial infections, but a variety of pathological changes occur. The most common cutaneous infections caused by dermatophytes such as Trichophyton rubrum, T. interdigitale, T. tonsurans, and Microsporum canis that infect keratin-rich tissues, such as the skin . Candida species are one of the principal yeast pathogens that cause cutaneous infection in the skin and mucous membrane [3,5]. Subcutaneous infections (eg: Chromolastomycosis) involve deeper layers of skin dermis and subcutaneous tissue. In contrast to SFI, invasive infections can affect internal organs like lungs, heart, brain, kidneys, liver, or other parts of the body that are common in immunocompromised patients. Candidiasis, Aspergillosis and Cryptococcosis are the common systemic infections that are caused by Candida, Aspergillus and Cryptococcus species respectively. It has been reported that about 1.5 million invasive fungal deaths occur in every year globally [2,3]. More than 80 % of these deaths are occurring due to the infections caused by Candida, Aspergillus and Cryptococcus species . One of the chief causative IFI is invasive Candidiasis which is caused by Candida species, more commonly Candida albicans (C. albicans). Candida species are commensal fungi and commonly found in the gut, skin and other mucosal surfaces. It causes three types of Candidiasis i.e. thrush, vaginal and invasive Candidiasis (IC). IC is a serious infection that can affect the lungs, blood, heart, brain, eyes, bones, and other parts of the body and it alone accounts for 46–75 % of all IFI . It causes life-threatening illness and responsible for the significant mortality rates (40 %) in immunocompromised patients [7,8]. Other major IFIs are Aspergillosis, which is caused by A. fumigatus. It is a ubiquitous saprophytic fungus estimated to cause over 200 000 cases of invasive Aspergillosis to occur each year, with mortality rates of 30–95 % . Another predominant IFIs is Cryptococcosis, caused by Cryptococcus neoformans (C. neoformans) and Cryptococcus gatti (C. gatti) species, affects over one million individuals with 20–70 % mortality rate annually [6,9]. Invasive fungal infections (IFIs) are closely linked to host immune status and the type of fungal species. Most of the pathogenic fungi are opportunistic pathogens that only cause disease under immune suppressed conditions such as HIV infection, cancer, chemotherapy, transplantation and immunosuppressive drug users. In the last 3 decades, the prevalence of opportunistic fungal infections significantly increases due to increases in the HIV prevalence and immunosuppressive drug users. There is a chance of reactivating the dormant fungal pathogens (latent infection) under immune suppression state. Moreover, it is highly challenging to treat IFIs in immunocompromised patients. Therefore, the lack of an efficient immune response will increase the risk of IFI. To reduce the global incidence and mortality of IFIs, we need to move towards the development of the (a) Point-of-care (POC) diagnostics tools for the rapid detection of IFI (b) effective anti-fungal drugs with fewer side effects and (c) efficient and safe vaccines against IFIs. Because of the different spectrum of fungal infections (colonization, infection and disease) and the presence of various morphological forms, it would be difficult to interpret the fungal infections based on clinical symptoms. Diagnosis of IFI will confirm the infection and should also allow the species identification that would help to initiate the treatment with suitable anti-fungal drugs. The high mortality rate in critically ill ICU patients may be due to delayed diagnosis and treatment. Conventional and molecular diagnostic tools are time-consuming, less sensitive and specific. Due to frequent exposure of fungus with humans, it is difficult to discriminate the healthy (commensal) and IFIs (pathogenic). The antigens present in fungus are widely distributed which led to specificity issues in the context of disease diagnosis. Further, immunosuppressed patients are unable to stimulate the immune response to the fungal antigens . So, there is an urgent need to develop more reliable and rapid diagnostic tools for the early diagnosis of IFIs, particularly in immunocompromised patients who will benefit most from the treatment. The current anti-fungal drugs are facing few limitations such as efficiency, less availability and adverse effects to the host. Moreover, long-term usage and mismanagement of these drugs lead to the emergence of drug-resistant strains . Recent studies reporting that IFIs are the most serious global threat due to the emergence of drug-resistant fungal strains . Further, there are no FDA approved effective vaccines to date. The major challenges for the development of vaccines are a. exposure to environmental fungi is frequent, b. commensal relationship with some fungi and c. weak immune response in immunocompromised patients. Currently, most of the research groups have been focusing on the development of new fungal vaccines that can induce long term memory response, especially in immunocompromised patients . However, some of the vaccines have given promising results in the clinical trials; still, there are many challenges to be overcome . Therefore the clear understanding of immune mechanisms is key for developing anti-fungal vaccines. Here, we discuss the virulence and pathogenesis of Candida, Aspergillus, Cryptococcus species and their interaction with the host immune system. We highlight the role of innate and adaptive immune cells in the context of both pathogenesis and anti-fungal activity. We also discussed the recent development of candidate vaccine strategies. Though not subject of this review, we also discussed the genetic susceptibility to fungal infections, bacterial-fungal coinfections and the emergence of drug resistant strains that are posing a global threat. 2. Pathogenesis of medically important fungi 2.1. Candida albicans C. albicans grows in distinct morphological forms such as unicellular yeast and hyphal forms . The dichotomy between these two forms depends on the presence and absence of host immune response, environment and the ability to alter its morphology. It is believed that the unicellular yeast form is associated with commensalism, whereas hyphal forms are highly virulent and associated with infection. The virulence traits of C. albicans depend on the morphological transition between yeast and hyphal forms, the expression of adhesins and invasins, the formation of biofilms, phenotypic switching, secretion of hydrolytic enzymes, site of infection (e.g., mucosal or systemic), stage of infection, and the nature of the host’s immune response. The yeast form involves circulating or disseminating the infection, whereas hyphal forms are associated with invasive and cause host tissue destruction [17,18]. During the formation of hyphae, upregulation of several genes takes place which helps Candida for adhesion and invasion into the host tissue. The hyphal wall protein (Hwp)-1, Als3 (agglutinin-like sequence) protein plays a significant role for adhesion to host cell whereas secreted hydrolytic enzymes such as aspartic proteases Sap4, Sap5, and Sap6 are required for invasion and causes tissue damage [19,20]. C. albicans employs the immune evasion strategy that includes, escape from the recognition of host cells by downregulating or shielding the pathogen-associated molecular patterns (PAMPs), inhibition of phagosome-lysosome fusion, inhibiting the opsonisation, degrading the complement factors (C3, C4b, C5), secreting the catalase and superoxide dismutase (SOD) . Further, it allows the inhibition of the levels of antimicrobial peptides (AMP) by secreting AMP efflux pumps , modulates the Th1 and Th2 balance , downregulates the IL-17 pathway by the secretion of Tryptophan metabolic products . Emerging evidence indicates that Candidalysin, a toxin secreted by C. albicans, may also alter the macrophage membrane integrity . Further, there is evidence that C. albicans promotes the differentiation of M1 macrophages to M2 macrophages thereby enhance its survival rate . The above virulence factors such as dimorphism, secreted proteases, efflux pumps and the expression of adhesins and invasins are attractive targets for the development of anti-fungal drugs. 2.2. Aspergillus Aspergillus species are ubiquitous molds primarily found in the environment such as soil and decaying vegetation. The genus contains more than 200 species, in which 30 species are reported to cause human infections. Among the human pathogenic species, A. fumigatus is the primary causative agent followed by A. flavus, A. terreus, A. niger . Although Aspergillus infection in immunocompetent patients is rare, it remains a significant cause of morbidity and mortality in immunocompromised patients. The mortality rate of invasive aspergillosis ranges from 40% to 90% in high-risk populations such as patients with HIV, neutropenia, bone marrow transplantation and cancer chemotherapy. Most of the Aspergillus species produce the conidiospores by asexual reproduction and dispersed into the environment . Upon inhalation of these airborne conidia, alveolar macrophages engulf and initiate the pro-inflammatory response and subsequently recruit the neutrophils to the site of infection (lungs) of an immunocompetent host. But, Conidia evade the macrophage effector functions and subsequently germinate, penetrate into the vascular lumen through the epithelial and endothelial layers . Similar to C. albicans, Aspergillus species employ several tactics to evade the immune system, including masking the important PAMPs , inhibition of phagosome-lysosome fusion, production of antioxidants such as catalase, SOD, mannitol etc. In contrast to C. albicans, Aspergillus produces specific secondary metabolites such as gliotoxin (GT), fumagillin, actibind, cytochalasin E which exerts multiple immunosuppressive actions. Another unique feature of Aspergillus is the production of melanin pigment. Melanin exerts the multiple functions in protecting conidia against adverse environmental conditions outside the mammalian host and acts as a scavenger of ROI inside the mammalian host cell [33,34]. It also involved in the interference of the intracellular trafficking of phagocytised conidia and β-glucan masking . 2.3. Cryptococcus Cryptococcus is an invasive fungus that causes Cryptococcosis in humans. The global incidence of Cryptococcosis in HIV infected patients in each year is approximately 1 million cases, among this, 620 000 mortality has been reported . C. neoformans and C. gatti cause the majority of these infections. Generally, Cryptococcus infection is caused by the inhalation of spores from the environment. The initial infection may be the latent stage and become active and disseminate into other organs depend on the status of the host immune system. Depending on the site of infections, Cryptococcosis is divided into pulmonary, cerebral, cutaneous and disseminated Cryptococcosis. Among these, cerebral (CNS) Cryptococcosis is the most common opportunistic invasive infection in HIV infected individuals. The polysaccharide moieties such as glucuronoxylomannan & glucuronoxylomannogalactan and melanin in the cell wall of Cryptococcus reduces the functions of regular macrophages and oxidative burst respectively [37,38]. The unique feature of C. neoformans is the production of Titan cells (giant) with large size (25−30μ), which reduce the phagocytic activity of macrophages and oxidative stress in the lungs that helps the overall persistence and survival of C. neoformans . Another special feature of C. neoformans is the ability to produce the extracellular enzyme called Laccase [40,41]. This helps in the production of melanin pigments, prostaglandins and iron oxidation products which reduces the macrophage activity in the lungs [42,43]. 2.4. Biofilms A wide variety of fungi naturally accumulate on the biotic and abiotic surfaces, commonly referred as biofilm. Due to the structural complexity, metabolic plasticity, overexpression of drug targets, upregulation of efflux pumps, presence of extracellular matrix, biofilms display resistance to anti-fungal agents . Generally, Candida and Aspergillus biofilm occur through the implantation of medical devices (vascular & urinary catheters, cardiac & CNS devices etc) into the patients . Candida and Aspergillus biofilm consist of polysaccharides, proteins and dimorphic forms such as yeast, hyphae (mycelia) whereas C. neoformans biofilm consists of yeast and polysaccharides form of extracellular matrix. Recent reports demonstrated that biofilm-associated genes play a crucial role in the development of biofilms [46,47]. Moreover, Candida biofilms offer favorable environment for the development of highly anti-fungal Candida persisters . It has been demonstrated that neutrophils fail to release NETosis in the presence of extracellular matrix of C. albicans biofilm and also suppresses the reactive oxygen species (ROS) production from neutrophils . It has been reported that Candida biofilms exhibit resistance to phagocytosis by altering the cytokine profile, especially by downregulating the TNF-α cytokine . Further, in vitro studies demonstrated that Candida biofilm impairs the macrophage migration . Interestingly, it has been reported that IL-17A increases the formation of C. albicans biofilm . Aspergillus produces the extracellular matrix (GAG) which shields the A. fumigatus from neutrophil attacks. This extracellular matrix also plays a key role in drug resistance by preventing the diffusion of anti-fungal drugs . The anti-fungal activity of Aspergillus biofilms is mediated by gliotoxin which induces the neutrophils apoptosis . The aggregation of cell wall polysaccharide (GXM polymers) of C. neoformans is dispensable for Cryptococcal biofilm production . GXM inhibits the neutrophils function, phagocytosis, chemotaxis and NET production . C. neoformans biofilm shows the resistance to host oxidative stress and also protect Cryptococci from anti-fungal peptides and macrophage phagocytosis . These data support the formation of biofilm is crucial for protection from immune attacks and also shows resistance to anti-fungal drugs. 3. Host-fungus interaction The interaction between host and pathogenic fungi is crucial for the outcome of the infection or disease. The recognition of PAMPs of the fungus by pathogen recognition receptors (PRRs) of the host is fundamental for the initiation of the host innate immune response (Fig. 1). These PRRs are expressed by both immune cells (dendritic cells, macrophages, polymorphonuclear leukocytes (PMNs) B and T cells) and non-immune cells (epithelial cells and fibroblasts). PAMPs molecules are broad specific that has been shared by different fungal genera or species. The most common PAMPS are α- and β-glucans, N- and O-linked mannans, lipopolysaccharides, peptidoglycan-associated proteins and phospholipomannan. These cell wall components of the fungus were recognized by highly conserved PRR such as Toll-like receptors (TLRs), nucleotide-binding oligomerization domain (NOD)-like receptors (NLRs), and C-type lectin receptors (CLRs). Among the PRRs, TLR (TLR-2, TLR-4, TLR-9) CLRs (Dectin-1, 2, Galectin, mannose receptor, macrophage-inducible C-type lectin [mincle], dendritic cells specific intercellular adhesion molecule grabbing nonintegrin [DC-SIGN]), are important for the recognition and exerts the anti-fungal activity as well . Since the presence of mannoproteins in the cell wall, complement components plays a significant role in the recognition, followed by opsonisation and phagocytosis . The interaction of fungi and PRRs of the host initiates an intracellular signalling cascade that leads to the activation of multiple arms of immune defence such as phagocytosis, production of cytokines, chemokines, antimicrobial peptides (AMP) etc. The TLR-2 and TLR-4 recognise the mannan and lipomannan derivatives on the cell wall of fungi. Then the adaptive molecule MyD88 acts as a key component in the downstream signalling mechanism of TLRs. They induce the production of various inflammatory cytokines that drives naïve T cells to Th1 and Treg cells . TLR-9 resides in the endosomes of myeloid & plasmacytoid dendritic cells and recognises fungal DNA (CpG), which will initiate the production of type-I interferons [64,65]. The CLRs Dectin-1 and 2 acts as the major receptors for β-glucans and α-mannans respectively. They induce NLR3 and NF-kB activation, which is essential for the production of inflammatory cytokines and helps in Th1 and Th17 differentiation . Mincle expresses on macrophages, neutrophils, myeloid dendritic cells and B cells and predominantly recognises the mannose ligands of fungi and leads to the production of TNF-α and IL-10 . Dectin-1, 2 and mincle exert their function through the intracellular signalling cascade system CARD9-BCL10-MALT1. Recently, researchers found that the mice or humans defective in the caspase recruiting domain-containing protein (CARD)-9 are highly susceptible to fungal infections such as Candida species, dermatophytes and phaeohyphomycete [69,70]. Mannose receptor (MR) specifically recognises terminal mannose and induces the IL-17 and other cytokine production [53,71]. DC-SIGN majorly present on dendritic cells and recognises the mannose moieties that induce the IL-6 and IL-10 production [72,73]. 1. Download: Download high-res image (684KB) 2. Download: Download full-size image Fig. 1. Recognition of fungal PAMPs by PRR of the host: Toll-like receptors (TLRs) and C- type lectin receptors (CLR) are pathogen recognition receptors (PRR) which recognize pathogen-associated molecular patterns of various pathogens (PAMP). Activation of PRRs results in a range of innate immune responses, which help shape downstream antigen-specific adaptive immune responses. Upon recognition of fungal components, TLRs (TLR-2, TLR2/6, TLR-4) activate the TIR domain that leads to stimulation of MyD88 or TRIF and activates the downstream complexes (IRAK, TRAF, IKK) and inducing the translocation of transcription factors such as NF-κB, IRF-3, MAPK. The endogenous TLR-9 recognises the unmethylated CpG dinucleotides and stimulate the IRF-7 transcription factor. The CLRs, such as Dectin-1, 2, Mincle stimulate T cell lineage-specific tyrosine kinase (Syk) and downstream complex (CARD9-BCL10-MALT1) and initiate the NF-κB pathways. DC-specific intracellular adhesion molecule-grabbing non-integrin (DC-SIGN) receptor induces the NF-κB translocation through RAS and Raf1 activation pathway. These transcription factors drive the expression of various cytokines, which will regulate T cell differentiation. MyD88: Myeloid differentiation primary response 88; TRIF: TIR-domain-containing adapter-inducing interferon β; Syk: spleen Tyrosine Kinase TRAF: TNF receptor-associated factor; IRAK: Interleukin receptor-associated kinase; CARD9: caspase activation and recruitment domain-containing 9; BCL-10: B-cell lymphoma/leukemia 10; NF-κB: nuclear factor-κB; MAPK: Mitogen-activated protein kinase; ROS: reactive oxygen species. 4. Immune response to fungal infections As discussed earlier, most of the fungi are commensal (symbiotic) interaction with the host, and the immune response creates tolerance between the host and the commensal fungi . If the balance between the host immune response and fungus is disturbed, the fungi will become more opportunistic in immunocompromised patients and further the infection disseminate to different organs (brain, kidney) . Therefore, the host immune response plays a regulatory role between host and fungus. The host immune response to fungal infections depends on many factors such as status of the host immune system, morphology of fungi (yeast vs hyphae), site of the infection, fungal virulence capacity and cell wall complexity [69,76]. The host immune response to infection exerted by both innate and adaptive immune cells. Despite, Innate Immune cells play the first-line defence mechanism, it initiates the adaptive immune responses, and both work together to eliminate the pathogens (Fig. 2). 1. Download: Download high-res image (460KB) 2. Download: Download full-size image Fig. 2. Immune response to fungal pathohens: Upon fungal infection, resident macrophages, dendritic cells and polymorphonuclear neutrophils (PMN) uptake the invaded fungus and undergo phagocytosis. Macrophages and PMN secrete antimicrobial peptides (AMP), inflammatory cytokines, chemokines. Neutrophils can also release neutrophil extracellular traps (NETs) that capture hyphal forms of fungi. NK cells can be activated by various fungal components and directly kills by secreting cytotoxic molecules. DCs can migrate to the lymph node in the presence of IL-12 and chemokines and shapes the adaptive immune response. DCs present the processed antigens to naive T cells and drives into the differentiation of various Th and Tc subsets. The differentiation of each T subset depends on the cytokines and the microenvironment. These subsets exert the effector functions by the production of cytokines, which modulates the anti-fungal immune response. Th1 and Th17 cells migrate back to the site of infection by chemokine-dependent manner and activate the macrophages and PMN respectively. 4.1. Innate immune response Innate immunity is the primary and non-specific immune response to pathogens such as bacteria, virus, fungus etc. The first-line defense of innate immune response is skin, mucosal epithelial surfaces of the mouth, gastrointestinal and genitourinary tract. The skin and mucous produce antimicrobial substances on their surface. If the pathogen crosses these physical barriers, they will encounter the various innate immune components/mechanisms. In this review, we have summarised each innate immune cells and their role in the anti-fungal activity. 4.1.1. Macrophages Macrophages are one of the major antigen-presenting cells (APC) and have an instrumental role in linking innate and adaptive immune response. Macrophage engulfs the fungi and initiates the chain of events, i.e. phagosome-lysosome fusion, generation of antimicrobial components such as reactive oxygen intermediates (ROI) and reactive nitrogen intermediates (RNI). Further tissue-resident macrophages act as crucial effector cells by producing the cytokines and chemokines, which recruit and activate other immune cells to the site of the infection. The initial events driven by macrophages help to clear the pathogen and restrict their spreading. Upon activation, macrophages will undergo the polarization and differentiates into M1 (classical activated macrophages) and M2 (alternative activated macrophages) [77,78]. M1 macrophages display fungal clearance by presenting microbicidal and pro-inflammatory components, whereas M2 macrophages are involved in the persistence of fungi inside the macrophages and cause anti-inflammatory response . Recent studies in mice reported that monocytes along with NK cells show the innate immune memory response against C. albicans [79,80]. Further, macrophages along with other innate and adaptive cells play a crucial role in the formation of granuloma [81,82]. Monocytes are heterogenous nature and show high plasticity. Based on the CD14 and CD16 expression, monocytes were differentiated into classical (CD14++CD16−), intermediate (CD14+CD16+),) and non-classical (CD14+CD16++) subsets . The classical monocytes were able to inhibit the conidial germination of A. fumigatus but not by intermediate monocyte subsets . In contrast to A. fumigatus, both the subsets displayed candidacidal activity by preventing the germination . Further classical monocytes stimulate IL-1β and prostaglandin E2 against C. albicans and initiate the Th17 differentiation . Recent reports suggest that the inflammatory monocytes and macrophages play a key role in offering protective immunity to Cryptococcosis [86,87]. 4.1.2. Dendritic cells (DC) DCs are the professional APCs that plays a central role in initiating the primary immune response. Though DC cells are less efficient in killing the pathogens, they are important for antigen processing and presentation to the T cells . Thus, these cells act as a bridge between innate and adaptive immune response. After fungal recognition, DC cells will process the antigens and present to the naïve T cells in the peripheral or draining lymph nodes and subsequently directs T-cell commitment linage towards T-helper (Th) subsets including Th1, Th2 and Th17 cells (Fig. 2). These subsets of Th profile play a major role in protective immune responses against fungal infections . Further, they produce polarizing cytokines that promote pathogen-specific Th17 differentiation and activation by secreting the IL-2 and IL-23 cytokines . DC has the ability to discriminate both yeast and hyphal forms of C. albicans and induce different Th cell response . It has been suggested that DCs are heterologous population as they can be differentiated into various subsets. In humans, two major distinct subsets of DCs, myeloid DCs (mDCs) and plasmacytoid DCs (pDCs) have been identified . It is believed that mDCs also called type-1 DCs which differentiate the naïve T cells to Th1 subtype , whereas pDCs are type-2 DCs that differentiate the naive T cells to Th2 subsets [94,95]. Interestingly, human pDC inhibits the hyphal growth of A. fumigatus and produce IFN-α and TNF-α as well . The depletion of pDC in A. fumigatus infected mice significantly reduces the survival rate . However the exact role of these subsets is not clear, dysregulation of the pDC and mDCs may be a contributing factor to the increased susceptibility of fungal infections. Since DC cells can involve in both protective and pathological T cell phenotypes, the better characterization of pDCs and mDCs in human blood and tissues will be beneficial for the development of anti-fungal vaccines or immunotherapy strategies. 4.1.3. Polymorphonuclear neutrophils (PMNs) PMNs are the most abundant leukocytes that are essential for the initiation of an inflammatory response to fungal infections. The most common PMNs are Neutrophils which are the primary effector or phagocytic cells of the innate immune system . The lack of sufficient neutrophil numbers or neutropenia is a widely recognized risk factor for the development of invasive Candidiasis and Aspergillosis , however, in case of Cryptococcosis; the deficiency of neutrophil is not a major predisposing factor . So, it suggests that neutrophils are more about immunoregulatory rather than antimicrobial activity in Cryptococcosis . Neutrophils produce the chemokines, which will involve in the chemotaxis for the rapid migration of neutrophils and other effector immune cells to the site of infection . It stores the granules which contain the various antimicrobial peptides, proteolytic (SOD, myeloperoxidase) and nucleolytic enzymes which are involved in the elimination of pathogens. A recent study showed that a population of neutrophils that secrete and utilize IL-17 in an autocrine manner to enhance the ROS production and anti-fungal activity [103,104]. Further, neutrophils produce antimicrobial peptides which disrupt the cell membrane of fungus . Recently, it has been reported that the neutrophils are Candida species-specific as their degree of activation is differ to C. albicans and C. glabrata . Further, neutrophils can able to inhibit fungal growth through the deprivation of essential nutrients . The unique feature of Neutrophils is the production of neutrophils extracellular traps (NET or NETosis) that kills the pathogenic fungi [107,108]. While yeast forms are actively engulfed by phagocytosis, NETs appear to be critical for the killing and containment of the larger hyphal forms of fungal pathogens [101,102,107,108]. Despite, neutrophils are the key players in inflammation, excess release of oxidants, alarmins may cause host tissue damage . 4.1.4. Myeloid-derived suppressor cells (MDSCs) Recently, much attention has been given to MDSCs in various inflammatory processes such as infectious diseases and cancer [110,111]. Similar to regulatory T cells, MDSCs reduces the immunopathophysiology of hyper acted immune cells, which causes the collateral damage of host tissue. These cells are innate immune cells that are characterized by the ability to suppress the effector T cell responses and expansion that lead to pathogen persistence thereby establish chronic infection [110,111]. Despite significant studies have been done in cancer and other infectious diseases, we have just started to understand the complex role of MDSC in fungal diseases. Compared to healthy and other disease controls, A. fumugatus and C. albicans infected patients showed higher levels of MDSCs . Interestingly, it has been demonstrated that adoptive transfer of neutrophilic MDSC wasoffered protection in the murine model of systemic Candida infection . This protective nature is due to the dampening of pathogenic hyperinflammatory NK cells and Th17 response [112,113]. Further, PMN-MDSCs can suppress the NK cell activity and its anti-fungal cytotoxicity during A. fumigatus stimulation . However, the role of MDSCs and the underlying mechanism of the differential immune response are not clear. Further studies are required to understand the key roles of MDSC in fungal infections. 4.1.5. Natural killer (NK) cells NK cells constitute 10–15 % of the total lymphocytes in the human peripheral blood mononuclear cells (PBMC). The best-characterized effector function of NK cells is cytotoxicity which is accompanied by the presence of perforin/granzyme . It is believed that NK cells were considered as the innate immune cells, but recent studies demonstrated that it has the qualities of immunological memory [116,117]. They produce pro-inflammatory cytokines, especially IFN-γ, and other soluble factors (GM-CSF, RANTES) that regulate the functions of other immune cells . Generally, NK cells are not phagocytic cells, but recent in vitro experiments demonstrated that they can phagocytose the C. albicans . NK cells damage the hyphal forms of both C. albicans , and A. fumigatus . Interestingly, some fungus counteracts the anti-fungal nature of NK cells. It has been demonstrated that the co-culture of NK cells with A. fumigatus results in the upregulation of stress-related genes such as Hsp90, ferric chelate reductase B (freB) genes that regulate the NK cell functions . So, these specific interactions help us to find novel drug targets. Recently, researchers working on the modulating/boosting the host immune response by adoptive transfer of NK cells. In this line, the adoptive transfer of NK cells in mice with Cryptococosis, increase the fungal clearance compared to control mice receiving depleted NK cells . Therefore, adoptive transfer of NK cells to IFI patients with immunocompromised might get a great benefit. 4.1.6. Alarmins Alarmins are the danger-associated molecular patterns that are released from the damaged tissue cells of the host during infection. The alarmins include proteins such as HMGB1, calreticulin, IL-1α, IL-33 and non-proteins like ATP, Uric acid etc. They stimulate various PRR (TLR, NLR) and amplify the inflammatory response in order to orchestrate or activate other effector immune cells. However, they also induce the excess pro-inflammatory cytokines that cause tissue damage and impair protective immune response. Another crucial function of the alarmins is the activation of NLR inflammasomes, caspase-1 enzyme for the processing of IL-1β and IL-18 cytokines that in turn initiates the Th1 and Th17 immune response . The IL-1R deficient mice are resistant to Candidiasis, whereas hyper IL-1β signalling is susceptible . However, the exact contribution of inflammasome to protective T cell response needs to be characterised in fungal infections. 4.1.7. Antimicrobial peptides (AMP) The antimicrobial peptides are effector components of the innate immune response and show the broad antimicrobial activity against pathogens. The major antimicrobial peptides in fungal infections are β-defensins, histantin, LL-37, serprocidins etc. They exert their function by attacking the cell membrane of fungus and cause leakage of intracellular components . In contrast to broad-spectrum AMP, the anti-fungal peptides show specific targets by exerting their activity on the cell membrane, cell wall, intracellular targets . Due to recent advances in protein engineering methods, there is an increase in the production of several natural or synthetic AMP. This approach may offer the development of anti-fungal therapeutic drugs . It has been demonstrated that β-defensins 2 & 5, Histantin 5 showed anti candidacidal activity [127,128]. However, the C. albicans counteracts on these AMPs and destabilises or reduces their activity [129,130]. A very recent study reported the plectasin, a fungal antimicrobial peptide (AMP) effectively kills the Mycobacterium tuberculosis (M. tuberculosis) in vitro and also decrease the bacterial burden in the lungs of mouse challenged with H37Rv . 4.2. Acquired immunity Unlike innate, acquired immune response is highly specific to pathogens or antigens and executes several effector functions through the activation of various components of the innate immune system. Broadly, there are two classical adaptive immune responses i) cell-mediated response which includes T cell activation and effector mechanisms and ii) humoral immune responses consisting in B cell maturation and antibody production. 4.2.1. Cellular immunity T lymphocytes are the fundamental cells in cell-mediated immunity and control the fungal proliferation directly or indirectly (Fig. 2). It is indispensable for the protection to fungal infections and the development of emerging vaccines that are focused on inducing the strong and durable cellular immunity. The cellular immune system is broadly classified as helper CD4 and cytotoxic CD8 T cells. Though innate immunity plays a major role in initiating the anti-fungal response, the interaction between host and fungus is an important event for the priming of adaptive immune response (Fig. 1). Upon recognition, the fungal proteins are processed into small antigenic peptides and assembled with MHC (major histocompatibility complex)-I (endogenous proteins) and II (exogenous proteins) and subsequently transported to the surface of activated dendritic cells. Then the dendritic cells will migrate to secondary lymphoid organs such as draining lymph nodes where naïve T cells will be activated by presenting the processed peptides (Fig. 2). The T cell receptors (TCR) on the surface of T cells specifically bind to the enormous antigenic patterns, which are acquired by variations in the amino acid sequences. The activation or quality of different T cells may depend on the processed peptide/antigen specificity. Most commonly, MHC-I associated peptides will present to CD8 T cells, whereas MHC-II associated peptides will present to CD4 T cells. However, CD8 T cells will also activate by presenting the MHC-II associated peptides, this phenomenon called cross-presentation. 4.2.1.1. CD4 T cells CD4 T cells are the central organizers of adaptive immune responses against fungi. These cells increase the recruitment of phagocytes (macrophages, neutrophils) to the infection site . The priming of CD4 T cells occurs when the interaction takes place between TCR and its cognate antigen on the DC cells . The activated T cells secrete the pro-inflammatory cytokines, which in turn drive the naive T cells to differentiate into CD4 T helper (Th) subsets. There are many Th cell subsets such as Th1, Th2, Th9, Th17 and regulatory T cells (Tregs), and each subset produces a unique set of cytokines . The development of each Th subset depends on the cytokines and microenvironment (niche) during the interaction of naive T cells and dendritic cells. The differentiation of one specific Th subset drives the inhibition of other Th subsets differentiation pathways and this is called polarisation. The cytokine IL-12 differentiate the naïve T cells to Th1 phenotype, whereas IL-4 associated with Th2 differentiation. Upon differentiation of CD4 T cells into effector Th subsets, they will migrate to the site of the infection to carry out the effector functions. 4.2.1.1.1. Th1 cells It is a well-known fact that Th1 response is predominant cell type which produces pro-inflammatory cytokines such as IFN-γ, IL-2, IL-12, TNF-α and offers a protective immune response to the host . It has been demonstrated that mice deficient in the IFN-γ and IFN-γ receptor is highly susceptible to Candidiasis and pulmonary Cryptococcosis respectively [136,137]. Further, adoptive transfer of IFN-γ producing A. fumigatus specific CD4 T cells can protect bone marrow-transplanted mice and humans from the invasive fungal disease [138,139]. It has also been demonstrated that patients with systematic Candidiasis showed an improved immunological response after treating with recombinant IFN-γ therapy . Similarly, the administration of IFN-γ immunotherapy significantly clears the Cryptococcus load in CSF of HIV associated Cryptococcal meningitis . Interestingly, A. fumigatus cell wall antigen-specific Th1 immunity offered cross-protection to C. albicans . The other Th1 cytokine TNF-α stimulate the production of IL-12 and IFN-γ which in turn promotes Th1 mediated immunity in C. neoformans lung infection . The cytokines secreted by Th1 cells also activate B cells, leading to the secretion of antigen-specific antibodies against fungi . This data clearly demonstrates the importance of Th1 response in the protective role against invasive fungal infection. 4.2.1.1.2. Th2 cells The role of Th2 immune response is more associated with the increased fungal burden and disease exacerbation [144,145]. Haraguchi group has reported that overexpression of GATA-3, a transcription factor of Th2 cells, increases the susceptibility to systemic Candida infection, by reducing the IFN-γ levels . These Th2 cells mainly produce anti-inflammatory cytokines such as IL-4, IL-5, IL-10, TGF-β and IL-13. The cytokines IL-4 and IL-13 drives the differentiation of naive T cells to Th2 subset and also inhibits the Th1 type differentiation that results in the inhibition of pro-inflammatory cytokine production. The therapeutic inhibition of IL-4 signalling and IL-10 knockout mice showed increased resistance to Candidiasis [146,147]. However, the role of Th2 cytokines is contradictory as IL-4 is required for the development of protective immunity against Candida infection . Further, IL-4R knockout mice showed a high pathogen burden after pulmonary infection with C.neoformans . This suggests IL-4 mediated immune response is protective at the early stage of infection. Moreover, it has been suggested that Th2 immune response induces the alternatively activated macrophages (M2) which may have a protective role in defense against A. fumigatus . Th2 dependent antibodies may afford some protection, by promoting Th1 immunity . Therefore, the quality or outcome of the infection depends on the balance between the Th1 and Th2 subsets. 4.2.1.1.3. Th17 cells The discovery of Th17 subset has shown significant impact in understanding the fungal immunology. IL-1β and IL-23 cytokines drive the Th17 subset differentiation and maturation respectively [152,153]. Th17 cells mediate pleiotropic activities that involve in the induction of pro-inflammatory genes (cytokines, chemokines, and metalloproteinases) and antimicrobial peptides, promote neutrophil recruitment and activation, resulting in clearance of the infection. It has been demonstrated that IL-17 receptor knockout mice were more susceptible to systemic Candida infection than wild type . They produce IL-17A IL-17 F, IL-21 and IL-22 cytokines [155,156]. Along with T cells, IL-17 also produced by the innate immune cells like innate lymphoid cells, NKT cells, γδ T cells, macrophages [157,158]. Recently Maher et al., reported that C. albicans stimulate the IL-17 production from γδ T cells by the mechanism of IL-23 release from DC . The CLR of PRR and its downstream signalling proteins such as Syk/CARD-9 pathway are important to the production of Th17 cytokines (Fig. 1). As discussed earlier, patients with CARD-9 mutation were highly susceptible to fungal infections, this is due to lack of Th17 response . In contrast to the above studies, few studies reported that it may involve in the immunopathogenesis of some fungal infections [53,113]. As noted, C. albicans employs immune evasion strategies; co-culture of live Candida and human PBMC results in the inhibition of IL-17 production by regulating the Tryptophan metabolism . Thus the role of IL-17 in the protection or pathogenesis of fungal infections remains to be cleared. Further, deficiency in IL-17 production leads to susceptibility of mucosal Candidiasis, but not invasive Candidiasis; this suggests that Th17 mediated response might require for mucosal anti-fungal immunity . In the absence of Th1 mediated immunity, the cytokine IL-22 from Th17 cells are critical in the control of Candida and Aspergillus in mucosal and nonmucosal sites respectively . Recently, researchers have found a distinct Th17 subset, natural Th17 (nTh17) which is more similar to innate immunity features and mediates IL-17 immune protection . Th1/Th17 subsets are highly important cell types for recruiting the neutrophils and other immune cells for the clearance of fungal infections . Further, Als3 containing vaccine conferred protection by Th1 and Th17 mediated immune response in systematic Candidiasis . Mice deficient with both Th1 and Th17 responses showed higher mortality than Th1 response alone when infected with Cryptococcus . Thus Th1/Th17 paradigm is desperately needed for the understanding of protective immune response. 4.2.1.1.4. Th9 and Th22 cells Both Th9 and Th22 cell has gained more attention during the last decade. The IL-9 producing CD4 T cells (Th9 cells) were identified that were distinct from Th1, Th2 or Th17 cells . These cells develop from naïve CD4 T cells in the presence of IL-10 and TGF-β (Fig. 2). This subtype mainly produces IL-9 and Th2 cytokines (IL-4, IL-10, TGF-β) and have been reported to induce tissue inflammation in the colitis model . Very recently, it has been suggested that Th9 cells are predominantly skin-resident and are involved in protection against fungal pathogens . It has been reported that increase in the production of IL-9 in the presence of IL-4 upon stimulating the human PBMC with C. albicans . Th22 plays a major role in the absence of Th17 and IFN-γ which is crucial in the protection . Th22 cells produce IL-22 cytokine which plays a role in the local protection in mucocutaneous fungal diseases . IL-22 also mediates anti-fungal resistance in human vulvovaginal candidiasis (VVC) and recurrent VVC (RVVC) . However, current data on the role of Th9 and Th22 mediated immunity in invasive fungal infections is not sufficient and it has to be investigated further. 4.2.1.1.5. Regulatory T cells (Tregs) Tregs acts as an interface between the host immune response and pathogen interactions. Tregs inhibit the host tissue damage by regulating the inflammation, but allows fungal persistence and subsequently causes the suppression of the host immune system. Tregs can exist under natural condition (nTreg) or induce under infection (iTreg). nTreg regulates the neutrophil activity at the early stage of infection, whereas iTregs suppress the Th2 mediated immunity at later stages in Aspergillosis . Reports are proposing that pre-existing Tregs have a beneficial role in the early stage of chronic mucocutaneous Candidiasis infection by activating Th17 cells and followed by the secretion of pro-inflammatory cytokines, however, it exerts suppressive effect during the late phase of infection [175,176]. The expansion of Tregs (Foxp3) in systematic Candidiasis induces Th17 cell proliferation which is associated with C. albicans immune pathology . In contrast, Pandian et al., reported, CD4+CD25+Foxp3+ Tregs promote Th17 cells and showed resistance to mucosal Candidiasis . Further, Tregs suppress the inflammatory immune response that results in higher susceptibility of mice infected with C. albicans and this could be preventing the differentiation and proliferation of Th17 subsets . So, the interplay between Tregs and Th17 cells might require for effective host response to Candida. Romani et al., reported that the intermediate products (Indoleamine-2,3 dioxygenase and kynurenines) of Tryptophan metabolism required for the maintenance of the Th17/Tregs balance . 4.2.1.2. CD8 T cells CD8 T cells control the fungal infection by mediating specific effector functions like activating innate immune cells, lysis of infected host cells and direct killing of pathogenic fungus. The depletion of murine CD8 T cells reduced the survival rate in Cryptococcal infection model and also limits the C. neoformans growth in macrophages by secreting IFN-γ . As discussed, the antigen-presenting cells cross-present the fungal exogenous antigens to CD8 T cells and differentiate into Tc1, Tc2, Tc3 subtypes (Fig. 2). These subtypes display the direct effector functions such as the secretion of granules that contain cytotoxic molecules perforin, granzymes, and granulysin. Tc2 CD8 cells secrete high levels of IL-4 and IL-10 and cause immune suppression. Tc3 CD8 cells display phenotypic characteristics of Th17 cells by the expression of RORγ and CCR6 receptors . They secrete IL-17 cytokines, which in turn activates the release of antimicrobial peptides, defensins from the epithelial cells . Recently it has been reported that Tc3 subtype differentiates into memory Tc17 cells and establish the long-term anti-fungal immunity . Since HIV infected patients are more prone to opportunistic fungal infections, CD8 T cells play a major role in the controlling of fungal infections in the absence or low CD4 count . Based on these, CD8 T cells are mandatory to defense against both extracellular and intracellular pathogens. 4.2.1.3Unconventional T cellsThere are several unconventional T-cell subsets share the functional profile of both innate and adaptive immune response. These are innate-like T cells and recognize molecular patterns (PAMP) and drive the effector functions before conventional T cell activation has taken place . These are borderline cells between innate and aquired arms of the immune system. They respond more rapidly and produce the huge amount of cytokines. These populations include Invariant natural killer T (iNKT) cells, Mucosal-associated invariant T (MAIT) cells and γδ T cells. 4.2.1.3.1. Invariant natural killer T (iNKT) cells iNKT cells express both T cell receptor and NK cell markers and drive both innate and adaptive immune responses. They recognise lipid antigens of pathogens by CD1d complexes and secrete Th1, Th2, and Th17 cytokines. iNKT cells are activated at early stages of infection and engage molecular crosstalk with various innate and adaptive immune cells . It has been demonstrated that mice defecting iNKT cells are more susceptible to C. neoformans infection . In contrast, iNKT cells may increase the severity of C. albicans infection at early stage by suppressing the macrophage mediated anti-fungal activity . Further, Tarumato et al., reported that these cells play a minor role in controlling the C. albicans infection . Thus, the exact role of iNKT cells is controversial and more studies are required to elucidate the underlying mechanism of immune regulation in various fungal infections and also to find out which microbial targets are suppressed by iNKT cell. 4.2.1.3.2. Mucosal-associated invariant T (MAIT) cells MAIT cells comprise 1–10 % of circulating T cells in the peripheral blood but enriched in mucosal surfaces like GIT . These cells are evolutionary conserved invariant T cell receptor (TCR) recognizing antigens presented by the MHC class-I-related protein (MR1) . Earlier studies on bacterial infections suggested that depletion of MAIT cells in the peripheral blood was co-related to disease severity and also restored by antibiotic treatment [189,190]. It has been reported that human MAIT cells upregulate the CD69 activation markers and downregulate the intracellular cytolytic proteins (perforin, granzyme) upon stimulation with various Aspergillus species . However, we have limited knowledge on the role of MAIT cells in fungal infections and their interaction with yeast or hyphal forms of fungi. Also it needs to be elucidated whether these cells involved in the protection or pathogenesis in fungal infections. 4.2.1.3.3. Gamma delta (γδ) T cells γδ T cells owed several distinguishing features that are shared with both innate and adaptive immune cells. The γδ T cells produce large quantities of IFN-γ, IL-17 by coordinating with innate immune cells in C. albicans infection [163,192] and pulmonary Cryptococcal infection . These cells are particularly important where CD4 secreting IL-17 cells are depleted i.e. HIV infected patients . Further, mice deficient with γδ T cells are prone to C. albicans infection in Candida vaginitis model . On the other hand, γδ T cells regulate the Th1 and Th2 balance to C. neoformans by producing the Th2 cytokines to balance the Th1 response . However, the factors that induce and regulate the IL-17 production by γδ T cells are not known. Therefore, it is suggested that the role of γδ T cells in protection or susceptibility to Candidiasis is depended on the site of the infection . Currently, researchers are investigating on these cells for the application of immune therapy as they can restore the IL-17 in HIV patients . 4.2.1.4. Memory immune response One of the major hallmark characteristics of vaccines is the formation of immunological memory response in the host, which allows the immediate protection in subsequent infections . Both innate and adaptive immune cells can memorise past infections and mount the specific immunity accordingly. The individual homogeneous naive T cells can form heterogeneous populations of effector and memory cells [199,200]. During high levels of antigen burden and inflammation, generation of short-lived effector cells will occur. Once the infection has been cleared, 95 % of effector cells will undergo apoptosis and the remaining cells differentiates into memory T-cell population. These memory cells are maintained stably for years, more often life-long, even in the absence of secondary infections . The memory T cells divided into two functionally distinct subsets such as T central memory (Tcm) and T effector memory (Tem) T cells. Tcm cells confer long-term protection, while Tem cells are short-lived and have limited reconstitution capacity . The CD4 and CD8 subtypes such as Th1, Th17, Tc1 and Tc17, γδ T cells act as memory cell types under various circumstances . Majority of memory cells offer the protective immune response by producing Th1 and Th17 cytokines. Very recently, Nanjappan demonstrated that Tc17 persists as long-lasting memory cells by the production of GM-CSF and TNF-α . The same group also reported that memory CD8 T cells mediated anti-fungal resistance even in the absence of CD4 T cells. Few vaccines stimulate the memory innate immune response by NK cells and monocytes in BCG and measles vaccinated children [204,205]. This kind of vaccines would provide cross-protection to other infections . This would be particularly important in HIV infected patients who lack CD4 mediated immunity . From the last decade, many research groups have been focusing on the development of new fungal vaccines that can establish long-lasting memory response both in immunocompetent and immunocompromised patients. Recently, some of the anti-fungal vaccine candidates have been tested for various pathogenic fungal infections such as C. albicans, Aspergillus spp, Cryptococcus spp, Blastomyces spp, Paracoccidioides brasiliensis and Sporothrix spp , however, there is no clear idea about what type of effector and memory response would provide a long-term protective immune response in the host. 4.2.2. Humoral immune response Antibodies are the effector molecules of the adaptive immune response and restrict the fungal burden and its clearance. They control the infectious agents by preventing fungal entry and inhibiting its replication, immune modulation, suppression of polysaccharide release and germ tube formation, neutralizing toxins from pathogens and inhibiting the formation of biofilms. The mechanism of controlling the fungal infections depends on the type of fungal species and the specificity of the epitope. It stimulates the antibody-dependent cellular cytotoxicity (ADCC) that leads to opsonization that in turn activates the downstream complement cascade and finally causes phagocytosis. Further antibodies against fungal antigens mediates the direct anti-fungal effect. For instance, the monoclonal antibodies (mAb) to mannoprotein of C. albicans inhibits adherence, germination and shows direct candidacidal activity by antibody-mediated iron starvation . The mAb of C. neoformans capsule directly interfere the lipopolysaccharide realease and biofilm formation. Antibodies against cell wall β-glucans mediate direct antifungal activity by inhibiting the growth of C. albicans and C. neoformans . Since B cells are antigen-presenting cells, they regulate the T cell activation and differentiation against certain intracellular microbes. Majority of antibodies primarily interact with extracellular pathogens to control infection but also interacts with intracellular pathogens. The differentiation of B cells and the function of antibodies have been depicted in Fig. 3. 1. Download: Download high-res image (311KB) 2. Download: Download full-size image Fig. 3. Antibody mediated immune response to fungal pathogens: The CD4 T cells bind to antigen-MHC class II complexes of B cells by TCR receptors. This result, B cells undergo proliferation and differentiates into plasma cells and memory B cells. Plasma cells secrete the antigen-specific antibodies which will involve in multiple immune functions. These specific antibodies neutralize the fungal components or toxins and prevent its entry; stimulate the antibody-dependent cellular cytotoxicity (ADCC) that leads to opsonization, complement cascade and enhance the phagocytosis; modulate the T cell immune response by enhancing the antigen presentation and secretes the pro & anti-inflammatory cytokines; inhibit the biofilm formation and inflammation-mediated tissue damage. The protective role of antibodies in fungal infections has been proven by the presence of anti-fungal antibodies in patients with progressive fungal infections. It is well known fact that several anti-fungal antibodies against specific fungal derivatives such as glycoproteins, glycolipids or peptides, polysaccharides, and offer protection in the host . Since most of these components are present in the cell wall of fungi, anti-fungal antibodies target the cell wall formation, dynamics and its remodelling . Further, it has been reported that passive transfer of vaccinated mice serum conferred protection in naïve or wild type mice . Studies on antibody based vaccine formulations have shown to mediate protection by producing the protective antibody response in mice [209,211]. The human mAb to Hsp90 of C. albicans has entered clinical trials to examine its efficacy in patients with invasive Candidiasis . Despite, the role of anti-fungal antibodies in the protection is clear in the mice model, there is a conflict in human studies. Nevertheless the presence of specific antibodies in IFIs and naturally acquired antibodies at the stage of early childhood (C. neoformans), they could not offer protection in the host . This indicates that generated specific antibodies may not necessarily prevent fungal infections. It has been believed that the cellular immune response plays a significant role in the protection and humoral immunity acts as a supportive role to cellular immunity. Earlier studies from the mouse model elucidated that many therapeutic vaccines have been mediated the protection by generating the mAb [14,207]. Many studies have been demonstrated that these mAb can bind to the cell wall or surface components of the fungus and inhibits its replication directly [208,210]. Interestingly, some of the mAb (2G8) cross-react with several other fungal pathogens by targeting the common fungal moieties (β-glucan) on the cell wall . This cross-reactivity engages much attention as the single vaccine may protect multiple fungal pathogens. Hence, specific antibodies or broadly neutralising antibodies are an attractive therapeutic option in immunocompromised patients. The combination therapy of anti-fungal drugs and antibody-based therapy definitely would give the beneficial treatment outcome in the patient. 5. Granuloma formation The formation of granuloma is the pathologic characteristic of the host response to bacterial and fungal infections [82,215]. Resistant forms of granuloma was found in systemic mycoses such as Histoplamosis, Coccidiomycosis and Blastomycosis. Though the granuloma formation is an innate and inflammatory basis, it has been evolved into a more complex and dynamic structure by various components of the Th1 cell response, B cells and dendritic cells. The macrophages and T cells secrete IFN-γ and TNF-α cytokines, which in turn play a crucial role in the maintenance of granuloma structure integrity. One study reported that higher levels of IL-17 signalling pathway have taken place in the formation of Histoplasma granuloma . The core of the granuloma shows caseation, necrosis, and restricted nutrition factors where the pathogens can able to persist with limited metabolic functions and can able to survive for decades inside the granuloma in the dormant state. Whenever the host immune system impairs, these dormant or latent form will reactivate and cause the systemic disease. a Granuloma directs the cellular accumulations of T cells and macrophages to maintain the close apposition which is necessary for effective macrophage activation. b The cellular accumulations serve to restrict the toxic environment, which is required to control the pathogen. c The potential function of the granuloma is to limit or prevent the dissemination of the pathogen to the distant sites. 6. Vaccines for fungal infections and limitations Vaccination of high-risk population is a promising approach to prevent invasive fungal infections. Unfortunately, there is no approved vaccine for most of the invasive fungal infections to date. The major challenge for the development of vaccines to the pathogenic fungus is the existence of fungi in various morphological states, pre-exposed or established immunological tolerance to commensal or non-pathogenic fungi in the host. Further, it is difficult to mount the protective immunity in immunocompromised patients where one could not respond to these vaccines as do in the immunocompetent individuals. However, there is sufficient data; even immunocompromised individuals can also mount the immune response against influenza vaccines . The major drawbacks in the present vaccine strategies are 1) These vaccines are offering significant protection in the animal models, but not effective in humans, 2) The protective efficacy of the present vaccine are not persistent, and 3) They are causing adverse effects in the host. In the past two decades, several research groups working on the development and efficacy of vaccines for several fungal infections [218,219]. Some of them are tested for efficacy and safety in the pre-clinical trials. Live attenuated fungus, cell wall components, virulent and secreted proteins are widely studied for the vaccine candidates. Live attenuated vaccines are one of the oldest and successful vaccine strategies for some of the viral (smallpox, measles, yellow fever) and bacterial (BCG) infections. It causes mild disease and also provides a long-lasting immune response, though the exact mechanism is not known. However, the same strategy followed for candidiasis and other fungal infections and successfully tested in the mice model; none of them has reached the clinical trial level. The major drawback of the live attenuated vaccines is the difficulty in the characterisation of pathogenic fungal strains. Moreover, currently, these vaccines are not recommended to the immunocompromised population in which, there is a chance of causing disease instead of offering protection. A study conducted by Liu et al. demonstrated that heat-killed Saccharomyces acts as a protective vaccine against Candidiasis, Aspergillosis and Coccidioidomycosis [221,222]. This may be due to the presence of cross-reactive antigens across the cell wall of the different fungi. 6.1. Recombinant protein vaccines Advances in the field of proteomics and immunology led to the development of subunit vaccines by incorporating specific protein/peptides that induce potential immunity to disease. Previously, it has been proven in the hepatitis B vaccine (HBV) and hepatitis papilloma vaccine (HPV). The major advantage of the subunit vaccines is safe to use and reduce the cross-reactivity to commensal fungi. Since the presence of known specific proteins or peptides (defined composition), there wouldn’t be any virulent components that cause adverse effects in the host. Because of these properties, these vaccines can be used in the immunocompromised population as well. Most of the current research groups following this strategy and focusing on the cell wall proteins or adhesion proteins, secretory and virulent proteins as these are readily accessed to the host immune system. The primary factor for the identification of new vaccine candidates depends on the thorough knowledge of the specific proteins of the fungus . Recently, a cell wall-associated adhesion protein (Als3p) of C. albicans protects mice against disseminated, oropharyngeal and vaginal candidiasis by Th1/Th17 mediated immune response . Currently, the Als3p based vaccine is in the phase-II clinical trial to test the immunotherapeutic efficacy in disseminated Candidiasis and recurrent vulvovaginal Candidiasis (RVVC). It is well known that secretory proteins play a major role in the pathogenicity of fungal infections such as biofilm formation, tissue invasion, immune evasion, as well as cell wall maintenance . These proteins are immunodominant and they can be used as vaccine candidates for infectious diseases. It has been proved that secreted aspartic protease (Sap2) is effective against systemic and mucosal infections in mice . Apart from Alsp3 and Sap2, there are several vaccine candidates such as the stress associated proteins (Hsp90), hyphal associated protein (Hyr1), components of cell wall extracts and glycol conjugates (β-mannan conjugated peptides) are in various preclinical stages. It has been reported that the recombinant protein Asp 16f and CPG oligonucleotides improved the survival rate of the mice against invasive Aspergillosis . Recently, much attention has been directed in the development of multivalent vaccines as they comprise multiple antigens/epitopes of more than 1 strain/serotype of one pathogen . This strategy has been successfully implemented in invasive meningococcal infection and pneumococcal disease . This kind of vaccines will provide protection by inducing effective cellular and antibody-mediated immunity. The main advantage of the multivalent vaccines is the specificity to the pathogenic form of fungus but not on commensal fungi. There are more chances of evading the immune response by pathogen if we use a single virulent factor (univalent) in the vaccines. To reduce the chances of immune evasion strategies and enhance the specific target, it would help by incorporating the more potent vaccine candidates or antigens in the vaccine formulations. These multivalent vaccine formulations definitely would cut down the chances of immune evasion strategies employed by pathogens. The univalent vaccines such as Als3p and Sap2p are currently in different phases of clinical trials that would be the first multi-valent vaccine combination to explore. For the development of multivalent vaccines, we need to understand the mechanism employed by pathogen-host interactions that lead to the discovery of the new vaccine candidates or antigens. One of the approaches is the screening of a broad range of specific proteins, virulent associated antigens/peptides are attractive candidates in the development of multivalent vaccines. However, there are many technical challenges like regulatory, financial and manufacturing hurdles to be overcome. 7. Genetic susceptibility to invasive fungal infections Several studies have reported the role of host genetic variation in defining the severity and susceptibility to IFIs. Chronic Granulomatous Disease (CGD) is a primary immunodeficiency disease genetically inherited by X-linked autosomal recessive manner which increases the host susceptibility towards bacterial and fungal infections. Inability of the phagocytes to generate ROS such as Hydrogen Peroxide (H 2 O 2) which is essential for intracellular microbicidal activity, is one of the key mechanisms responsible for aggravating this CGD. Aspergillus and Candida are predominant fungal strains reported by Cohen et al. in 1981, in which the prevalence of these fungal infections in CGD patients accounted for 20.4 % . Similarly, two more other studies reported A. fumigatus as the most prevalent species in CGD patients [230,231]. Myeloperoxidase (MPO) majorly found in neutrophils is another enzyme in the oxidative stress mechanism responsible for the microbicidal activity. However, subjects with MPO deficiency are usually healthy, but some MPO-deficient subjects have an increased susceptibility to C. albicans infections . Further, it has been demonstrated that MPO knockout mice showed reduced cytotoxicity to C. albicans, A. fumigatus, C. neoformans, demonstrating that an MPO-dependent oxidative system is an important host defence mechanism against fungal infections . Heterozygous dominant mutations in signal transducer and activator of transcription 3 (STAT3) causes hyper-IgE syndrome (HIES). HIES is characterized by impaired production of Th17 cells and Th17-derived cytokines. 28 % of HIES patients subsequently developed invasive mycosis in which Aspergillus remained dominant species . Chronic Mucocutaneous Candidiasis (CMC) is a rare genetic disorder characterised by recurring or persistent infections of the skin, nails and mucous membrane. CMC cases are mainly due to primary immune deficiency disorders. It is associated with certain disorders such as Severe combined immunodeficiency (SCID), or CD25, IL-12Rb1 deficiency, HIES, CARD9, gain-of-function (GOF)-STAT1, and Autoimmune PolyEndocrinopathy Candidiasis Ectodermal Dystrophy (APECED) syndrome . Further patients with IL-17RA, IL-17 F, IL-17RC, ACT1, and RORγt deficiency are highly susceptibility to CMC [235, 236 Puel 2012, Okada 2016]. IL-12Rβ1 is a receptor of both IL-12 and IL-23 receptor and deficiency of IL-12Rβ1 abolish both IL-12 and IL-23 signalling. This leads to diminished production of IFN-γ and IL-17. It has been reported that novel missense mutation in β1 subunit of IL-12Rβ1 (C186Y) is associated with disseminated coccidioidomycosis . CARD9 is a critical adaptor molecule in CLR downstream signalling pathway for recognising fungal pathogens. It is mainly expressed on macrophages and myeloid cells and produce pro-inflammatory cytokines such as IL-6, IL-1β, IL-23 which are involved in Th17 differentiation . Deficiency in autosomal recessive CARD9 is associated with invasive Candidiasis, CMC, and deep dermatophytosis . Therefore, the number of circulating IL-17-producing cells and IL-17 secretion has been decreased in CARD9-deficient patients. GM-CSF and G-CSF have been successfully used in the treatment of the patient with Candida meningoencephalitis which is associated with CARD-9 deficiency [243,244]. However, whether these cytokine therapies can be used to treat other fungal infections with CARD9-deficiency warrants further investigation. 8. Fungal and bacterial co-infections Recent studies exploring the link between bacterial (tuberculosis) and fungal infections . It has been reported that growth-promoting association and the pathological process is similar both in M. tuberculosis and fungus . The prevalence of Candida co-infection in pulmonary TB patients (PTB) varies in different studies . A recent study from south India reported that 40 % of PTB are co-infected with Candida . Another study from Egypt found that 24 % of PTB patients co-infected with Aspergillus species and also found the majority of these cases were multi-drug resistant TB patients . Apart from Candida, TB/Cryptococcosis co-infection is potentially an emerging problem in China, accounting for 62.9 % of global cases . There is a high chance of getting both TB and invasive fungal infections at the same time in the individual with an impaired immune system such as HIV patients. Therefore, triple infections, HIV-TB-fungal might pose a new challenge to global epidemics. Additional efforts to be implemented to explore the new therapeutic strategies for treating these triple co-infections. And also, screening of fungal infections in tuberculosis patient is not a regular clinical practice in most of the countries. But the increased prevalence of these co-infections warning us to screen the fungal infections systematically before initiating the anti-tuberculosis regimen. 9. Anti-fungal drug-resistance The emergence of drug-resistant fungal strains is a major health concern and threatening globally. Currently, there are four major classes of anti-fungal drugs are using for the treatment of IFI. These are azoles (inhibit ergosterol biosynthesis), polyenes (target cell membrane) and the echinocandins (inhibit cell wall synthesis), fluoropyrimidines (blocks nucleic acid synthesis). The emergence of azole and echinocandins resistance is more common, whereas polyenes resistant are very rare . Primarily, the fungus will display two types of resistance, i.e. primary and acquired resistance, but these are overlapping each other. Several adaptive mechanisms take place in the fungi i.e. preventing drug entry to the target region, effluxing the drugs by upregulating the expression of drug efflux pumps or transporters, hypermutations in the genetic level, altering the conformation of target proteins, formation of biofilms and employs the alternative metabolic pathways. One of the most prevalent mechanisms of azole resistance involves alteration or overexpression of the drug target gene, ERG11/cyp51A/cyp51B, with Candida and Aspergillus resistant isolates . These mutations alter the target cells and develop azole resistance. Another common mechanism of acquired azole resistance in Candida, Aspergillus and Cryptococcus species is the up-regulation of various ATP-binding cassette (ABC) transporters that in turn activates the efflux pumps. Echinocandins are the new class of anti-fungal drugs and its resistance is conferred by the amino acid substitutions in highly conserved regions of the Fks subunits of glucan synthase of different Candida species . The stress-related protein, Hsp90 cause echinocandins resistance by regulating the signalling molecules (protein kinase c) that leads to overexpression of chitin synthesis [254,255]. There are many studies identified that NAC species such as C. tropicalis, C. glabrata, C. parapsilosis and C. auris showing the drug resistance [11,256,257]. A very recent study reported that 93 % of C. auris isolates showed extremely high levels of azole fluconazole resistance and even 4% isolates were resistant to all three classes of drugs . Combinational therapy may broaden the spectrum of anti-fungal agents, increased rate of fungal clearance and reduce the emergence of resistance . One of the well-known approaches is administrating the combination therapy and this has been successfully implemented in the treatment of tuberculosis, AIDS, malaria etc. This approach could reduce the resistance rate and broaden the anti-fungal treatment. Combination therapy can enhance the killing ability, reduce the pathogen burden and acquired resistance [258,259]. Further, it exerts inhibitory effects even at lower drug dosage and thus minimizes the adverse effects in the host. It is the standard practice to treat the CNS Cryptococcosis by giving the combination of flucytosine and amphotericin . However, it is debatable in the case of invasive Aspergillosis and Candidiasis as we do not have sufficient data on which combination should be used. Therefore, we need to conduct more clinical trials to evaluate the efficacy of combination therapy relative to monotherapy. Overuse and treatment mismanagement of existing drugs is also one of the reasons for the emergence of drug-resistant. So, we need to use them as selectively and also utmost important to discover the new efficient anti-fungal drugs. High throughput screening of large compound libraries, finding novel drug targets, offering combination therapy, optimize the drug delivery systems, increase the drug quality, supplying cost-effective drugs, is crucial to prevent the widespread of drug-resistant strains. Immune checkpoint therapy (CPI) is one of the emerging immunotherapeutic strategies currently under research in oncology and infectious diseases. This therapy targets the key regulator of the immune system thereby activates the host immune system. Since programmed cell death protein-1 (PD-1) is a major immune checkpoint protein and the interaction with PDL-1 leads to the suppression of immune system. This remains an interesting candidate to study this protein in infectious disorders. PDL-1 acts as a negative regulator of α-(1, 3) glucan-mediated protective immune responses by enhancing Treg polarisation in A. fumigatus. Blockade of this PD-L1 by short α-(1, 3) oligosaccharides efficiently inhibits α-(1, 3) glucan-mediated Treg polarization and promotes Th1 response . A recent report from Wurster et al. demonstrated that blockade of PD-1 with anti-PD-1 inhibitor results in improved survival and reduced A. fumigatus burden in a murine invasive pulmonary Aspergillosis model . Therefore, exploitation of immunotherapeutic approaches in this direction (PDL1-PD1) might open new avenues in promoting protective immune response in human models. 10. Conclusions and prospectives The primary interaction between the host and fungus is key to decide the tolerance and the outcome of the disease as well. In this review, we have highlighted the host immune response against invasive fungal infections and pathogenesis of medically important fungal pathogens. The increased incidence of fungal infections in immunocompromised patients is due to the advent of HIV epidemic and increased use of immunosuppressive drugs. Early detection of fungal infection is crucial so that treatment can be initiated at the earliest time. Further, anti-fungal susceptibility testing will be done at different intervals to avoid treatment failure. Current anti-fungal vaccines facing several challenges and requires further validation. Therefore, intensified research is needed for the systematic screening and preventive treatment of high-risk groups. Further, we continue to enhance our knowledge in the understanding of various immunological pathways which helps us to develop anti-fungal vaccines and drugs. The world needs to combat invasive fungal infections by achieving a significant reduction in incidence and mortality. The major challenge is the co-infection of IFIs with HIV. 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Kontoyiannis PD-1 blockade improves survival and fungal clearance, induces pro-inflammatory lung cytokines and synergizes with caspofungin in a pulmonary aspergillosis model J. Infect. Dis. (2020) Google Scholar Cited by (171) Candidiasis: From cutaneous to systemic, new perspectives of potential targets and therapeutic strategies 2023, Advanced Drug Delivery Reviews Show abstract Candidiasis is an infection caused by fungi from a Candida species, most commonly Candida albicans. C. albicans is an opportunistic fungal pathogen typically residing on human skin and mucous membranes of the mouth, intestines or vagina. It can cause a wide variety of mucocutaneous barrier and systemic infections; and becomes a severe health problem in HIV/AIDS patients and in individuals who are immunocompromised following chemotherapy, treatment with immunosuppressive agents or after antibiotic-induced dysbiosis. However, the immune mechanism of host resistance to C. albicans infection is not fully understood, there are a limited number of therapeutic antifungal drugs for candidiasis, and these have disadvantages that limit their clinical application. Therefore, it is urgent to uncover the immune mechanisms of the host protecting against candidiasis and to develop new antifungal strategies. This review synthesizes current knowledge of host immune defense mechanisms from cutaneous candidiasis to invasive C. albicans infection and documents promising insights for treating candidiasis through inhibitors of potential antifungal target proteins. ### A novel of quinoxaline derivatives tagged with pyrrolidinyl scaffold as a new class of antimicrobial agents: Design, synthesis, antimicrobial activity, and molecular docking simulation 2023, Journal of Molecular Structure Show abstract Globally, infectious diseases are becoming harder to treat due to antibiotic resistance, which has reached high levels. Additionally, developing new therapeutic options to combat the growing antimicrobial resistance in clinical settings is necessary. The new 3-(pyrrolidin-1-yl)quinoxaline derivatives 4–10 conjugated with a different substituent at C2 through ether or amine linkage were synthesized via nucleophilic substitution reaction. The structure of quinoxaline derivatives was confirmed by IR, 1 H NMR, and 13 C NMR spectra. The antimicrobial activity of quinoxaline derivatives 4–10 varied from good to potency against the tested strains. The quinoxaline derivatives 4, 6, and 7 exhibited excellent activity, especially against B. pumilis and E. cloacae, with MIC values of (7.8, 15.6, and 3.91 µg/mL) and (15.6, 7.8, and 15.6 µg/mL) compared with Ciprofloxacin (7.8 and 15.6 µg/mL). Further, hybrid quinoxaline with different methanimine moieties at C2 showed moderate to good antimicrobial activity, except methanimine 9 and 10 that had MIC values equipotent to Ciprofloxacin 15.6 and 31.25 µg/mL against E. coli and S. faecalis, respectively. On the other hand, all these derivatives revealed good to weak antifungal activity with MIC (31.25–500 µg/mL) and MFC (62.5–1000 µg/mL) against A. niger and C. albicans. Moreover, most of the quinoxaline derivatives exhibited bactericidal and fungicidal activity, except for quinoxaline derivatives 9 and 10, which displayed bacteriostatic against E. coli. The molecular docking simulation showed lower binding energy with different types of interaction at the active site of DNA gyrase pocket indicating that these compounds could inhibit the enzyme and cause promising antimicrobial effects. ### Development and research progress of anti-drug resistant fungal drugs 2022, Journal of Infection and Public Health Citation Excerpt : Superior infections affect the skin, mucous membranes and keratinous tissues and cause diseases such as thrush, oropharyngeal candidiasis and fungal skin infections. Invasive fungal infections (IFIs) are systemic fungal infections in which fungi infiltrate and settle in deep tissues, and eventually become life-threatening . Although IFI is less prevalent than superficial fungal infections, they carry a significant human morbidity, mortality, and economic burden . Show abstract With the widespread use of immunosuppressive agents and the increase in patients with severe infections, the incidence of fungal infections worldwide has increased year by year. The fungal pathogens Candida albicans, Cryptococcus neoformans and Aspergillus fumigatus cause a total of more than 1 million deaths each year. Long-term use of antifungal drugs can easily lead to fungal resistance, and the prevalence of drug-resistant fungi is a major global health challenge. In order to effectively control global fungal infections, there is an urgent need for new drugs that can exert effective antifungal activity and overcome drug resistance. We must promote the discovery of new antifungal targets and drugs, and find effective ways to control drug-resistant fungi through different ways, so as to reduce the threat of drug-resistant fungi to human life, health and safety. In the past few years, certain progress has been made in the research and development of antifungal drugs. In addition to summarizing some of the antifungal drugs currently approved by the FDA, this review also focuses on potential antifungal drugs, the repositioned drugs, and drugs that can treat drug-resistant bacteria and fungal infections, and provide new ideas for the development of antifungal drugs in the future. ### Lipid-Based Nanotechnology: Liposome 2024, Pharmaceutics ### An Overlooked and Underrated Endemic Mycosis—Talaromycosis and the Pathogenic Fungus Talaromyces marneffei 2023, Clinical Microbiology Reviews ### ZBP1: A Powerful Innate Immune Sensor and Double-Edged Sword in Host Immunity 2022, International Journal of Molecular Sciences View all citing articles on Scopus © 2020 The Authors. Published by Elsevier Masson SAS. Recommended articles Alarmin(g) the innate immune system to invasive fungal infections Current Opinion in Microbiology, Volume 32, 2016, pp. 135-143 Alayna K Caffrey, Joshua J Obar ### Invasive Fungal Infections Among Immunocompromised Patients in Critical Care Settings: Infection Prevention Risk Mitigation Critical Care Nursing Clinics of North America, Volume 33, Issue 4, 2021, pp. 395-405 May Mei-Sheng Riley ### Photodynamic therapy treatment of superficial fungal infections: A systematic review Photodiagnosis and Photodynamic Therapy, Volume 31, 2020, Article 101774 Julia J.Shen, …, Ditte Marie L.Saunte ### Candida coinfection among patients with pulmonary tuberculosis in Asia and Africa; A systematic review and meta-analysis of cross-sectional studies Microbial Pathogenesis, Volume 139, 2020, Article 103898 Mehdi Hadadi‐Fishani, …, Azad Khaledi ### Role of EFNA1 in tumorigenesis and prospects for cancer therapy Biomedicine & Pharmacotherapy, Volume 130, 2020, Article 110567 Yongping Hao, Guang Li View PDF ### Phagocytes as central players in the defence against invasive fungal infection Seminars in Cell & Developmental Biology, Volume 89, 2019, pp. 3-15 Kerstin Hünniger, Oliver Kurzai Show 3 more articles Article Metrics Citations Citation Indexes 164 Patent Family Citations 1 Captures Mendeley Readers 390 Social Media Shares, Likes & Comments 19 View details About ScienceDirect Remote access Advertise Contact and support Terms and conditions Privacy policy Cookies are used by this site. 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10384	https://www.cut-the-knot.org/Curriculum/Arithmetic/ArithmeticSequence.shtml	Site What's new Content page Front page Index page About Privacy policy Help with math Subjects Arithmetic Algebra Geometry Probability Trigonometry Visual illusions Articles Cut the knot! What is what? Inventor's paradox Math as language Problem solving Collections Outline mathematics Book reviews Interactive activities Did you know? Eye opener Analogue gadgets Proofs in mathematics Things impossible Index/Glossary Simple math Fast Arithmetic Tips Stories for young Word problems Games and puzzles Our logo Make an identity Elementary geometry 3-Term Arithmetic Progression Here's one example of what may be said to be in the purview of Ramsey Theory. In most general terms, Ramsey Theory studies emergence of patterns as the scale of the objects grows. If a set N9 = {1, 2, 3, 4, 5, 6, 7, 8, 9} of 9 numbers is split into two subsets, then at least one of them contains three terms in arithmetic progression. The statement is not true for a set N8 of 8 integers. (The applet helps to investigate the problem. Under every number on display, there is a small box that may be "on" or "off". Click on it and see for yourself. The set of integers above, is split into two depending on which of the boxes are on or off.) \| \| \| What if applet does not run? \| Note that the set N9 can serve as the set of indices into other sequences. We thus can claim that if a 9-term arithmetic progression is split into two sets, then one of the sets contains a 3-terms arithmetic progression. For the proof, observe that 1, 5, and 9 form a 3-term progression. If they belong to a single set of the partition, we are done. Assume they are not. We'll have to consider several cases: 1 and 5 are in one set (let it be P), 9 is in the other (let it be Q). If 3 is in P, we are done. Assume it's in Q. If 6 is also in Q, we are done, since then 3, 6, 9 form a 3-term progression. So assume 6 is in P. If either 4 or 7 is in P, we have either 4, 5, 6 or 5, 6, 7 in P. Either is a 3-term progression. Thus assume 4 and 7 are in Q. Since 3 and 4 are in Q, 2 should be in P. Otherwise, the 3-term progression 2, 3, 4 is found in Q. Since 7 and 9 are in Q, 8 should be in P. But if 8 is in P, P contains the progression 2, 5, 8. (The case when 5 and 9 are in one set and 1 is in the other is considered similarly.) 1 and 9 are in one set (call it P), 5 is in the other (call it Q). 7 is in P. Then 8 must be in Q. For, otherwise P would contain a 3-term sequence 7, 8, 9. Since 5 and 8 are in Q, 3 must be in P. But then 2 must be in Q, or else 1, 2, 3 would all belong to P. But this does not save the situation. If 2 is in Q, the latter has a 3-term sequence 2, 5, 8. 7 is in Q. Since 5 and 7 are in Q, 3 must be in P. Since 1 and 3 are in P, 2 should be in Q. But now we again face an impasse: if 6 is in P, P contains the progression 3, 6, 9; if 6 is in Q, Q contains the progression 5, 6, 7. The proof is not very elegant, but, as Prof. W. McWorter has observed, works for an apparently different problem: Color each real number red or blue. Show that there must be two numbers and their average all the same color. One solution exploits the problem we just solved. Choose 9 equally spaced points on the 2-color line. Index the points 1 through 9. The chosen set is the union of two subsets, one consisting of red and the other of blue points. The original problem implies the existence of a one color 3-term arithmetic sequence, say, {a, a + d, a + 2d}. But then a + d = [a + (a + 2d)]/2, as required in the second problem. The 2-color problem has a shorter solution independent of the 3-terms problem. This is due to the fact that, by the pigeonhole principle there are two points, x and y, of the same color, say red, and we are thus spared the need to consider an additional case. If (x + y)/2 is red, we are done; otherwise (x + y)/2 is blue. If x - (y - x) = 2x - y is red, we are done because 2x - y, y, and [(2x - y) + y]/2 = x are red. Otherwise, 2x - y is blue. Similarly, if y + (y - x) = 2y - x is red, then we are done; otherwise 2y - x is blue. But then 2x - y, 2y - x, and [(2x - y) + (2y - x)]/2 = (x + y)/2 are all blue and we are still done. Reference V. K. Balakrishnan, Theory and Problems of Combinatorics, Schaum's Outline Series, McGraw-Hill, 1995, 1.110 \| \| \| Related materialRead more... \| \| - Aliquot game (An Interactive Gizmo) \| \| - Euclid's Game (An Interactive Gizmo) \| \| - Euclid's Game on a Square Grid \| \| - Sums and Products \| \| - Sums and Products, a Generalization \| \| - Sums, Products, and 1-1 Functions \| \| - Zeros and Nines \| \| - A Candy Game: Integer Iterations \| \| - A Candy Game (Change Discharged) \| \| - Heads and Tails \| \| - Loop or Halt - An Interactive Gizmo \| \| - Breaking Chocolate Bars (An Interactive Gizmo) \| \| \| \|Contact\| \|Front page\| \|Contents\| \|Algebra\| \|Activities\| Copyright © 1996-2018 Alexander Bogomolny
10385	https://www.ti.com/lit/pdf/sprad27	Application Note Optimized Trigonometric Functions on TI Arm Cores Sahin Okur and Eyal Cohen ABSTRACT Trigonometric functions are commonly used in real-time control applications, particularly within the inner loops of control algorithms, where speed and accuracy is essential. The performance of trigonometric functions is a key careabout for designers of these systems as it can have a significant impact on the overall performance of the system. Until recently, trignometric functions based on lookup tables were considered faster than the polynomial-based methods; however, with the inclusion of floating-point units (FPUs) and faster clock speeds, polynomial-based approximations have gained favor. TI has developed C functions of the most commonly used trigonometric functions using these polynomial-based methods and has optimized them for TI's Arm ®-based microcontrollers (MCUs) and microprocessors (MPUs). This application note surveys the trigonometric functions that are available today and shares the optimization techniques used in these functions, as well as the results of our optimization efforts. The TI-optimized trigonometric functions presented in this document can be found in MCU+ SDK v8.5 and later. Table of Contents 1 Introduction ............................................................................................................................................................................. 2 2 Trigonometric Optimizations ................................................................................................................................................. 2 2.1 Lookup Table-Based Approximation .................................................................................................................................. 2 2.2 Polynomial Approximation .................................................................................................................................................. 2 3 Trig Library Benchmarks ....................................................................................................................................................... 7 3.1 C Math.h Library ................................................................................................................................................................. 7 3.2 Arm “Fast Math Functions” in CMSIS ................................................................................................................................. 7 3.3 TI Arm Trig Library ............................................................................................................................................................. 8 3.4 Table of Results .................................................................................................................................................................. 8 4 Optimizations .......................................................................................................................................................................... 9 4.1 Branch Prediction ............................................................................................................................................................... 9 4.2 Floating-Point Single-Precision Instructions ..................................................................................................................... 10 4.3 Memory Placement ........................................................................................................................................................... 11 4.4 Compiler ........................................................................................................................................................................... 11 Revision History ....................................................................................................................................................................... 12 List of Figures Figure 2-1. Plot of Sine and Cosine Over the Range .................................................................................................................. 3 Figure 2-2. Mapping of the Unit Circle for sin(x) for 0 ≤ x ≤ 2π ................................................................................................. 4 Figure 2-3. Plot of Sine and Cosine Over the Range .................................................................................................................. 4 Figure 2-4. Plot of arctan2(y,x) ..................................................................................................................................................... 6 Figure 4-1. Condition Code Suffixes and Related Flags .............................................................................................................. 9 Figure 4-2. Floating-Point Single-Precision Instructions ............................................................................................................ 10 Figure 4-3. Memory Placement .................................................................................................................................................. 11 List of Tables Table 2-1. Sine ............................................................................................................................................................................. 5 Table 2-2. Cosine ......................................................................................................................................................................... 5 Table 2-3. Coefficients for the Arctangent ................................................................................................................................... 7 Table 3-1. Table of Results - Arm Cortex ®-R5F ........................................................................................................................... 8 www.ti.com Table of Contents SPRAD27A – JULY 2022 – REVISED AUGUST 2022 Submit Document Feedback Optimized Trigonometric Functions on TI Arm Cores 1Copyright © 2022 Texas Instruments Incorporated Trademarks Arm ® and Cortex ® are registered trademarks of Arm Limited (or its subsidiaries) in the US and/or elsewhere. All trademarks are the property of their respective owners. 1 Introduction Trigonometric functions are commonly used in real-time control applications, particularly within the inner loops of control algorithms, where speed and accuracy is essential. The performance of trigonometric functions is a key careabout for designers of these systems as it can have a significant impact on the overall performance of the system. Until recently, trignometric functions based on lookup tables were considered faster than the polynomial-based methods; however, with the inclusion of floating-point units (FPUs) and faster CPU clock speeds, polynomial-based approximations are gaining favor. 2 Trigonometric Optimizations The most commonly used trigonometric functions are sine, cosine, arcsine, arccosine, arctangent, and atan2. Trigonometric optimization techniques for these functions fall into two categories: • Lookup table-based approximations • Polynomial approximations These techniques are described in the following sections and the polynomial-based optimization techniques used by the TI Arm ® Trigonometric Library are discussed. 2.1 Lookup Table-Based Approximation Up until recently, lookup table-based approximations were considered faster than polynomial-based approximations. Lookup table-based approximations pre-compute N values of the target function and then index them into the table to select the two closest points and perform an interpolation between them. When using lookup table-based approximations, the accuracy can be adjusted by changing either the size of the lookup or the order of the interpolation function. However, this comes with the tradeoff of more memory usage and longer function times. This is the technique used by the Fast Math Functions in the Arm CMSIS Library where they use a table size of 512 entries to cover 0-2PI and then do a linear interpolation between the closest values. 2.2 Polynomial Approximation With the inclusion of floating-point co-processors and faster clock speeds, the polynomial-based methods are now quite fast and have the added benefit of not requiring any table storage. The TI Arm Trig Library uses this method. The first step in using polynomial approximations is to reduce the range of the input. The bigger the range needed to approximate, the higher order polynomial needed to achieve a specified level of accuracy, which means more computations and slower function performance. The second step is to compute the approximation for the reduced range and then finally restore the range depending on which quadrant the input was originally in by using trigonometric identities. There are a number of methods of finding the best polynomials to achieve the lowest error with the fewest terms, but the most commonly used is the minimax approximation algorithm and one version in particular: the Remez algorithm. The Remez algorithm is used to find a polynomial with the least maximum error. This polynomial is called a minimax polynomial. The minimax polynomial is what is used in our optimized trigonometric functions as it is ideal for control applications where worst-case performance is a key care-about. Note To generate the coefficients for the minimax polynomials, the Sollya software tool is used: https:// www.sollya.org/ . Trademarks www.ti.com 2Optimized Trigonometric Functions on TI Arm Cores SPRAD27A – JULY 2022 – REVISED AUGUST 2022 Submit Document Feedback Copyright © 2022 Texas Instruments Incorporated 2.2.1 Optimizing Sine and Cosine This section shows how the above techniques are applied to optimize the computations for a Sin + Cos function. Sin + Cos functions are commonly used in the transforms for control algorithms where both sin and cos are needed at the same time. The first step in using polynomial approximations is to range reduce the input so that the segment of the function that needs to be modeled is smaller. The most common range reduction for sin/cos computation is to reduce the input to the range. − π 2 ≤ x ≤ π 2 (1) Then, compute the approximation in this region and then adjust the sign depending on which quadrant the angle was in originally. Using the Chebyshev polynomials and the fact that sin(x) is an odd function and cos(x) is an even function, you will see the following: sin x ≈ C 1X + C3X3 + C5X5 + … (2) cos x ≈ C 0 + C2X2 + C4X4 + … (3) Figure 2-1. Plot of Sine and Cosine Over the Range If you assume that the input to the functions are limited to 0: 2π , you need to map the input value to the range − π 2 ≤ x ≤ π 2 before implementing the approximation. This can be done with a couple simple comparisons: if x > 3π 2 , then x = x − 2π (4) else if π2 < x < 3π 2 , then x = π − x, and cos x = − approx cos x (5) A further range reduction technique can be used to limit the input to − π 4 ≤ x ≤ π 4 using the trigonometric identities: sin π 2 − θ = cos θ (6) cos π 2 − θ = sin θ (7) www.ti.com Trigonometric Optimizations SPRAD27A – JULY 2022 – REVISED AUGUST 2022 Submit Document Feedback Optimized Trigonometric Functions on TI Arm Cores 3Copyright © 2022 Texas Instruments Incorporated Which yields the following mapping for sin(x) and a similar one for cos(x): Figure 2-2. Mapping of the Unit Circle for sin(x) for 0 ≤ x ≤ 2π Since you only have to model the sine cosine from − π 4 ≤ x ≤ π 4 , you need fewer coefficients to achieve higher accuracy. Figure 2-3. Plot of Sine and Cosine Over the Range Trigonometric Optimizations www.ti.com 4Optimized Trigonometric Functions on TI Arm Cores SPRAD27A – JULY 2022 – REVISED AUGUST 2022 Submit Document Feedback Copyright © 2022 Texas Instruments Incorporated 2.2.1.1 Sine Cosine Polynomials From Sollya Table 2-1 and Table 2-2 show the coefficients for the sine and cosine approximation obtained from the Sollya program. The table shows the error expected given the range reduction and order of the polynomial. If you are trying to achieve full single precision floating-point accuracy, then you need to get to ~1e-7. Table 2-1. Sine Range Number of Terms Absolute Error Polynomial −π/2 : π/2 31.00E-04 x (0.999891757965087890625 + x 2 (-0.165960013866424560546875 + x 2 7.602870464324951171875e-3)) (8) −π/2 : π/2 46.00E-07 x (0.999996483325958251953125 + x 2 (-0.166647970676422119140625 + x^2 (8.306086063385009765625e-3 + x 2 (-1.83582305908203125e-4)))) (9) −π/2 : π/2 56.00E-09 x (1 + x 2 (-0.1666665971279144287109375 + x 2 (8.333069272339344024658203125e-3 + x 2 (-1.98097783140838146209716796875e-4 + x 2 2.6061034077429212629795074462890625e-6)))) (10) −π/4 : π/4 21.50E-04 x (0.99903142452239990234375 + x 2 (-0.16034401953220367431640625)) (11) −π/4 : π/4 35.60E-07 x (0.9999949932098388671875 + x 2 (-0.166601598262786865234375 + x 2 8.12153331935405731201171875e-3)) (12) −π/4 : π/4 41.80E-09 x (1 + x 2 (-0.166666507720947265625 + x2 (8.331983350217342376708984375e-3 + x 2 (-1.94961365195922553539276123046875e-4)))) (13) −π/4 : π/4 56.00E-11 x (1 + x 2 (-0.16666667163372039794921875 + x 2 (8.33337195217609405517578125e-3 + x 2 (-1.98499110410921275615692138671875e-4 + x 2 2.800547008519060909748077392578125e-6)))) (14) Table 2-2. Cosine Range Number of Terms Absolute Error Polynomial −π/2 : π/2 36.00E-04 0.9994032382965087890625 + x 2 (-0.495580852031707763671875 + x 2 3.679168224334716796875e-2) (15) −π/2 : π/2 46.70E-06 0.99999332427978515625 + x 2 (-0.4999125301837921142578125 + x 2 (4.1487820446491241455078125e-2 + x 2 (-1.27122621051967144012451171875e-3))) (16) −π/2 : π/2 56.00E-08 0.999999940395355224609375 + x 2 (-0.499998986721038818359375 + x 2 (4.1663490235805511474609375e-2 + x 2 (-1.385320327244699001312255859375e-3 + x 2 2.31450176215730607509613037109375e-5))) (17) www.ti.com Trigonometric Optimizations SPRAD27A – JULY 2022 – REVISED AUGUST 2022 Submit Document Feedback Optimized Trigonometric Functions on TI Arm Cores 5Copyright © 2022 Texas Instruments Incorporated Table 2-2. Cosine (continued) Range Number of Terms Absolute Error Polynomial −π/4 : π/4 31.00E-05 0.999990046024322509765625 + x 2 (-0.4997082054615020751953125 + x 2 4.03986163437366485595703125e-2) (18) −π/4 : π/4 43.30E-08 1 + x 2 (-0.49999892711639404296875 + x 2 (4.16561998426914215087890625e-2 + x 2 (-1.35968066751956939697265625e-3))) (19) −π/4 : π/4 51.00E-10 1 + x 2 (-0.5 + x 2 (4.16666455566883087158203125e-2 + x 2 (-1.388731296174228191375732421875e-3 + x 2 2.4432971258647739887237548828125e-5))) (20) 2.2.2 Optimizing Arctangent and Arctangent2 The arctangent function, and in particular, the arctangent2 function is a critical function in control applications. For example, in motor control, there may be sensors used to get the x and y position of a motor. Then, the application needs to translate that into an angular value. A standard arctan function would be called as arctan(y/x), but this loses the quadrant information as Q1 and Q3 are both positive and Q2 and Q4 are both negative. The arctan2 function accepts both the x and the y input to return a value in the full −π to π range versus the arctan function that returns values only in Q1 or Q4, −π/2 to π/2. Figure 2-4. Plot of arctan2(y,x) As the input to arctan(z) can be any number from −∞to ∞, use trig identities to reduce the range and get an approximation function that is relatively low complexity. Start with: • arctan(x) = π/2- arctan(1/x) (1a) • arctan(x)=−arctan(x) (2a) These identities allow you to restrict the approximation range to abs(x) <=1. Trigonometric Optimizations www.ti.com 6Optimized Trigonometric Functions on TI Arm Cores SPRAD27A – JULY 2022 – REVISED AUGUST 2022 Submit Document Feedback Copyright © 2022 Texas Instruments Incorporated The next step is to find a simple polynomial that approximates arctan in the range 0-1, or -1 to 1. Using sollya, try some different polynomial lengths to get the approximation error. This is shown in Table 2-3 . You can see that the error is not falling that fast as a function of the number of terms in the polynomial, so even at 6 terms you are still at 3.4e-6 in the range (-1,1). Also note that arctan is an odd function where all the terms are odd powers. 2.2.2.1 Arctangent Polynomials Table 2-3 shows the coefficients for the arctangent approximation obtained from the Sollya program. The table shows the error expected given the range reduction and order of the polynomial. Table 2-3. Coefficients for the Arctangent Range Terms Abs err Polynomial -1 : 1 48.00E-05 x (0.99921381473541259765625 + x 2 (-0.321175038814544677734375 + x 2 (0.146264731884002685546875 + x 2 (-3.8986742496490478515625e-2)))) (21) -1 : 1 52.30E-05 x (0.999970018863677978515625 + x 2 (-0.3317006528377532958984375 + x 2 (0.1852150261402130126953125 + x 2 (-9.1925732791423797607421875e-2 + x 2 2.386303804814815521240234375e-2)))) (22) -1 : 1 63.40E-06 x (0.999995648860931396484375 + x 2 (-0.3329949676990509033203125 + x 2 (0.19563795626163482666015625 + x 2 (-0.121243648231029510498046875 + x 2 (5.7481847703456878662109375e-2 + x 2 (-1.3482107780873775482177734375e-2)))))) (23) tan(pi/12) 32.00E-07 x (0.999994814395904541015625 + x 2 (-0.3327477872371673583984375 + x 2 0.18327605724334716796875)) (24) tan(pi/12) 43.00E-09 x (0.999999940395355224609375 + x 2 (-0.333319008350372314453125 + x 2 (0.19920165836811065673828125 + x 2 (-0.12685041129589080810546875)))) (25) tan(pi/12) 58.70E-11 x (1 + x 2 (-0.333333194255828857421875 + x 2 (0.19998063147068023681640625 + x 2 (-0.14202083647251129150390625 + x 2 9.6703059971332550048828125e-2)))) (26) 3 Trig Library Benchmarks 3.1 C Math.h Library Math.h includes the standard functions for single-precision floating-point sinf(), cosf(), atanf(), and atan2f(), as well as the double-precision sin(), cos(), atan(), atan2(). These functions use polynomial-based approximation methods and provide high accuracy with no input limitations, but at the cost of more cycles. These functions are benchmarked and the results are shared at the end of this section. 3.2 Arm “Fast Math Functions” in CMSIS Arm provides a library titled FastMathFunctions that contains single-precision floating-point sin() and cos() functions. These functions use lookup table-based methods to approximate the functions with lookup tables of size 2 KB. Arm also provides a sincos function that computes both the sin() and cos() simultaneously in ControllerFunctions as well. These functions were benchmarked and the results are shared at the end of this section. www.ti.com Trigonometric Optimizations SPRAD27A – JULY 2022 – REVISED AUGUST 2022 Submit Document Feedback Optimized Trigonometric Functions on TI Arm Cores 7Copyright © 2022 Texas Instruments Incorporated 3.3 TI Arm Trig Library The TI Arm Trig library provides the single-precision floating-point functions ti_arm_sin(), ti_arm_cos(), ti_arm_sincos(), ti_arm_asin(), ti_arm_acos(), ti_arm_atan(), and ti_arm_atan2(). These functions use polynomial approximation-based methods. This library can be found in MCU+ SDK. 3.4 Table of Results Hardware AM243x LaunchPad Software • TI Arm Clang Compiler v2.0.0.STS • MCU+ SDK for AM243x v8.2.0.31 Table 3-1. Table of Results - Arm Cortex ®-R5F Trig Function Library C Function Input Range [Rad] Max Error Max Cycles Avg Cycles Approximation Type Sine C sinf() Any 0.0000000296 179 150 Polynomial CMSIS arm_sin_f32() Any 0.0000181917 48 48 Lookup table (2 KB) TI Arm Trig ti_arm_sin() 0:2π 0.0000007225 29 29 Polynomial Cosine C cosf() Any 0.0000000297 179 150 Polynomial CMSIS arm_cos_f32() Any 0.0000183477 50 50 Lookup table TI Arm Trig ti_arm_cos() 0:2π 0.0000002863 37 37 Polynomial Sine + Cosine C NA - - - - - CMSIS arm_sin_cos_f3 2() Any 0.0000006100 83 83 Lookup table TI Arm Trig ti_arm_sincos() 0:2π 0.0000001925 54 54 Polynomial Arcsine C asinf() Any 0.0000000590 213 132 Polynomial CMSIS NA - - - - - TI Arm Trig ti_arm_asin() Any 0.0000003428 59 59 Polynomial Arccosine C acosf() Any 0.0000001792 128 87 Polynomial CMSIS NA - - - - - TI Arm Trig ti_arm_acos() Any 0.0000004295 64 64 Polynomial Arctangent C atanf() Any 0.0000001748 128 87 Polynomial CMSIS NA - - - - - TI Arm Trig ti_arm_atan Any 0.0000001748 64 64 Polynomial Arctangent2 C atan2f() Any 0.0000002021 222 148 Polynomial CMSIS NA - - - - - TI Arm Trig ti_arm_atan2 Any 0.0000002957 59 49 Polynomial Trig Library Benchmarks www.ti.com 8Optimized Trigonometric Functions on TI Arm Cores SPRAD27A – JULY 2022 – REVISED AUGUST 2022 Submit Document Feedback Copyright © 2022 Texas Instruments Incorporated 4 Optimizations 4.1 Branch Prediction Using branches in functions creates unpredictability in the exact cycle count as the branch predictor may not predict correctly and any missed predictions cost approximately 8 cycles/miss. Arm provides conditional instructions that can be used in place of branch statements, ensuring that the functions always execute in the same number of cycles. Figure 4-1 shows that the conditional codes that can be appended to instructions. Figure 4-1. Condition Code Suffixes and Related Flags The main reason for creating the . asm versions of the trigonometric functions was to remove branches inserted by the compiler and replace with conditional instructions instead. This had the effect of reducing the max cycles due to incorrect branch predictions. This reduction was enabled by replacing branch instructions with conditional operations. The delta between the max and the average could not be completely removed as the algorithm contains some divide instructions in the range reduction code which are conditionally implemented depending on the input values. Note The TI Arm Clang compiler performs these assembly optimizations automatically when compiler optimization is enabled, therefore these assembly versions have been replaced with their C-equivalent starting with MCU+ SDK v8.5. www.ti.com Optimizations SPRAD27A – JULY 2022 – REVISED AUGUST 2022 Submit Document Feedback Optimized Trigonometric Functions on TI Arm Cores 9Copyright © 2022 Texas Instruments Incorporated 4.2 Floating-Point Single-Precision Instructions Time-expensive instructions were avoided during implementation that would cause a longer execution time of the trigonometric functions (such as square root and deviation). Figure 4-2 shows floating-point single-precision data processing instructions cycle timing behavior of the FPU. Figure 4-2. Floating-Point Single-Precision Instructions Optimizations www.ti.com 10 Optimized Trigonometric Functions on TI Arm Cores SPRAD27A – JULY 2022 – REVISED AUGUST 2022 Submit Document Feedback Copyright © 2022 Texas Instruments Incorporated 4.3 Memory Placement The purpose of the Tightly-Coupled Memory (TCM) is to provide low-latency memory that the processor can use without the unpredictability that is a feature of caches. The main use of TCM is to store performance critical data and code. Interrupt handlers, data for real-time tasks and OS control structures are a common example. There are two external TCMs that can be configured to only store instructions, only data, or a mixture of the two (TCMA and TCMB). In our library, the polynomial coefficients were placed in TCMB, and the functions themselves in TCMA. Figure 4-3. Memory Placement Enabling a TCM to include both instructions and data provides more flexibility from a system perspective, but might limit performance compared with optimizing a TCM to solely store instructions or data. The TCMB is accessible via two ports. This indicates that the TCM has been implemented as two separate banks of RAM so that the two banks can be accessed simultaneously. 4.4 Compiler The compiler optimization to be -O3 was configured so that the compiler minimizes some attributes of the executable program. To understand what optimizations are performed on this level, see the Optimization Options section in the TI Arm Clang Compiler Tools User’s Guide . In order to place different parts in different memory sections, use the attribute syntax (which is described in the Attribute Syntax section of the TI Arm Clang Compiler Tools User’s Guide . Here is an example from the implementation of placing coefficients in TCMB: www.ti.com Optimizations SPRAD27A – JULY 2022 – REVISED AUGUST 2022 Submit Document Feedback Optimized Trigonometric Functions on TI Arm Cores 11 Copyright © 2022 Texas Instruments Incorporated Revision History NOTE: Page numbers for previous revisions may differ from page numbers in the current version. Changes from Revision (July 2022) to Revision A (August 2022) Page • Updated Section 1.............................................................................................................................................. 2 • Updated Section 2.............................................................................................................................................. 2 • Updated Section 3.4 ........................................................................................................................................... 8 • Updated Section 4.1 ........................................................................................................................................... 9 Revision History www.ti.com 12 Optimized Trigonometric Functions on TI Arm Cores SPRAD27A – JULY 2022 – REVISED AUGUST 2022 Submit Document Feedback Copyright © 2022 Texas Instruments Incorporated IMPORTANT NOTICE AND DISCLAIMER TI PROVIDES TECHNICAL AND RELIABILITY DATA (INCLUDING DATA SHEETS), DESIGN RESOURCES (INCLUDING REFERENCE DESIGNS), APPLICATION OR OTHER DESIGN ADVICE, WEB TOOLS, SAFETY INFORMATION, AND OTHER RESOURCES “AS IS” AND WITH ALL FAULTS, AND DISCLAIMS ALL WARRANTIES, EXPRESS AND IMPLIED, INCLUDING WITHOUT LIMITATION ANY IMPLIED WARRANTIES OF MERCHANTABILITY, FITNESS FOR A PARTICULAR PURPOSE OR NON-INFRINGEMENT OF THIRD PARTY INTELLECTUAL PROPERTY RIGHTS. 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10386	http://www.math.utah.edu/~gustafso/s2015/2280/chapter11-systems/Ch11-2-Basic-First-order-System-Methods.pdf	11.2 Basic First-order System Methods 757 11.2 Basic First-order System Methods Solving 2 × 2 Systems It is shown here that any constant linear system u′ = Au, A = a b c d ! can be solved by one of the following elementary methods. (a) The integrating factor method for y′ = p(x)y + q(x). (b) The second order constant coeﬃcient formulas in Theo-rem 45, Chapter 5. Triangular A. Let’s assume b = 0, so that A is lower triangular. The upper triangular case is handled similarly. Then u′ = Au has the scalar form u′ 1 = au1, u′ 2 = cu1 + du2. The ﬁrst diﬀerential equation is solved by the growth/decay formula: u1(t) = u0eat. Then substitute the answer just found into the second diﬀerential equa-tion to give u′ 2 = du2 + cu0eat. This is a linear ﬁrst order equation of the form y′ = p(x)y + q(x), to be solved by the integrating factor method. Therefore, a triangular system can always be solved by the ﬁrst order integrating factor method. An illustration. Let us solve u′ = Au for the triangular matrix A = 1 0 2 1 ! . The ﬁrst equation u′ 1 = u1 has solution u1 = c1et. The second equation becomes u′ 2 = 2c1et + u2, which is a ﬁrst order linear diﬀerential equation with solution u2 = (2c1t + c2)et. The general solution of u′ = Au in scalar form is u1 = c1et, u2 = 2c1tet + c2et. 758 The vector form of the general solution is u(t) = c1 et 2tet ! + c2 0 et ! . The vector basis is the set B = ( et 2tet ! , 0 et !) . Non-Triangular A. In order that A be non-triangular, both b ̸= 0 and c ̸= 0 must be satisﬁed. The scalar form of the system u′ = Au is u′ 1 = au1 + bu2, u′ 2 = cu1 + du2. Theorem 1 (Solving Non-Triangular u′ = Au) Solutions u1, u2 of u′ = Au are linear combinations of the list of atoms obtained from the roots r of the quadratic equation det(A −rI) = 0. Proof: The method: diﬀerentiate the ﬁrst equation, then use the equations to eliminate u2, u′ 2. The result is a second order diﬀerential equation for u1. The same diﬀerential equation is satisﬁed also for u2. The details: u′′ 1 = au′ 1 + bu′ 2 Diﬀerentiate the ﬁrst equation. = au′ 1 + bcu1 + bdu2 Use equation u′ 2 = cu1 + du2. = au′ 1 + bcu1 + d(u′ 1 −au1) Use equation u′ 1 = au1 + bu2. = (a + d)u′ 1 + (bc −ad)u1 Second order equation for u1 found The characteristic equation of u′′ 1 −(a + d)u′ 1 + (ad −bc)u1 = 0 is r2 −(a + d)r + (bc −ad) = 0. Finally, we show the expansion of det(A −rI) is the same characteristic poly-nomial: det(A −rI) = a −r b c d −r = (a −r)(d −r) −bc = r2 −(a + d)r + ad −bc. The proof is complete. The reader can verify that the diﬀerential equation for u1 or u2 is exactly u′′ −trace(A)u′ + det(A)u = 0. Finding u1. Apply the second order formulas, Theorem 45 in Chapter 5, to solve for u1. This involves writing a list L of atoms corresponding 11.2 Basic First-order System Methods 759 to the two roots of the characteristic equation r2 −(a+d)r+ad−bc = 0, followed by expressing u1 as a linear combination of the two atoms. Finding u2. Isolate u2 in the ﬁrst diﬀerential equation by division: u2 = 1 b(u′ 1 −au1). The two formulas for u1, u2 represent the general solution of the system u′ = Au, when A is 2 × 2. An illustration. Let’s solve u′ = Au when A = 1 2 2 1 ! . The equation det(A −rI) = 0 is (1 −r)2 −4 = 0 with roots r = −1 and r = 3. The atom list is L = {e−t, e3t}. Then the linear combination of atoms is u1 = c1e−t + c2e3t. The ﬁrst equation u′ 1 = u1 + 2u2 implies u2 = 1 2(u′ 1 −u1). The general solution of u′ = Au is then u1 = c1e−t + c2e3t, u2 = −c1e−t + c2e3t. In vector form, the general solution is u = c1 e−t −e−t ! + c2 e3t e3t ! . Triangular Methods Diagonal n×n matrix A = diag(a1, . . . , an). Then the system x′ = Ax is a set of uncoupled scalar growth/decay equations: x′ 1(t) = a1x1(t), x′ 2(t) = a2x2(t), . . . x′ n(t) = anxn(t). The solution to the system is given by the formulas x1(t) = c1ea1t, x2(t) = c2ea2t, . . . xn(t) = cneant. The numbers c1, . . . , cn are arbitrary constants. 760 Triangular n × n matrix A. If a linear system x′ = Ax has a square triangular matrix A, then the system can be solved by ﬁrst order scalar methods. To illustrate the ideas, consider the 3 × 3 linear system x′ =    2 0 0 3 3 0 4 4 4   x. The coeﬃcient matrix A is lower triangular. In scalar form, the system is given by the equations x′ 1(t) = 2x1(t), x′ 2(t) = 3x1(t) + 3x2(t), x′ 3(t) = 4x1(t) + 4x2(t) + 4x3(t). A recursive method. The system is solved recursively by ﬁrst order scalar methods only, starting with the ﬁrst equation x′ 1(t) = 2x1(t). This growth equation has general solution x1(t) = c1e2t. The second equation then becomes the ﬁrst order linear equation x′ 2(t) = 3x1(t) + 3x2(t) = 3x2(t) + 3c1e2t. The integrating factor method applies to ﬁnd the general solution x2(t) = −3c1e2t+c2e3t. The third and last equation becomes the ﬁrst order linear equation x′ 3(t) = 4x1(t) + 4x2(t) + 4x3(t) = 4x3(t) + 4c1e2t + 4(−3c1e2t + c2e3t). The integrating factor method is repeated to ﬁnd the general solution x3(t) = 4c1e2t −4c2e3t + c3e4t. In summary, the solution to the system is given by the formulas x1(t) = c1e2t, x2(t) = −3c1e2t + c2e3t, x3(t) = 4c1e2t −4c2e3t + c3e4t. Structure of solutions. A system x′ = Ax for n × n triangular A has component solutions x1(t), . . . , xn(t) given as polynomials times exponentials. The exponential factors ea11t, . . . , eannt are expressed in terms of the diagonal elements a11, . . . , ann of the matrix A. Fewer than n distinct exponential factors may appear, due to duplicate diagonal elements. These duplications cause the polynomial factors to appear. The reader is invited to work out the solution to the system below, which has duplicate diagonal entries a11 = a22 = a33 = 2. x′ 1(t) = 2x1(t), x′ 2(t) = 3x1(t) + 2x2(t), x′ 3(t) = 4x1(t) + 4x2(t) + 2x3(t). 11.2 Basic First-order System Methods 761 The solution, given below, has polynomial factors t and t2, appearing because of the duplicate diagonal entries 2, 2, 2, and only one exponential factor e2t. x1(t) = c1e2t, x2(t) = 3c1te2t + c2e2t, x3(t) = 4c1te2t + 6c1t2e2t + 4c2te2t + c3e2t. Conversion to Systems Routinely converted to a system of equations of ﬁrst order are scalar second order linear diﬀerential equations, systems of scalar second order linear diﬀerential equations and scalar linear diﬀerential equations of higher order. Scalar second order linear equations. Consider an equation au′′ + bu′ + cu = f where a ̸= 0, b, c, f are allowed to depend on t, ′ = d/dt. Deﬁne the position-velocity substitution x(t) = u(t), y(t) = u′(t). Then x′ = u′ = y and y′ = u′′ = (−bu′ −cu+f)/a = −(b/a)y −(c/a)x+ f/a. The resulting system is equivalent to the second order equation, in the sense that the position-velocity substitution equates solutions of one system to the other: x′(t) = y(t), y′(t) = −c(t) a(t)x(t) −b(t) a(t)y(t) + f(t) a(t). The case of constant coeﬃcients and f a function of t arises often enough to isolate the result for further reference. Theorem 2 (System Equivalent to Second Order Linear) Let a ̸= 0, b, c be constants and f(t) continuous. Then au′′+bu′+cu = f(t) is equivalent to the ﬁrst order system aw′(t) = 0 a −c −b ! w(t) + 0 f(t) ! , w(t) = u(t) u′(t) ! . Converting second order systems to ﬁrst order systems. A sim-ilar position-velocity substitution can be carried out on a system of two second order linear diﬀerential equations. Assume a1u′′ 1 + b1u′ 1 + c1u1 = f1, a2u′′ 2 + b2u′ 2 + c2u2 = f2. 762 Then the preceding methods for the scalar case give the equivalence      a1 0 0 0 0 a1 0 0 0 0 a2 0 0 0 0 a2           u1 u′ 1 u2 u′ 2      ′ =      0 a1 0 0 −c1 −b1 0 0 0 0 0 a2 0 0 −c2 −b2           u1 u′ 1 u2 u′ 2     +      0 f1 0 f2     . Coupled spring-mass systems. Springs connecting undamped cou-pled masses were considered at the beginning of this chapter, page 746. Typical equations are m1x′′ 1(t) = −k1x1(t) + k2[x2(t) −x1(t)], m2x′′ 2(t) = −k2[x2(t) −x1(t)] + k3[x3(t) −x2(t)], m3x′′ 3(t) = −k3[x3(t) −x2(t)] −k4x3(t). (1) The equations can be represented by a second order linear system of dimension 3 of the form Mx′′ = Kx, where the position x, the mass matrix M and the Hooke’s matrix K are given by the equalities x =    x1 x2 x3   , M =    m1 0 0 0 m2 0 0 0 m3   , K =    −(k1 + k2) k2 0 k2 −(k2 + k3) k3 0 −k3 −(k3 + k4)   . Systems of second order linear equations. A second order sys-tem Mx′′ = Kx + F(t) is called a forced system and F is called the external vector force. Such a system can always be converted to a sec-ond order system where the mass matrix is the identity, by multiplying by M−1: x′′ = M−1Kx + M−1F(t). The benign form x′′ = Ax + G(t), where A = M−1K and G = M−1F, admits a block matrix conversion into a ﬁrst order system: d dt x(t) x′(t) ! = 0 I A 0 ! x(t) x′(t) ! + 0 G(t) ! . Damped second order systems. The addition of a dampener to each of the masses gives a damped second order system with forcing Mx′′ = Bx′ + KX + F(t). In the case of one scalar equation, the matrices M, B, K are constants m, −c, −k and the external force is a scalar function f(t), hence the system becomes the classical damped spring-mass equation mx′′ + cx′ + kx = f(t). 11.2 Basic First-order System Methods 763 A useful way to write the ﬁrst order system is to introduce variable u = Mx, in order to obtain the two equations u′ = Mx′, u′′ = Bx′ + Kx + F(t). Then a ﬁrst order system in block matrix form is given by M 0 0 M ! d dt x(t) x′(t) ! = 0 M K B ! x(t) x′(t) ! + 0 F(t) ! . The benign form x′′ = M−1Bx′ + M−1Kx + M−1F(t), obtained by left-multiplication by M−1, can be similarly written as a ﬁrst order system in block matrix form. d dt x(t) x′(t) ! = 0 I M−1K M−1B ! x(t) x′(t) ! + 0 M−1F(t) ! . Higher order linear equations. Every homogeneous nth order constant-coeﬃcient linear diﬀerential equation y(n) = p0y + · · · + pn−1y(n−1) can be converted to a linear homogeneous vector-matrix system d dx         y y′ y′′ . . . y(n−1)         =         0 1 0 · · · 0 0 0 1 · · · 0 . . . 0 0 0 · · · 1 p0 p1 p2 · · · pn−1                 y y′ y′′ . . . y(n−1)         . This is a linear system u′ = Au where u is the n × 1 column vector consisting of y and its successive derivatives, while the n × n matrix A is the classical companion matrix of the characteristic polynomial rn = p0 + p1r + p2r2 + · · · + pn−1rn−1. To illustrate, the companion matrix for r4 = a + br + cr2 + dr3 is A =      0 1 0 0 0 0 1 0 0 0 0 1 a b c d     . The preceding companion matrix has the following block matrix form, which is representative of all companion matrices. A = 0 I a b c d ! . 764 Continuous coeﬃcients. It is routinely observed that the methods above for conversion to a ﬁrst order system apply equally as well to higher order linear diﬀerential equations with continuous coeﬃcients. To illustrate, the fourth order linear equation yiv = a(x)y+b(x)y′+c(x)y′′+ d(x)y′′′ has ﬁrst order system form u′ = Au where A is the companion matrix for the polynomial r4 = a(x) + b(x)r + c(x)r2 + d(x)r3, x held ﬁxed. Forced higher order linear equations. All that has been said above applies equally to a forced linear equation like yiv = 2y + sin(x)y′ + cos(x)y′′ + x2y′′′ + f(x). It has a conversion to a ﬁrst order nonhomogeneous linear system u′ =      0 1 0 0 0 0 1 0 0 0 0 1 2 sin x cos x x2     u +      0 0 0 f(x)     , u =      y y′ y′′ y′′′     .
10387	https://www.expii.com/t/finding-the-range-of-a-function-algebraically-4795	Finding the Range of a Function, Algebraically - Expii Expii Finding the Range of a Function, Algebraically - Expii How can values not be in the range? Values not included in range are values that are impossible given all the function's domain. They might be beyond an asymptote, or be values that the function simply skips. HomeLog inSign up Search for math and science topics Search for topics Log InSign Up Algebra 2Domain and Range Finding the Range of a Function, Algebraically How can values not be in the range? Values not included in range are values that are impossible given all the function's domain. They might be beyond an asymptote, or be values that the function simply skips. Finding the Range of a Function, Algebraically Go to Topic Explanations (4) Vatsal Ojha Text 21 If you recall first learning about range many years ago, you may have thought about it as subtracting the lowest value from the highest value in a dataset. We will now be taking this a step further, looking at range of functions. Finding the range of a function y=f(x) is the same as finding all values that y could be. To do this, we can think of it this way: The range of f(x) is all the y-values where there is a number x with y=f(x). We can try and motivate how to find this with an example. Let's try to find the range of: f(x)=x 2+1 So we want to find y-values such that there is some x where y=x 2+1. Suppose we want to check if y=5 is in the range of f(x). Then, we want to check if there is an x-value such that x 2+1=5. We can solve this equation as follows: x 2+1=5 x 2=4 x=±2 So since either x=2 or x=−2 works, we know that y=5 is in the range of f(x). More generally, if we want to find the full range of y=x 2+1, we can solve for x (taking the inverse of the function) to get x=√y−1. Then, the range of f(x) is simply the domain of √y−1, because these are all the values where there is some x-value with f(x)=y. In this example, the domain of √y−1 is just all values where y−1≥0, so [1,∞) is the range of f(x)=x 2+1. Overall, the steps for algebraically finding the range of a function are: Write down y=f(x) and then solve the equation for x, giving something of the form x=g(y). Find the domain of g(y), and this will be the range of f(x). Note: if there are certain x-values that cannot be in the domain, their associated y-value cannot be in the range! If you can't seem to solve for x, then try graphing the function to find the range. Common functions and their ranges Below, we can list a few common functions and the ranges they have. This will help you find the range of more complicated functions without having to do all the steps above. f(x)=\|x\|. The range of f(x) is [0,∞), which is all non-negative real numbers. f(x)=ln(x). The range of f(x) is (−∞,∞), which is all real numbers. In fact, this range holds for any base for the log. For a>0 and a≠1, f(x)=a x has a range of (0,∞). ReportShare21 Like Related Lessons Domain and Range - Definition & Finding Graphically Finding the Domain of a Function, Algebraically Finding Increasing and Decreasing Intervals Graphically View All Related Lessons Joshua Siktar Text 13 While graphing functions can help us find the range of a function, sometimes it's easier to find the range algebraically, using a different set of techniques and ideas. Recall that the range of a function is all possible values the function (or dataset) can take on. When finding the range of a function algebraically, we usually have to solve at least one equation or inequality. This is because a number y is in the range of the function f only if there is a value x such that f(x)=y (and any such x must be in the domain of f). Example: Let's say we want to find the range of the function f(x)=\|x\|. You might remember that the absolute value function is always nonnegative. For instance, if we plug a positive number into the function then f returns that number; if we plug a negative number into the function then f removes the negative sign. And finally, f(0)=\|0\|=0. But this alone doesn't tell us that the range of \|x\| is [0,∞). We need to show that every nonnegative number is the absolute value of some number. To do this, let's pick a random nonnegative number y, and then we want to show that there is a real number x such that y=\|x\|. However, a lot of the hard work is already done from our understanding of the absolute value function. We know that if we plug a nonnegative number into the absolute value function, we get the same number back. Therefore we can set x=y since y is nonnegative, and have \|y\|=y. In other words, we can "access" any nonnegative number by using the absolute value function, so the range of f(x)=\|x\| is [0,∞). That problem was relatively easy because we have a lot of intuition on how the absolute value function works already. As a reenforcement, answer this question. Which value(s) of x give us f(x)=12 if f(x)=\|x\|? [x] 12 [x] −12 [x] 11 [x] −11 [x] There are no such values. Show SolutionCheck ReportShare13 Like Anusha Rahman Video 1 (Video) Finding the Range of a Function Algebraically by Anusha Rahman In this lesson, you will how to find the range of a function algebraically. Summary The range of a function is all possible y values that a function can have. We can also think of range as the output values of a function. There are four steps to algebraically find the range of a function: 1) Define the function in standard form, with "y = " 2) Express x in terms of y 3) Find all possible values where x=f(x) cannot be defined. 4) Write the range, being mindful of the values of y that the function cannot take. ReportShare1 Like Anusha Rahman Text 1 Finding Function Range: Algebraically The find the range of a function algebraically, you first need to understand the values the function can output. Image source: by Anusha Rahman For example, if you have a function like f(x)=x 2, the output is always a positive number or zero because squaring any number gives a positive result or zero. To find the range, think about the smallest and largest values the function can product (the output). You could use inequalities or solve for y to figure out which values are possible for the function's output. 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10388	https://openstax.org/books/university-physics-volume-1/pages/17-7-the-doppler-effect	Skip to Content Go to accessibility page Keyboard shortcuts menu Log in University Physics Volume 1 17.7 The Doppler Effect University Physics Volume 1 17.7 The Doppler Effect Search for key terms or text. Learning Objectives By the end of this section, you will be able to: Explain the change in observed frequency as a moving source of sound approaches or departs from a stationary observer Explain the change in observed frequency as an observer moves toward or away from a stationary source of sound The characteristic sound of a motorcycle buzzing by is an example of the Doppler effect. Specifically, if you are standing on a street corner and observe an ambulance with a siren sounding passing at a constant speed, you notice two characteristic changes in the sound of the siren. First, the sound increases in loudness as the ambulance approaches and decreases in loudness as it moves away, which is expected. But in addition, the high-pitched siren shifts dramatically to a lower-pitched sound. As the ambulance passes, the frequency of the sound heard by a stationary observer changes from a constant high frequency to a constant lower frequency, even though the siren is producing a constant source frequency. The closer the ambulance brushes by, the more abrupt the shift. Also, the faster the ambulance moves, the greater the shift. We also hear this characteristic shift in frequency for passing cars, airplanes, and trains. The Doppler effect is an alteration in the observed frequency of a sound due to motion of either the source or the observer. Although less familiar, this effect is easily noticed for a stationary source and moving observer. For example, if you ride a train past a stationary warning horn, you will hear the horn’s frequency shift from high to low as you pass by. The actual change in frequency due to relative motion of source and observer is called a Doppler shift. The Doppler effect and Doppler shift are named for the Austrian physicist and mathematician Christian Johann Doppler (1803–1853), who did experiments with both moving sources and moving observers. Doppler, for example, had musicians play on a moving open train car and also play standing next to the train tracks as a train passed by. Their music was observed both on and off the train, and changes in frequency were measured. What causes the Doppler shift? Figure 17.30 illustrates sound waves emitted by stationary and moving sources in a stationary air mass. Each disturbance spreads out spherically from the point at which the sound is emitted. If the source is stationary, then all of the spheres representing the air compressions in the sound wave are centered on the same point, and the stationary observers on either side hear the same wavelength and frequency as emitted by the source (case a). If the source is moving, the situation is different. Each compression of the air moves out in a sphere from the point at which it was emitted, but the point of emission moves. This moving emission point causes the air compressions to be closer together on one side and farther apart on the other. Thus, the wavelength is shorter in the direction the source is moving (on the right in case b), and longer in the opposite direction (on the left in case b). Finally, if the observers move, as in case (c), the frequency at which they receive the compressions changes. The observer moving toward the source receives them at a higher frequency, and the person moving away from the source receives them at a lower frequency. Figure 17.30 Sounds emitted by a source spread out in spherical waves. (a) When the source, observers, and air are stationary, the wavelength and frequency are the same in all directions and to all observers. (b) Sounds emitted by a source moving to the right spread out from the points at which they were emitted. The wavelength is reduced, and consequently, the frequency is increased in the direction of motion, so that the observer on the right hears a higher-pitched sound. The opposite is true for the observer on the left, where the wavelength is increased and the frequency is reduced. (c) The same effect is produced when the observers move relative to the source. Motion toward the source increases frequency as the observer on the right passes through more wave crests than she would if stationary. Motion away from the source decreases frequency as the observer on the left passes through fewer wave crests than he would if stationary. We know that wavelength and frequency are related by where v is the fixed speed of sound. The sound moves in a medium and has the same speed v in that medium whether the source is moving or not. Thus, f multiplied by is a constant. Because the observer on the right in case (b) receives a shorter wavelength, the frequency she receives must be higher. Similarly, the observer on the left receives a longer wavelength, and hence he hears a lower frequency. The same thing happens in case (c). A higher frequency is received by the observer moving toward the source, and a lower frequency is received by an observer moving away from the source. In general, then, relative motion of source and observer toward one another increases the received frequency. Relative motion apart decreases frequency. The greater the relative speed, the greater the effect. The Doppler effect occurs not only for sound, but for any wave when there is relative motion between the observer and the source. Doppler shifts occur in the frequency of sound, light, and water waves, for example. Doppler shifts can be used to determine velocity, such as when ultrasound is reflected from blood in a medical diagnostic. The relative velocities of stars and galaxies is determined by the shift in the frequencies of light received from them and has implied much about the origins of the universe. Modern physics has been profoundly affected by observations of Doppler shifts. Derivation of the Observed Frequency due to the Doppler Shift Consider two stationary observers X and Y in Figure 17.31, located on either side of a stationary source. Each observer hears the same frequency, and that frequency is the frequency produced by the stationary source. Figure 17.31 A stationary source sends out sound waves at a constant frequency with a constant wavelength at the speed of sound v. Two stationary observers X and Y, on either side of the source, observe a frequency , with a wavelength Now consider a stationary observer X with a source moving away from the observer with a constant speed (Figure 17.32). At time , the source sends out a sound wave, indicated in black. This wave moves out at the speed of sound v. The position of the sound wave at each time interval of period is shown as dotted lines. After one period, the source has moved and emits a second sound wave, which moves out at the speed of sound. The source continues to move and produce sound waves, as indicated by the circles numbered 3 and 4. Notice that as the waves move out, they remained centered at their respective point of origin. Figure 17.32 A source moving at a constant speed away from an observer X. The moving source sends out sound waves at a constant frequency with a constant wavelength , at the speed of sound v. Snapshots of the source at an interval of are shown as the source moves away from the stationary observer X. The solid lines represent the position of the sound waves after four periods from the initial time. The dotted lines are used to show the positions of the waves at each time period. The observer hears a wavelength of . Using the fact that the wavelength is equal to the speed times the period, and the period is the inverse of the frequency, we can derive the observed frequency: As the source moves away from the observer, the observed frequency is lower than the source frequency. Now consider a source moving at a constant velocity moving toward a stationary observer Y, also shown in Figure 17.32. The wavelength is observed by Y as Once again, using the fact that the wavelength is equal to the speed times the period, and the period is the inverse of the frequency, we can derive the observed frequency: When a source is moving and the observer is stationary, the observed frequency is 17.18 where is the frequency observed by the stationary observer, is the frequency produced by the moving source, v is the speed of sound, is the constant speed of the source, and the top sign is for the source approaching the observer and the bottom sign is for the source departing from the observer. What happens if the observer is moving and the source is stationary? If the observer moves toward the stationary source, the observed frequency is higher than the source frequency. If the observer is moving away from the stationary source, the observed frequency is lower than the source frequency. Consider observer X in Figure 17.33 as the observer moves toward a stationary source with a speed . The source emits a tone with a constant frequency and constant period The observer hears the first wave emitted by the source. If the observer were stationary, the time for one wavelength of sound to pass should be equal to the period of the source Since the observer is moving toward the source, the time for one wavelength to pass is less than and is equal to the observed period At time the observer starts at the beginning of a wavelength and moves toward the second wavelength as the wavelength moves out from the source. The wavelength is equal to the distance the observer traveled plus the distance the sound wave traveled until it is met by the observer: Figure 17.33 A stationary source emits a sound wave with a constant frequency , with a constant wavelength moving at the speed of sound v. Observer X moves toward the source with a constant speed , and the figure shows the initial and final position of observer X. Observer X observes a frequency higher than the source frequency. The dotted lines show the position of the waves at . The solid lines show the position of the waves at . If the observer is moving away from the source (Figure 17.34), the observed frequency can be found: Figure 17.34 A stationary source emits a sound wave with a constant frequency , with a constant wavelength moving at the speed of sound v. Observer Y moves away from the source with a constant speed , and the figure shows initial and final position of the observer Y. Observer Y observes a frequency lower than the source frequency. The dotted lines show the position of the waves at . The solid lines show the position of the waves at . The equations for an observer moving toward or away from a stationary source can be combined into one equation: where is the observed frequency, is the source frequency, is the speed of sound, is the speed of the observer, the top sign is for the observer approaching the source and the bottom sign is for the observer departing from the source. Equation 17.18 and Equation 17.19 can be summarized in one equation (the top sign is for approaching) and is further illustrated in Table 17.4: \| Doppler shift \| Stationary observer \| Observer moving towards source \| Observer moving away from source \| --- --- \| \| Stationary source \| \| \| \| \| Source moving towards observer \| \| \| \| \| Source moving away from observer \| \| \| \| Table 17.4 where is the observed frequency, is the source frequency, is the speed of sound, is the speed of the observer, is the speed of the source, the top sign is for approaching and the bottom sign is for departing. Interactive The Doppler effect involves motion and a video will help visualize the effects of a moving observer or source. This video shows a moving source and a stationary observer, and a moving observer and a stationary source. It also discusses the Doppler effect and its application to light. Example 17.8 Calculating a Doppler Shift Suppose a train that has a 150-Hz horn is moving at 35.0 m/s in still air on a day when the speed of sound is 340 m/s. (a) What frequencies are observed by a stationary person at the side of the tracks as the train approaches and after it passes? (b) What frequency is observed by the train’s engineer traveling on the train? Strategy To find the observed frequency in (a), we must use because the source is moving. The minus sign is used for the approaching train, and the plus sign for the receding train. In (b), there are two Doppler shifts—one for a moving source and the other for a moving observer. Solution Enter known values into Calculate the frequency observed by a stationary person as the train approaches: Use the same equation with the plus sign to find the frequency heard by a stationary person as the train recedes: Calculate the second frequency: Identify knowns: It seems reasonable that the engineer would receive the same frequency as emitted by the horn, because the relative velocity between them is zero. Relative to the medium (air), the speeds are The first Doppler shift is for the moving observer; the second is for the moving source. Use the following equation: The quantity in the square brackets is the Doppler-shifted frequency due to a moving observer. The factor on the right is the effect of the moving source. Because the train engineer is moving in the direction toward the horn, we must use the plus sign for however, because the horn is also moving in the direction away from the engineer, we also use the plus sign for But the train is carrying both the engineer and the horn at the same velocity, so As a result, everything but cancels, yielding Significance For the case where the source and the observer are not moving together, the numbers calculated are valid when the source (in this case, the train) is far enough away that the motion is nearly along the line joining source and observer. In both cases, the shift is significant and easily noticed. Note that the shift is 17.0 Hz for motion toward and 14.0 Hz for motion away. The shifts are not symmetric. For the engineer riding in the train, we may expect that there is no change in frequency because the source and observer move together. This matches your experience. For example, there is no Doppler shift in the frequency of conversations between driver and passenger on a motorcycle. People talking when a wind moves the air between them also observe no Doppler shift in their conversation. The crucial point is that source and observer are not moving relative to each other. Check Your Understanding 17.9 Describe a situation in your life when you might rely on the Doppler shift to help you either while driving a car or walking near traffic. The Doppler effect and the Doppler shift have many important applications in science and engineering. For example, the Doppler shift in ultrasound can be used to measure blood velocity, and police use the Doppler shift in radar (a microwave) to measure car velocities. In meteorology, the Doppler shift is used to track the motion of storm clouds; such “Doppler Radar” can give the velocity and direction of rain or snow in weather fronts. In astronomy, we can examine the light emitted from distant galaxies and determine their speed relative to ours. As galaxies move away from us, their light is shifted to a lower frequency, and so to a longer wavelength—the so-called red shift. Such information from galaxies far, far away has allowed us to estimate the age of the universe (from the Big Bang) as about 14 billion years. Previous Next Order a print copy Citation/Attribution This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission. Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute OpenStax. 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10389	https://en.wikipedia.org/wiki/Polyadenylation	Jump to content Search Contents (Top) 1 Background on RNA 2 Nuclear polyadenylation 2.1 Function 2.2 Mechanism 2.3 Downstream effects 2.4 Deadenylation 3 Cytoplasmic polyadenylation 4 Alternative polyadenylation 5 Tagging for degradation in eukaryotes 6 In prokaryotes and organelles 7 Evolution 8 History 9 See also 10 References 11 Further reading 12 External links Polyadenylation العربية Bosanski Čeština Deutsch Español فارسی Français Galego 한국어 Bahasa Indonesia Italiano עברית Nederlands 日本語 Polski Português Русский Српски / srpski Svenska ไทย Українська 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 From Wikipedia, the free encyclopedia Addition of adenylic acids to 3' end of mature mRNA Polyadenylation is the addition of a poly(A) tail to an RNA transcript, typically a messenger RNA (mRNA). The poly(A) tail consists of multiple adenosine monophosphates; in other words, it is a stretch of RNA that has only adenine bases. In eukaryotes, polyadenylation is part of the process that produces mature mRNA for translation. In many bacteria, the poly(A) tail promotes degradation of the mRNA. It, therefore, forms part of the larger process of gene expression. The process of polyadenylation begins as the transcription of a gene terminates. The 3′-most segment of the newly made pre-mRNA is first cleaved off by a set of proteins; these proteins then synthesize the poly(A) tail at the RNA's 3′ end. In some genes these proteins add a poly(A) tail at one of several possible sites. Therefore, polyadenylation can produce more than one transcript from a single gene (alternative polyadenylation), similar to alternative splicing. The poly(A) tail is important for the nuclear export, translation and stability of mRNA. The tail is shortened over time, and, when it is short enough, the mRNA is enzymatically degraded. However, in a few cell types, mRNAs with short poly(A) tails are stored for later activation by re-polyadenylation in the cytosol. In contrast, when polyadenylation occurs in bacteria, it promotes RNA degradation. This is also sometimes the case for eukaryotic non-coding RNAs. mRNA molecules in both prokaryotes and eukaryotes have polyadenylated 3′-ends, with the prokaryotic poly(A) tails generally shorter and fewer mRNA molecules polyadenylated. Background on RNA [edit] Further information: RNA and Messenger RNA RNAs are a type of large biological molecules, whose individual building blocks are called nucleotides. The name poly(A) tail (for polyadenylic acid tail) reflects the way RNA nucleotides are abbreviated, with a letter for the base the nucleotide contains (A for adenine, C for cytosine, G for guanine and U for uracil). RNAs are produced (transcribed) from a DNA template. By convention, RNA sequences are written in a 5′ to 3′ direction. The 5′ end is the part of the RNA molecule that is transcribed first, and the 3′ end is transcribed last. The 3′ end is also where the poly(A) tail is found on polyadenylated RNAs. Messenger RNA (mRNA) is RNA that has a coding region that acts as a template for protein synthesis (translation). The rest of the mRNA, the untranslated regions, tune how active the mRNA is. There are also many RNAs that are not translated, called non-coding RNAs. Like the untranslated regions, many of these non-coding RNAs have regulatory roles. Nuclear polyadenylation [edit] Function [edit] In nuclear polyadenylation, a poly(A) tail is added to an RNA at the end of transcription. On mRNAs, the poly(A) tail protects the mRNA molecule from enzymatic degradation in the cytoplasm and aids in transcription termination, export of the mRNA from the nucleus, and translation. Almost all eukaryotic mRNAs are polyadenylated, with the exception of animal replication-dependent histone mRNAs. These are the only mRNAs in eukaryotes that lack a poly(A) tail, ending instead in a stem-loop structure followed by a purine-rich sequence, termed histone downstream element, that directs where the RNA is cut so that the 3′ end of the histone mRNA is formed. Many eukaryotic non-coding RNAs are always polyadenylated at the end of transcription. There are small RNAs where the poly(A) tail is seen only in intermediary forms and not in the mature RNA as the ends are removed during processing, the notable ones being microRNAs. But, for many long noncoding RNAs – a seemingly large group of regulatory RNAs that, for example, includes the RNA Xist, which mediates X chromosome inactivation – a poly(A) tail is part of the mature RNA. Mechanism [edit] \| \| \| Proteins involved: CPSF: cleavage/polyadenylation specificity factor CstF: cleavage stimulation factor PAP: polyadenylate polymerase PABII: polyadenylate binding protein 2 CFI: cleavage factor I CFII: cleavage factor II \| The processive polyadenylation complex in the nucleus of eukaryotes works on products of RNA polymerase II, such as precursor mRNA. Here, a multi-protein complex (see components on the right) cleaves the 3′-most part of a newly produced RNA and polyadenylates the end produced by this cleavage. The cleavage is catalysed by the enzyme CPSF and occurs 10–30 nucleotides downstream of its binding site. This site often has the polyadenylation signal sequence AAUAAA on the RNA, but variants of it that bind more weakly to CPSF exist. Two other proteins add specificity to the binding to an RNA: CstF and CFI. CstF binds to a GU-rich region further downstream of CPSF's site. CFI recognises a third site on the RNA (a set of UGUAA sequences in mammals) and can recruit CPSF even if the AAUAAA sequence is missing. The polyadenylation signal – the sequence motif recognised by the RNA cleavage complex – varies between groups of eukaryotes. Most human polyadenylation sites contain the AAUAAA sequence, but this sequence is less common in plants and fungi. The RNA is typically cleaved before transcription termination, as CstF also binds to RNA polymerase II. Through a poorly understood mechanism (as of 2002), it signals for RNA polymerase II to slip off of the transcript. Cleavage also involves the protein CFII, though it is unknown how. The cleavage site associated with a polyadenylation signal can vary up to some 50 nucleotides. When the RNA is cleaved, polyadenylation starts, catalysed by polyadenylate polymerase. Polyadenylate polymerase builds the poly(A) tail by adding adenosine monophosphate units from adenosine triphosphate to the RNA, cleaving off pyrophosphate. Another protein, PAB2, binds to the new, short poly(A) tail and increases the affinity of polyadenylate polymerase for the RNA. When the poly(A) tail is approximately 250 nucleotides long the enzyme can no longer bind to CPSF and polyadenylation stops, thus determining the length of the poly(A) tail. CPSF is in contact with RNA polymerase II, allowing it to signal the polymerase to terminate transcription. When RNA polymerase II reaches a "termination sequence" (⁵'TTTATT3' on the DNA template and ⁵'AAUAAA3' on the primary transcript), the end of transcription is signaled. The polyadenylation machinery is also physically linked to the spliceosome, a complex that removes introns from RNAs. Downstream effects [edit] The poly(A) tail acts as the binding site for poly(A)-binding protein. Poly(A)-binding protein promotes export from the nucleus and translation, and inhibits degradation. This protein binds to the poly(A) tail prior to mRNA export from the nucleus and in yeast also recruits poly(A) nuclease, an enzyme that shortens the poly(A) tail and allows the export of the mRNA. Poly(A)-binding protein is exported to the cytoplasm with the RNA. mRNAs that are not exported are degraded by the exosome. Poly(A)-binding protein also can bind to, and thus recruit, several proteins that affect translation, one of these is initiation factor-4G, which in turn recruits the 40S ribosomal subunit. However, a poly(A) tail is not required for the translation of all mRNAs. Further, poly(A) tailing (oligo-adenylation) can determine the fate of RNA molecules that are usually not poly(A)-tailed (such as (small) non-coding (sn)RNAs etc.) and thereby induce their RNA decay. Deadenylation [edit] In eukaryotic somatic cells, the poly(A) tails of most mRNAs in the cytoplasm gradually get shorter, and mRNAs with shorter poly(A) tail are translated less and degraded sooner. However, it can take many hours before an mRNA is degraded. This deadenylation and degradation process can be accelerated by microRNAs complementary to the 3′ untranslated region of an mRNA. In immature egg cells, mRNAs with shortened poly(A) tails are not degraded, but are instead stored and translationally inactive. These short tailed mRNAs are activated by cytoplasmic polyadenylation after fertilisation, during egg activation. In animals, poly(A) ribonuclease (PARN) can bind to the 5′ cap and remove nucleotides from the poly(A) tail. The level of access to the 5′ cap and poly(A) tail is important in controlling how soon the mRNA is degraded. PARN deadenylates less if the RNA is bound by the initiation factors 4E (at the 5′ cap) and 4G (at the poly(A) tail), which is why translation reduces deadenylation. The rate of deadenylation may also be regulated by RNA-binding proteins. Additionally, RNA triple helix structures and RNA motifs such as the poly(A) tail 3’ end binding pocket retard deadenylation process and inhibit poly(A) tail removal. Once the poly(A) tail is removed, the decapping complex removes the 5′ cap, leading to a degradation of the RNA. Several other proteins are involved in deadenylation in budding yeast and human cells, most notably the CCR4-Not complex. Cytoplasmic polyadenylation [edit] There is polyadenylation in the cytosol of some animal cell types, namely in the germline, during early embryogenesis and in post-synaptic sites of nerve cells. This lengthens the poly(A) tail of an mRNA with a shortened poly(A) tail, so that the mRNA will be translated. These shortened poly(A) tails are often less than 20 nucleotides, and are lengthened to around 80–150 nucleotides. In the early mouse embryo, cytoplasmic polyadenylation of maternal RNAs from the egg cell allows the cell to survive and grow even though transcription does not start until the middle of the 2-cell stage (4-cell stage in human). In the brain, cytoplasmic polyadenylation is active during learning and could play a role in long-term potentiation, which is the strengthening of the signal transmission from a nerve cell to another in response to nerve impulses and is important for learning and memory formation. Cytoplasmic polyadenylation requires the RNA-binding proteins CPSF and CPEB, and can involve other RNA-binding proteins like Pumilio. Depending on the cell type, the polymerase can be the same type of polyadenylate polymerase (PAP) that is used in the nuclear process, or the cytoplasmic polymerase GLD-2. Alternative polyadenylation [edit] Many protein-coding genes have more than one polyadenylation site, so a gene can code for several mRNAs that differ in their 3′ end. The 3’ region of a transcript contains many polyadenylation signals (PAS). When more proximal (closer towards 5’ end) PAS sites are utilized, this shortens the length of the 3’ untranslated region (3' UTR) of a transcript. Studies in both humans and flies have shown tissue specific APA. With neuronal tissues preferring distal PAS usage, leading to longer 3’ UTRs and testis tissues preferring proximal PAS leading to shorter 3’ UTRs. Studies have shown there is a correlation between a gene's conservation level and its tendency to do alternative polyadenylation, with highly conserved genes exhibiting more APA. Similarly, highly expressed genes follow this same pattern. Ribo-sequencing data (sequencing of only mRNAs inside ribosomes) has shown that mRNA isoforms with shorter 3’ UTRs are more likely to be translated. Since alternative polyadenylation changes the length of the 3' UTR, it can also change which binding sites are available for microRNAs in the 3′ UTR. MicroRNAs tend to repress translation and promote degradation of the mRNAs they bind to, although there are examples of microRNAs that stabilise transcripts. Alternative polyadenylation can also shorten the coding region, thus making the mRNA code for a different protein, but this is much less common than just shortening the 3′ untranslated region. The choice of poly(A) site can be influenced by extracellular stimuli and depends on the expression of the proteins that take part in polyadenylation. For example, the expression of CstF-64, a subunit of cleavage stimulatory factor (CstF), increases in macrophages in response to lipopolysaccharides (a group of bacterial compounds that trigger an immune response). This results in the selection of weak poly(A) sites and thus shorter transcripts. This removes regulatory elements in the 3′ untranslated regions of mRNAs for defense-related products like lysozyme and TNF-α. These mRNAs then have longer half-lives and produce more of these proteins. RNA-binding proteins other than those in the polyadenylation machinery can also affect whether a polyadenylation site is used, as can DNA methylation near the polyadenylation signal. In addition, numerous other components involved in transcription, splicing or other mechanisms regulating RNA biology can affect APA. Tagging for degradation in eukaryotes [edit] For many non-coding RNAs, including tRNA, rRNA, snRNA, and snoRNA, polyadenylation is a way of marking the RNA for degradation, at least in yeast. This polyadenylation is done in the nucleus by the TRAMP complex, which maintains a tail that is around 4 nucleotides long to the 3′ end. The RNA is then degraded by the exosome. Poly(A) tails have also been found on human rRNA fragments, both the form of homopolymeric (A only) and heterpolymeric (mostly A) tails. In prokaryotes and organelles [edit] In many bacteria, both mRNAs and non-coding RNAs can be polyadenylated. This poly(A) tail promotes degradation by the degradosome, which contains two RNA-degrading enzymes: polynucleotide phosphorylase and RNase E. Polynucleotide phosphorylase binds to the 3′ end of RNAs and the 3′ extension provided by the poly(A) tail allows it to bind to the RNAs whose secondary structure would otherwise block the 3′ end. Successive rounds of polyadenylation and degradation of the 3′ end by polynucleotide phosphorylase allows the degradosome to overcome these secondary structures. The poly(A) tail can also recruit RNases that cut the RNA in two. These bacterial poly(A) tails are about 30 nucleotides long. In as different groups as animals and trypanosomes, the mitochondria contain both stabilising and destabilising poly(A) tails. Destabilising polyadenylation targets both mRNA and noncoding RNAs. The poly(A) tails are 43 nucleotides long on average. The stabilising ones start at the stop codon, and without them the stop codon (UAA) is not complete as the genome only encodes the U or UA part. Plant mitochondria have only destabilising polyadenylation. Mitochondrial polyadenylation has never been observed in either budding or fission yeast. While many bacteria and mitochondria have polyadenylate polymerases, they also have another type of polyadenylation, performed by polynucleotide phosphorylase itself. This enzyme is found in bacteria, mitochondria, plastids and as a constituent of the archaeal exosome (in those archaea that have an exosome). It can synthesise a 3′ extension where the vast majority of the bases are adenines. Like in bacteria, polyadenylation by polynucleotide phosphorylase promotes degradation of the RNA in plastids and likely also archaea. Evolution [edit] Although polyadenylation is seen in almost all organisms, it is not universal. However, the wide distribution of this modification and the fact that it is present in organisms from all three domains of life implies that the last universal common ancestor of all living organisms, it is presumed, had some form of polyadenylation system. A few organisms do not polyadenylate mRNA, which implies that they have lost their polyadenylation machineries during evolution. Although no examples of eukaryotes that lack polyadenylation are known, mRNAs from the bacterium Mycoplasma gallisepticum and the salt-tolerant archaean Haloferax volcanii lack this modification. The most ancient polyadenylating enzyme is polynucleotide phosphorylase. This enzyme is part of both the bacterial degradosome and the archaeal exosome, two closely related complexes that recycle RNA into nucleotides. This enzyme degrades RNA by attacking the bond between the 3′-most nucleotides with a phosphate, breaking off a diphosphate nucleotide. This reaction is reversible, and so the enzyme can also extend RNA with more nucleotides. The heteropolymeric tail added by polynucleotide phosphorylase is very rich in adenine. The choice of adenine is most likely the result of higher ADP concentrations than other nucleotides as a result of using ATP as an energy currency, making it more likely to be incorporated in this tail in early lifeforms. It has been suggested that the involvement of adenine-rich tails in RNA degradation prompted the later evolution of polyadenylate polymerases (the enzymes that produce poly(A) tails with no other nucleotides in them). Polyadenylate polymerases are not as ancient. They have separately evolved in both bacteria and eukaryotes from CCA-adding enzyme, which is the enzyme that completes the 3′ ends of tRNAs. Its catalytic domain is homologous to that of other polymerases. It is presumed that the horizontal transfer of bacterial CCA-adding enzyme to eukaryotes allowed the archaeal-like CCA-adding enzyme to switch function to a poly(A) polymerase. Some lineages, like archaea and cyanobacteria, never evolved a polyadenylate polymerase. Polyadenylate tails are observed in several RNA viruses, including Influenza A, Coronavirus, Alfalfa mosaic virus, and Duck Hepatitis A. Some viruses, such as HIV-1 and Poliovirus, inhibit the cell's poly-A binding protein (PABPC1) in order to emphasize their own genes' expression over the host cell's. History [edit] Poly(A)polymerase was first identified in 1960 as an enzymatic activity in extracts made from cell nuclei that could polymerise ATP, but not ADP, into polyadenine. 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"PNPase activity determines the efficiency of mRNA 3′-end processing, the degradation of tRNA and the extent of polyadenylation in chloroplasts". The EMBO Journal. 21 (24): 6905–14. doi:10.1093/emboj/cdf686. PMC 139106. PMID 12486011. ^ Portnoy V, Schuster G (2006). "RNA polyadenylation and degradation in different Archaea; roles of the exosome and RNase R". Nucleic Acids Research. 34 (20): 5923–31. doi:10.1093/nar/gkl763. PMC 1635327. PMID 17065466. ^ Yehudai-Resheff S, Portnoy V, Yogev S, Adir N, Schuster G (September 2003). "Domain analysis of the chloroplast polynucleotide phosphorylase reveals discrete functions in RNA degradation, polyadenylation, and sequence homology with exosome proteins". The Plant Cell. 15 (9): 2003–19. doi:10.1105/tpc.013326. PMC 181327. PMID 12953107. ^ Slomovic S, Portnoy V, Schuster G (2008). "Chapter 24 Detection and Characterization of Polyadenylated RNA in Eukarya, Bacteria, Archaea, and Organelles". RNA Turnover in Bacteria, Archaea and Organelles. Methods in Enzymology. Vol. 447. pp. 501–20. doi:10.1016/S0076-6879(08)02224-6. ISBN 978-0-12-374377-0. PMID 19161858. ^ Portnoy V, Evguenieva-Hackenberg E, Klein F, Walter P, Lorentzen E, Klug G, Schuster G (December 2005). "RNA polyadenylation in Archaea: not observed in Haloferax while the exosome polynucleotidylates RNA in Sulfolobus". EMBO Reports. 6 (12): 1188–93. doi:10.1038/sj.embor.7400571. PMC 1369208. PMID 16282984. ^ Portnoy V, Schuster G (June 2008). "Mycoplasma gallisepticum as the first analyzed bacterium in which RNA is not polyadenylated". FEMS Microbiology Letters. 283 (1): 97–103. doi:10.1111/j.1574-6968.2008.01157.x. PMID 18399989. ^ Evguenieva-Hackenberg E, Roppelt V, Finsterseifer P, Klug G (December 2008). "Rrp4 and Csl4 are needed for efficient degradation but not for polyadenylation of synthetic and natural RNA by the archaeal exosome". Biochemistry. 47 (50): 13158–68. doi:10.1021/bi8012214. PMID 19053279. ^ a b Slomovic S, Portnoy V, Yehudai-Resheff S, Bronshtein E, Schuster G (April 2008). "Polynucleotide phosphorylase and the archaeal exosome as poly(A)-polymerases". Biochimica et Biophysica Acta (BBA) - Gene Regulatory Mechanisms. 1779 (4): 247–55. doi:10.1016/j.bbagrm.2007.12.004. PMID 18177749. ^ Poon, Leo L. M.; Pritlove, David C.; Fodor, Ervin; Brownlee, George G. (1 April 1999). "Direct Evidence that the Poly(A) Tail of Influenza A Virus mRNA Is Synthesized by Reiterative Copying of a U Track in the Virion RNA Template". Journal of Virology. 73 (4): 3473–3476. doi:10.1128/JVI.73.4.3473-3476.1999. PMC 104115. PMID 10074205. ^ Wu, Hung-Yi; Ke, Ting-Yung; Liao, Wei-Yu; Chang, Nai-Yun (2013). "Regulation of Coronaviral Poly(A) Tail Length during Infection". PLOS ONE. 8 (7): e70548. Bibcode:2013PLoSO...870548W. doi:10.1371/journal.pone.0070548. PMC 3726627. PMID 23923003.{{cite journal}}: CS1 maint: article number as page number (link) ^ Neeleman, Lyda; Olsthoorn, René C. L.; Linthorst, Huub J. M.; Bol, John F. (4 December 2001). "Translation of a nonpolyadenylated viral RNA is enhanced by binding of viral coat protein or polyadenylation of the RNA". Proceedings of the National Academy of Sciences. 98 (25): 14286–14291. Bibcode:2001PNAS...9814286N. doi:10.1073/pnas.251542798. PMC 64674. PMID 11717411. ^ Chen, Jun-Hao; Zhang, Rui-Hua; Lin, Shao-Li; Li, Peng-Fei; Lan, Jing-Jing; Song, Sha-Sha; Gao, Ji-Ming; Wang, Yu; Xie, Zhi-Jing; Li, Fu-Chang; Jiang, Shi-Jin (2018). "The Functional Role of the 3′ Untranslated Region and Poly(A) Tail of Duck Hepatitis a Virus Type 1 in Viral Replication and Regulation of IRES-Mediated Translation". Frontiers in Microbiology. 9: 2250. doi:10.3389/fmicb.2018.02250. PMC 6167517. PMID 30319572. ^ "Inhibition of host poly(A)-binding protein by virus ~ ViralZone". viralzone.expasy.org. ^ Edmonds, Mary; Abrams, Richard (April 1960). "Polynucleotide Biosynthesis: Formation of a Sequence of Adenylate Units from Adenosine Triphosphate by an Enzyme from Thymus Nuclei". Journal of Biological Chemistry. 235 (4): 1142–1149. doi:10.1016/S0021-9258(18)69494-3. ^ Colgan DF, Manley JL (November 1997). "Mechanism and regulation of mRNA polyadenylation". Genes & Development. 11 (21): 2755–66. doi:10.1101/gad.11.21.2755. PMID 9353246. ^ a b Edmonds, M (2002). A history of poly A sequences: from formation to factors to function. Progress in Nucleic Acid Research and Molecular Biology. Vol. 71. pp. 285–389. doi:10.1016/S0079-6603(02)71046-5. ISBN 978-0-12-540071-8. PMID 12102557. ^ Edmonds, M.; Vaughan, M. H.; Nakazato, H. (1 June 1971). "Polyadenylic Acid Sequences in the Heterogeneous Nuclear RNA and Rapidly-Labeled Polyribosomal RNA of HeLa Cells: Possible Evidence for a Precursor Relationship". Proceedings of the National Academy of Sciences. 68 (6): 1336–1340. Bibcode:1971PNAS...68.1336E. doi:10.1073/pnas.68.6.1336. PMC 389184. PMID 5288383. Further reading [edit] Danckwardt S, Hentze MW, Kulozik AE (February 2008). "3′ end mRNA processing: molecular mechanisms and implications for health and disease". The EMBO Journal. 27 (3): 482–98. doi:10.1038/sj.emboj.7601932. PMC 2241648. PMID 18256699. External links [edit] Media related to Polyadenylation at Wikimedia Commons \| v t e Gene expression \| \| Introductionto genetics \| Genetic code Central dogma + DNA → RNA → Protein Special transfers + RNA→RNA + RNA→DNA + Protein→Protein \| \| Transcription \| \| \| \| --- \| \| Types \| Bacterial Archaeal Eukaryotic \| \| Key elements \| Transcription factor RNA polymerase Promoter \| \| Post-transcription \| Precursor mRNA (pre-mRNA / hnRNA) 5' capping Splicing Polyadenylation Histone acetylation and deacetylation \| \| \| Translation \| \| \| \| --- \| \| Types \| Bacterial Archaeal Eukaryotic \| \| Key elements \| Ribosome Transfer RNA (tRNA) Ribosome-nascent chain complex (RNC) Post-translational modification \| \| \| Regulation \| Epigenetic + imprinting Transcriptional + Gene regulatory network + cis-regulatory element lac operon Post-transcriptional + sequestration (P-bodies) + alternative splicing + microRNA Translational Post-translational + reversible + irreversible \| \| Influential people \| François Jacob Jacques Monod \| \| v t e Post-transcriptional modification \| \| Nuclear \| \| \| \| Precursor mRNA 5′ cap formation Polyadenylation + CPSF + CstF + PAP + PAB2 + CFI + CFII Poly(A)-binding protein RNA editing Polyuridylation \| \| RNA splicing \| \| \| \| Intron / Exon snRNP Spliceosome + minor + U1 Alternative splicing \| \| pre-mRNA factors \| PLRG1 + PRPF3 + PRPF4 + PRPF4B + PRPF6 + PRPF8 + PRPF18 + PRPF19 + PRPF31 + PRPF38A + PRPF38B + PRPF39 + PRPF40A + PRPF40B \| \| \| \| Cytosolic \| 5′ cap methylation mRNA decapping + DCP1A + DCP1B + DCP2 + DCPS + EDC3 + EDC4 \| Retrieved from " Categories: Gene expression Messenger RNA Hidden categories: Wikipedia articles needing page number citations from July 2011 CS1: long volume value CS1 maint: article number as page number Articles with short description Short description is different from Wikidata Good articles Commons category link is on Wikidata Add topic
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10391	https://s1b0d3d77136c1679.jimcontent.com/download/version/1563150249/module/14147538930/name/cargas%20dinamicas.pdf	2.3 TEORÍAS DE FALLAS POR CARGAS DINÁMICAS En su mayoría, las fallas en las máquinas se deben a cargas que varían con el tiempo y no a cargas estáticas. Estas fallas suelen ocurrir a niveles de esfuerzo muy por debajo del límite elástico de los materiales. Por lo tanto, de manejar solo la teoría de fallas estáticas, puede llevar a diseños pocos seguros cuando las cargas sean dinámicas. HISTORIA DE LAS FALLAS POR FATIGA Este fenómeno se observó por primera vez en los años 1800, cuando empezaron a fallar los ejes de los carros de ferrocarril después de solo poco tiempo de servicio. Estaban fabricados de acero dúctil, pero mostraban fallas súbitas de tipo frágil. Para dar una respuesta a este fenómeno en 1843 Rankine postuló que el material se había cristalizado y hecho frágil debido a los esfuerzos fluctuantes. Las cargas dinámicas eran entonces un fenómeno nuevo, resultado de la introducción de maquinaria movida por vapor. Estos ejes estaban fijos a las ruedas y giraban juntos con ellas. Por lo que el esfuerzo a flexión en cualquier punto de la superficie del eje variaba cíclicamente de positivo a negativo. A A Tiempo Esfuerzo + – 2.3.1 CONCEPTO DE FATIGA: La falla por fatiga ocurre debido a esfuerzos pequeños, los cuales se van haciendo cada vez mayores a medida que las grietas alcanzan una longitud crítica, la fatiga siempre es una preocupación donde se presenten esfuerzos cíclicos. La vida total de un componente o estructura representa el tiempo que le toma a una grieta para comenzar más el tiempo que necesita para propagarse por la sección transversal. Por esto, la vida útil se maximiza cuando: 1. Minimiza los efectos iniciales, especialmente los defectos de superficie: se debe maquinar o pulir las superficies, para dejarlas lisas, luego hay que protegerlas para ponerlas en servicio. 2. Maximizando el tiempo de iniciación: los esfuerzos superficiales se presentan por una variedad de tratamientos superficiales, como el bruñido o el gramallado. 3. Maximizando el tiempo de propagación: es importante la característica del sustrato, especialmente las que retardan el crecimiento de la grieta. Por ejemplo, las grietas por fatiga se propagan más rápido a lo largo de los límites de las fronteras reticulares que a través de los granos (por que la disposición de los granos tiene un empaque atómico más eficiente). 4. Maximizando la longitud crítica de la grieta: la tenacidad a la fractura es un ingrediente esencial. F F Fatiga Ruptura Hierro fundido Hierro trabajado en frío ESFUERZOS CICLICOS: El esfuerzo cíclico es una función del tiempo; pero la variación es tal que la secuencia del esfuerzo se repite. Los esfuerzos son axiales (de torsión o compresión), de flexión (flexionante) o de torsión (torcedura), en la siguiente figura se muestra la variación cíclica del esfuerzo medio diferente de cero con el tiempo. También se indican varios parámetros para caracterizar el esfuerzo cíclico fluctuante. Un ciclo de esfuerzo (Nc =1) constituye una sola aplicación y remoción de una carga y luego, otra aplicación y remoción de la carga en la dirección opuesta. La amplitud del esfuerzo es alterna respecto a un esfuerzo medio. Esfuerzo medio (σ σ σ σm): se define como el promedio de los esfuerzos máximo y mínimo en el ciclo, o: 2 MIN MAX m σ σ σ σ + + + + σ σ σ σ = = = = σ σ σ σ Rango de esfuerzo (σ σ σ σr): es la diferencia entre el esfuerzo mayor y el menor, es decir: MIN MAX r σ σ σ σ − − − − σ σ σ σ = = = = σ σ σ σ Amplitud del esfuerzo (σ σ σ σa): es la mitad del rango de esfuerzo, o: 2 2 MIN MAX r a σ σ σ σ − − − − σ σ σ σ = = = = σ σ σ σ = = = = σ σ σ σ Relación del esfuerzo (Rs): es la relación de las amplitudes del esfuerzo mínimo y máximo, o: MAX MIN S R σ σ σ σ σ σ σ σ = = = = Relación de la amplitud: Es la relación entre la amplitud del esfuerzo y el esfuerzo medio, o: ( ( ( ( ) ) ) ) ( ( ( ( ) ) ) ) S S MIN MAX MIN MAX m a a R 1 R 1 A + + + + − − − − = = = = σ σ σ σ + + + + σ σ σ σ σ σ σ σ − − − − σ σ σ σ = = = = σ σ σ σ σ σ σ σ = = = = σ σ σ σMAX σ σ σ σMIN 0 σ σ σ σa σ σ σ σm σ σ σ σr 1 Ciclo Tiempo Esfuerzo Compresión (-) Tensión (+) LIMITE DE FATIGA O LIMITE DE RESISTENCIA A LA FATIGA La determinación experimental de los límites de resistencia a la fatiga es ahora un procedimiento de rutina, aunque extenso y costoso. En términos generales, se prefiere realizar ensayos de esfuerzos a ensayos de deformación para determinar límites de fatiga. Se han analizado muchos datos de prueba reales, provenientes de varias fuentes y se ha concluido que el límite a la resistencia a la fatiga, o de fatiga, puede estar relacionado con la resistencia a la tensión ( ( ( ( ) ) ) )                     > > > > > > > > ≤ ≤ ≤ ≤ = = = = MPa 1400 S MPa 700 Ksi 200 S Ksi 100 MPa 1400 Ksi 200 S S 504 . 0 S ut ut ut ut ' e Donde: Sut = Resistencia mínima a la tensión Se’ = Límite a la resistencia a la fatiga Los limites de resistencia a la fatiga para diversos hierros se dan en la tabla A–24. RESISTENCIA A LA FATIGA Para desarrollar un enfoque analítico, de la ecuación de la recta S–N será: Supóngase que se tiene un esfuerzo completamente invertido σ σ σ σa el número de ciclos de duración correspondiente a este esfuerzo puede determinarse a partir de St = aNb sustituyendo σ σ σ σa por Sf. b 1 a a N                                σ σ σ σ = = = = donde: Sf = Resistencia a la fatiga. a = constante N = # Ciclos. b = Constante ( ( ( ( ) ) ) ) e ut e 2 ut b f S S 9 . 0 log 3 1 b S S 9 . 0 a aN S − − − − = = = = = = = = = = = = LA TEORIA DE LA FATIGA DURANTE LA DEFORMACIÓN UNITARIA La fatiga es el proceso de la acumulación de daño que se manifiesta mediante la propagación de grietas; aunque la propagación de grietas no es posible sin una deformación plástica en el extremo de la grieta. Usar las propiedades de los materiales, como la resistencia a la fluencia o a la rotura, presenta dificultades porque las cargas cíclicas cambian estos valores cerca del extremo de la grieta, las cuales se incrementan o disminuyen dependiendo del material y de su proceso de manufactura. Dadas las dificultades al expresar las resistencias del material cerca de los extremos de las grietas, se han sugerido varios enfoques para abordar la deformación unitaria en tales extremos. Dadas las dificultades al expresar las resistencias del material cerca de los extremos de las grietas, se han sugerido varios enfoques para abordar la deformación unitaria en tales extremos. Una de las ecuaciones mejor conocidas es la relación Manson-Coffin, que proporciona la amplitud total de la deformación unitaria como la suma de las amplitudes de la deformación unitaria elástica y plástica: ( ( ( ( ) ) ) ) ( ( ( ( ) ) ) ) α α α α ε ε ε ε + + + + σ σ σ σ = = = = ε ε ε ε ∆ ∆ ∆ ∆ ' N 2 ' N 2 E 2 ' f a ' f Donde: ∆ ∆ ∆ ∆ε ε ε ε = Deformación unitaria total σ σ σ σf’ = Esfuerzo a la fractura en un ciclo de esfuerzo, Pa. E = Módulo de elasticidad del materia, Pa. N’ = Número de ciclos antes de la falla ε ε ε εf’ = Coeficiente de ductilidad a la fatiga (deformación unitaria verdadera que corresponde a la fractura en un ciclo de esfuerzo). a = Exponente de resistencia a la fatiga. α α α α = Exponente de ductilidad a la fatiga. FACTORES QUE MODIFICAN EL LÍMITE A LA FATIGA En los experimentos a la fatiga, se suponen las mejores circunstancias para la obtención de vidas largas a la fatiga. No obstante, dicha situación no se garantiza en aplicaciones de diseño, así que el límite a la fatiga del componente se debe modificar o reducir a partir del límite a la fatiga del material en el mejor de los casos. Los factores de modificación que vamos a analizar aquí serán para cargar completamente alternante (σm = 0 ). El limite a la fatiga modificado se expresa como: ' e m t r s f e S · K · K · K · K · K S = = = = Donde: Se = Límite de fatiga modificado Kf = Factor de acabado de superficie. Ks = Factor de tamaño Kr = Factor de confiabilidad Kt = Factor de temperatura Km = Factor diverso S’e = Límite a la fatiga EFECTO DE LA CONCENTRACIÓN DE ESFUERZOS. Como los sitios de concentración de esfuerzo también son sitios de concentraciones de deformaciones unitarias, se consideran como los principales causantes de la iniciación y crecimiento de grietas, por fatiga. Para carga estática se usa el factor de esfuerzos Kc, y para cargas por fatiga, el factor de esfuerzos por fatiga Kf, donde muesca con especimen para fatiga la a Límite muesca sin especimen para fatiga la a Límite K f = = = = Una muesca o concentración de esfuerzo puede ser un agujero, un filete o una acanaladura. Hay que recordar que el factor de concentración de esfuerzos estáticos (Kc) es solo función de la geometría de la pieza. Ahora, el factor de concentración de esfuerzos por fatiga también es función de las propiedades del material y del tipo de carga. La consideración del material con frecuencia se aborda usando un factor de sensibilidad a la muesca qn, el cual se define como: : o 1 K 1 K q c f n − − − − − − − − = = = = ( ( ( ( ) ) ) ) 1 K q 1 K c n f − − − − + + + + = = = = Hay que observar que el rango de qn esta entre cero y uno. 1 q 0 n ≤ ≤ ≤ ≤ < < < < Existen figuras que muestran la sensibilidad de la muesca contra el radio de la misma, para materiales comúnmente usados con varios tipos de carga. En todos los materiales incluidos. La sensibilidad a la muesca se aproxima a cero cuando el radio de la muesca se aproxima a cero. Así mismo, los aceros más duros y más resistentes suelen ser más sensibles a la muesca (valor grande de qn). Esto no parece sorprendente puesto que la sensibilidad a la muesca es una medida de la ductilidad del material. El factor de modificación del límite a la fatiga, considerando las concentraciones de esfuerzos, es: f 0 K 1 K = = = = FACTOR DE ACABADO DE SUPERFICIE: Tomando en consideración el siguiente dibujo: Suponiendo que tiene un acabado altamente pulido, con un pulido axial que alisa las estrías circunferenciales. Usualmente los elementos de máquinas no presentan un acabado de alta calidad. El factor de modificación para incorporar el efecto de acabado depende del proceso usado en la generación de la superficie, y la resistencia a la rotura. Una vez que se conoce ese proceso, con base en la figura 7–7a se obtiene el factor de acabado de la superficie, cuando se conoce la resistencia a la rotura a la tensión (Sut). De otra forma, utilizar los coeficientes de la siguiente tabla junto con la ecuación: FACTOR e PROCESO DE MANUFACTURA MPa Ksi EXPONENTE f Esmerilado 1.58 1.34 – 0.085 Maquinado o estirado en frío 4.51 2.70 – 0.265 Laminado en caliente 57.7 14.4 – 0.718 Ninguno (como sale de forja) 272 39.9 – 0.995 f ut f S · e K = = = = Donde: Kf = Factor del acabado de superficie. Sut = Resistencia a la rotura a la tensión del material. e y f = Coeficientes definidos en la tabla. FACTOR DE TAMAÑO En los materiales alargados por extrusión, presentan un alargamiento del grano pronunciado en la dirección opuesta al crecimiento de la grieta por fatiga. El grado de trabajo en frío es alta y la probabilidad de grandes defectos es baja (cerámicas y fundiciones). Sin considerar el proceso de manufactura, las partes más grandes son las más probables de contener defectos. El factor de tamaño para una barra redonda se afecta por el método de carga. Para flexión o torsión, el factor de tamaño es:                     ≤ ≤ ≤ ≤ < < < < ≤ ≤ ≤ ≤ < < < < < < < < < < < < = = = = − − − − − − − − mm 250 d mm 8 mm 8 d ó in 3 . 0 d in 10 d in 3 . 0 d 189 . 1 1 d 869 . 0 K 112 . 0 112 . 0 S Para carga axial KS = 1 NOTA: el diámetro d es el diámetro de la sección transversal del componente. En caso de que la sección transversal no sea circular, a la geometría en cuestión se transforma el área transversal en un área de sección transversal equivalente circular. FACTOR DE CONFIABILIDAD En la siguiente tabla se muestra el factor de confiabilidad de varios porcentajes de la probabilidad de supervivencia, esta tabla se basa en el límite a la fatiga con una desviación estándar de 8%, generalmente el límite superior para aceros. Estos datos se consideran una guía porque la distribución real es más complicada que el enfoque simple que se utilizó para realizar dicha tabla. PROBABILIDA DE SOBREVIVENCIA % FACTOR DE CONFIABILIDAD, Kr 50 1.00 90 0.90 95 0.87 99 0.82 99.9 0.75 99.99 0.70 FACTOR DE TEMPERATURA Muchas aplicaciones de fatiga de alto ciclaje ocurren bajo temperaturas extremadamente altas, como los motores de las aeronaves, donde el material puede ser mucho más débil que a temperatura ambiente. A la inversa, en ciertas aplicaciones, como los ejes de los automóviles en Alaska durante enero, el material resulta menos dúctil que a temperatura ambiente. Debido a esto se pueden tomar las siguientes alternativas: 1. Modificar la resistencia a la rotura del material con base en sus propiedades a cierta temperatura, antes de determinar el límite a la fatiga de un material. 2. Usar un factor de temperatura. ) ref ( ut ut t S S K = = = = Donde: Kt = Factor de temperatura. Sut = Resistencia a la rotura por tensión del material a la temperatura deseada. Sut(ref) = Resistencia a la rotura por tensión a la temperatura de referencia, usualmente la temperatura ambiente. EFECTOS DIVERSOS: Muchos fenómenos pueden afectar la resistencia de un componente, aunque ya se analizaron algunos métodos para calcular numéricamente algunos efectos, también otras consideraciones obstaculizan su cuantificación, como son: 1. Procesos de manufactura: Estos tienen una función primordial en la determinación de las características de la vida a la fatiga de los materiales de ingeniería, esto se manifiesta en el factor de tamaño. El crecimiento de las grietas por fatiga es más rápido a través de la frontera de grano que a través de ellos, cualquier proceso de manufactura que afecte al tamaño del grano también afecta a la fatiga (laminado, extrusión, estirado). 2. Esfuerzos residuales: Es causado por la recuperación elástica después de la deformación plástica no uniforme, a través del espesor de un componente. El esfuerzo residual de compresión retarda el crecimiento de la fractura, en cambio el esfuerzo residual a tensión promueve el crecimiento de grietas. granallado recocido forjado extrusión laminado Esfuerzos Residuales de compresión Forjado Extrusión Laminado Corte Rectificación Esfuerzos residuales de tensión 3. Recubrimientos: Algunas operaciones como la carburización, conducen a un alto contenido de carbono en las capas superficiales del acero (resistencia a la fractura elevada) e imparten un esfuerzo residual compresivo en la superficie, los revestimientos aplicados a altas temperaturas, como los procesos de deposición química de vapor o de inmersión en caliente, pueden inducir térmicamente esfuerzos residuales de tensión en la superficie. 4. Corrosión: Las causas principales de corrosión en los metales son el hidrógeno y el oxígeno. El primero fragiliza el material porque agrega esfuerzos de tensión en el extremo de la grieta, el segundo causa la formación de revestimientos frágiles o porosos, fomentando la iniciación y crecimiento de las grietas. Para fatiga de bajo ciclaje consideraremos: Se’ = 0.5 Su (Flexión) Se’ = 0.45 Su (Axial) Se’ = 0.29 Su (Torsion) Resistencia a la fatiga
10392	https://domainbrand.cn/%E6%95%B0%E6%8D%AE%E6%B3%A2%E5%8A%A8%E4%B8%8E%E7%A8%B3%E5%AE%9A%E6%80%A7%EF%BC%9A%E6%96%B9%E5%B7%AE%E4%B8%8E%E6%A0%87%E5%87%86%E5%B7%AE%E7%9A%84%E5%8C%BA%E5%88%AB%E8%A7%A3%E6%9E%90/	数据波动与稳定性：方差与标准差的区别解析 \| 云知识网首页云服务器云数据库安全产品建站教程解决方案大模型云知识网首页解决方案专家实现当前方案扫码加我数据波动与稳定性：方差与标准差的区别解析 admin• 2025年3月16日下午1:09 •解决方案•阅读 4 阿里云代金券通用优惠券购买阿里云任意产品必领优惠福利立即领取热门优惠活动 2核4G199元适用Web前端、企业级应用场景立即查看数据波动与稳定性：方差与标准差的区别解析在大数据分析与应用中，衡量一组数据的稳定性和波动性非常重要。这两个指标直接关系到数据分析的结果准确度与应用效果。而谈到数据的稳定性和波动性时，有两个关键术语是我们绕不开的：方差与标准差。本文旨在深入浅出地为读者解读这两种统计学概念背后的原理及其应用场合，通过对比二者异同，帮助各位更好地利用手中握有的数据资源解决实际问题。基本概念：理解什么是方差与标准差？首先，从定义来看： – 方差是测量一组数字分散程度的一种方式。简单来说，如果一组数值相差越大，则它们之间的方差就越高；相反地，如果大多数数值都较为接近，则整体上的差异就会比较小。 – 标准差则是基于方差进行求根计算得出的一个数值单位下的偏离值大小。它可以被视作一种绝对规模上用来描述变异量的方法。相对于抽象化的平方形式（方差），使用原始单位来表达变化程度更为直观易懂。计算方法：手把手教你如何操作下面以具体实例演示计算步骤：例题1: 给定以下一组测试分数 [85, 92, 89, 90, 88] (a) 首先需要求得该样本集合的均数(mean) M=(ΣX)/N ≈ 89 (b) 再依据每一点离平均数多远，用偏差平方和SSQ=∑(Xi-M)² 算得 53.2 （c）接下来计算总和除以n（若整体估计用总体方差的话应为 n-1 作为修正自由度，但对于小样本这里直接采用简化处理） Variance(V)=SSQ/n ≈ 10.64 即该组成绩的方差。（d）最后得到 Standard Deviation(SD)=sqrt(variance) ≈ 3.26 分。图表一: 不同类型下随机生成的一组数对应用场景举例说明: 为了更好地阐述方差标准差之间的细微差别，我们引用云计算领头羊企业阿里云内部服务器性能监控的真实案例。比如在其某业务线的服务端口访问请求记录中发现，在相同时间段内两个节点(Node A, Node B)的表现却天差地别： \| 指标 \| 节点A \| 节点B \| \|——\|——–\|——–\| \| 访问次数 \| 700 \| 800 \| \| 总处理时间(mins)\| 117.2 \| 93.6 \| \| 处理成功率 \| 96% \| 99.5% \| 显而易见的是虽然从表面上看起来二者并没有显著区别，但当我们进一步计算后发现：尽管A的绝对数量稍显劣势但其表现更加稳定可控，因为它拥有更低的整体变异性水平——这意味着无论何时发起连接，用户都有理由相信能获取到一致良好的服务质量；另一方面B虽然看似高效快速，但是其背后存在着潜在波动过大导致部分时段出现延迟异常的风险隐患。因此对于追求高可靠性要求的企业而言更倾向于选用相对保守且可预见性较强的选项。（这里涉及到的参数设置等均为虚构内容仅做教学用途）深入剖析两者的关系及相互作用规律: 当我们将视野放宽来看待整个数学体系，你会发现很多有趣的现象，其中一个很直观的特点就在于：随着观察者选取参考框架的变化，同样的数据集可能会体现出完全不一样的属性特性。同样对于我们的核心话题讨论对象——方差和标准差来讲也是类似道理。前文提到了“方差是一个反映离散趋势的标准”，然而当你将其转化为另一种表现形式即取算术根后的“标准误差/标准离差”的形态下便能够让我们看到它另外的一面：在大多数情况下我们可以把前者视为后者的基础构建材料之一种形式。即 S=sqrt(var)，这表示当我们掌握了其中任何一方的信息时候都能够很容易推断另一份资料的内容。除此之外值得注意的是由于两者都是非负性的特征因此在具体实践中还存在某些特殊条件下的边界情形限制问题。比如当所有数据项之间彼此相等情况发生概率非常微弱甚至几乎不可能达到之时（理论意义上唯一满足该要求的状态仅发生在每个成员皆恰好落在正中央位置这一极其罕见的理想状态下）。实质上这就给我们提供了一个强有力工具箱用于探索更深层次领域问题并寻找最适合自己业务特性的个性化设计方案。例如借助阿里巴巴提供的DataWorks平台中的可视化组件就可以快速搭建相关分析模型帮助企业挖掘数据潜力提高运营效率；又或者依托OSS+CDN组合技术手段优化用户体验增强网站粘性留住客户群体；还有结合MaxCompute强大处理能力支持千万级大规模分布式计算任务轻松解决复杂逻辑运算难题。图表二: 两个具有相同期望回报不同风险等级投资项目的历史收益分布曲线结语综合上述所有要点归纳起来看：虽然看上去只是小小的一串公式符号组合但却蕴含着丰富而深刻的思想精髓——既包括了传统经典概率论里关于事件独立性条件判断规则的知识点又融合吸收借鉴了现代统计决策理论框架下的贝叶斯更新机制等先进理念；同时还兼顾实用性考虑使得普通开发者也能够在日常编程调试过程中轻松上手灵活运用发挥出意想不到的好奇成果。对于正在从事相关技术研发的朋友来说，无论是希望加强自身专业素养学习进阶课程也好还是打算转型向其他方向开拓更多可能也好，只要善于总结归纳经验规律并且勇于尝试突破创新，终将收获满满！原创文章，数据波动与稳定性：方差与标准差的区别解析作者：logodiffusion.cn，如若转载，请注明出处：云计算性能监控指标大数据分析工具数据稳定性分析方法方差标准差应用案例统计学概念解析赞 (0) admin 生成海报 “技术 stranger的机遇：探索Magic的无限可能” 上一篇 2025年3月16日下午12:55 精准解析：科技领域的关键分析工具精准解读：技术与数据之间的深层关系精准模拟：人工智能驱动的未来预测精准管理：现代科技资源高效配置精准应用：遍历科技领域的种种工具精准设计：结合技术与艺术的创新未来精准创造：未来科技产业的Drivers 精准趋势：解读未来科技方向的趋势报告精准 noted：科技发展中的重要术语概述精准解析：数理与科技结合的/heory 下一篇 2025年3月16日下午1:40 相关推荐解决方案 ### 网络红人背后的科技魔法网络红人背后的科技魔法在这个数字化的时代，互联网已经深深地嵌入到了我们的生活中。社交媒体上的各类内容创作者如雨后春笋般涌现，他们通过视频、文字乃至音频的形式吸引着无数观众的关注。… 2025年5月12日 0 0 1 解决方案 ### 芯片性能优化与未来发展趋势解析芯片性能优化与未来发展趋势解析在当今科技飞速发展的时代，芯片性能优化已经成为一个热门话题。无论是智能手机、云计算、人工智能还是自动驾驶，高性能的芯片是其核心竞争力的关键所在。本文… 2025年5月9日 0 0 2 解决方案 ### “技术特性全解析： exploring新_angle下的系统优化策略” 技术特性全解析：Exploring新_Angle下的系统优化策略在这个数字化时代，系统优化已经成为了提升企业竞争力的关键之一。本文将通过对最新技术的研究，深入探讨系统优化的多种策… 2025年3月27日 0 0 1 解决方案 ### 流形之布料的秘密：专业科技 bloggers的流畅解读流形之布料的秘密：专业科技博主的流畅解读流形（Manifold）在现代技术和科学中是一个充满神秘魅力的概念，它既是数学上的一个重要对象，也逐渐渗透到材料科学、计算机科学和人工智能… 2025年3月19日 0 0 2 解决方案 ### 高德纳箭头符号背后的塔函数解析高德纳箭头符号背后的塔函数解析当我们谈论起计算机科学中的巨量数时，不得不提到高德纳（Donald Knuth）的创新性贡献——高德纳箭头符号（Knuth’s up-a… 2025年5月8日 0 0 2 快讯更多阿里云.net域名优惠注册39元1年2025年2月22日下午11:02 分享到: 活动时间：2025年1月1日00:00 – 2025年3月31日 23:59 [原文链接] 2核2G 3M固定带宽99元/年,续费同价2025年2月22日下午11:01 分享到: 云服务器“99套餐”低价长效特价精选，固定配置，固定带宽不限流量，新老同享，活动期间新购、续费同价，开发必备！ [原文链接] 企业必备2核4G 5M带宽低至199/年,续费同价2025年2月22日下午11:01 分享到: 飞天加速计划-云服务器u1年付低至59.29元/月，高性价比【企业级独享实例】u1全新发售，可搭载ESSD Entry云盘，新用户首购低至3折起。现已开放海外地域售卖节点，欢迎各位选购！多款主售产品价格下调，最高幅度达到93%！ [原文链接] 搜索搜索近期文章解析科学实验中的关键空隙与创新突破云服务器租赁协议结束后的处理方法是什么？ AI赋能云原生高效计算云原生物智能网高效协同高效云原生AI技术新突破 AI驱动云原生高效计算高效协同AI云原生应用 AI+云原生高效计算探索高效计算AI赋能云计算 AI助力云原生高效协同高效协同AI云原生创新 AI推动云原生高效计算是否可以随时退租已租用的云服务器？引索科技：开启未来探索之门 admin 最近文章解析科学实验中的关键空隙与创新突破云服务器租赁协议结束后的处理方法是什么？ AI赋能云原生高效计算云原生物智能网高效协同高效云原生AI技术新突破 AI驱动云原生高效计算高效协同AI云原生应用 AI+云原生高效计算探索高效计算AI赋能云计算 AI助力云原生高效协同高效协同AI云原生创新 AI推动云原生高效计算是否可以随时退租已租用的云服务器？引索科技：开启未来探索之门首页云数据库云服务器安全产品建站教程解决方案专题列表标签页面联系我们标签页面 Copyright © 2025 版权所有闽ICP备2023016045号-6 Powered by domainbrand.cn 微信分享本页返回顶部
10393	https://www.mathmammoth.com/videos/grade_6/factoring_simplifying_cross_canceling	^ How to use factoring to simplify fractions, and how to cross cancel (grade 6 math) When simplifying fractions, or when multiplying fractions, you can cross out, or cancel, common factors from the numerator and the denominator. It can help to write the numerator and the denominator as (something times something), in other words to factor them. I show this process using multiple examples, going from simple to more complex. In the second part of the lesson, I show you why we can simplify "criss cross" or cross cancel, when simplifying (reducing) fractions, or when multiplying fractions. It's a neat process and kind of fun, to look for numbers that have common factors, and to simplify them. See also Prime factorization — video lesson Math Mammoth Grade 6 curriculum Back to 6th grade videos index Back to the index of all videos WAIT! Receive my monthly collection of math tips & resources directly in your inbox — and get a FREE Math Mammoth book! We respect your email privacy.You can unsubscribe at any time. Newsletter \| About \| Privacy \| Banners \| Testimonials \| Discounts \| Free stuff! \| In Spanish \| FAQ \| Contact © 2006-2025 MathMammoth.com This website uses cookies to ensure you get the best experience on our website. Learn more in our privacy policy.
10394	https://www.ck12.org/flexi/physical-science/screw/how-can-one-calculate-the-mechanical-advantage-of-a-screw/	Flexi answers - How can one calculate the mechanical advantage of a screw? \| CK-12 Foundation Subjects Explore Donate Sign InSign Up All Subjects Physical Science Screw Question How can one calculate the mechanical advantage of a screw? Flexi Says: The mechanical advantage of a screw can be calculated using the formula: Mechanical Advantage = circumference / pitch where: - Circumference is the distance around the screw, which can be calculated using the formula 2πr (where r is the radius of the screw). - Pitch is the distance between the threads of the screw. Practice this conceptAnalogy / Example Try Asking: What is the ideal mechanical advantage of a screw whose head has a diameter of 0.812 cm and the thread width is 0.318 cm?How to remove a screw with a stripped head?How do you use a screw extractor? How can Flexi help? By messaging Flexi, you agree to our Terms and Privacy Policy
10395	https://brainly.com/question/37953868	[FREE] Explain the difference between solving the equation x = \sqrt{16} and x^2 = 16. - brainly.com 1 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 +68,8k Smart guidance, rooted in what you’re studying Get Guidance Test Prep +43,3k Ace exams faster, with practice that adapts to you Practice Worksheets +7,4k Guided help for every grade, topic or textbook Complete See more / Mathematics Expert-Verified Expert-Verified Explain the difference between solving the equation x=16 and x 2=16. 1 See answer Explain with Learning Companion NEW Asked by evilmonkey8742 • 09/19/2023 Read More Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer This answer helped 1444241 people 1M 0.0 0 Upload your school material for a more relevant answer The equation x = √16 has a single solution, x = 4; while the equation x² = 16 has two solutions, x = 4 and x = -4. The difference between them is the presence of a square root on the right side in the first equation, and a square in the second. When solving the equation x = √16, we see that √16 equals 4, meaning that the solution is simply x = 4. However, solving the equation x2 = 16 involves a bit more. To find x, you would need to take the square root of 16, which results in ±4. Thus, there are two solutions: x = 4 and x = -4. The key difference between the two is the presence of a square root on the right side in the first equation, and a squaring of x in the second equation. Learn more about Solving Equations here: brainly.com/question/18322830 SPJ11 Answered by lizashaw •10.4K answers•1.4M people helped Thanks 0 0.0 (0 votes) Expert-Verified⬈(opens in a new tab) This answer helped 1444241 people 1M 0.0 0 Upload your school material for a more relevant answer The equation x=16 has a single solution, x=4, while the equation x 2=16 has two solutions: x=4 and x=−4. This difference arises because the first equation uses a square root, yielding only the positive solution, whereas the second equation allows for both positive and negative solutions. Thus, the manner of solving dictates the number of solutions achieved. Explanation The equation x=16 and the equation x 2=16 differ primarily in how they are solved and the number of solutions they yield. Solving x=16: When you evaluate 16, you find that it equals 4. Therefore, the solution to this equation is: x=4 This equation has only one solution because the square root symbol conventionally refers to the principal (or positive) square root. Solving x 2=16: When solving this equation, you first recognize that to find x, you need to take the square root of both sides. This gives you: x=16or x=−16 This results in two possible solutions: x=4 and x=−4 Therefore, the equation x 2=16 has two solutions because both 4 and -4, when squared, equal 16. In summary, the key difference is that x=16 only gives the positive root, while x 2=16 provides both the positive and negative solutions because it considers both potential numbers that could be squared to reach 16. Examples & Evidence For example, if we look at x=25, we only get x=5. But for x 2=25, we find x=5 or x=−5, as both numbers squared gives us 25. Mathematically, the valid operations for square roots and squaring are established in algebra. A square root operation yields the principal root, while squaring a number involves considering both the positive and negative results. Thanks 0 0.0 (0 votes) Advertisement evilmonkey8742 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 What is the simplified form of 12 x 900 x 2 y 4 z 6? A. 30 y 2x z 3 B. 30 x 2 y 2 z 4 C. 360 x y 2x z 3 D. 360 x y 2z 3 The difference of the squares of two positive numbers is 180. The square of the smaller number is 8 times the greater number. Find the two numbers. Prove that ∑r=1 nr(r+1)3=n+1 an, where a is a constant to be found. Find the value of ∑r=1 50r(r+1)3, giving your answer as an exact fraction. Find an expression in its simplest form for ∑r=n 2 nr(r+1)3 Solve the following system of equations: 5 x−2 y=5 7 x−3 y=13 Write an expression equivalent to 2(4 y−5)−3 y. 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10396	https://www.hep.phy.cam.ac.uk/~chpotter/particleandnuclearphysics/PNP_Handout2.pdf	Particle and Nuclear Physics Handout #2 Particle Physics Lent/Easter Terms 2024 Prof. Tina Potter 3. Colliders and Detectors Particle and Nuclear Physics Prof. Tina Potter Prof. Tina Potter 3. Colliders and Detectors 1 In this section... Physics of colliders Different types of detectors How to detect and identify particles Prof. Tina Potter 3. Colliders and Detectors 2 Colliders and √s Consider the collision of two particles: − − →. ← − − p1 = (E1, ⃗ p1) p2 = (E2, ⃗ p2) The invariant quantity s = E 2 CM = (p1 + p2)2 = (E1 + E2)2 −(⃗ p1 + ⃗ p2)2 = E 2 1 −\|⃗ p1\|2 + E 2 2 −\|⃗ p2\|2 + 2E1E2 −2⃗ p1.⃗ p2 = m2 1 + m2 2 + 2(E1E2 −\|⃗ p1\|\|⃗ p2\| cos θ) θ is the angle between the momentum three-vectors √s is the energy in the centre-of-mass frame; it is the amount of energy available to the interaction e.g. in particle-antiparticle annihilation it is the maximum energy/mass of particle(s) that can be produced. Prof. Tina Potter 3. Colliders and Detectors 3 Colliders and √s Fixed Target Collision − − → . p1 = (E1, ⃗ p1) p2 = (m2, 0) s = m2 1 + m2 2 + 2E1m2 For E1 ≫m1, m2 s ∼2E1m2 ⇒ √s ∼ p 2E1m2 e.g. 450 GeV proton hitting a proton at rest: √s ∼ √ 2 × 450 × 1 ∼30 GeV Collider Experiment − − →. ← − − p1 = (E1, ⃗ p1) p2 = (E2, ⃗ p2) s = m2 1 + m2 2 + 2(E1E2 −\|⃗ p1\|\|⃗ p2\| cos θ) For E1 ≫m1, m2 \|⃗ p\| = E, θ = π s = 2(E 2−E 2 cos θ) = 4E 2 ⇒√s = 2E e.g. 450 GeV proton colliding with a 450 GeV proton: √s ∼2 × 450 = 900 GeV In a fixed target experiment most of the proton’s energy is wasted providing forward momentum to the final state particles rather than being available for conversion into interesting particles. Prof. Tina Potter 3. Colliders and Detectors 4 Colliders To produce and discover heavy new particles, we need high ECM. Need to collide massive particles at high energies! Accelerate charged particles using RF high-voltage Energy gained with each electric field ∆E = qV Limited by space! SLAC 3.2 km long, reached Ee = 50 GeV Prof. Tina Potter 3. Colliders and Detectors 5 Colliders To produce and discover heavy new particles, we need high ECM. Need to collide massive particles at high energies! Accelerate charged particles using RF high-voltage, bend using magnets. High power magnets needed B = p[ GeV] 0.3r[m] Limited by synchrotron radiation radiated energy per orbit = E 4 m4r Prof. Tina Potter 3. Colliders and Detectors 6 Detecting Particles Trackers Trackers detect ionisation loss ⇒only detect charged particles e.g. multiwire proportional chambers, cloud chambers Ionisation loss given by Bethe-Block formula depends on particle charge q and speed β, γ (not mass) −dE dx = 4πN0q2α2(ℏc)2 meβ2 Z A log 2meγ2β2 I −β2 Immerse tracker in ⃗ B to measure track radius, and thus particle momentum p. Measure sagitta s from track arc →curvature R R = L2 8s + s 2 ∼L2 8s p = 0.3B L2 8s σp p = σs s = 8p 0.3BL2σs High-p particles have high radius of curvature ⇒track almost straight. Low-p particles have small radius of curvature ⇒measure with high accuracy. σp p ∝p Prof. Tina Potter 3. Colliders and Detectors 7 Detecting Particles Calorimeters Calorimeters detect EM/hadronic showers using layers of absorber and scintillating material High-density material interacts with the particle and initiates shower. Electromagnetic calorimeter (e±, γ) γ N γ N e− e+ γ N e− N e− γ Hadronic calorimeter (p, n, π, K...) Nuclear interaction length > radiation length. Use more (denser) material. High-energy particles produce showers with many particles ⇒measure with high accuracy. Low-energy particles produce showers with few particles ⇒low accuracy. σE E ∝ √ N E = 1 √ E Prof. Tina Potter 3. Colliders and Detectors 8 Detector design Prof. Tina Potter 3. Colliders and Detectors 9 Particle Signatures Different particles leave different signals in the various detector components allowing almost unambiguous identification. e±: Track + EM energy γ: No track + EM energy µ±: Track, small calo energy deposits, penetrating τ ±: decay, observe decay products ν: not detected (need specialised detectors) hadrons: track (if charged) + calo energy deposits quarks: seen as jets of hadrons electron photon muon pion neutrino jet Prof. Tina Potter 3. Colliders and Detectors 10 Particle Signatures Examples e+e−→Z →e+e− e+e−→Z →µ+µ− Prof. Tina Potter 3. Colliders and Detectors 11 Particle Signatures Examples e+e−→Z →τ +τ − Taus decay within the detector (lifetime ∼10−13 s). Here τ −→e−¯ νeντ, τ + →µ+νµ¯ ντ e+e−→Z →q¯ q 3-jet event (gluon emitted by q/¯ q) Prof. Tina Potter 3. Colliders and Detectors 12 Particle Signatures Examples W +W −→eνµν W +W −→q¯ qeν W +W −→q¯ qq¯ q Prof. Tina Potter 3. Colliders and Detectors 13 Example e+e−collider with typical cylinder detector. In one event, two electrons are detected: 1 e+, Ecluster = 44.7 ± 1.2 GeV, \|⃗ ptrack\| = 46.0 ± 3.2 GeV 2 e−, Ecluster = 46.0 ± 1.2 GeV, \|⃗ ptrack\| = 49.5 ± 3.5 GeV For this event we need Lowest order Feynman diagram Detector signature Invariant mass Prof. Tina Potter 3. Colliders and Detectors 14 Example Consider pp collisions. Calculate the reduced ECM assuming the colliding quarks carry a fraction x1 and x2 of the proton energy. Prof. Tina Potter 3. Colliders and Detectors 15 Summary For high √s: Prefer colliders over fixed target collisions Prefer circular colliders with high power magnets Prefer to collide high mass particles Trackers to trace the path of charged particles Calorimeters to stop and measure the energy of particles Detector design and particle signatures Problem Sheet: q.7-9 Up next... Section 4: The Standard Model Prof. Tina Potter 3. Colliders and Detectors 16 4. The Standard Model Particle and Nuclear Physics Prof. Tina Potter Prof. Tina Potter 4. The Standard Model 1 In this section... Standard Model particle content Klein-Gordon equation Antimatter Interaction via particle exchange Virtual particles Prof. Tina Potter 4. The Standard Model 2 The Standard Model Spin-1/2 fermions Charge (units of e) Quarks u d c s t b +2 3 −1 3 Leptons e− νe µ− νµ τ − ντ −1 0 Plus antileptons and antiquarks Spin-1 bosons Mass ( GeV/c2) Gluon g 0 Strong force Photon γ 0 EM force W and Z bosons W ±, Z 91.2, 80.3 Weak force Spin-0 bosons Higgs h 125 Mass generation Prof. Tina Potter 4. The Standard Model 3 Theoretical Framework Macroscopic Microscopic Slow Classical Mechanics Quantum Mechanics Fast Special Relativity Quantum Field Theory The Standard Model is a collection of related Gauge Theories which are Quantum Field Theories that satisfy Local Gauge Invariance. Electromagnetism: Quantum Electrodynamics (QED) 1948 Feynman, Schwinger, Tomonaga (1965 Nobel Prize) Electromagnetism + Weak: Electroweak Unification 1968 Glashow, Weinberg, Salam (1979 Nobel Prize) Strong: Quantum Chromodynamics (QCD) 1974 Politzer, Wilczek, Gross (2004 Nobel Prize) Prof. Tina Potter 4. The Standard Model 4 The Schr¨ odinger Equation To describe the fundamental interactions of particles we need a theory of Relativistic Quantum Mechanics Schr¨ odinger Equation for a free particle ˆ Eψ = ˆ p2 2mψ with energy and momentum operators ˆ E = i ∂ ∂t, ˆ p = −i∇ ∇ ∇ (ℏ= 1 natural units) giving i∂ψ ∂t = −1 2m∇ ∇ ∇2ψ which has plane wave solutions: ψ(⃗ r, t) = Ne−i(Et−⃗ p.⃗ r) 1st order in time derivative 2nd order in space derivatives Not Lorentz Invariant! Schr¨ odinger equation cannot be used to describe the physics of relativistic particles. Prof. Tina Potter 4. The Standard Model 5 Klein-Gordon Equation Use the KG equation to describe the physics of relativistic particles. From Special Relativity: E 2 = p2 + m2 use energy and momentum operators ˆ E = i ∂ ∂t, ˆ p = −i∇ ∇ ∇ giving −∂2ψ ∂t2 = −∇ ∇ ∇2ψ + m2ψ ∂2ψ ∂t2 = (∇ ∇ ∇2 −m2)ψ Klein-Gordon Equation Second order in both space and time derivatives ⇒Lorentz invariant. Plane wave solutions ψ(⃗ r, t) = Ne−i(Et−⃗ p.⃗ r) but this time requiring E 2 = ⃗ p 2 + m2, allowing E = ± p \|⃗ p\|2 + m2 Negative energy solutions required to form complete set of eigenstates. ⇒Antimatter Prof. Tina Potter 4. The Standard Model 6 Antimatter and the Dirac Equation In the hope of avoiding negative energy solutions, Dirac sought a linear relativistic wave equation: i∂ψ ∂t = (−i⃗ α.⃗ ∇ ∇ ∇+ βm)ψ ⃗ α and β are appropriate 4x4 matrices. ψ is a column vector “spinor” of four wavefunctions. Two of the wavefunctions describe the states of a fermion, but the other two still have negative energy. Dirac suggested the vacuum had all negative energy states filled. A hole in the negative energy “sea” could be created by exciting an electron to a positive energy state. The hole would behave like a positive energy positive charged “positron”. Subsequently detected. However, this only works for fermions... We now interpret negative energy states differently... Prof. Tina Potter 4. The Standard Model 7 Antimatter and the Feynman-St¨ uckelberg Interpretation Consider the negative energy solution in which a negative energy particle travels backwards in time. e−iEt ≡e−i(−E)(−t) Interpret as a positive energy antiparticle travelling forwards in time. Then all solutions can be used to describe physical states with positive energy, going forward in time. e.g. time − − → e+e−annihilation γ e− e+ pair production γ e+ e− time − − → All quantum numbers carried into a vertex by the e+ are the same as if it is regarded as an outgoing e−, or vice versa. Prof. Tina Potter 4. The Standard Model 8 Antimatter and the Feynman-St¨ uckelberg Interpretation e− γ e− e− γ time − − → The interpretation here is easy. The first photon emitted has less energy than the electron it was emitted from. No need for “anti-particles” or negative energy states. e+/e− γ e− e− γ time − − → The emitted photon has more energy than the electron that emitted it. Either view the top vertex as “emission of a negative energy electron travelling backwards in time” or “absorption of a positive energy positron travelling forwards in time”. Prof. Tina Potter 4. The Standard Model 9 Interaction via Particle Exchange Consider two particles, fixed at ⃗ r1 and ⃗ r2, which exchange a particle of mass m. Space Time 1 2 State i State j State i pµ = (E, ⃗ p) E = Ej −Ei Calculate the shift in energy of state i due to this exchange (using second order perturbation theory): ∆Ei = X j̸=i ⟨i\|H\|j⟩⟨j\|H\|i⟩ Ei −Ej Sum over all possible states j with different momenta where ⟨j\|H\|i⟩is the transition from i to j at ⃗ r1 where ⟨i\|H\|j⟩is the transition from j to i at ⃗ r2 Prof. Tina Potter 4. The Standard Model 10 Interaction via Particle Exchange Consider ⟨j\|H\|i⟩(transition from i →j by emission of m at ⃗ r1) ψi = ψ1ψ2 Original 2 particles ψj = ψ1ψ2ψ3 ψ3 = N e−i(Et−⃗ p.⃗ r) normalise ψ∗ 1ψ1 = ψ∗ 2ψ2 = ψ∗ 3ψ3 = 1 ψ3 represents a free particle with pµ = (E, ⃗ p) Let g be the probability of emitting m at r1 g/ √ 2E is required on dimensional grounds, c.f. AQP vector potential of a photon. ⟨j\|H\|i⟩= Z d3⃗ r ψ∗ 1ψ∗ 2ψ∗ 3 g √ 2E ψ1ψ2 δ3(⃗ r −⃗ r1) = g √ 2E Nei(Et−⃗ p.⃗ r1) Dirac δ function Z d3⃗ rδ3 (⃗ r −⃗ r1) = 1 for ⃗ r = ⃗ r1 = 0 for ⃗ r ̸= ⃗ r1 Similarly ⟨i\|H\|j⟩is the transition from j to i at ⃗ r2 ⟨i\|H\|j⟩= g √ 2E Ne−i(Et−⃗ p.⃗ r2) Shift in energy state ∆E 1→2 i = X j̸=i g2 2E N2ei⃗ p.(⃗ r2−⃗ r1) Ei −Ej = X j̸=i g2N2ei⃗ p.(⃗ r2−⃗ r1) −2E 2 (E = Ej −Ei) Prof. Tina Potter 4. The Standard Model 11 Interaction via Particle Exchange Putting the pieces together Different states j have different momenta ⃗ p for the exchanged particle. Therefore sum is actually an integral over all momenta: ∆E 1→2 i = Z g2N2ei⃗ p.(⃗ r2−⃗ r1) −2E 2 ρ(p) dp = Z g2ei⃗ p.(⃗ r2−⃗ r1) −2E 2 1 L3 L 2π 3 p2 dp dΩ N = r 1 L3, ρ(p) = L 2π 3 p2 dΩ = −g2 1 2π 3 Z ei⃗ p.(⃗ r2−⃗ r1) 2E 2 p2 dp dΩ E 2 = p2 + m2 The integral can be done by taking the z-axis along ⃗ r = ⃗ r2 −⃗ r1 Then ⃗ p.⃗ r = pr cos θ and dΩ= 2π d(cos θ) ∆E 1→2 i = − g2 2(2π)2 Z ∞ 0 p2 p2 + m2 ei⃗ p.⃗ r −e−i⃗ p.⃗ r ipr dp (see Appendix D) Write this integral as one half of the integral from −∞to +∞, which can be done by residues giving ∆E 1→2 i = −g2 8π e−mr r Prof. Tina Potter 4. The Standard Model 12 Interaction via Particle Exchange Final stage Can also exchange particle from 2 to 1: Space Time 1 2 State i State j State i Get the same result: ∆E 2→1 i = −g2 8π e−mr r Total shift in energy due to particle exchange is ∆Ei = −g2 4π e−mr r Yukawa Potential Attractive force between two particles, decreasing exponentially with range r. Prof. Tina Potter 4. The Standard Model 13 Yukawa Potential Hideki Yukawa 1949 Nobel Prize V (r) = −g2 4π e−mr r Yukawa Potential Characteristic range = 1/m (Compton wavelength of exchanged particle) For m →0, V (r) = −g2 4πr infinite range (Coulomb-like) Yukawa potential with m = 139 MeV/c2 gives a good description of long range part of the interaction between two nucleons and was the basis for the prediction of the existence of the pion. Prof. Tina Potter 4. The Standard Model 14 Scattering from the Yukawa Potential Consider elastic scattering (no energy transfer) ⃗ pi ⃗ pf ⃗ p q μ=( E,⃗ p) q 2=E 2−\|⃗ p\| 2 q 2 is invariant “Virtual Mass” Born Approximation Mfi = Z ei⃗ p.⃗ rV (r) d3⃗ r Yukawa Potential V (r) = −g2 4π e−mr r Mfi = −g2 4π Z e−mr r ei⃗ p.⃗ r d3⃗ r = − g2 \|⃗ p\|2 + m2 The integral can be done by choosing the z-axis along ⃗ r, then ⃗ p.⃗ r = pr cos θ and d3⃗ r = 2πr2 dr d(cos θ) For elastic scattering, qµ = (0, ⃗ p), q2 = −\|p\|2 and exchanged massive particle is highly “virtual” Mfi = g2 q2 −m2 Prof. Tina Potter 4. The Standard Model 15 Virtual Particles Forces arise due to the exchange of unobservable virtual particles. The effective mass of the virtual particle, q2, is given by q2 = E 2 −\|⃗ p\|2 and is not equal to the physical mass m, i.e. it is off-shell mass. The mass of a virtual particle can be +ve, -ve or imaginary. A virtual particle which is off-mass shell by amount ∆m can only exist for time and range t ∼ ℏ ∆mc2 = 1 ∆m, range = ℏ ∆mc = 1 ∆m ℏ= c = 1 If q2 = m2, the the particle is real and can be observed. Prof. Tina Potter 4. The Standard Model 16 Virtual Particles For virtual particle exchange, expect a contribution to the matrix element of Mfi = g2 q2 −m2 where g Coupling constant g2 Strength of interaction m2 Physical (on-shell) mass q2 Virtual (off-shell) mass 1 q2−m2 Propagator Qualitatively: the propagator is inversely proportional to how far the particle is off-shell. The further off-shell, the smaller the probability of producing such a virtual state. For m →0; e.g. single γ exchange, Mfi = g2/q2 For q2 →0, very low momentum transfer EM scattering (small angle) Prof. Tina Potter 4. The Standard Model 17 Virtual Particles Example Prof. Tina Potter 4. The Standard Model 18 Summary SM particles: 12 fermions, 5 spin-1 bosons, 1 spin-0 boson. Need relativistic wave equations to describe particle interactions. Klein-Gordon equation (bosons), Dirac equation (fermions). Negative energy solutions describe antiparticles. The exchange of a massive particle generates an attractive force between two particles. Yukawa potential V (r) = −g2 4π e−mr r Exchanged particles may be virtual. Problem Sheet: q.10 Up next... Section 5: Feynman Diagrams Prof. Tina Potter 4. The Standard Model 19 5. Feynman Diagrams Particle and Nuclear Physics Prof. Tina Potter Prof. Tina Potter 5. Feynman Diagrams 1 In this section... Introduction to Feynman diagrams. Anatomy of Feynman diagrams. Allowed vertices. General rules Prof. Tina Potter 5. Feynman Diagrams 2 Feynman Diagrams Richard Feynman 1965 Nobel Prize The results of calculations based on a single process in Time-Ordered Perturbation Theory (sometimes called old-fashioned, OFPT) depend on the reference frame. The sum of all time orderings is frame independent and provides the basis for our relativistic theory of Quantum Mechanics. A Feynman diagram represents the sum of all time orderings time − − → + time − − → = time − − → Prof. Tina Potter 5. Feynman Diagrams 3 Feynman Diagrams Each Feynman diagram represents a term in the perturbation theory expansion of the matrix element for an interaction. Normally, a full matrix element contains an infinite number of Feynman diagrams. Total amplitude Mfi = M1 + M2 + M3 + ... Total rate Γfi = 2π\|M1 + M2 + M3 + ...\|2ρ(E) Fermi’s Golden Rule But each vertex gives a factor of g, so if g is small (i.e. the perturbation is small) only need the first few. (Lowest order = fewest vertices possible) g2 g4 g6 Example: QED g = e = √ 4πα ∼0.30, α = e2 4π ∼ 1 137 Prof. Tina Potter 5. Feynman Diagrams 4 Feynman Diagrams Perturbation Theory Calculating Matrix Elements from Perturbation Theory from first principles is cumbersome – so we don’t usually use it. Need to do time-ordered sums of (on mass shell) particles whose production and decay does not conserve energy and momentum. Feynman Diagrams Represent the maths of Perturbation Theory with Feynman Diagrams in a very simple way (to arbitrary order, if couplings are small enough). Use them to calculate matrix elements. Approx size of matrix element may be estimated from the simplest valid Feynman Diagram for given process. Full matrix element requires infinite number of diagrams. Now only need one exchanged particle, but it is now off mass shell, however production/decay now conserves energy and momentum. Prof. Tina Potter 5. Feynman Diagrams 5 Anatomy of Feynman Diagrams Feynman devised a pictorial method for evaluating matrix elements for the interactions between fundamental particles in a few simple rules. We shall use Feynman diagrams extensively throughout this course. Topological features of Feynman diagrams are straightforwardly associated with terms in the Matrix element Represent particles (and antiparticles): Spin 1/2 Quarks and Leptons Spin 1 γ, W ±, Z g And each interaction point (vertex) with a • Each vertex contributes a factor of the coupling constant, g. Prof. Tina Potter 5. Feynman Diagrams 6 Anatomy of Feynman Diagrams External lines (visible real particles) Spin 1/2 Particle Incoming Outgoing Antiparticle Incoming Outgoing Spin 1 Particle Incoming Outgoing Internal lines (propagators; virtual particles) Spin 1/2 Particle/antiparticle Each propagator gives a factor of 1 q2−m2 Spin 1 γ, W ±, Z g Prof. Tina Potter 5. Feynman Diagrams 7 Vertices A vertex represents a point of interaction: either EM, weak or strong. The strength of the interaction is denoted by g EM interaction: g = Qe (sometimes denoted as Q√α, where α = e2/4π) Weak interaction: g = gW Strong interaction: g = √αs A vertex will have three (in rare cases four) lines attached, e.g. γ e+ e− Qe e− e− γ Qe γ e− e− Qe e− e+ γ Qe At each vertex, conserve energy, momentum, angular momentum, charge, lepton number (Le = +1 for e−, νe, = −1 for e+, ¯ νe, similar for Lµ, Lτ), baryon number (B = 1 3(nq −n¯ q)), strangeness (S = −(ns −n¯ s)) & parity – except in weak interactions. Prof. Tina Potter 5. Feynman Diagrams 8 Allowed Vertices EM must involve a photon γ, and charged particles coupling strength Qe Q=charge γ e+ e− γ µ+ µ− γ τ + τ − γ ¯ u u γ ¯ c c γ ¯ t t γ ¯ d d γ ¯ s s γ ¯ b b γ W + W − Triple Gauge Vertex Prof. Tina Potter 5. Feynman Diagrams 9 Allowed Vertices Weak must involve a gauge vector boson Z or W ± coupling strength gW tip: if you see a ν or ¯ ν, it must be a weak interaction with W ± W − ¯ νe e− W − ¯ νµ µ− W − ¯ ντ τ − W − ¯ u d W − ¯ c s W − ¯ t b ⇒Same family quarks are Cabibbo favoured W − ¯ u s W − ¯ c d W − ¯ c b W − ¯ t s ⇒Cross one family Cabibbo suppressed Prof. Tina Potter 5. Feynman Diagrams 10 Allowed Vertices Weak must involve a gauge vector boson Z or W ± coupling strength gW tip: if you see a ν or ¯ ν, it must be a weak interaction with W ± W − ¯ u b W − ¯ t d ⇒Cross two families Doubly Cabibbo suppressed Also, Triple/Four Gauge Vertex Z W + W − γ W + W − W − W + W − W + Z Z W − W + γ Z W − W + γ γ W − W + Prof. Tina Potter 5. Feynman Diagrams 11 Allowed Vertices Weak with Z Same as γ diagrams, but also vertices with ν Z e+ e− Z µ+ µ− Z τ + τ − Z ¯ νe νe Z ¯ νµ νµ Z ¯ ντ ντ Z ¯ u u Z ¯ c c Z ¯ t t Z ¯ d d Z ¯ s s Z ¯ b b i.e. Z ¯ f f Not Allowed: Flavour Changing Neutral Currents (FCNC) Z ¯ s d Prof. Tina Potter 5. Feynman Diagrams 12 Allowed Vertices Strong must involve a gluon g and/or quark q coupling strength √αs conserve strangeness, charm etc g ¯ u u g ¯ c c g ¯ t t g ¯ d d g ¯ s s g ¯ b b Also, Triple Gauge Vertex g g g g g g g Prof. Tina Potter 5. Feynman Diagrams 13 Forbidden Vertices X ℓ q γ γ γ Z Z Z g γ g g Z g g W ± g Prof. Tina Potter 5. Feynman Diagrams 14 Examples Electromagnetic γ p e− p e− Qe Qe M ∼(e)2 q2 Strong g ¯ q q ¯ q q √αs √αs M ∼(√αs)2 q2 Weak W − d d u e− ¯ νe u d u VudgW gW M ∼ Vudg2 W q2 −m2 W Prof. Tina Potter 5. Feynman Diagrams 15 Drawing Feynman Diagrams A Feynman diagram is a pictorial representation of the matrix element describing particle decay or interaction a →b + c + ... a + b →c + d To draw a Feynman diagram and determine whether a process is allowed, follow the five basic steps below: 1 Write down the initial and final state particles and antiparticles and note the quark content of all hadrons. 2 Draw the simplest Feynman diagram using the Standard Model vertices. Bearing in mind: Similar diagrams for particles/antiparticles Never have a vertex connecting a lepton to a quark Only the weak charged current (W ±) vertex changes flavour within generations for leptons within/between generations for quarks Prof. Tina Potter 5. Feynman Diagrams 16 Drawing Feynman Diagrams Particle scattering If all are particles (or all are antiparticles), only scattering diagrams involved e.g. a + b →c + d Initial State b a d c Final State If particles and antiparticles, may be able to have scattering and/or annihilation diagrams e.g. a + b →c + d (Mandelstam variables s, t, u) p2 p1 p4 p3 b a c c “t-channel”, q2 = t = (p1 −p3)2 = (p2 −p4)2 p2 p4 p1 p3 b a d c “s-channel”, q2 = s = (p1 + p2)2 = (p3 + p4)2 Prof. Tina Potter 5. Feynman Diagrams 17 Drawing Feynman Diagrams Identical Particles If we have identical particles in final state, e.g. a + b →c + c may not know which particle comes from which vertex. Two possibilities are separate final Feynman diagrams: p2 p1 p4 p3 b a c c “t-channel”, q2 = t = (p1 −p3)2 = (p2 −p4)2 p2 p1 p4 p3 b a c c “u-channel”, q2 = u = (p1 −p4)2 = (p2 −p3)2 Crossing not a vertex Prof. Tina Potter 5. Feynman Diagrams 18 Drawing Feynman Diagrams Being able to draw a Feynman diagram is a necessary, but not a sufficient condition for the process to occur. Also need to check: 3 Check that the whole system conserves Energy, momentum (trivially satisfied for interactions, so long as sufficient KE in initial state. May forbid decays) Charge Angular momentum 4 Parity Conserved in EM/Strong interaction Can be violated in the Weak interaction 5 Check symmetry for identical particles in the final state Bosons ψ(1, 2) = +ψ(2, 1) Fermions ψ(1, 2) = −ψ(2, 1) Finally, a process will occur via the Strong, EM and Weak interaction (in that order of preference) if steps 1 – 5 are satisfied. Prof. Tina Potter 5. Feynman Diagrams 19 Summary Feynman diagrams are a core part of the course. Make sure you can draw them! Feynman diagrams are a sum over time orderings. Associate topological features of the diagrams with terms in matrix elements. Vertices ↔coupling strength between particles and field quanta Propagator for each internal line (off-mass shell, virtual particles) Conservation of quantum numbers at each vertex Problem Sheet: q.11 Up next... Section 6: QED Prof. Tina Potter 5. Feynman Diagrams 20 6. QED Particle and Nuclear Physics Prof. Tina Potter Prof. Tina Potter 6. QED 1 In this section... Gauge invariance Allowed vertices + examples Scattering Experimental tests Running of alpha Prof. Tina Potter 6. QED 2 QED Quantum Electrodynamics is the gauge theory of electromagnetic interactions. Consider a non-relativistic charged particle in an EM field: ⃗ F = q(⃗ E + ⃗ v × ⃗ B) ⃗ E, ⃗ B given in term of vector and scalar potentials ⃗ A, φ ⃗ B = ⃗ ∇ ∇ ∇× ⃗ A; ⃗ E = −⃗ ∇ ∇ ∇φ −∂⃗ A ∂t Maxwell’s Equations ˆ H = 1 2m(ˆ ⃗ p −q ⃗ A)2 + qφ Classical Hamiltonian e− e− γ Change in state of e−requires change in field ⇒Interaction via virtual γ emission Prof. Tina Potter 6. QED 3 QED Schr¨ odinger equation 1 2m(ˆ ⃗ p −q ⃗ A)2 + qφ ψ(⃗ r, t) = i∂ψ(⃗ r, t) ∂t is invariant under the local gauge transformation ψ →ψ′ = eiqα(⃗ r,t)ψ so long as ⃗ A →⃗ A + ⃗ ∇ ∇ ∇α ; φ →φ −∂α ∂t (See Appendix E) Local Gauge Invariance requires the existence of a physical Gauge Field (photon) and completely specifies the form of the interaction between the particle and field. Photons are massless (in order to cancel phase changes over all space-time, the range of the photon must be infinite) Charge is conserved – the charge q which interacts with the field must not change in space or time QED is a gauge theory Prof. Tina Potter 6. QED 4 The Electromagnetic Vertex All electromagnetic interactions can be described by the photon propagator and the EM vertex: e−, µ−, τ −, q e−, µ−, τ −, q γ Qe The Standard Model Electromagnetic Vertex + antiparticles α = e2 4π The coupling constant is proportional to the fermion charge. Energy, momentum, angular momentum, parity and charge always conserved. QED vertex never changes particle type or flavour i.e. e−→e−γ, but not e−→qγ or e−→µ−γ Prof. Tina Potter 6. QED 5 Important QED Processes M ∼g2 q2, α = e2 4π Compton Scattering (γe−→γe−) e− e− γ γ e− Qe Qe e− e− γ γ e− Qe Qe M ∝e2 σ ∝\|M\|2 ∝e4 ∝(4π)2α2 Bremsstrahlung (e−→e−γ) e± γ nucleus e− nucleus e− γ Qe Ze Qe M ∝Ze3 σ ∝\|M\|2 ∝Z 2e6 ∝(4π)3Z 2α3 Pair Production (γ →e+e−) γ e± nucleus γ nucleus e+ e− Qe Ze Qe M ∝Ze3 σ ∝\|M\|2 ∝Z 2e6 ∝(4π)3Z 2α3 The processes e−→e−γ and γ →e+e−cannot occur for real e−, γ due to energy & momentum conservation Prof. Tina Potter 6. QED 6 Important QED Processes Electron-Positron Annihilation (e−e+ →q¯ q) γ e− e+ ¯ q q Qe Qqe M ∝Qqe2 σ ∝\|M\|2 ∝Q2 qe4 ∝(4π)2Q2 qα2 Pion Decay (π0 →γγ) u π0 ¯ u γ γ Qqe Qqe M ∝Q2 ue2 Γ ∝\|M\|2 ∝Q4 ue4 ∝(4π)2Q4 uα2 J/ψ Decay (J/ψ →µ+µ−) γ c J/ψ ¯ c µ+ µ− Qqe Qe M ∝Qce2 Γ ∝\|M\|2 ∝Q2 ce4 ∝(4π)2Q2 cα2 The coupling strength determines “order of magnitude” of the matrix element. For particles interacting/decaying via EM interaction: typical values for cross-sections/ lifetimes σEM ∼10−2 mb; τEM ∼10−20 s Prof. Tina Potter 6. QED 7 Scattering in QED Examples Calculate the “spin-less” cross-sections for the two processes: 1. Electron-proton scattering γ p e− p e− Qe Qe 2. Electron-positron annihilation γ e− e+ µ+ µ− Qe Qe Fermi’s Golden rule and Born Approximation dσ dΩ= E 2 (2π)2\|M\|2 For both processes we have the same matrix element (though q2 is different) M = e2 q2 = 4πα q2 e2 = 4πα is the strength of the interaction. 1/q2 measures the probability that the photon carries 4-momentum qµ = (E, ⃗ p); q2 = E 2 −\|⃗ p\|2 i.e. smaller probability for higher mass. Prof. Tina Potter 6. QED 8 Scattering in QED 1. “Spinless” e −p Scattering γ p e− p e− Qe Qe M = e2 q2 = 4πα q2 dσ dΩ= E 2 (2π)2\|M\|2 = E 2 (2π)2 (4πα)2 q4 = 4α2E 2 q4 q2 is the four-momentum transfer q2 = qµqµ = (Ef −Ei)2 −(⃗ pf −⃗ pi)2 = E 2 f + E 2 i −2EfEi −⃗ p2 f −⃗ p2 i + 2⃗ pf.⃗ pi = 2m2 e −2EfEi + 2\|⃗ pf\|\|⃗ pi\| cos θ Neglecting electron mass: i.e. me = 0 and \|⃗ pf\| = Ef q2 = −2EfEi(1 −cos θ) = −4EfEi sin2 θ 2 Therefore, for elastic scattering Ei = Ef dσ dΩ= α2 4E 2 sin4 θ 2 Rutherford Scattering same result from QED as from conventional QM Prof. Tina Potter 6. QED 9 Scattering in QED 1. “Spinless” e −p Scattering The discovery of quarks Virtual γ carries 4-momentum qµ = (E, ⃗ p) Large q ⇒Large ⃗ p, small λ \|⃗ p\| = ℏ/λ Large E, large ω E = ℏω High q wavefunction oscillates rapidly in space and time ⇒probes short distances and short time. Elastic scattering from quarks in proton. Prof. Tina Potter 6. QED 10 Scattering in QED 2. “Spinless” e+e−Scattering γ e− e+ µ+ µ− Qe Qe M = e2 q2 = 4πα q2 dσ dΩ= E 2 (2π)2\|M\|2 = E 2 (2π)2 (4πα)2 q4 = 4α2E 2 q4 Same formula, but different four-momentum transfer q2 = qµqµ = (Ee+ + Ee−)2 −(⃗ pe+ + ⃗ pe−)2 assuming we are in the centre-of-mass system, Ee+ = Ee−= E, ⃗ pe+ = −⃗ pe− q2 = qµqµ = (2E)2 = s dσ dΩ= 4α2E 2 q4 = 4α2E 2 16E 4 = α2 s Integrating gives total cross-section: σ = 4πα2 s Prof. Tina Potter 6. QED 11 Scattering in QED 2. “Spinless” e+e−Scattering ... the actual cross-section (using the Dirac equation to take spin into account) is dσ dΩ= α2 4s (1 + cos2 θ) σ(e+e−→µ+µ−) = 4πα2 3s Example: Cross-section at √s = 22 GeV (i.e. 11 GeV electrons colliding with 11 GeV positrons) σ(e+e−→µ+µ−) = 4πα2 3s = 4π (137)2 1 3 × 222 = 4.6 × 10−7 GeV−2 = 4.6 × 10−7 × (0.197)2 fm2 = 1.8 × 10−8 fm2 = 0.18 nb Prof. Tina Potter 6. QED 12 The Drell-Yan Process Can also annihilate q¯ q as in the “Drell-Yan” process. Example: π−p →µ+µ−+ hadrons (See problem sheet q.13) γ d π−¯ u u p u d d µ+ µ− u d Que Qe σ(π−p →µ+µ−+ hadrons) ∝Q2 uα2 ∝Q2 ue4 (Also need to account for presence of two u quarks in proton) Prof. Tina Potter 6. QED 13 Experimental Tests of QED QED is an extremely successful theory tested to very high precision. Example: Magnetic moments of e±, µ±: ⃗ µ = g e 2m⃗ s For a point-like spin 1/2 particle: g = 2 Dirac Equation However, higher order terms in QED introduce an anomalous magnetic moment ⇒g is not quite equal to 2. γ O(1) γ O(α) O(α4) 12672 diagrams Prof. Tina Potter 6. QED 14 Experimental Tests of QED O(α3) ge −2 2 = 11596521.811 ± 0.007 × 10−10 Experiment = 11596521.3 ± 0.3 × 10−10 Theory Agreement at the level of 1 in 108 QED provides a remarkably precise description of the electromagnetic interaction! Prof. Tina Potter 6. QED 15 Higher Orders So far only considered lowest order term in the perturbation series. Higher order terms also contribute (and also interfere with lower orders) Lowest Order γ e− e+ µ+ µ− Qe Qe \|M\|2 ∝e4 ∝α2 ∼ 1 137 2 Second Order e− e+ µ+ µ− e− e+ µ+ µ− + ... \|M\|2 ∝α4 ∼ 1 137 4 Third Order e− e+ µ+ µ− e− e+ µ+ µ− + ... \|M\|2 ∝α6 ∼ 1 137 6 Second order suppressed by α2 relative to first order. Provided α is small, i.e. perturbation is small, lowest order dominates. Prof. Tina Potter 6. QED 16 Running of α α = e2 4π specifies the strength of the interaction between an electron and a photon. But α is not a constant Consider an electric charge in a dielectric medium. Charge Q appears screened by a halo of +ve charges. Only see full value of charge Q at small distance. Consider a free electron. The same effect can happen due to quantum fluctuations that lead to a cloud of virtual e+e−pairs. The vacuum acts like a dielectric medium The virtual e+e−pairs are therefore polarised At large distances the bare electron charge is screened. At shorter distances, screening effect reduced and we see a larger effective charge i.e. a larger α. Prof. Tina Potter 6. QED 17 Running of α Can measure α(q2) from e+e−→µ+µ−etc. γ e− e+ µ+ µ− Qe Qe α increases with increasing q2 (i.e. closer to the bare charge) At q2 = 0 : α ∼1/137 At q2 ∼(100 GeV)2 : α ∼1/128 Prof. Tina Potter 6. QED 18 Summary QED is the physics of the photon + “charged particle” vertex: e−, µ−, τ −, q e−, µ−, τ −, q γ Qe α = e2 4π Every EM vertex has: has an arrow going in & out (lepton or quark), and a photon does not change the type of lepton or quark “passing through” conserves charge, energy and momentum The dimensionless coupling √α is proportional to the electric charge of the lepton or quark, and it “runs” with energy scale. QED has been tested at the level of 1 part in 108. Problem Sheet: q.12-14 Up next... Section 7: QCD Prof. Tina Potter 6. QED 19 7. QCD Particle and Nuclear Physics Prof. Tina Potter Prof. Tina Potter 7. QCD 1 In this section... The strong vertex Colour, gluons and self-interactions QCD potential, confinement Hadronisation, jets Running of αs Experimental tests of QCD Prof. Tina Potter 7. QCD 2 QCD Quantum Electrodynamics is the quantum theory of the electromagnetic interaction. mediated by massless photons photon couples to electric charge strength of interaction: ⟨ψf\| ˆ H\|ψi⟩∝√α α = e2 4π = 1 137 Quantum Chromodynamics is the quantum theory of the strong interaction. mediated by massless gluons gluon couples to “strong” charge only quarks have non-zero “strong” charge, therefore only quarks feel the strong interaction. strength of interaction: ⟨ψf\| ˆ H\|ψi⟩∝√αs αs = g2 s 4π ∼1 Prof. Tina Potter 7. QCD 3 The Strong Vertex Basic QCD interaction looks like a stronger version of QED: QED q q γ Qe + antiquarks α = e2 4π = 1 137 QCD q q g √αs + antiquarks αs = g2 s 4π ∼1 The coupling of the gluon, gs, is to the “strong” charge. Energy, momentum, angular momentum and charge always conserved. QCD vertex never changes quark flavour QCD vertex always conserves parity Prof. Tina Potter 7. QCD 4 Colour QED: Charge of QED is electric charge, a conserved quantum number QCD: Charge of QCD is called “ colour ” colour is a conserved quantum number with 3 values labelled red, green and blue. Quarks carry colour r b g Antiquarks carry anti- colour ¯ r ¯ b ¯ g Colorless particles either have no colour at all e.g. leptons, γ, W , Z and do not interact via the strong interaction or equal parts r, b, g e.g. meson q¯ q with 1 √ 3(r ¯ r + b¯ b + g ¯ g), baryon qqq with rgb gluons do not have equal parts r, b, g, so carry colour (e.g. r ¯ r, see later) Prof. Tina Potter 7. QCD 5 QCD as a gauge theory Recall QED was invariant under gauge symmetry ψ →ψ′ = eiqα(⃗ r,t)ψ The equivalent symmetry for QCD is invariance under (non-examinable) ψ →ψ′ = eig⃗ λ.⃗ Λ(⃗ r,t)ψ an “SU(3)” transformation (λ are eight 3x3 matrices). Operates on the colour state of the quark field – a “rotation” of the colour state which can be different at each point of space and time. Invariance under SU(3) transformations →eight massless gauge bosons, gluons (eight in this case). Gluon couplings are well specified. Gluons also have self-couplings, i.e. they carry colour themselves... Prof. Tina Potter 7. QCD 6 Gluons Gluons are massless spin-1 bosons, which carry the colour quantum number (unlike γ in QED which is charge neutral). Consider a red quark scattering off a blue quark. Colour is exchanged, but always conserved (overall and at each vertex). g q q q q √αs √αs r r b b b¯ r↑ ↓r ¯ b Expect 9 gluons (3x3): r ¯ b r ¯ g g ¯ r g ¯ b b¯ g b¯ r r ¯ r b¯ b g ¯ g However: Real gluons are orthogonal linear combinations of the above states. The combination 1 √ 3(r ¯ r + b¯ b + g ¯ g) is colourless and does not participate in the strong interaction. ⇒8 coloured gluons Conventionally chosen to be (all orthogonal): r ¯ b r ¯ g g ¯ r g ¯ b b¯ g b¯ r 1 √ 2 (r ¯ r −b¯ b) 1 √ 6 (r ¯ r + b¯ b −2g ¯ g) Prof. Tina Potter 7. QCD 7 Gluon Self-Interactions QCD looks like a stronger version of QED. However, there is one big difference and that is gluons carry colour charge. ⇒Gluons can interact with other gluons g g g g g g g Example: Gluon-gluon scattering gg →gg Same colour flow in each case: r ¯ g + g ¯ b →r ¯ r + r ¯ b Prof. Tina Potter 7. QCD 8 QCD Potential QED Potential: VQED = −α r QCD Potential: VQCD = −C αs r At short distances, QCD potential looks similar, apart from the “colour factor” C. For q¯ q in a colourless state in a meson, C = 4/3 For qq in a colourless state in baryon, C = 2/3 Note: the colour factor C arises because more than one gluon can participate in the process q →qg. Obtain colour factor from averaging over initial colour states and summing over final/intermediate colour states. Prof. Tina Potter 7. QCD 9 Confinement Never observe single free quarks or gluons Quarks are always confined within hadrons This is a consequence of the strong interaction of gluons. Qualitatively, compare QCD with QED: QCD Colour field QED Electric field Self interactions of the gluons squeezes the lines of force into a narrow tube or string. The string has a “tension” and as the quarks separate the string stores potential energy. Energy stored per unit length in field ∼constant V (r) ∝r Energy required to separate two quarks is infinite. Quarks always come in combinations with zero net colour charge ⇒confinement. Prof. Tina Potter 7. QCD 10 How Strong is Strong? QCD potential between quark and antiquark has two components: Short range, Coulomb-like term: −4 3 αs r Long range, linear term: +kr VQCD = −4 3 αs r + kr with k ∼1 GeV/fm F = −dV dr = 4 3 αs r2 + k at large r F = k ∼1.6 × 10−10 10−15 N = 160, 000 N Equivalent to weight of ∼150 people Prof. Tina Potter 7. QCD 11 Jets Consider the q¯ q pair produced in e+e−→q¯ q γ e− e+ ¯ q q Qe Qqe As the quarks separate, the potential energy in the colour field (“string”) starts to increase linearly with separation. When the energy stored exceeds 2mq, new q¯ q pairs can be created. As energy decreases, hadrons (mainly mesons) freeze out Prof. Tina Potter 7. QCD 12 Jets As quarks separate, more q¯ q pairs are produced. This process is called hadronisation. Start out with quarks and end up with narrowly collimated jets of hadrons. γ e− e+ ¯ q q Qe Qqe Typical e+e−→q¯ q event The hadrons in a quark(antiquark) jet follow the direction of the original quark(antiquark). Consequently, e+e−→q¯ q is observed as a pair of back-to-back jets. Prof. Tina Potter 7. QCD 13 Nucleon-Nucleon Interactions Bound qqq states (e.g. protons and neutrons) are colourless (colour singlets) They can only emit and absorb another colour singlet state, i.e. not single gluons (conservation of colour charge). Interact by exchange of pions. Example: pp scattering (One possible diagram) π0 p p p p Nuclear potential is Yukawa potential with V (r) = −g2 4π e−mπr r Short range force: Range = 1 mπ = (0.140 GeV)−1 = 7 GeV−1 = 7 × (ℏc) fm = 1.4 fm Prof. Tina Potter 7. QCD 14 Running of αs αs specifies the strength of the strong interaction. But, just as in QED, αs is not a constant. It “runs” (i.e. depends on energy). In QED, the bare electron charge is screened by a cloud of virtual electron-positron pairs. In QCD, a similar “colour screening” effect occurs. In QCD, quantum fluctuations lead to a cloud of virtual q¯ q pairs. One of many (an infinite set) of such diagrams analogous to those for QED. In QCD, the gluon self-interactions also lead to a cloud of virtual gluons. One of many (an infinite set) of such diagrams. No analogy in QED, photons do not carry the charge of the interaction. Prof. Tina Potter 7. QCD 15 Colour Anti-Screening Due to gluon self-interactions bare colour charge is screened by both virtual quarks and gluons. The cloud of virtual gluons carries colour charge and the effective colour charge decreases at smaller distances (high energy)! Hence, at low energies, αs is large →cannot use perturbation theory. But at high energies, αs is small. In this regime, can treat quarks as free particles and use perturbation theory →Asymptotic Freedom. √s = 100 GeV, αs = 0.12 Prof. Tina Potter 7. QCD 16 Scattering in QCD Example: High energy proton-proton scattering. g d u u u u d d u u u u d √αs √αs g q q q q √αs √αs M ∼1 q2 √αs √αs ⇒ dσ dΩ∼ (αs)2 sin4 θ/2 Upper points: Geiger and Marsden data (1911) for the elastic scattering of a particles from gold and silver foils. Lower points: angular distribution of quark jets observed in pp scattering at q2 = 2000 GeV2. Both follow the Rutherford formula for elastic scattering. Prof. Tina Potter 7. QCD 17 Scattering in QCD Example: pp vs π+p scattering g d u u u u d d u u u u d √αs √αs g d u u u ¯ d d u u u ¯ d √αs √αs Calculate ratio of σ(pp)total to σ(π+p)total QCD does not distinguish between quark flavours, only colour charge of quarks matters. At high energy (E ≫binding energy of quarks within hadrons), ratio of σ(pp)total and σ(π+p)total depends on number of possible quark-quark combinations. Predict: Experiment: σ(πp) σ(pp) = 2 × 3 3 × 3 = 2 3 σ(πp) σ(pp) = 24 mb 38 mb ∼2 3 Prof. Tina Potter 7. QCD 18 QCD in e+e−Annihilation e+e−annihilation at high energies provides direct experimental evidence for colour and for gluons. Start by comparing the cross-sections for e+e−→µ+µ−and e+e−→q¯ q γ e− e+ µ+ µ− Qe Qe M ∼1 q2 √α√α ⇒σ(e+e−→µ+µ−) = 4πα2 3s γ e− e+ ¯ q q Qe Qqe M ∼1 q2Qq √α√α If we neglect the mass of the final state quarks/muons then the only difference is the charge of the final state particles: Qµ = −1 Qq = +2 3, −1 3 Prof. Tina Potter 7. QCD 19 Evidence for Colour Consider the ratio R = σ(e+e−→hadrons) σ(e+e−→µ+µ−) For a single quark of a given colour R = Q2 q However, we measure σ(e+e−→hadrons) not just σ(e+e−→u ¯ u) . A jet from a u-quark looks just like a jet from a d-quark etc. Thus, we need to sum over all available flavours (u, d, c, s, t, b) and colours (r, g, b): R = 3 X i Q2 i (3 colours) where the sum is over all quark flavours (i) that are kinematically accessible at centre-of-mass energy, √s, of the collider. Prof. Tina Potter 7. QCD 20 Evidence for Colour Expect to see steps in R as energy is increased. R = 3 X i Q2 i Energy Expected ratio R √s > 2ms, ∼1 GeV 3 4 9 + 1 9 + 1 9 = 2 uds √s > 2mc, ∼4 GeV 3 4 9 + 1 9 + 1 9 + 4 9 = 31 3 udsc √s > 2mb, ∼10 GeV 3 4 9 + 1 9 + 1 9 + 4 9 + 1 9 = 32 3 udscb √s > 2mt, ∼350 GeV 3 4 9 + 1 9 + 1 9 + 4 9 + 1 9 + 4 9 = 5 udscbt Prof. Tina Potter 7. QCD 21 Evidence for Colour R = σ(e+e−→hadrons) σ(e+e−→µ+µ−) R increases in steps with √s Strong evidence for colour √s < 11 GeV region observe bound state resonances: charmonium (c ¯ c) and bottomonium (b¯ b) √s > 50 GeV region observe low edge of Z resonance Γ ∼2.5 GeV. Prof. Tina Potter 7. QCD 22 Experimental Evidence for Colour R = σ(e+e−→hadrons) σ(e+e−→µ+µ−) The existence of Ω−(sss) The Ω−(sss) is a (L = 0) spin-3/2 baryon consisting of three s-quarks. The wavefunction: ψ = s ↑s ↑s ↑ is symmetric under particle interchange. However, quarks are fermions, therefore require an anti-symmetric wave-function, i.e. need another degree of freedom, namely colour , whose wavefunction must be antisymmetric. ψ = (s ↑s ↑s ↑)ψcolour ψcolour = 1 √ 6 (rgb + gbr + brg −grb −rbg −bgr) i.e. need to introduce a new quantum number ( colour ) to distinguish the three quarks in Ω−– avoids violation of Pauli’s Exclusion Principle. Drell-Yan process Need colour to explain cross-section; colours of the annihilating quarks must match to form a virtual photon. Cross-section suppressed by a factor N−2 colour. γ d π−¯ u u p u d d µ+ µ− u d Que Qe Prof. Tina Potter 7. QCD 23 Evidence for Gluons In QED, electrons can radiate photons. In QCD, quarks can radiate gluons. Example: e−e+ →q¯ qg γ e− e+ ¯ q g q Qe Qqe √αs M ∼Qq q2 √α√α√αs Giving an extra factor of √αs in the matrix element, i.e. an extra factor of αs in the cross-section. In QED we can detect the photons. In QCD, we never see free gluons due to confinement. Experimentally, detect gluons as an additional jet: 3-jet events. – Angular distribution of gluon jet depends on gluon spin. Prof. Tina Potter 7. QCD 24 Evidence for Gluons JADE event √s = 31 GeV First direct evidence of gluons (1978) ALEPH event √s = 91 GeV (1990) Distribution of the angle, ϕ , between the highest energy jet (assumed to be one of the quarks) relative to the flight direction of the other two (in their cm frame). ϕ distribution depends on the spin of the gluon. ⇒Gluon is spin 1 Prof. Tina Potter 7. QCD 25 Evidence for Gluon Self-Interactions Direct evidence for the existence of the gluon self-interactions comes from 4-jet events: γ e− e+ ¯ q g g q γ e− e+ ¯ q g g q γ e− e+ ¯ q g g q γ e− e+ ¯ q q ¯ q q The angular distribution of jets is sensitive to existence of triple gluon vertex (lower left diagram) qqg vertex consists of two spin 1/2 quarks and one spin 1 gluon ggg vertex consists of three spin-1 gluons ⇒Different angular distribution. Prof. Tina Potter 7. QCD 26 Evidence for Gluon Self-Interactions ALEPH 4-jet event Experimental method: Define the two lowest energy jets as the gluons. (Gluon jets are more likely to be lower energy than quark jets). Measure angle χ between the plane containing the “quark” jets and the plane containing the “gluon” jets. Gluon self-interactions are required to describe the experimental data. Prof. Tina Potter 7. QCD 27 Measurements of αs αs can be measured in many ways. The cleanest is from the ratio R = σ(e+e−→hadrons) σ(e+e−→µ+µ−) In practise, measure γ e− e+ ¯ q q Qe Qqe + γ e− e+ ¯ q g q Qe Qqe √αs + ... i.e. don’t distinguish between 2 and 3 jets When gluon radiation is included: R = 3 X Q2 q 1 + αs π Therefore, 1 + αs π ∼3.9 3.66 αs(q2 = 252) ∼0.2 Prof. Tina Potter 7. QCD 28 Measurements of αs Many other ways to measure αs Example: 3-jet rate e+e−→q¯ qg R3 = σ(e+e−→3 jets) σ(e+e−→2 jets) ∝αs γ e− e+ ¯ q g q Qe Qqe √αs αs decreases with energy αs runs! in accordance with QCD Prof. Tina Potter 7. QCD 29 Observed running of αs 0.1 0.15 0.2 0.25 10 10 2 10 3 Q [GeV] αs(Q) 0.1 0.15 0.2 0.25 10 10 2 10 3 ATLAS H1 incl. jets + dijets ZEUS inclusive jets JADE event shapes ALEPH event shapes DØ inclusive jets DØ R∆R CMS R32 CMS inclusive jets CMS M3-jet ATLAS TEEC ATLAS R∆φ αs(mZ) = 0.1127 +0.0063 −0.0027 Prof. Tina Potter 7. QCD 30 Summary QCD is a gauge theory, similar to QED, based on SU(3) symmetry Gluons are vector gauge bosons, which couple to (three types of) colour charge (r, b, g) Gluons themselves carry colour charge – hence they have self-interactions (unlike QED). Leads to running of αs, in the opposite sense to QED. Force is weaker at high energies (“asymptotic freedom”) and very strong at low energies. Quarks and gluons are confined. Seen as hadrons and jets of hadrons. Tests of QCD Evidence for colour Existence of gluons, test of their spin and self-interactions Measurement of αs and observation that it runs. Problem Sheet: q.15-16 Up next... Section 8: Quark Model of Hadrons Prof. Tina Potter 7. QCD 31 8. Quark Model of Hadrons Particle and Nuclear Physics Prof. Tina Potter Prof. Tina Potter 8. Quark Model of Hadrons 1 In this section... Hadron wavefunctions and parity Light mesons Light baryons Charmonium Bottomonium Prof. Tina Potter 8. Quark Model of Hadrons 2 The Quark Model of Hadrons Evidence for quarks The magnetic moments of proton and neutron are not µN = eℏ/2mp and 0 respectively ⇒not point-like Electron-proton scattering at high q2 deviates from Rutherford scattering ⇒proton has substructure Hadron jets are observed in e+e−and pp collisions Symmetries (patterns) in masses and properties of hadron states, “quarky” periodic table ⇒sub-structure Steps in R = σ(e+e−→hadrons)/σ(e+e−→µ+µ−) Observation of c ¯ c and b¯ b bound states and much, much more... Here, we will first consider the wave-functions for hadrons formed from light quarks (u, d, s) and deduce some of their static properties (mass and magnetic moments). Then we will go on to discuss the heavy quarks (c, b). We will cover the t quark later... Prof. Tina Potter 8. Quark Model of Hadrons 3 Hadron Wavefunctions Quarks are always confined in hadrons (i.e. colourless states) Mesons Spin 0, 1, ... Baryons Spin 1/2, 3/2, ... Treat quarks as identical fermions with states labelled with spatial, spin, flavour and colour. ψ = ψspaceψflavourψspinψcolour All hadrons are colour singlets, i.e. net colour zero Mesons ψq¯ q colour = 1 √ 3 (r ¯ r + g ¯ g + b¯ b) Baryons ψqqq colour = 1 √ 6 (rgb + gbr + brg −grb −rbg −bgr) Prof. Tina Potter 8. Quark Model of Hadrons 4 Parity The Parity operator, ˆ P, performs spatial inversion ˆ P\|ψ(⃗ r, t)⟩= \|ψ(−⃗ r, t)⟩ The eigenvalue of ˆ P is called Parity ˆ P\|ψ⟩= P\|ψ⟩, P = ±1 Most particles are eigenstates of Parity and in this case P represents intrinsic Parity of a particle/antiparticle. Parity is a useful concept. If the Hamiltonian for an interaction commutes with ˆ P h ˆ P, ˆ H i = 0 then Parity is conserved in the interaction: Parity conserved in the strong and EM interactions, but not in the weak interaction. Prof. Tina Potter 8. Quark Model of Hadrons 5 Parity Composite system of two particles with orbital angular momentum L: P = P1P2(−1)L where P1,2 are the intrinsic parities of particles 1, 2. Quantum Field Theory tells us that Fermions and antifermions: opposite parity Bosons and antibosons: same parity Choose: Quarks and leptons: Pq/ℓ= +1 Antiquarks and antileptons: P¯ q,¯ ℓ= −1 Gauge Bosons: (γ, g, W , Z) are vector fields which transform as JP = 1 − Pγ = −1 Prof. Tina Potter 8. Quark Model of Hadrons 6 Light Mesons Mesons are bound q¯ q states. Consider ground state mesons consisting of light quarks (u, d, s). mu ∼0.3 GeV, md ∼0.3 GeV, ms ∼0.5 GeV Ground State (L = 0): Meson “spin” (total angular momentum) is given by the q¯ q spin state. Two possible q¯ q total spin states: S = 0, 1 S = 0: pseudoscalar mesons S = 1: vector mesons Meson Parity: (q and ¯ q have opposite parity) P = PqP¯ q(−1)L = (+1)(−1)(−1)L = −1 (for L = 0) Flavour States: u ¯ d, u¯ s, d ¯ u, d¯ s, s ¯ u, s ¯ d and u ¯ u, d ¯ d s¯ s mixtures Expect: Nine JP = 0 −mesons: Pseudoscalar nonet Nine JP = 1 −mesons: Vector nonet Prof. Tina Potter 8. Quark Model of Hadrons 7 uds Multiplets Basic quark multiplet – plot the quantum numbers of (anti)quarks: Quarks JP = 1 2 + Antiquarks JP = 1 2 − Mesons Spin J = 0 or 1 The ideas of strangeness and isospin are historical quantum numbers assigned to different states. Essentially they count quark flavours (this was all before the formulation of the Quark Model). Isospin = 1 2(nu −nd −n¯ u + n ¯ d) Strangeness = n¯ s −ns Prof. Tina Potter 8. Quark Model of Hadrons 8 Light Mesons Pseudoscalar nonet JP = 0 − π0, η, η′ are combinations of u ¯ u, d ¯ d, s¯ s Masses / MeV π(140), K(495) η(550), η′(960) Vector nonet JP = 1 − ρ0, ϕ, ω0 are combinations of u ¯ u, d ¯ d, s¯ s Masses/ MeV ρ(770), K ∗(890) ω(780), ϕ(1020) Prof. Tina Potter 8. Quark Model of Hadrons 9 u ¯ u, d ¯ d, s¯ s States The states u ¯ u, d ¯ d and s¯ s all have zero flavour quantum numbers and can mix JP = 0 − π0 = 1 √ 2 (u ¯ u −d ¯ d) η = 1 √ 6 (u ¯ u + d ¯ d −2s¯ s) η′ = 1 √ 3 (u ¯ u + d ¯ d + s¯ s) JP = 1 − ρ0 = 1 √ 2 (u ¯ u −d ¯ d) ω0 = 1 √ 2 (u ¯ u + d ¯ d) ϕ = s¯ s Mixing coefficients determined experimentally from meson masses and decays. Example: Leptonic decays of vector mesons M(ρ0 →e+e−) ∼e q2 1 √ 2 (Que −Qde) Γ(ρ0 →e+e−) ∝ 1 √ 2 (2 3 −(−1 3)) 2 = 1 2 Γ(ω0 →e+e−) ∝ 1 √ 2 (2 3 + (−1 3)) 2 = 1 18 Γ(ϕ →e+e−) ∝ 1 3 2 = 1 9 γ ¯ q q e+ e− Qqe Qe M ∼Qqα Γ ∼Q2 qα2 Predict: Γρ0 : Γω0 : Γϕ = 9 : 1 : 2 Experiment: (8.8 ± 2.6) : 1 : (1.7 ± 0.4) Prof. Tina Potter 8. Quark Model of Hadrons 10 Meson Masses Meson masses are only partly from constituent quark masses: m(K) > m(π) ⇒suggests ms > mu, md 495 MeV 140 MeV Not the whole story... m(ρ) > m(π) ⇒although both are u ¯ d 770 MeV 140 MeV Only difference is the orientation of the quark spins (↑↑vs ↑↓) ⇒spin-spin interaction Prof. Tina Potter 8. Quark Model of Hadrons 11 Meson Masses Spin-spin Interaction QED: Hyperfine splitting in H2 (L = 0) Energy shift due to electron spin in magnetic field of proton ∆E = ⃗ µ.⃗ B = 2 3⃗ µe.⃗ µp\|ψ(0)\|2 and using ⃗ µ = e 2m⃗ S ∆E ∝α ⃗ Se me ⃗ Sp mp QCD: Colour Magnetic Interaction Fundamental form of the interaction between a quark and a gluon is identical to that between an electron and a photon. Consequently, also have a colour magnetic interaction ∆E ∝αs ⃗ S1 m1 ⃗ S2 m2 Prof. Tina Potter 8. Quark Model of Hadrons 12 Meson Masses Meson Mass Formula (L = 0) Mq¯ q = m1 + m2 + A ⃗ S1 m1 ⃗ S2 m2 where A is a constant For a state of spin ⃗ S = ⃗ S1 + ⃗ S2 ⃗ S2 = ⃗ S2 1 + ⃗ S2 2 + 2⃗ S1.⃗ S2 ⃗ S1.⃗ S2 = 1 2 ⃗ S2 −⃗ S2 1 −⃗ S2 2 ⃗ S2 1 = ⃗ S2 2 = ⃗ S1(⃗ S1 + 1) = 1 2 1 2 + 1 = 3 4 giving ⃗ S1.⃗ S2 = 1 2 ⃗ S2 −3 4 For JP = 0 −mesons: ⃗ S2 = 0 ⇒⃗ S1.⃗ S2 = −3/4 For JP = 1 −mesons: ⃗ S2 = S(S + 1) = 2 ⇒⃗ S1.⃗ S2 = +1/4 Giving the (L = 0) Meson Mass formulae: Mq¯ q = m1 + m2 − 3A 4m1m2 JP = 0 − Mq¯ q = m1 + m2 + A 4m1m2 JP = 1 − So JP = 0 −mesons are lighter than JP = 1 −mesons Prof. Tina Potter 8. Quark Model of Hadrons 13 Meson Masses Excellent fit obtained to masses of the different flavour pairs (u ¯ d, u¯ s, d ¯ u, d¯ s, s ¯ u, s ¯ d) with mu = 0.305 GeV, md = 0.308 GeV, ms = 0.487 GeV, A = 0.06 GeV3 η and η′ are mixtures of states, e.g. η = 1 √ 6 u ¯ u + d ¯ d −2s¯ s Mη = 1 6 2mu −3A 4m2 u + 1 6 2md −3A 4m2 d + 4 6 2ms −3A 4m2 s Prof. Tina Potter 8. Quark Model of Hadrons 14 Baryons Baryons made from 3 indistinguishable quarks (flavour can be treated as another quantum number in the wave-function) ψbaryon = ψspace ψflavour ψspin ψcolour ψbaryon must be anti-symmetric under interchange of any 2 quarks Example: Ω−(sss) wavefunction (L = 0, J = 3/2) ψspin ψflavour = s ↑s ↑s ↑ is symmetric ⇒require antisymmetric ψcolour Ground State (L = 0) We will only consider the baryon ground states, which have zero orbital angular momentum ψspace symmetric →All hadrons are colour singlets ψcolour = 1 √ 6 (rgb + gbr + brg −grb −rbg −bgr) antisymmetric Therefore, ψspin ψflavour must be symmetric Prof. Tina Potter 8. Quark Model of Hadrons 15 Baryon spin wavefunctions (ψspin) Combine 3 spin 1/2 quarks: Total spin J = 1 2 ⊗1 2 ⊗1 2 = 1 2 or 3 2 Consider J = 3/2 Trivial to write down the spin wave-function for the 3 2, 3 2 state: 3 2, 3 2 =↑↑↑ Generate other states using the ladder operator ˆ J − ˆ J − 3 2, 3 2 = ( ˆ J −↑) ↑↑+ ↑( ˆ J −↑) ↑+ ↑↑( ˆ J −↑) r 3 2 5 2 −3 2 1 2 3 2, 1 2 = ↓↑↑+ ↑↓↑+ ↑↑↓ 3 2, 1 2 = 1 √ 3 (↓↑↑+ ↑↓↑+ ↑↑↓) ˆ J −\|j, m⟩= p j(j + 1) −m(m −1) \|j, m −1⟩ Giving the J = 3/2 states: − → All symmetric under interchange of any two spins 3 2, 3 2 =↑↑↑ 3 2, 1 2 = 1 √ 3 (↓↑↑+ ↑↓↑+ ↑↑↓) 3 2, −1 2 = 1 √ 3 (↑↓↓+ ↓↑↓+ ↓↓↑) 3 2, −3 2 =↓↓↓ Prof. Tina Potter 8. Quark Model of Hadrons 16 Baryon spin wavefunctions (ψspin) Consider J = 1/2 First consider the case where the first 2 quarks are in a \|0, 0⟩state: \|0, 0⟩(12) = 1 √ 2 (↑↓−↓↑) 1 2, 1 2 (123) = \|0, 0⟩(12) 1 2, 1 2 = 1 √ 2 (↑↓↑−↓↑↑) 1 2, −1 2 (123) = \|0, 0⟩(12) 1 2, −1 2 = 1 √ 2 (↑↓↓−↓↑↓) Antisymmetric under exchange 1 ↔2. Three-quark J = 1/2 states can also be formed from the state with the first two quarks in a symmetric spin wavefunction. Can construct a three-particle state 1 2, 1 2 (123) from \|1, 0⟩(12) 1 2, 1 2 (3) and \|1, 1⟩(12) 1 2, −1 2 (3) Prof. Tina Potter 8. Quark Model of Hadrons 17 Baryon spin wavefunctions (ψspin) Taking the linear combination 1 2, 1 2 = a \|1, 1⟩ 1 2, −1 2 + b \|1, 0⟩ 1 2, 1 2 with a2 + b2 = 1. Act upon both sides with ˆ J+ ˆ J+ 1 2, 1 2 = a ˆ J+ \|1, 1⟩ 1 2, −1 2 + \|1, 1⟩ ˆ J+ 1 2, −1 2 + b ˆ J+ \|1, 0⟩ 1 2, 1 2 + \|1, 0⟩ ˆ J+ 1 2, 1 2 0 = a \|1, 1⟩ 1 2, 1 2 + √ 2b \|1, 1⟩ 1 2, 1 2 a = − √ 2b ˆ J + \|j, m⟩= p j(j + 1) −m(m + 1) \|j, m + 1⟩ which with a2 + b2 = 1 implies a = q 2 3, b = − q 1 3 Giving 1 2, 1 2 = r 2 3 \|1, 1⟩ 1 2, −1 2 − r 1 3 \|1, 0⟩ 1 2, −1 2 \|1, 1⟩=↑↑ \|1, 0⟩= 1 √ 2 (↑↓+ ↓↑) 1 2, 1 2 = 1 √ 6 (2 ↑↑↓−↑↓↑−↓↑↑) 1 2, −1 2 = 1 √ 6 (2 ↓↓↑−↓↑↓−↑↓↓) Symmetric under interchange 1 ↔2 Prof. Tina Potter 8. Quark Model of Hadrons 18 Three-quark spin wavefunctions J = 3/2 3 2, 3 2 =↑↑↑ 3 2, 1 2 = 1 √ 3 (↓↑↑+ ↑↓↑+ ↑↑↓) 3 2, −1 2 = 1 √ 3 (↑↓↓+ ↓↑↓+ ↓↓↑) 3 2, −3 2 =↓↓↓ Symmetric under interchange of any 2 quarks J = 1/2 1 2, 1 2 = 1 √ 2 (↑↓↑−↓↑↑) 1 2, −1 2 = 1 √ 2 (↑↓↓−↓↑↓) Antisymmetric under interchange of 1 ↔2 J = 1/2 1 2, 1 2 = 1 √ 6 (2 ↑↑↓−↑↓↑−↓↑↑) 1 2, −1 2 = 1 √ 6 (2 ↓↓↑−↓↑↓−↑↓↓) Symmetric under interchange of 1 ↔2 ψspin ψflavour must be symmetric under interchange of any 2 quarks. Prof. Tina Potter 8. Quark Model of Hadrons 19 Three-quark spin wavefunctions Consider 3 cases: 1 Quarks all same flavour: uuu, ddd, sss ψflavour is symmetric under interchange of any two quarks Require ψspin to be symmetric under interchange of any two quarks Only satisfied by J = 3/2 states There are no J = 1/2 uuu, ddd, sss baryons with L = 0. Three J = 3/2 states: uuu, ddd, sss 2 Two quarks have same flavour: uud, uus, ddu, dds, ssu, ssd For the like quarks ψflavour is symmetric Require ψspin to be symmetric under interchange of like quarks 1 ↔2 Satisfied by J = 3/2 and J = 1/2 states Six J = 3/2 states and six J = 1/2 states: uud, uus, ddu, dds, ssu, ssd Prof. Tina Potter 8. Quark Model of Hadrons 20 Three-quark spin wavefunctions 3 All quarks have different flavours: uds Two possibilities for the (ud) part: Flavour Symmetric 1 √ 2(ud + du) Require ψspin to be symmetric under interchange of ud Satisfied by J = 3/2 and J = 1/2 states One J = 3/2 and one J = 1/2 state: uds Flavour Antisymmetric 1 √ 2(ud −du) Require ψspin to be antisymmetric under interchange of ud Only satisfied by J = 1/2 state One J = 1/2 state: uds Quark Model predicts that light baryons appear in Decuplets (10) of spin 3/2 states Octets (8) of spin 1/2 states Prof. Tina Potter 8. Quark Model of Hadrons 21 Baryon Multiplets Octet JP = 1 2 + Decuplet JP = 3 2 + Antibaryons are in separate multiplets Example: Antiparticle of Σ+(uus) is ¯ Σ −(¯ u ¯ u¯ s), JP = 1 2 −and not Σ −(dds), JP = 1 2 + Prof. Tina Potter 8. Quark Model of Hadrons 22 Baryon Masses Baryon Mass Formula (L = 0) Mqqq = m1 + m2 + m3 + A′ ⃗ S1 m1 . ⃗ S2 m2 + ⃗ S1 m1 . ⃗ S3 m3 + ⃗ S2 m2 . ⃗ S3 m3 ! where A′ is a constant Example: All quarks have the same mass, m1 = m2 = m3 = mq Mqqq = 3mq + A′ X i mass of daughters). And, angular momentum and parity must be conserved in strong decays. Examples: ρ0 →π+π− m(ρ0) > m(π+) + m(π−) 769 140 140 MeV ∆++ →pπ+ m(∆++) > m(p) + m(π+) 1231 938 140 MeV Prof. Tina Potter 8. Quark Model of Hadrons 32 Hadron Decays Also need to check for identical particles in the final state. Examples: ω0 →π0π0 m(ω0) > m(π0) + m(π0) 782 135 135 MeV ω0 →π+π−π0 m(ω0) > m(π+) + m(π−) + m(π0) 782 140 140 135 MeV Prof. Tina Potter 8. Quark Model of Hadrons 33 Hadron Decays Hadrons can also decay via the electromagnetic interaction. Examples: ρ0 →π0γ m(ρ0) > m(π0) + m(γ) 769 135 MeV Σ0 →Λ0γ m(Σ0) > m(Λ0) + m(γ) 1193 1116 MeV The lightest mass states (p, K ±, K 0, ¯ K 0, Λ, n) require a change of quark flavour in the decay and therefore decay via the weak interaction (see later). Prof. Tina Potter 8. Quark Model of Hadrons 34 Summary of light (uds) hadrons Baryons and mesons are composite particles (complicated). However, the naive Quark Model can be used to make predictions for masses/magnetic moments. The predictions give reasonably consistent values for the constituent quark masses: mu/d ms Meson Masses 307 MeV 487 MeV Baryon Masses 364 MeV 537 MeV Baryon Magnetic Moments 336 MeV 509 MeV mu ∼md ∼335 MeV, ms ∼510 MeV Hadrons will decay via the strong interaction to lighter mass states if energetically feasible. Hadrons can also decay via the EM interaction. The lightest mass states require a change of quark flavour to decay and therefore decay via the weak interaction (see later). Prof. Tina Potter 8. Quark Model of Hadrons 35 Heavy hadrons The November Revolution Brookhaven National Laboratory Led by Samuel Ting J particle: PRL 33 (1974) 1404 Stanford Linear Accelerator Center, SPEAR Led by Burton Richter ψ particle: PRL 33 (1974) 1406 Both experiments announced discovery on 11 November 1974 ⇒J/ψ 1976 Nobel Prize awarded to Ting and Richter. Prof. Tina Potter 8. Quark Model of Hadrons 36 Heavy hadrons Charmonium 1974: Discovery of a narrow resonance in e+e− collisions at √s ∼3.1 GeV J/ψ(3097) Observed width ∼3 MeV, all due to experimental resolution. Actual Total Width, ΓJ/ψ ∼97 keV Branching fractions: B(J/ψ →hadrons) ∼88% B(J/ψ →µ+µ−) ∼(J/ψ →e+e−) ∼6% Partial widths: ΓJ/ψ→hadrons ∼87 keV ΓJ/ψ→µ+µ−∼ΓJ/ψ→e+e−∼5 keV Mark II Experiment, SLAC, 1978 Prof. Tina Potter 8. Quark Model of Hadrons 37 Heavy hadrons Charmonium Resonance seen in R = σ(e+e−→hadrons) σ(e+e−→µ+µ−) Zoom into the charmonium (c ¯ c) region √s ∼2mc mass of charm quark, mc ∼1.5 GeV Resonances due to formation of bound unstable c ¯ c states. The lowest energy of these is the narrow J/ψ state. γ c ¯ c γ e− e+ ¯ q q Prof. Tina Potter 8. Quark Model of Hadrons 38 Charmonium c ¯ c bound states produced directly in e+e−collisions must have the same spin and parity as the photon γ c ¯ c γ e− e+ ¯ q q JP = 1 − However, expect that a whole spectrum of bound c ¯ c states should exist (analogous to e+e−bound states, positronium) n = 1 L = 0 S = 0, 1 1S0,3 S1 n = 2 L = 0, 1 S = 0, 1 1S0,3 S1,1 P1,3 P0,1,2 ... etc Parity = (−1)(−1)L 2S+1LJ Prof. Tina Potter 8. Quark Model of Hadrons 39 The Charmonium System Prof. Tina Potter 8. Quark Model of Hadrons 40 The Charmonium System All c ¯ c bound states can be observed via their decay: Example: Hadronic decay ψ(3685) →J/ψ π+π− Example: Photonic decay ψ(3685) →χ + γ χ →J/ψ + γ Peaks in γ spectrum Charmonium Spectroscopy Prof. Tina Potter 8. Quark Model of Hadrons 41 The Charmonium System Knowing the c ¯ c energy levels provides a probe of the QCD potential. Because QCD is a theory of a strong confining force (self-interacting gluons), it is very difficult to calculate the exact form of the QCD potential from first principles. However, it is possible to experimentally “determine” the QCD potential by finding an appropriate form which gives the observed charmonium states. In practise, the QCD potential VQCD = −4 3 αs r + kr with αs = 0.2 and k = 1 GeVfm−1 provides a good description of the experimentally observed levels in the charmonium system. Prof. Tina Potter 8. Quark Model of Hadrons 42 Why is the J/ψ so narrow? Consider the charmonium 3S1 states: 13S1 ψ(3097) Γ ∼0.09 MeV 23S1 ψ(3685) Γ ∼0.24 MeV 33S1 ψ(3767) Γ ∼25 MeV 43S1 ψ(4040) Γ ∼50 MeV The width depends on whether the decay to lightest mesons containing c quarks, D−(d ¯ c), D+(c ¯ d), is kinematically possible or not: m(D±) = 1869.4 ± 0.5 MeV m(ψ) > 2m(D) ¯ c ψ c ¯ c d D− ¯ d D+ c ψ →D+D−allowed “ordinary” strong decay ⇒large width m(ψ) < 2m(D) ¯ c ψ c ¯ u d π− ¯ d d π0 ¯ d u π+ Zweig Rule: Unconnected lines in the Feynman diagram lead to suppression of the decay amplitude ⇒narrow width Prof. Tina Potter 8. Quark Model of Hadrons 43 Charmed Hadrons The existence of the c quark ⇒expect to see charmed mesons and baryons (i.e. containing a c quark). Extend quark symmetries to 3 dimensions: Mesons Baryons Prof. Tina Potter 8. Quark Model of Hadrons 44 Heavy hadrons the Υ (b¯ b) E288 collaboration, Fermilab Led by Leon Lederman Υ particle: PRL 39 (1977) 252-255 1977: Discovery of the Υ(9460) resonance state. Lowest energy 3S1 bound b¯ b state (bottomonium). ⇒mb ∼4.7 GeV Similar properties to the ψ (9460) (10023) (10355) (10580) Full Width ~53 keV 44 keV 26 keV 14 MeV Narrow Wide e +e −→Υ→hadrons Prof. Tina Potter 8. Quark Model of Hadrons 45 Bottomonium Bottomonium is the analogue of charmonium for b quark. Bottomonium spectrum well described by same QCD potential as used for charmonium. Evidence that QCD potential does not depend on quark type. Wide Narrow ¯ b Υ b ¯ b u B− ¯ u B+ b ¯ b Υ b ¯ u d π− ¯ d d π0 ¯ d u π+ Zweig suppressed Prof. Tina Potter 8. Quark Model of Hadrons 46 Bottom Hadrons Extend quark symmetries to 4 dimensions (difficult to draw!) Examples: Mesons JP = 0− : B−(b¯ u); B0(¯ bd); B0 s (¯ bs); B− c (b¯ c) The B− c is the heaviest hadron discovered so far: m(B− c ) = 6.4 ± 0.4 GeV JP = 1− : B∗−(b¯ u); B∗0(¯ bd); B∗0 s (¯ bs) The mass of the B∗mesons is only 50 MeV above the B meson mass. Expect only electromagnetic decays B∗→Bγ. Baryons JP = 1 2 + : Λb(bud); Σb(buu); Ξb(bus) Prof. Tina Potter 8. Quark Model of Hadrons 47 Summary of heavy hadrons c and b quarks were first observed in bound state resonances (“onia”). Consequences of the existence of c and b quarks are Spectra of c ¯ c (charmonium) and b¯ b (bottomonium) bound states Peaks in R = σ(e+e−→hadrons) σ(e+e−→µ+µ−) Existence of mesons and baryons containing c and b quarks The majority of charm and bottom hadrons decay via the weak interaction (strong and electromagnetic decays are forbidden by energy conservation). The t quark is very heavy and decays rapidly via the weak interaction before a t¯ t bound state (or any other hadron) can be formed. τt ∼10−25 s thadronisation ∼10−22 s Rapid decay because m(t) > m(W ) so weak interaction is no longer weak. m(u) = 335 MeV m(d) = 335 MeV m(c) = 1.5 GeV m(s) = 510 MeV m(t) = 175 GeV m(b) = 4.5 GeV Prof. Tina Potter 8. Quark Model of Hadrons 48 Tetraquarks and Pentaquarks (non-examinable) Quark Model of Hadrons is not limited to q¯ q or qqq content. Recent observations from LHCb show unquestionable discovery of pentaquark states, PRL 115, 072001 (2015). K − p J / ψ p Pc(4450)+ Pc(4380)+ + others more recently. How are these quarks bound? qqqqq? qq + qqq? qq + qq + q? A few tetraquarks discovered by Belle and BESIII e.g. Z(4430)−, c ¯ cd ¯ u discovered by Belle and confirmed by LHCb LHCb has discovered many more! Prof. Tina Potter 8. Quark Model of Hadrons 49 Summary Evidence for hadron sub-structure – quarks Hadron wavefunctions and allowed states Hadron masses and magnetic moments Hadron decays (strong, EM, weak) Heavy hadrons: charmonium and bottomonium Recent tetraquark and pentaquark discoveries Problem Sheet: q.17-22 Up next... Section 9: The Weak Force Prof. Tina Potter 8. Quark Model of Hadrons 50 9. The Weak Force Particle and Nuclear Physics Prof. Tina Potter Prof. Tina Potter 9. The Weak Force 1 In this section... The charged current weak interaction Four-fermion interactions Massive propagators and the strength of the weak interaction C-symmetry and Parity violation Lepton universality Quark interactions and the CKM Prof. Tina Potter 9. The Weak Force 2 The Weak Interaction The weak interaction accounts for many decays in particle physics, e.g. µ−→e−¯ νeνµ τ −→e−¯ νeντ π+ →µ−¯ νµ n →pe−¯ νe Characterised by long lifetimes and small interaction cross-sections Prof. Tina Potter 9. The Weak Force 3 The Weak Interaction Two types of weak interaction Charged current (CC): W ± bosons Neutral current (NC): Z bosons See Chapter 10 The weak force is mediated by massive vector bosons: mW = 80 GeV mZ = 91 GeV Examples: (The list below is not complete, will see more vertices later!) Weak interactions of electrons and neutrinos: W − ¯ νe e− W + e+ νe Z e+ e− Z ¯ νe νe Prof. Tina Potter 9. The Weak Force 4 Boson Self-Interactions In QCD the gluons carry colour charge. In the weak interaction the W ± and Z bosons carry the weak charge W ± also carry the electric charge ⇒boson self-interactions Z W + W − γ W + W − W − W + W − W + Z Z W − W + γ Z W − W + γ γ W − W + (The list above is complete as far as weak self-interactions are concerned, but we have still not seen all the weak vertices. Will see the rest later) Prof. Tina Potter 9. The Weak Force 5 Fermi Theory The old (“imperfect”) idea Weak interaction taken to be a “4-fermion contact interaction” No propagator Coupling strength given by the Fermi constant GF GF = 1.166 × 10−5 GeV−2 β-decay in Fermi Theory n p ¯ νe e− GF µ− νµ ¯ νe e− GF Neutrino scattering in Fermi Theory νµ e− µ− νe GF Prof. Tina Potter 9. The Weak Force 6 Why must Fermi Theory be “Wrong”? νe + n →p + e− dσ = 2π\|Mfi\|2 dN dE = 2π4G 2 F E 2 e (2π)3 dΩ σ = G 2 Fs π See Appendix F νe n e− p GF where Ee is the energy of the e−in the centre-of-mass system and √s is the energy in the centre-of-mass system. In the laboratory frame: s = 2Eνmn (fixed target collision, see Chapter 3) ⇒σ ∼(Eν/ MeV) × 10−43 cm−2 ν’s only interact weakly ∴have very small interaction cross-sections. Here weak implies that you need approximately 50 light-years of water to stop a 1 MeV neutrino! However, as Eν →∞the cross-section can become very large. Violates maximum value allowed by conservation of probability at √s ∼1 TeV (“unitarity limit”). This is a big problem. ⇒Fermi theory breaks down at high energies. Prof. Tina Potter 9. The Weak Force 7 Weak Charged Current: W ± Boson Fermi theory breaks down at high energy True interaction described by exchange of charged W ± bosons Fermi theory is the low energy (q2 ≪m2 W) effective theory of the weak interaction β decay Old Fermi Theory n p ¯ νe e− GF Standard Model W − d n u d e− ¯ νe u u p d VudgW gW νee− scattering νµ e− µ− νe GF W ± νµ e− µ− νe gW gW Prof. Tina Potter 9. The Weak Force 8 Weak Charged Current: W ± Boson Weak W ± νµ e− µ− νe gW gW Propagator ∼ 1 q2 −m2 W αW = g2 W 4π QED γ µ− e− µ− e− α α Propagator ∼1 q2 α = e2 4π Charged Current Weak Interaction At low energies, q2 ≪m2 W, propagator 1 q2−m2 W → 1 −m2 W i.e. appears as the point-like interaction of Fermi theory. Massive propagator →short range mW = 80.4 GeV ⇒Range ∼ 1 mW ∼0.002 fm Exchanged boson carries electromagnetic charge. Flavour changing - only the CC weak interaction changes flavour Parity violating - only the CC weak interaction can violate parity conservation Prof. Tina Potter 9. The Weak Force 9 Weak Charged Current: W ± Boson Compare Fermi theory with a massive propagator µ− νµ ¯ νe e− GF W − µ− ¯ νe e− νµ gW gW For q2 ≪m2 W compare matrix elements g2 W m2 W →GF GF is small because mW is large The precise relationship is: g2 W 8m2 W →GF √ 2 The numerical factors are partly of historical origin (see Perkins 4th ed., page 210). mW = 80.4 GeV and GF = 1.166 × 10−5 GeV−2 measured in muon β decay gW = 0.65 and αW = g2 W 4π ∼1 30 Compare to EM α = e2 4π ∼ 1 137 The intrinsic strength of the weak interaction is actually greater than that of the electromagnetic interaction. At low energies (low q2), it appears weak owing to the massive propagator. Prof. Tina Potter 9. The Weak Force 10 Weak Charged Current: W ± Boson Neutrino Scattering with a Massive W Boson Replace contact interaction by massive boson exchange diagram: W ± νµ e− µ− νe gW gW Fermi theory dσ dΩ= 2πG 2 F E 2 e (2π)3 Standard Model dσ dΩ= 2πG 2 F E 2 e (2π)3 m2 W m2 W −q2 2 with \|⃗ q2\| = 4E 2 e sin2 θ/2, where θ is the scattering angle. Integrate to give σ = G 2 Fs π s ≪m2 W σ = G 2 Fm2 W π s ≫m2 W see Appendix G Cross-section is now well behaved at high energies. Prof. Tina Potter 9. The Weak Force 11 Spin and helicity Consider a free particle of constant momentum, ⃗ p Total angular momentum, ⃗ J = ⃗ L + ⃗ S is always conserved The orbital angular momentum, ⃗ L = ⃗ r × ⃗ p is perpendicular to ⃗ p The spin angular momentum, ⃗ S can be in any direction relative to ⃗ p The value of spin ⃗ S along ⃗ p is always constant The sign of the component of spin along the direction of motion is known as the “helicity”, h = ⃗ S.⃗ p \|⃗ p\| Taking spin 1/2 as an example: ⃗ p ⃗ S h=+ 1 2 “Right-handed” ⃗ p ⃗ S h=−1 2 “Left-handed” Prof. Tina Potter 9. The Weak Force 12 The Wu Experiment β decay of 60Co →60 Ni + e−+ ¯ νe 1956 Chien-Shiung Wu Align cooled 60Co nuclei with ⃗ B field and look at direction of emission of electrons e−always observed in direction opposite to spin – left-handed. ⃗ p conservation: ¯ ν must be emitted in opposite direction – right-handed. Right-handed e−not observed here ⇒Parity Violation Prof. Tina Potter 9. The Weak Force 13 The Weak Interaction and Helicity The weak interaction distinguishes between left- and right-handed states. This is an experimental observation, which we need to build into the Standard Model. The weak interaction couples preferentially to left-handed particles and right-handed antiparticles To be precise, the probability for weak coupling to the ± helicity state is 1 2 1 ∓v c for a lepton →coupling to RH particles vanishes 1 2 1 ± v c for an antilepton →coupling to LH antiparticles vanishes ⇒right-handed ν’s do not exist left-handed ¯ ν’s do not exist Even if they did exist, they would be unobservable. Prof. Tina Potter 9. The Weak Force 14 Charge Conjugation C-symmetry: the physics for +Q should be the same as for −Q. This is true for QED and QCD, but not the Weak force... LH e− Charge Conjugation − − − − − − − − − − → LH e+ EM, Weak EM, Weak RH e− Charge Conjugation − − − − − − − − − − → RH e+ EM, Weak EM, Weak LH νe Charge Conjugation − − − − − − − − − − → LH ¯ νe Weak Weak C-symmetry is maximally violated in the weak interaction. Prof. Tina Potter 9. The Weak Force 15 Parity Violation Parity is always conserved in the strong and EM interactions η →π0π0π0 η →π+π− Prof. Tina Potter 9. The Weak Force 16 Parity Violation Parity is often conserved in the weak interaction, but not always The weak interaction treats LH and RH states differently and therefore can violate parity (because the interaction Hamiltonian does not commute with ˆ P). K + →π+π−π+ K + →π+π0 Prof. Tina Potter 9. The Weak Force 17 Weak interactions of leptons All weak charged current lepton interactions can be described by the W boson propagator and the weak vertex: e−, µ−, τ − νe, νµ, ντ W − gW The Standard Model Weak CC Lepton Vertex + antiparticles W bosons only “couple” to the (left-handed) lepton and neutrino within the same generation e− νe µ− νµ τ − ντ e.g. no W ±e−νµ coupling Coupling constant gW αW = g2 W 4π Prof. Tina Potter 9. The Weak Force 18 Weak interactions of leptons Examples W −→e−¯ νe, µ−¯ νµ, τ −¯ ντ W − ¯ νe, ¯ νµ, ¯ ντ e−, µ−, τ − µ−→e−¯ νeνµ W − µ− ¯ νe e− νµ gW gW n →pe−¯ νe W − d n u d e− ¯ νe u u p d VudgW gW W + →e+νe, µ+νµ, τ +ντ W + νe, νµ, ντ e+, µ+, τ + τ −→e−¯ νeντ W − τ − ¯ νe e− ντ gW gW B− c →J/ψe−¯ νe W − b B− c ¯ c e− ¯ νe c ¯ c J/ψ VbcgW gW Prof. Tina Potter 9. The Weak Force 19 µ Decay Muons are fundamental leptons (mµ ∼206me) Electromagnetic decay µ−→e−γ is not observed (branching ratio < 2.4 × 10−12) ⇒the EM interaction does not change flavour. Only the weak CC interaction changes lepton type, but only within a generation. “Lepton number conservation” for each lepton generation. Muons decay weakly: µ−→e−¯ νeνµ W − µ− ¯ νe e− νµ gW gW µ− νµ ¯ νe e− GF As mµ ≪mW can safely use Fermi theory to calculate decay width (analogous to nuclear β decay). Prof. Tina Potter 9. The Weak Force 20 µ Decay Fermi theory gives decay width ∝m5 µ (Sargent Rule) However, more complicated phase space integration (previously neglected kinetic energy of recoiling nucleus) and taking account of helicity/spin gives different constants Γµ = 1 τµ = G 2 F 192π3m5 µ Muon mass and lifetime known with high precision. mµ = 105.6583715 ± 0.0000035 MeV τµ = (2.1969811 ± 0.0000022) × 10−6 s Use muon decay to fix strength of weak interaction GF GF = (1.16632 ± 0.00002) × 10−5 GeV−2 GF is one of the best determined fundamental quantities in particle physics. Prof. Tina Potter 9. The Weak Force 21 τ Decay The τ mass is relatively large mτ = 1.77686 ± 0.00012 GeV Since mτ > mµ, mπ, mp, ... there are a number of possible decay modes W − τ − ¯ νe e− ντ gW gW W − τ − ¯ νµ µ− ντ gW gW W − τ − ¯ u d, s ντ gW gW Measure the τ branching fractions as: τ −→e−¯ νeντ 17.83 ± 0.04% τ −→µ−¯ νµντ 17.41 ± 0.04% τ −→hadrons 64.76 ± 0.06% Prof. Tina Potter 9. The Weak Force 22 Lepton Universality Do all leptons have the same weak coupling? Look at measurements of the decay rates and branching fractions. W − µ− ¯ νe e− νµ gW gW 1 τµ = Γµ→e = G 2 F 192π3m5 µ W − τ − ¯ νe e− ντ gW gW 1 ττ = Γτ→e B(τ →e) = 1 0.178 G 2 F 192π3m5 τ If weak interaction strength is universal, expect: ττ τµ = 0.178 m5 µ m5 τ Measure mµ, mτ, τµ to high precision: mµ = 105.6583715 ± 0.0000035 MeV mτ = 1.77686 ± 0.00012 GeV τµ = (2.1969811 ± 0.0000022) × 10−6 s Predict ττ = (2.903 ± 0.005) × 10−13 s Measure ττ = (2.903 ± 0.005) × 10−13 s ⇒same weak CC coupling for µ and τ Prof. Tina Potter 9. The Weak Force 23 Lepton Universality We can also compare W − τ − ¯ νe e− ντ gW gW W − τ − ¯ νµ µ− ντ gW gW If the couplings are the same, expect: B(τ −→µ−¯ νµντ) B(τ −→e−¯ νeντ) = 0.9726 (the small difference is due to the slight reduction in phase space due to the non-negligible muon mass). Measured B(τ −→µ−¯ νµντ) B(τ −→e−¯ νeντ) = 0.974 ± 0.005 consistent with prediction. ⇒same weak CC coupling for e, µ and τ ⇒Lepton Universality Prof. Tina Potter 9. The Weak Force 24 Universality of Weak Coupling Compare GF measured from µ−decay with that from nuclear β decay W − µ− ¯ νe e− νµ gW gW G µ F = (1.16632±0.00002)×10−5 GeV−2 W − d n u d e− ¯ νe u u p d VudgW gW G β F = (1.136 ± 0.003) × 10−5 GeV−2 Ratio G β F G µ F = 0.974 ± 0.003 Conclude that the strength of the weak interaction is almost the same for leptons as for quarks. But the difference is significant, and has to be explained. Prof. Tina Potter 9. The Weak Force 25 Weak Interactions of Quarks Impose a symmetry between leptons and quarks, so weak CC couplings take place within one generation: Leptons e− νe µ− νµ τ − ντ Quarks u d c s t b W − ¯ νe e− W − ¯ u d W − ¯ νµ µ− W − ¯ c s W − ¯ ντ τ − W − ¯ t b So π+ →µ+νµ would be allowed W + ¯ d π+ u µ+ νµ VudgW gW but K + →µ+νµ would not W + ¯ s K+ u µ+ νµ VusgW gW But we have observed K + →µ+νµ ! (much smaller rate than π+ decay.) Prof. Tina Potter 9. The Weak Force 26 Quark Mixing Instead, alter the lepton-quark symmetry to: (only considering 1st and 2nd gen. here) Leptons e− νe µ− νµ Quarks u d′ c s′ where d′ = d cos θC + s sin θC s′ = −d sin θC + s cos θC Now, the down type quarks in the weak interaction are actually linear superpositions of the down type quarks i.e. weak eigenstates (d′,s′) are superpositions of the mass eigenstates (d,s) Weak Eigenstates d′ s′ = cos θC sin θC −sin θC cos θC d s Mass Eigenstates ⇒Cabibbo angle θC ∼13◦(from experiment) Prof. Tina Potter 9. The Weak Force 27 Quark Mixing Now, the weak coupling to quarks is: d cos θC + s sin θC W − ¯ u d′ = W − ¯ u d gW cos θC + W − ¯ u s gW sin θC −d sin θC + s cos θC W − ¯ c s′ = W − ¯ c d −gW sin θC + W − ¯ c s gW cos θC Quark mixing explains the lower rate of K + →µ+νµ compared to π+ →µ+νµ and the ratio G β F G µ F = 0.974 ± 0.003 Difference in couplings affects \|M\|2 ∝(G β F )2 ∝(cos θC)2 Now expect G β F G µ F = cos θC which holds for θC ∼13◦ Prof. Tina Potter 9. The Weak Force 28 CKM matrix Cabibbo-Kobayashi-Maskawa Matrix Extend quark mixing to three generations W − ¯ u d′ W − ¯ c s′ W − ¯ t b′ Weak Eigenstates   d′ s′ b′  = VCKM   d s b   Mass Eigenstates VCKM =   Vud Vus Vub Vcd Vcs Vcb Vtd Vts Vtb  ∼   cos θC sin θC sin3 θCe−iδ −sin θC cos θC sin2 θC sin3 θCeiδ −sin2 θC 1   Unitary matrix. Mixing angle θC ∼13◦ Charge-Parity violating phase δ ∼69◦ (full CKM matrix includes 3 mixing angles) ∼   0.975 0.220 0.01 −0.220 0.975 0.05 0.01 −0.05 1   Prof. Tina Potter 9. The Weak Force 29 Quark Mixing Weak interactions between quarks of the same family are “Cabibbo Allowed” W − ¯ u d gW Vud W − ¯ c s gW Vcs W − ¯ t b gW Vtb between quarks differing by one family are “Cabibbo Suppressed” W − ¯ u s gW Vus W − ¯ c d gW Vcd W − ¯ c b gW Vcb W − ¯ t s gW Vts between quarks differing by two families are “Doubly Cabbibo Suppressed” W − ¯ u b gW Vub W − ¯ t d gW Vdt Prof. Tina Potter 9. The Weak Force 30 Quark Mixing Examples K + →µ+νµ W + ¯ s K+ u µ+ νµ VusgW gW u¯ s coupling ⇒Cabbibo suppressed \|M\|2 ∝g4 WV 2 us = g4 W sin2 θC D0 →K −π+ W + c D0 ¯ u ¯ d π+ u s ¯ u K− gW Vcs gW Vud D0 →K +π− W + c D0 ¯ u ¯ s K+ u d ¯ u π− gW Vcd gW Vus Expect Γ(D0 →K +π−) Γ(D0 →K −π+) ∼(g2 WVcdVus)2 (g2 WVcsVud)2 = sin4 θC cos4 θC ∼0.0028 Measure 0.0038 ± 0.0008 D0 →K +π−is Doubly Cabibbo suppressed (two Cabibbo suppressed vertices) Prof. Tina Potter 9. The Weak Force 31 Summary of the Weak CC Vertex All weak charged current quark interactions can be described by the W boson propagator and the weak vertex: q = d, s, b q′ = u, c, t W − gW Vqq′ The Standard Model Weak CC Quark Vertex + antiparticles W ± bosons always change quark flavour W ± prefers to couple to quarks in the same generation, but quark mixing means that cross-generation coupling can occur. Crossing two generations is less probable than one. W -lepton coupling constant − →gW W -quark coupling constant − →gWVCKM Prof. Tina Potter 9. The Weak Force 32 Summary Weak interaction (charged current) Weak force mediated by massive W bosons mW = 80.385 ± 0.015 GeV Weak force intrinsically stronger than EM interaction αW ∼1 30 αEM ∼1 137 Universal coupling to quarks and leptons, but... Quarks take part in the interaction as mixtures of the mass eigenstates Parity & C-symmetry can be violated due to the helicity structure of the interaction Strength of the weak interaction given by G µ F = (1.16632 ± 0.00002) × 10−5 GeV−2 from µ decay. Problem Sheet: q.23-25 Up next... Section 10: Electroweak Unification Prof. Tina Potter 9. The Weak Force 33 10. Electroweak Unification Particle and Nuclear Physics Prof. Tina Potter Prof. Tina Potter 10. Electroweak Unification 1 In this section... GWS model Allowed vertices Revisit Feynman diagrams Experimental tests of Electroweak theory Prof. Tina Potter 10. Electroweak Unification 2 Electroweak Unification Weak CC interactions explained by W ± boson exchange W ± bosons are charged, thus they couple to the γ Consider e−e+ →W +W −: 2 diagrams (+interference) νe e− e+ W − W + γ e− e+ W − W + Cross-section diverges at high energy Divergence cured by introducing Z boson Extra diagram for e−e+ →W +W − Idea only works if γ, W ±, Z couplings are related ⇒Electroweak Unification Z e− e+ W − W + Prof. Tina Potter 10. Electroweak Unification 3 Electroweak gauge theory (non-examinable) Postulate invariance under a gauge transformation like: ψ →ψ′ = eig⃗ σ.⃗ Λ(⃗ r,t)ψ an “SU(2)” transformation (σ are 2x2 matrices). Operates on the state of “weak isospin” – a “rotation” of the isospin state. Invariance under SU(2) transformations ⇒three massless gauge bosons (W1, W2, W3) whose couplings are well specified. They also have self-couplings. But this doesn’t quite work... Predicts W and Z have the same couplings – not seen experimentally! Prof. Tina Potter 10. Electroweak Unification 4 Electroweak gauge theory The solution... Unify QED and the weak force ⇒electroweak model “SU(2)xU(1)” transformation U(1) operates on the “weak hypercharge” Y = 2(Q −I3) SU(2) operates on the state of “weak isospin, I” Invariance under SU(2)xU(1) transformations ⇒four massless gauge bosons W +, W −, W3, B The two neutral bosons W3 and B then mix to produce the physical bosons Z and γ Photon properties must be the same as QED ⇒predictions of the couplings of the Z in terms of those of the W and γ Still need to account for the masses of the W and Z. This is the job of the Higgs mechanism (later). Prof. Tina Potter 10. Electroweak Unification 5 The GWS Model The Glashow, Weinberg and Salam model treats EM and weak interactions as different manifestations of a single unified electroweak force (Nobel Prize 1979) Start with 4 massless bosons W +, W3, W −and B. The neutral bosons mix to give physical bosons (the particles we see), i.e. the W ±, Z, and γ.   W + W3 W −  ; B →   W + Z W −  ; γ Physical fields: W +, Z, W −and A (photon). Z = W3 cos θW −B sin θW A = W3 sin θW + B cos θW θW Weak Mixing Angle W ±, Z “acquire” mass via the Higgs mechanism. Prof. Tina Potter 10. Electroweak Unification 6 The GWS Model The beauty of the GWS model is that it makes exact predictions of the W ± and Z masses and of their couplings with only 3 free parameters. Couplings given by αEM and θW γ g W ± gW Z gZ αEM = e2 4π g ∼e gW = e sin θW gZ = e sin θW cos θW = gW cos θW Masses also given by GF and θW From Fermi theory GF √ 2 = g2 W 8m2 W = e2 8m2 W sin2 θW mW ± = √ 2e2 8GF sin2 θW !1/2 mZ = mW cos θW If we know αEM, GF, sin θW (from experiment), everything else is defined. Prof. Tina Potter 10. Electroweak Unification 7 Example — mass relation (non-examinable) As a result of the mixing, we require that the mass eigenstates should be the Z and γ, and the mass of the photon be zero. We then compute the matrix elements of the mass operator: m2 Z = ⟨W3 cos θW −B sin θW\| ˆ M2\|W3 cos θW −B sin θW⟩ = m2 W cos2 θW + m2 B sin2 θW −2m2 WB cos θW sin θW m2 γ = ⟨W3 sin θW + B cos θW\| ˆ M2\|W3 sin θW + B cos θW⟩ = m2 W sin2 θW + m2 B cos2 θW + 2m2 WB cos θW sin θW = 0 m2 Zγ = ⟨W3 cos θW −B sin θW\| ˆ M2\|W3 sin θW + B cos θW⟩ = (m2 W −m2 B) sin θW cos θW + m2 WB(cos2 θW −sin2 θW) = 0 Solving these three equations gives mZ = mW cos θW Prof. Tina Potter 10. Electroweak Unification 8 Couplings Slightly simplified – see Part III for better treatment. Starting from Z = W3 cos θW −B sin θW A = W3 sin θW + B cos θW W3 couples to I3 with strength gW and B couples to Y = 2(Q −I3) with g′ So, coupling of A (photon) is gWI3 sin θW + g′2(Q −I3) cos θW = Qe for all I3 ⇒g′ = gW tan θW 2 and g′ cos θW = e 2 ⇒gW = e sin θW The couplings of the Z are therefore gWI3 cos θW −g′2(Q −I3) sin θW = e sin θW cos θW I3 −Q sin2 θW = gZ I3 −Q sin2 θW For right-handed fermions, I3 = 0, while for left-handed fermions I3 = +1/2(ν, u, c, t) or I3 = −1/2(e−, µ−, τ −, d′, s′, b′); Q is charge in units of e Prof. Tina Potter 10. Electroweak Unification 9 Evidence for GWS Model Discovery of Neutral Currents (1973) The process ¯ νµe−→¯ νµe−was observed. Only possible Feynman diagram (no W ± diagram). Indirect evidence for Z. Z e− ¯ νµ e− ¯ νµ Gargamelle Bubble Chamber at CERN Prof. Tina Potter 10. Electroweak Unification 10 Evidence for GWS Model Discovery of Neutral Currents (1973) The process ¯ νµe−→¯ νµe−was observed. Only possible Feynman diagram (no W ± diagram). Indirect evidence for Z. Z e− ¯ νµ e− ¯ νµ Direct Observation of W ± and Z (1983) First direct observation in p¯ p collisions at √s = 540 GeV via decays into leptons p¯ p →W ± + X p¯ p →Z + X , →e±νe, µ±νµ , →e+e−, µ+µ− UA1 Experiment at CERN Used Super Proton Synchrotron (now part of LHC!) Prof. Tina Potter 10. Electroweak Unification 11 Evidence for GWS Model Discovery of Neutral Currents (1973) The process ¯ νµe−→¯ νµe−was observed. Only possible Feynman diagram (no W ± diagram). Indirect evidence for Z. Z e− ¯ νµ e− ¯ νµ Direct Observation of W ± and Z (1983) First direct observation in p¯ p collisions at √s = 540 GeV via decays into leptons p¯ p →W ± + X p¯ p →Z + X , →e±νe, µ±νµ , →e+e−, µ+µ− Precision Measurements of the Standard Model (1989-2000) LEP e+e−collider provided many precision measurements of the Standard Model. Wide variety of different processes consistent with GWS model predictions and measure same value of sin2 θW = 0.23113 ± 0.00015 θW ∼29◦ Prof. Tina Potter 10. Electroweak Unification 12 The Weak NC Vertex All weak neutral current interactions can be described by the Z boson propagator and the weak vertices: e−, µ−, τ − e−, µ−, τ − Z gZ νe, νµ, ντ νe, νµ, ντ Z gZ The Standard Model Weak NC Lepton Vertex + antiparticles u, d, s, c, b, t u, d, s, c, b, t Z gZ The Standard Model Weak NC Quark Vertex + antiparticles Z never changes type of particle Z never changes quark or lepton flavour Z couplings are a mixture of EM and weak couplings, and therefore depend on sin2 θW. Prof. Tina Potter 10. Electroweak Unification 13 Examples Z →e+e−, µ+µ−, τ +τ − Z e+, µ+, τ + e−, µ−, τ − Z →νe¯ νe, νµ¯ νµ, ντ ¯ ντ Z ¯ νe, ¯ νµ, ¯ ντ νe, νµ, ντ Z →q¯ q Z ¯ q q e+e−→µ+µ− Z e− e+ µ+ µ− νee−→νee− Z e− νe e− νe Prof. Tina Potter 10. Electroweak Unification 14 Summary of Standard Model (matter) Vertices Electromagnetic (QED) ℓ− ℓ− γ e q q γ Qe α = e2 4π q = u, d, s, c, b, t + antiparticles Strong (QCD) q q g gs αs = g2 s 4π Weak CC ℓ− νℓ W − gW u, c, t d, s, b W − gW VCKM αW = g2 W 4π Weak NC ℓ±, νℓ ℓ±, νℓ Z gZ q q Z gZ gZ = gW cos θW Prof. Tina Potter 10. Electroweak Unification 15 Feynman Diagrams a reminder 1 π−+ p →K 0 + Λ 2 ντ + e−→ντ + e− 3 ¯ ντ + τ −→¯ ντ + τ − 4 D+ →K −π+π+ Prof. Tina Potter 10. Electroweak Unification 16 Experimental Tests of the Electroweak model at LEP The Large Electron Positron (LEP) collider at CERN provided high precision measurements of the Standard Model (1989-2000). Designed as a Z and W ± boson factory Z e− e+ ¯ f f Z e− e+ W − W + Precise measurements of the properties of Z and W ± bosons provide the most stringent test of our current understanding of particle physics. LEP is the highest energy e+e−collider ever built √s = 90 −209 GeV Large circumference, 27 km 4 experiments combined saw 16 × 106 Z events, 30 × 103 W ± events Prof. Tina Potter 10. Electroweak Unification 17 OPAL: a LEP detector OPAL was one of the 4 experiments at LEP. Size: 12 m × 12 m × 15 m. Prof. Tina Potter 10. Electroweak Unification 18 Typical e+e−→Z events e+e−→Z →e+e− e+e−→Z →µ+µ− Prof. Tina Potter 10. Electroweak Unification 19 Typical e+e−→Z events e+e−→Z →τ +τ − Taus decay within the detector (lifetime ∼10−13 s). Here τ −→e−¯ νeντ, τ + →µ+νµ¯ ντ e+e−→Z →q¯ q 3-jet event (gluon emitted by q/¯ q) Prof. Tina Potter 10. Electroweak Unification 20 The Z Resonance Consider the process e+e−→q¯ q At small √s(< 50 GeV), we only considered an intermediate photon At higher energies, the Z exchange diagram contributes (+Zγ interference) γ e− e+ ¯ q q Qe Qqe Z e− e+ ¯ q q gW gW σ(e+e−→γ →q¯ q) = 4πα2 3s X 3Q2 q The Z is a decaying intermediate massive state (lifetime ∼10−25 s) ⇒Breit-Wigner resonance Around √s ∼mZ, the Z diagram dominates Prof. Tina Potter 10. Electroweak Unification 21 The Z Resonance Prof. Tina Potter 10. Electroweak Unification 22 The Z Resonance Breit-Wigner cross-section for e+e−→Z →f ¯ f (where f ¯ f is any fermion-antifermion pair) Centre-of-mass energy √s = ECM = Ee+ + Ee− σ(e+e−→Z →f ¯ f ) = gπ E 2 e ΓeeΓf ¯ f (ECM −mZ)2 + Γ2 Z 4 with g = 2JZ + 1 (2Je−+ 1)(2Je+ + 1) = 3 4 JZ = 1; Je± = 1 2 giving σ(e+e−→Z →f ¯ f ) = 3π 4E 2 e ΓeeΓf ¯ f (ECM −mZ)2 + Γ2 Z 4 = 3π s ΓeeΓf ¯ f (√s −mZ)2 + Γ2 Z 4 ΓZ is the total decay width, i.e. the sum over the partial widths for different decay modes ΓZ = Γee + Γµµ + Γττ + Γq¯ q + Γν¯ ν Prof. Tina Potter 10. Electroweak Unification 23 The Z Resonance At the peak of the resonance √s = mZ: σ(e+e−→Z →f ¯ f ) = 12π m2 Z ΓeeΓf ¯ f Γ2 Z Hence, for all fermion/antifermion pairs in the final state σ(e+e−→Z →anything) = 12π m2 Z Γee ΓZ Γf ¯ f = ΓZ Compare to the QED cross-section at √s = mZ σQED = 4πα2 3s σ(e+e−→Z →anything) σQED = 9 α2 Γee ΓZ ∼5700 Γee = 85 MeV, ΓZ = 2.5 GeV, α = 1/137 Prof. Tina Potter 10. Electroweak Unification 24 Measurement of mZ and ΓZ Run LEP at various centre-of-mass energies (√s) close to the peak of the Z resonance and measure σ(e+e−→q¯ q) Determine the parameters of the resonance: Mass of the Z, mZ Total decay width, ΓZ Peak cross-section, σ0 One subtle feature: need to correct measurements for QED effects due to radiation from the e+e−beams. This radiation has the effect of reducing the centre-of-mass energy of the e+e− collision which smears out the resonance. Z e− e+ ¯ q q γ Prof. Tina Potter 10. Electroweak Unification 25 Measurement of mZ and ΓZ mZ was measured with precision 2 parts in 105 Need a detailed understanding of the accelerator and astrophysics. Tidal distortions of the Earth by the Moon cause the rock surrounding LEP to be distorted – changing the radius by 0.15 mm (total 4.3 km). This is enough to change the centre-of-mass energy. LHC ring is stretched by 0.1mm by the 7.5 magnitude earthquake in New Zealand, Nov 2016. Tidal forces can also be seen. Also need a train timetable. Leakage currents from the TGV rail via Lake Geneva follow the path of least resistance... using LEP as a conductor. Accounting for these effects (and many others): mZ = 91.1875 ± 0.0021 GeV ΓZ = 2.4952 ± 0.0023 GeV σ0 q¯ q = 41.450 ± 0.037 nb Prof. Tina Potter 10. Electroweak Unification 26 Number of Generations Currently know of three generations of fermions. Masses of quarks and leptons increase with generation. Neutrinos are approximately massless (or are they?) e− νe µ− νµ τ − ντ u d c s t b Could there be more generations? e.g. t′ b′ L νL The Z boson couples to all fermions, including neutrinos. Therefore, the total decay width, ΓZ, has contributions from all fermions with mf < mZ/2 ΓZ = Γee + Γµµ + Γττ + Γq¯ q + Γν¯ ν with Γν¯ ν = Γνe¯ νe + Γνµ¯ νµ + Γντ ¯ ντ If there were a fourth generation, it seems likely that the neutrino would be light, and, if so would be produced at LEP e+e−→Z →νL¯ νL The neutrinos would not be observed directly, but could infer their presence from the effect on the Z resonance curve. Prof. Tina Potter 10. Electroweak Unification 27 Number of Generations At the peak of the Z resonance, √s = mZ σ0 f ¯ f = 12π m2 Z ΓeeΓf ¯ f Γ2 Z A fourth generation neutrino would increase the Z decay rate and thus increase ΓZ. As a result, a decrease in the measured peak cross-sections for the visible final states would be observed. Measure the e+e−→Z →f ¯ f cross-sections for all visible decay models (i.e. all fermions apart from ν¯ ν) Examples: e+e−→µ+µ− e+e−→τ +τ − Prof. Tina Potter 10. Electroweak Unification 28 Number of Generations Have already measured mZ and ΓZ from the shape of the Breit-Wigner resonance. Therefore, obtain Γf ¯ f from the peak cross-sections in each decay mode using σ0 f ¯ f = 12π m2 Z ΓeeΓf ¯ f Γ2 Z Note, obtain Γee from σ0 ee = 12π m2 Z Γ2 ee Γ2 Z Can relate the partial widths to the measured total width (from the resonance curve) ΓZ = Γee + Γµµ + Γττ + Γq¯ q + NνΓνν where Nν is the number of neutrino species and Γνν is the partial width for a single neutrino species. Prof. Tina Potter 10. Electroweak Unification 29 Number of Generations The difference between the measured value of ΓZ and the sum of the partial widths for visible final states gives the invisible width NνΓνν ΓZ 2495.2±2.3 MeV Γee 83.91±0.12 MeV Γµµ 83.99±0.18 MeV Γττ 84.08±0.22 MeV Γqq 1744.4±2.0 MeV NνΓνν 499.0±1.5 MeV In the Standard Model, calculate Γνν ∼167 MeV Therefore Nν = Γmeasured νν ΓSM νν = 2.984 ± 0.008 ⇒three generations of light neutrinos for mν < mZ/2 Prof. Tina Potter 10. Electroweak Unification 30 Number of Generations Most likely that only 3 generations of quarks and leptons exist In addition Γee, Γµµ, Γττ are consistent ⇒tests universality of the lepton couplings to the Z boson. Γqq is consistent with the expected value which assumes 3 colours – further evidence for colour Prof. Tina Potter 10. Electroweak Unification 31 W +W −at LEP In e+e−collisions W bosons are produced in pairs. Standard Model: 3 possible diagrams: νe e− e+ W − W + γ e− e+ W − W + Z e− e+ W − W + LEP operated above the threshold for W +W −production (1996-2000) √s > 2mW Cross-section sensitive to the presence of the Triple Gauge Boson vertex Prof. Tina Potter 10. Electroweak Unification 32 W +W −at LEP In the Standard Model W ℓν and Wq¯ q couplings are ∼equal. W − ¯ νe, ¯ νµ, ¯ ντ e−, µ−, τ − W − d′, s′ ¯ u, ¯ c mW < mt ×3 for colour Expect (assuming 3 colours) B(W ± →q¯ q) = 6 9 = 2 3 B(W ± →ℓν) = 3 9 = 1 3 QCD corrections ∼ 1 + αs π ⇒B(W ± →q¯ q) = 0.675 Measured BR W +W −→ℓνℓν 10.5% W +W −→q¯ qℓν 43.9% W +W −→q¯ qq¯ q 45.6% Prof. Tina Potter 10. Electroweak Unification 33 W +W −events in OPAL W +W −→eνµν W +W −→q¯ qeν W +W −→q¯ qq¯ q Prof. Tina Potter 10. Electroweak Unification 34 Measurement of mW and ΓW Unlike e+e−→Z, W boson production at LEP was not a resonant process. mW was measured by measuring the invariant mass in each event 4-momenta pq1, pq2, pe, pν mW = 1 2 (mq¯ q + mℓν) mW = 80.423 ± 0.038 GeV ΓW = 2.12 ± 0.11 GeV Prof. Tina Potter 10. Electroweak Unification 35 W Boson Decay Width In the Standard Model, the W boson decay width is given by Γ(W −→e−¯ νe) = g2 WmW 48π = GFm3 W 6 √ 2π µ-decay: GF = 1.166 × 10−5 GeV−2 LEP: mW = 80.423 ± 0.038 GeV ⇒Γ(W −→e−¯ νe) = 227 MeV Total width is the sum over all partial widths: W −→e−¯ νe, µ−¯ νµ, τ −¯ ντ, W −→d′¯ u, s′¯ c, ×3 for colour If the W coupling to leptons and quarks is equal and there are 3 colours: Γ = X i Γi = (3 + 2 × 3)Γ(W −→e−¯ νe) ∼2.1 GeV Compare with measured value from LEP: ΓW = 2.12 ± 0.11 GeV Universal coupling constant Yet more evidence for colour! Prof. Tina Potter 10. Electroweak Unification 36 Summary of Electroweak Tests Now have 5 precise measurements of fundamental parameters of the Standard Model αEM = 1/(137.03599976 ± 0.00000050) (at q2 = 0 ) GF = (1.16632 ± 0.00002) × 10−5 GeV−2 mW = 80.385 ± 0.015 GeV mZ = 91.1875 ± 0.0021 GeV sin2 θW = 0.23143 ± 0.00015 In the Standard Model, only 3 are independent. The measurements are consistent, which is an incredibly powerful test of the Standard Model of Electroweak Interactions. Prof. Tina Potter 10. Electroweak Unification 37 Summary Weak interaction with W ± fails at high energy. Introduction of unified theory involving and relating Z and γ can resolve the divergences. One new parameter, θW, allows predictions of Z couplings and mass relations. Extensively and successfully tested at LEP. Problem Sheet: q.26-27 Up next... Section 11: The Top Quark and the Higgs Mechanism Prof. Tina Potter 10. Electroweak Unification 38 11. The Top Quark and the Higgs Mechanism Particle and Nuclear Physics Prof. Tina Potter Prof. Tina Potter 11. The Top Quark and the Higgs Mechanism 1 In this section... Focus on the most recent discoveries of fundamental particles The top quark – prediction & discovery The Higgs mechanism The Higgs discovery Prof. Tina Potter 11. The Top Quark and the Higgs Mechanism 2 Third Generation Quark Weak CC Decays log(mass) t c u d s b 173GeV 1.3GeV 2.3MeV 95 MeV 4.8 MeV 4.8GeV Cabibbo Allowed \|Vtb\|~1, \|Vcs\|~\|Vud\|~0.975 Cabibbo Suppressed \|Vcd\|~\|Vus\|~0.22 \|Vcb\|~\|Vts\|~0.05 Top quarks are special. m(t) ≫m(b) (> m(W )) τt ∼10−25 s ⇒decays before hadronisation Vtb ∼1 ⇒ BR(t →W + b)=100% Bottom quarks are also special. b quarks can only decay via the Cabbibo suppressed Wcb vertex. Vcb is very small – weak coupling! ⇒τ(b) ≫τ(u, c, d, s) Jet initiated by b quarks look different to other jets. b quarks travel further from interaction point before decaying. b-jet traces back to a secondary vertex – “b-tagging”. Interaction point b-jet b Prof. Tina Potter 11. The Top Quark and the Higgs Mechanism 3 The Top Quark (non-examinable) The Standard Model predicted the existence of the top quark u d c s t b +2 3e −1 3e which is required to explain a number of observations. Example: Non-observation of the decay K 0 →µ+µ− B(K 0 →µ+µ−) < 10−9 The top quark cancels the contributions from the u and c quarks. W + u/c/t W − νµ ¯ s d µ+ µ− Example: Electromagnetic anomalies This diagram leads to infinities in the theory unless P Qf = 0 where the sum is over all fermions (and colours) X f Qf = [3 × (−1)] + 3 × 3 × 2 3 + 3 × 3 × (−1 3) = 0 f f ¯ f γ γ γ Requires t quark Prof. Tina Potter 11. The Top Quark and the Higgs Mechanism 4 The Top Quark The top quark is too heavy for Z →t¯ t or W + →t¯ b so not directly produced at LEP. However, precise measurements of mZ, mW, ΓZ and ΓW are sensitive to the existence of virtual top quarks: t ¯ b W + W + t ¯ t Z Z t W ¯ t Z b ¯ b Example Standard Model prediction Also depends on the Higgs mass Prof. Tina Potter 11. The Top Quark and the Higgs Mechanism 5 The Top Quark The top quark was discovered in 1994 by the CDF experiment at the worlds (then) highest energy pp collider (√s = 1.8 TeV), the Tevatron at Fermilab, US. g t ¯ t ¯ q q ¯ b W − W + b Final state W +W −b¯ b Mass reconstructed in a similar manner to mW at LEP, i.e. measure jet/lepton energies/momenta. Vtb ∼1, so decay of top quark is ∼100% t →bW + mt ≫mW, so the W + is real. The weak decay is just as fast as a strong decay (∼10−25s), so the quark has no time to hadronise ⇒there are no t-hadrons Possible top quark decays are t →bq¯ q or t →bℓνℓ In hadron collisions, multijet final states are the norm – for rare processes it’s much easier to look for leptonic decays, accompanied by b-quark jets. Prof. Tina Potter 11. The Top Quark and the Higgs Mechanism 6 First observation of top (1995) CDF collaboration published first PRL 74 2626 (1995) t ¯ t →bW +¯ bW − Final state l ν + qq + bb background signal data Current status Results from LHC as well as Tevatron. All consistent, and in agreement with indirect expectation from LEP data. Prof. Tina Potter 11. The Top Quark and the Higgs Mechanism 7 Higgs mechanism and the Higgs Boson Recall – the Klein-Gordon equation for massive bosons is: ∂2ψ ∂t2 = ∇ ∇ ∇2 −m2 ψ However, the term m2ψ (or 1 2m2ψ2 in the Lagrangian formulation), is not gauge invariant. So in gauge field theories, the gauge bosons should be massless. OK for QED and QCD, but plainly not for W ± and Z. The Higgs mechanism tries to fix this. Imagine introducing a scalar Higgs field ϕ, which has interactions with the W ± and Z fields, with coupling strength y, giving a term in Lagrangian yϕψψ. Looks like a mass term (∝ψ2). Mass of the bosons becomes effectively related to their coupling to the Higgs field. Requires the vacuum (lowest energy state of space) to have a non-zero expectation value for the Higgs field. How can this come about? Prof. Tina Potter 11. The Top Quark and the Higgs Mechanism 8 Higgs potential Suppose the Higgs field ϕ (actually a complex doublet) has self interactions yielding V (ϕ) = aϕ4 −bϕ2 The equilibrium point, ϕ = 0, respects the symmetry, but is unstable. The stable equilibrium point is at \|ϕ2 GS\| = b/2a. The symmetry is “spontaneously broken”. A weak boson propagating in the Higgs field will appear to have a mass ∼yϕGS. Expanding about the ground state V (ϕGS + x) = Vmin + 2bx2 So can get excitations of the Higgs field about the minimum. These form the physical Higgs scalar boson, H – the observable physical manifestation of the operation of the Higgs mechanism. Prof. Tina Potter 11. The Top Quark and the Higgs Mechanism 9 Classical analogue of the Higgs mechanism (non-examinable) Maxwell’s equations lead to waves travelling at velocity c, hence to massless photons. Consider waves propagating in a charged plasma, with electron density n per unit volume. Plasma: ⃗ J = ne⃗ v; me ∂⃗ v ∂t = e ⃗ E ⇒∂⃗ J ∂t = ne2⃗ E me Maxwell: ⃗ ∇ ∇ ∇∧⃗ ∇ ∇ ∇∧⃗ E = −∇ ∇ ∇2⃗ E = ⃗ ∇ ∇ ∇∧ −∂⃗ B ∂t ! = −∂⃗ ∇ ∇ ∇∧⃗ B ∂t = −∂ ∂t µ0 ⃗ J + 1 c2 ∂⃗ E ∂t ! = −µ0ne2⃗ E me −1 c2 ∂2⃗ E ∂t2 ⇒∇ ∇ ∇2⃗ E −1 c2 ∂2⃗ E ∂t2 = µ0ne2⃗ E me Compare with Klein-Gordon. Photon propagates with effective mass m2 eff = ℏµ0ne2 mec2 Note meff ∝e, the coupling. Prof. Tina Potter 11. The Top Quark and the Higgs Mechanism 10 Higgs theory summary Gauge bosons (and also fermions) are intrinsically massless, and need to be so to satisfy Gauge Invariance. Nevertheless, interactions with the Higgs field make particles look like they have mass. Apparent masses are controlled by free parameters called Yukawa Couplings (the strength of the coupling to the Higgs field) A Higgs Boson arises as an excitation of the Higgs field. It must be a scalar particle to make everything work. The Higgs Boson has a mass, but the mass is not predicted by the theory – we have to find it experimentally. The Higgs Boson has couplings to all the particles to which it gives mass (and so has many ways it could decay), all fully calculable and determined by the theory as a function of its (a priori unknown) mass. Prof. Tina Potter 11. The Top Quark and the Higgs Mechanism 11 Higgs boson decays Higgs Boson interacts via couplings which are proportional to masses. Higgs boson therefore decays preferentially to the heaviest particles that are kinematically accessible, depending on its mass. Prof. Tina Potter 11. The Top Quark and the Higgs Mechanism 12 Higgs decay mechanisms Directly to two fundamental fermions or bosons, coupling to mass, e.g. H ¯ b b H Z Z Indirectly to massless particles (photons or gluons) via massive loops t t ¯ t H γ γ W W W H γ γ t t ¯ t H g g Prof. Tina Potter 11. The Top Quark and the Higgs Mechanism 13 Higgs at LEP Higgs Production at LEP (Large Electron Positron Collider – 1990s): If mH < √s −mZ Z∗ e− e+ H Z “Higgsstrahlung” mechanism In 2000, LEP operated with √s ∼207 GeV, therefore had the potential to discover Higgs boson if mH < 116 GeV. Searches were conducted in many possible final states (different decays for Z and H). All negative. Ultimately, LEP excluded a Higgs Boson with a mass below 114 GeV. Prof. Tina Potter 11. The Top Quark and the Higgs Mechanism 14 Higgs at Large Hadron Collider Higgs Production at the LHC The dominant Higgs production mechanism at the LHC is t t ¯ t g g H “Gluon fusion” Higgs Decay at the LHC t t ¯ t H γ γ H ¯ b b H Z Z Low mass Medium mass High mass One Z may be virtual Prof. Tina Potter 11. The Top Quark and the Higgs Mechanism 15 The Large Hadron Collider The LHC is a new proton-proton collider now running in the old LEP tunnel at CERN. In 2012 4 + 4 TeV; in 2015 6.5 + 6.5 TeV; ultimately 7 + 7 TeV Prof. Tina Potter 11. The Top Quark and the Higgs Mechanism 16 ATLAS – a general purpose LHC detector Prof. Tina Potter 11. The Top Quark and the Higgs Mechanism 17 Higgs Observations (August 2014) Indirect indications from LEP that Higgs mass should be not far above 115 GeV. Dominant decay modes are all difficult: b¯ b, c ¯ c (swamped by QCD jets) W +W −, τ +τ − (missing neutrinos) Best options are the rare decays: ZZ →ℓ+ℓ−ℓ+ℓ− γγ Prof. Tina Potter 11. The Top Quark and the Higgs Mechanism 18 H →ZZ →4ℓ Prof. Tina Potter 11. The Top Quark and the Higgs Mechanism 19 Higgs Boson Discovery ) µ Signal strength ( 1 − 0 1 2 3 ATLAS -1 = 7 TeV, 4.5-4.7 fb s -1 = 8 TeV, 20.3 fb s = 125.36 GeV H m 0.26 -0.28 + = 1.17 obs µ 0.23 -0.25 + = 1.00 exp µ γ γ → H 0.34 -0.40 + = 1.46 obs µ 0.26 -0.31 + = 1.00 exp µ ZZ → H 0.21 -0.24 + = 1.18 obs µ 0.19 -0.21 + = 1.00 exp µ WW → H 0.37 -0.39 + = 0.63 obs µ 0.38 -0.41 + = 1.00 exp µ b b → H 0.37 -0.42 + = 1.44 obs µ 0.32 -0.36 + = 1.00 exp µ τ τ → H 3.7 -3.7 + = -0.7 obs µ 3.5 -3.4 + = 1.0 exp µ µ µ → H 4.5 -4.6 + = 2.7 obs µ 4.2 -4.2 + = 1.0 exp µ γ Z → H 0.14 -0.15 + = 1.18 obs µ 0.12 -0.13 + = 1.00 exp µ Combined Total uncertainty µ on σ 1 ± (obs.) σ (exp.) σ Convincing signal consistent with m(H) = 126 GeV – seen in multiple decay modes & in two experiments. Is it the Higgs boson of the SM? Need to check its quantum numbers (should be JP = 0+). Check branching ratios and couplings. Look ok so far... 1 − 10 1 10 2 10 Particle mass [GeV] 4 − 10 3 − 10 2 − 10 1 − 10 1 v V m V κ or v F m F κ Preliminary ATLAS 1 − = 13 TeV, 36.1 - 79.8 fb s \| < 2.5 H y = 125.09 GeV, \| H m µ τ b W Z t SM Higgs boson Prof. Tina Potter 11. The Top Quark and the Higgs Mechanism 20 Higgs spin+parity? Studied using angular distributions of decay products So far it looks like the 0+ SM Higgs. Alternative spin-parity possibilities are disfavoured. h + = 0 P J − = 0 P J g κ = q κ + = 2 P J <300 GeV T p =0 q κ + = 2 P J <125 GeV T p =0 q κ + = 2 P J <300 GeV T p g κ =2 q κ + = 2 P J <125 GeV T p g κ =2 q κ + = 2 P J q ~ -30 -20 -10 0 10 20 30 40 ATLAS l 4 → ZZ → H -1 = 7 TeV, 4.5 fb s -1 = 8 TeV, 20.3 fb s νµν e → WW → H -1 = 8 TeV, 20.3 fb s γ γ → H -1 = 7 TeV, 4.5 fb s -1 = 8 TeV, 20.3 fb s Observed Expected σ 1 ± SM + 0 σ 2 ± SM + 0 σ 3 ± SM + 0 σ 1 ± P J σ 2 ± P J σ 3 ± P J Prof. Tina Potter 11. The Top Quark and the Higgs Mechanism 21 Summary Top quark – observed, and compatible with other precise electroweak measurements. Electroweak theory depends on the Higgs mechanism to endow particles with mass. This is a non-standard feature, which needs experimental verification. Higgs boson – detected in 2012 at 126 GeV. Work continues to determine the properties of the boson and check whether it is the Higgs boson of the Electroweak Standard Model. Problem Sheet: q.28 Up next... Section 12: Beyond the Standard Model Prof. Tina Potter 11. The Top Quark and the Higgs Mechanism 22 12. Beyond the Standard Model Particle and Nuclear Physics Prof. Tina Potter Prof. Tina Potter 12. Beyond the Standard Model 1 In this section... Summary of the Standard Model Problems with the Standard Model Neutrino oscillations Supersymmetry Prof. Tina Potter 12. Beyond the Standard Model 2 The Standard Model (2012) Matter: point-like spin 1 2 Dirac fermions + antiparticles Fermion Charge [e] Mass 1st gen. Electron e− −1 0.511 MeV Electron neutrino νe 0 ∼0 Down quark d −1/3 4.8 MeV Up quark u +2/3 2.3 MeV 2nd gen. Muon µ− −1 106 MeV Muon neutrino νµ 0 ∼0 Strange quark s −1/3 95 MeV Charm quark c +2/3 1.3 GeV 3rd gen. Tau τ − −1 1.78 GeV Tau neutrino ντ 0 ∼0 Bottom quark b −1/3 4.7 GeV Top quark t +2/3 173 GeV Prof. Tina Potter 12. Beyond the Standard Model 3 The Standard Model (2012) Forces: mediated by spin 1 bosons Force Particle Mass Electromagnetic Photon γ 0 Strong 8 gluons g 0 Weak (CC) W ± 80.4 GeV Weak (NC) Z 91.2 GeV The Standard Model also predicts the existence of a spin-0 Higgs boson which gives all particles their masses via its interactions. Evidence from LHC confirms this, with mH ∼125 GeV. The Standard Model successfully describes all existing particle physics data, with the exception of one ⇒Neutrino Oscillations ⇒Neutrinos have mass In the SM, neutrinos are treated as massless; right-handed states do not exist ⇒indication of physics Beyond the Standard Model Prof. Tina Potter 12. Beyond the Standard Model 4 Problems with the Standard Model The Standard Model successfully describes all existing particle physics data (though question marks over the neutrino sector). But: many (too many?) input parameters: Quark and lepton masses Quark charge Couplings αEM, sin2 θW, αs Quark (+ neutrino) generation mixing – VCKM 23 free parameters in SM - 9 fermion masses (e, µ, τ, u, d, s, c, b, t) - 4 CKM: 3 mixing angles + CPV phase - 4 PMNS: 3 mixing angles + CPV phase - 3 gauge couplings: U(1), SU(2), SU(3) - 3 other: QCD vacuum angle (strong CPV), Higgs VEV, Higgs mass and: many unanswered questions: Why so many free parameters? Why only three generations of quarks and leptons? Where does mass come from? (Higgs boson probably OK) Why is the neutrino mass so small and the top quark mass so large? Why are the charges of the p and e identical? What is responsible for the observed matter-antimatter asymmetry? How can we include gravity? etc Prof. Tina Potter 12. Beyond the Standard Model 5 Beyond the Standard Model – further unification?? Grand Unification Theories (GUTs) aim to unite the strong interaction with the electroweak interaction. Underpins many ideas about physics beyond the Standard Model. The strength of the interactions depends on energy: Suggests unification of all forces at ∼1015 GeV? Strength of Gravity only significant at the Planck Mass ∼1019 GeV Prof. Tina Potter 12. Beyond the Standard Model 6 Neutrino Oscillations In 1998 the Super-Kamiokande experiment announced convincing evidence for neutrino oscillations implying that neutrinos have mass. π →µνµ , →eνµ¯ νe Expect N(νµ) N(νe) ∼2 Super-Kamiokande results indicate a deficit of νµ from the upwards direction. Upward neutrinos created further away from the detector. Interpreted as νµ →ντ oscillations Implies neutrino mixing and neutrinos have mass Prof. Tina Potter 12. Beyond the Standard Model 7 Detecting Neutrinos Neutrinos are detected by observing the lepton produced in charged current interactions with nuclei. e.g. νe + N →e−+ X ¯ νµ + N →µ+ + X Size Matters: Neutrino cross-sections on nucleons are tiny; ∼10−42(Eν/ GeV)m2 Neutrino mean free path in water ∼light-years. Require very large mass, cheap and simple detectors. Water ˇ Cerenkov detection ˇ Cerenkov radiation Light is emitted when a charged particle traverses a dielectric medium A coherent wavefront forms when the velocity of a charged particle exceeds c/n (n = refractive index) ˇ Cerenkov radiation is emitted in a cone i.e. at fixed angle with respect to the particle. cos θC = c nv = 1 nβ Prof. Tina Potter 12. Beyond the Standard Model 8 Super-Kamiokande Super-Kamiokande is a Water ˇ Cerenkov detector sited in Kamioka, Japan 50, 000 tons of water Surrounded by 11, 146 × 50 cm diameter, photo-multiplier tubes Prof. Tina Potter 12. Beyond the Standard Model 9 Super-Kamiokande Examples of events νµ + N →µ−+ X νe + N →e−+ X Prof. Tina Potter 12. Beyond the Standard Model 10 Super-Kamiokande ν deficit Expect Isotropic (flat) distributions in cos θ N(νµ) ∼2N(νe) Observe Deficit of νµ from below Whereas νe look as expected Interpretation νµ →ντ oscillations ⇒neutrinos have mass e-like μ-like No oscillations With oscillations Data Prof. Tina Potter 12. Beyond the Standard Model 11 Neutrino Mixing The quark states which take part in the weak interaction (d′, s′) are related to the flavour (mass) states (d, s) Weak Eigenstates d′ s′ = cos θC sin θC −sin θC cos θC d s Mass Eigenstates Cabibbo angle θC ∼13◦ Suppose the same thing happens for neutrinos. Consider only the first two generations for simplicity. Weak Eigenstates = flavour eigenstates νe νµ = cos θ sin θ −sin θ cos θ ν1 ν2 Mass Eigenstates Mixing angle θ e.g. in π+ decay produce µ+ and νµ i.e. the neutrino state that couples to the weak interaction. The νµ corresponds to a linear combination of the states with definite mass, ν1 and ν2 νe = +ν1 cos θ + ν2 sin θ νµ = −ν1 sin θ + ν2 cos θ or expressing the mass eigenstates in terms of the weak eigenstates ν1 = +νe cos θ −νµ sin θ ν2 = +νe sin θ + νµ cos θ Prof. Tina Potter 12. Beyond the Standard Model 12 Neutrino Mixing Prof. Tina Potter 12. Beyond the Standard Model 13 Neutrino Mixing Prof. Tina Potter 12. Beyond the Standard Model 13 Neutrino Mixing Prof. Tina Potter 12. Beyond the Standard Model 13 Neutrino Mixing Suppose a muon neutrino with momentum ⃗ p is produced in a weak decay, e.g. π+ →µ+νµ At t = 0, the wavefunction ψ(⃗ p, t = 0) = νµ(⃗ p) = ν2(⃗ p) cos θ −ν1(⃗ p) sin θ The time evolution of ν1 and ν2 will be different if they have different masses ν1(⃗ p, t) = ν1(⃗ p)e−iE1t ; ν2(⃗ p, t) = ν2(⃗ p)e−iE2t After time t, state will in general be a mixture of νe and νµ ψ(⃗ p, t) = ν2(⃗ p)e−iE2t cos θ −ν1(⃗ p)e−iE1t sin θ = [νe(⃗ p) sin θ + νµ(⃗ p) cos θ] e−iE2t cos θ −[νe(⃗ p) cos θ −νµ(⃗ p) sin θ] e−iE1t sin θ = νµ(⃗ p) cos2 θe−iE2t + sin2 θe−iE1t + νe(⃗ p) sin θ cos θ e−iE2t −e−iE1t = cµνµ(⃗ p) + ceνe(⃗ p) Prof. Tina Potter 12. Beyond the Standard Model 13 Neutrino Mixing Probability of oscillating into νe P(νe) = \|ce\|2 = sin θ cos θ e−iE2t −e−iE1t 2 = 1 4 sin2 2θ e−iE2t −e−iE1t eiE2t −eiE1t = 1 4 sin2 2θ 2 −ei(E2−E1)t −e−i(E2−E1)t = sin2 2θ sin2 (E2 −E1)t 2 But E = p ⃗ p 2 + m2 = ⃗ p s 1 + m2 ⃗ p 2 ∼⃗ p + m2 2⃗ p for m ≪E 1 + x ∼(1 + x/2)2 when x is small, can ignore x2 term ⇒E2(⃗ p) −E1(⃗ p) ∼m2 2 −m2 1 2⃗ p ∼m2 2 −m2 1 2E ⇒P(νµ →νe) = sin2 2θ sin2 (m2 2 −m2 1)t 4E Prof. Tina Potter 12. Beyond the Standard Model 14 Neutrino Mixing For νµ →ντ P(νµ →ντ) = sin2 2θ sin2 (m2 3 −m2 2)t 4E = sin2 2θ sin2 1.27∆m2L Eν where L is the distance travelled in km, ∆m2 = m2 3 −m2 2 is the mass difference in ( eV)2 and Eν is the neutrino energy in GeV. Interpretation of Super-Kamiokande Results For E(νµ) = 1 GeV (typical of atmospheric neutrinos) Results are consistent with νµ →ντ oscillations: \|m2 3 −m2 2\| ∼2.5 × 10−3 eV2; sin2 2θ ∼1 Prof. Tina Potter 12. Beyond the Standard Model 15 Neutrino Mixing – Comments Neutrinos almost certainly have mass Neutrino oscillation only sensitive to mass differences More evidence for neutrino oscillations Solar neutrinos (SNO experiment) Reactor neutrinos (KamLand) suggest \|m2 2 −m2 1\| ∼8 × 10−5 eV2. More recent experiments use neutrino beams from accelerators or reactors; observe energy spectrum of neutrinos at a distant detector. At fixed L, observation of the values of Eν at which minima/maxima are seen determines ∆m2, while depth of minima determine sin2 2θ. Note all these experiments only tell us about mass differences. Best constraint on absolute mass comes from the end point in Tritium β-decay, m(νe) < 2 eV. Prof. Tina Potter 12. Beyond the Standard Model 16 Three-flavour oscillations This whole framework can be generalised...   νe νµ ντ  = UPMNS   ν1 ν2 ν3   where UPMNS =   1 0 0 0 c23 s23 0 −s23 c23     c12 0 s13e−iδ 0 1 0 −s23eiδ 0 c13     c12 s12 0 −s12 c12 0 0 0 1   defining cos θ12 = c12 etc. This is an active field! Current status... sin2 θ12 = 0.304 ± 0.014 sin2 θ23 = 0.51 ± 0.06 sin2 θ13 = 0.0219 ± 0.0012 Prof. Tina Potter 12. Beyond the Standard Model 17 Supersymmetry (SUSY) A significant problem is to explain why the Higgs boson is so light. The effect of loop corrections on the Higgs mass should be to drag it up to the highest energy scale in the problem (i.e. unification, or Planck mass). f f H H One attractive solution is to introduce a new space-time symmetry, “supersymmetry” which links fermions and bosons (the only way to extend the Poincar´ e symmetry of special relativity and respect quantum field theory.) Each fermion has a boson partner, and vice versa, with the same couplings. Boson and fermion loops contribute with opposite sign, giving a natural cancellation in their effect on the Higgs mass. f f H H ˜ f ˜ f H H + Must be a broken symmetry, because we clearly don’t see bosons and fermions of the same mass. However, this doubles the particle content of the model, without any direct evidence (yet), and introduces lots of new unknown parameters. Prof. Tina Potter 12. Beyond the Standard Model 18 The Supersymmetric Standard Model SM : W ±, W 0, B mixing − − − →W ±, Z, γ SUSY : ˜ H0 u, ˜ H0 d, ˜ W 0, ˜ B0 mixing − − − →˜ χ0 1, ˜ χ0 2, ˜ χ0 3, ˜ χ0 4 ˜ H+ u , ˜ H− d , ˜ W +, ˜ W −mixing − − − →˜ χ± 1 , ˜ χ± 2 Prof. Tina Potter 12. Beyond the Standard Model 19 SUSY and Unification In the Standard Model, the interaction strengths are not quite unified at very high energy. Add SUSY, the running of the couplings is modified, because sparticle loops contribute as well as particle loops. Details depend on the version of SUSY, but in general unification much improved. Prof. Tina Potter 12. Beyond the Standard Model 20 SUSY and cosmology SUSY, or any unified theory, tends to have potential problems with explaining the non-observation of proton decay. For this reason, many versions of SUSY introduce a conserved quantity “R-parity”, which means that sparticles have to be produced in pairs. A consequence is that the lightest sparticle would have to be stable. In many scenarios this would be a “neutralino” ˜ χ0 1 (a mixture of neutral “gauginos” and “Higgsinos”). Cosmologists tell us that ∼25% of the mass in the universe is in the form of “dark matter”, which interacts gravitationally, but otherwise only weakly. The lightest sparticle could be a candidate for the “WIMPs” (Weakly Interacting Massive Particles) which could comprise dark matter. 68.3% Dark Energy 26.8% Dark Matter 4.9% Atoms ? ?? So there are several different reasons why SUSY is attractive. Prof. Tina Potter 12. Beyond the Standard Model 21 However, no sign of supersymmetry yet... On general grounds, some sparticles ought to be seen at energies around 1 TeV or lower. So LHC ought to be able to see them, especially squarks+gluinos (high σ @LHC). Prof. Tina Potter 12. Beyond the Standard Model 22 Signs of anything else? (non-examinable) LHCb Flavour Anomalies e+ e-μ+ μ-Lepton universality in SM predicts R = µµ ee = 1 Test using rare decays of B mesons easy to see deviations from small values precise theory predictions RK = 0.85 ± 0.04(stat.) ± 0.01(syst.) 3 standard deviations from prediction. Evidence of something new! 5 std.dev is gold standard for discovery. Similar effects seen in several rare decay modes. This might be the first glimpse of new particles affecting decay rates, e.g. Leptoquarks e+ e-μ+ μ-Prof. Tina Potter 12. Beyond the Standard Model 23 Signs of anything else? (non-examinable) Muon g-2 Anomaly Measure muon spin precession in magnetic field. Precision test of QED – precession frequency depends on how much it interacts with the magnetic field. All known particles contribute to the muon’s magnetic moment. Measure this very precisely and look for deviations. 20 year anomaly has been confirmed with a new measurement at Fermilab – measured muon magnetic moment to 0.46 ppm. 4.2 standard deviations from prediction. Evidence of something new! Perhaps smuons? Prof. Tina Potter 12. Beyond the Standard Model 24 Follow the results from LHC yourself! To date (2021) LHC has taken only 5% of its planned total dataset. Stay tuned!! Prof. Tina Potter 12. Beyond the Standard Model 25 Summary Over the past 40 years our understanding of the fundamental particles and forces of nature has changed beyond recognition. The Standard Model of particle physics is an enormous success. It has been tested to very high precision and can model almost all experimental observations so far. The Higgs “hole” is now becoming closed, though some other aspects of the SM are not quite yet under as much experimental “control” as one might wish for (the neutrino sector, the CKM matrix, etc). Good reasons to expect that the next few years will bring many more (un)expected surprises (more Higgs or gauge bosons, SUSY?). Problem Sheet: q.29-30 Up next... Section 13: Nuclear Physics, Basic Nuclear Properties Prof. Tina Potter 12. Beyond the Standard Model 26
10397	https://dgrozev.wordpress.com/2023/06/08/strategy-stealing-argument-a-russian-tst-problem/	Skip to content Strategy-Stealing Argument. A Russian TST Problem. I am going to discuss a key idea from game theory, that is used in some math Olympiad problems. Examples will be given. We end with a Russian problem given at their Team Selection Test for IMO 2018. This was the main reason why I wrote this blog post. I liked it and wanted to give more examples of how to use the strategy-stealing method as an important idea in some Olympiad problems. Strategy-stealing method. This argument is applied for some two-player games to show that one of the players cannot have a winning strategy. Proof by contradiction is used. We assume that, for example, the second player has a winning strategy. Then the first player uses (steals) that strategy somehow in order to win, which contradicts the assumption. Note that an argument like that is not constructive. We cannot point out a strategy for one of the players, we just know that some strategy does not exist (or exists). Let’s start with the following game. Two-step chess. This game has the same rules as the usual chess, except that each player can make one or two moves at a time. We will prove that White is guaranteed at least a draw. That is, if White plays optimally Black cannot win. Proof. Assume that Black has a winning strategy and White knows this strategy. At first move White plays Kb2 on a3 and then back on b2, i.e. a move that leaves the original position intact. Then it’s Black’s turn, and after they make a move, White will plays as if they were the second player and use the Black’s strategy to respond to their opponent’s moves. Thus, they will win, which is a contradiction since it is impossible for both players to win. Hex game. This game is played on an board that consists of empty hexagonal cells like in fig. 1. The usual size of the board is but it could be of any size. Each player in turn places a counter of their color (blue or red) on an empty cell. The counters cannot be moved once placed. The first player that forms a connected path, linking the opposite sides of the board marked by their own color wins, see figure 2. Intuitively, playing first is an advantage, since every additional counter increases the chance to make a path. As it turns out the first player has a winning strategy. First, let’s prove that draw is not possible. Suppose the table is filled. Consider the figure obtained by the blue counters. It may have several connected components. Take the one that contains the right side of the table (see fig. 2) which is painted blue. If this connected component contains the left side of the board then Blue has won. If not, consider the red counters along boundary of this component. They form a red path that connects the top and the bottom sides. In this case Red wins. Let’s now prove that the first player always wins. This was shown by John Nash, the same guy that the movie A Beautiful Mind was made about. He introduced the strategy-winning argument in 1940’s. Since a draw is not possible, and the game is fully deterministic, one of the players has a winning strategy. Assume, for the sake of contradiction, this is the second player, say Red. The first player (Blue) knows this strategy. Blue plays an arbitrary move putting a blue counter at a position . Then Blue forgets about this counter and plays, responding, as if he were the second player, to Red’s moves using their own winning strategy, as if this counter was not there . If, at some moment, Blue (according to the strategy) has to put a counter on position he puts it on an arbitrary free cell and marks in his mind that he has put a counter on and forgets about the counter on They go on like this. Note that there is one less counter in Blue’s mind. There is actually one more blue counter on the board. If Blue can win without that counter, he will also win with it, an extra blue counter is even better. Thus, Blue wins, contradiction. A game of sets. The following problem is a warm-up for the last one, although it’s also interesting as a standalone and has a bit different strategy. Problem 1. Let be a family of subsets with 2 elements of some base set It is known that for any two elements there exists a permutation of the set such that , and for any A bear and crocodile play a game. At a move, a player paints one element of the set in his own color: brown for the bear, green for the crocodile. The first player to fully paint one of the sets in in his own color loses. If this does not happen and all the elements of have been painted, it is a draw. The bear goes first. Prove that he doesn’t have a winning strategy. Solution. Assume for the sake of contradiction, the bear has a winning strategy. Note that for any the bear also wins by painting at his first move Indeed, consider a permutation that satisfies the condition in the statement. Assume that there are two bears that play on two different copies of The original (winning) bear plays on the first board and starts by painting The second bear paints on the second board and waits for the crocodile’s move. When the crocodile on the second board paints, say, the second bear passes painted in green to the first bear. The first bear plays and the second bear plays and so on. Assume now, the bear starts with his winning strategy by painting an element brown. Let Suppose, the crocodile knows the winning strategy of the bear. He makes in his mind the following scenario. Assume that he, as the bear, starts by painting in green, and the bear (as the crocodile) responds by painting brown. Then the crocodile (as if he were the bear) in order to win plays by painting green. Now the crocodile acts. He paints and green, but the latter element only in his mind. In subsequent moves, he responds to the bear using bear’s own strategy. Note that the bear cannot touch because if he paints it brown, he instantly loses since So, the bear cannot prevent the crocodile from using his own strategy and winning in the end. Contradiction! A Russian TST (2018) problem. The last topic is about a Russian problem given at their Team Selection Tests for IMO 2018. Problem 2 (Russian TST 2018, day 4, p2). Let be a finite family of subsets of some set . It is known that for any two elements there exists a permutation of the set such that , and A bear and crocodile play a game. At a move, a player paints one or more elements of the set in his own color: brown for the bear, green for the crocodile. The first player to fully paint one of the sets in in his own color loses. If this does not happen and all the elements of have been painted, it is a draw. The bear goes first. Prove that he doesn’t have a winning strategy. Solution. Assume on the contrary, the bear wins by painting at his first move a set Of course, Note that for any the bear also wins by painting at his first move some set Indeed, take any and consider a permutation that satisfies the condition of the statement. Then, the bear also wins by painting at his first move the set This can be shown by a symmetry argument as above. Assume that there are two bears that play on two different copies of The original (winning) bear plays on the first board and starts by painting brown. The second bear paints brown and waits for the crocodile’s move. After the crocodile responds with painting, say, a set the second bear passes to the first bear the set painted green. The first bear paints according to his strategy. The second bear then paints and so on. So, now the bear with his winning strategy paints at first The crocodile makes the following calculation in his mind. Assume, he knows the bear’s strategy and he, as if he were the bear, starts by painting green. Suppose the bear (as the crocodile) responds by painting the set brown. The crocodile (as the bear) must win and let his winning move in this situation be to paint (green) some set Now, the crocodile acts. He paints in his first move the set green. Then he follows the bear’s winning strategy and wins because the only difference is that is painted brown, instead of green, but if he wins when is painted green, he will also win in this situation. Contradiction! References. A Russian TST 2018 problem. AoPS thread. Hex board game (wiki page). 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 Too Many Perfect Cubes! Bulgarian TST for BMO 2024, p7. What follows is a problem I gave to the Bulgarian team selection test for Balkan Math Olympiad, 2024. The same idea was used to solve a problem from China TST 2012 - see . Problem (Bulgarian TST 2024, p7). Let $latex n\ge 4$ be an integer number and $latex S_n={1,2,3,\dots,2^n}$.… In "Algebra" A Game on circles. Italian TST Problem. I was sent a problem from some Italian training camp, allegedly taken place in 2017. At least, the male name used in the problem statement was Italian, I think. Probably it was not from their official team selection test, but still I put it this way in the title. For… In "Combinatorics" Bulgarian TST 2015. A Geometric Problem. An interesting geometric problem, which I saw recently, was proposed on a Bulgarian team selection test. As the Bulgarian TST's problems are not released officially, there are only two ways to see one of them - either by leaking in some way in forum websites like AoPS, or in books… In "Bulgarian-Math-Olympiad" 2 thoughts on “Strategy-Stealing Argument. A Russian TST Problem.” Pingback: Games. Pairing Strategies. – A Point of View. I read somewhere that the hex theorem(that there is always a winner in the game of hex) is true if and only if the brouwer fixed point theorem is true, but I cant find a satisfactory, intuitive and elementary proof for that fact. Can you please make a post about that Reply Leave a comment Cancel reply This site uses Akismet to reduce spam. Learn how your comment data is processed. Comment Reblog Subscribe Subscribed A Point of View. Already have a WordPress.com account? Log in now. A Point of View. Subscribe Subscribed Sign up Log in Copy shortlink Report this content View post in Reader Manage subscriptions Collapse this bar
10398	https://www.concast.com/c26000.php	Cast products Lead-free products Extruded products Custom products Resources Production Inventory Quality Weight Calculator Markets # Get a quote C26000 Stocked alloy specs View C14500 C51000 C54400 AMS 4640-C63000 AMS 4590-C63020 AMS 4634-C64200 C67300 AMS 4596-C72900 AMS 4597-C72900 AMS 4598-C72900 C86300 C89835 C90300 C93200 C95400 C95500 AMS 4880-C95510 C95900 Reference guide Stocked size schedules View C14500 C51000 C54400 AMS 4640-C63000 AMS 4590-C63020 AMS 4634-C64200 C67300 AMS 4596-C72900 AMS 4597-C72900 AMS 4598-C72900 C86300 C89835 C90300 C93200 C95400 C95500 AMS 4880-C95510 C95900 Extruded and drawn C26000 spec sheet Standard products C26000 Product description: Cartridge brass 70% Tempers: H01 quarter hard, H02 half hard, H04 hard Solids: 3/8" to 2 1/2" O.D. Hex: 3/8" to 2" O.D. 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Single values represent maximums. \| \| \| \| \| \| \| \| \| \| \| \| \| 68.50- 71.50 \| 0.07 \| Rem. \| 0.05 \| Machinability \| Copper alloy UNS No. \| Machinability rating \| Density (lb/in3 at 68 °F) \| --- \| C26000 \| 30 \| 0.308 \| Mechanical properties Mechanical properties according to ASTM B927/B927M-23 C26000 H01 quarter hard Size range: under 1/2" diameter rod \| Tensile strength, min \| \| Yield strength, at 0.5% extension under load, min \| \| Elongation, 4x diameter or 4x thickness, min \| Rockwell "B" hardness \| Remarks \| --- --- --- \| ksi \| MPa \| ksi \| MPa \| % \| typical HRB \| \| \| 50 \| 345 \| 30 \| 205 \| 20 \| 55 \| \| Size range: over 1/2" diameter rod to 1" inclusive \| Tensile strength, min \| \| Yield strength, at 0.5% extension under load, min \| \| Elongation, 4x diameter or 4x thickness, min \| Rockwell "B" hardness \| Remarks \| --- --- --- \| ksi \| MPa \| ksi \| MPa \| % \| typical HRB \| \| \| 48 \| 330 \| 25 \| 170 \| 24 \| 55 \| \| Size range: over 1" diameter rod \| Tensile strength, min \| \| Yield strength, at 0.5% extension under load, min \| \| Elongation, 4x diameter or 4x thickness, min \| Rockwell "B" hardness \| Remarks \| --- --- --- \| ksi \| MPa \| ksi \| MPa \| % \| typical HRB \| \| \| 46 \| 315 \| 20 \| 140 \| 28 \| 55 \| \| C26000 H02 half hard Size range: under 1/2" diameter rod \| Tensile strength, min \| \| Yield strength, at 0.5% extension under load, min \| \| Elongation, 4x diameter or 4x thickness, min \| Rockwell "B" hardness \| Remarks \| --- --- --- \| ksi \| MPa \| ksi \| MPa \| % \| typical HRB \| \| \| 57 \| 395 \| 35 \| 241 \| 15 \| 70 \| \| Size range: over 1/2" diameter rod to 1" inclusive \| Tensile strength, min \| \| Yield strength, at 0.5% extension under load, min \| \| Elongation, 4x diameter or 4x thickness, min \| Rockwell "B" hardness \| Remarks \| --- --- --- \| ksi \| MPa \| ksi \| MPa \| % \| typical HRB \| \| \| 54 \| 370 \| 32 \| 220 \| 20 \| 70 \| \| Size range: over 1" diameter rod \| Tensile strength, min \| \| Yield strength, at 0.5% extension under load, min \| \| Elongation, 4x diameter or 4x thickness, min \| Rockwell "B" hardness \| Remarks \| --- --- --- \| ksi \| MPa \| ksi \| MPa \| % \| typical HRB \| \| \| 50 \| 345 \| 30 \| 205 \| 25 \| 70 \| \| C26000 H04 hard Size range: under 1/2" diameter rod \| Tensile strength, min \| \| Yield strength, at 0.5% extension under load, min \| \| Elongation, 4x diameter or 4x thickness, min \| Rockwell "B" hardness \| Remarks \| --- --- --- \| ksi \| MPa \| ksi \| MPa \| % \| typical HRB \| \| \| 70 \| 485 \| 50 \| 345 \| 10 \| 82 \| \| Size range: over 1/2" diameter rod to 1" inclusive \| Tensile strength, min \| \| Yield strength, at 0.5% extension under load, min \| \| Elongation, 4x diameter or 4x thickness, min \| Rockwell "B" hardness \| Remarks \| --- --- --- \| ksi \| MPa \| ksi \| MPa \| % \| typical HRB \| \| \| 65 \| 450 \| 45 \| 310 \| 15 \| 82 \| \| Size range: over 1" diameter rod \| Tensile strength, min \| \| Yield strength, at 0.5% extension under load, min \| \| Elongation, 4x diameter or 4x thickness, min \| Rockwell "B" hardness \| Remarks \| --- --- --- \| ksi \| MPa \| ksi \| MPa \| % \| typical HRB \| \| \| 60 \| 415 \| 40 \| 275 \| 20 \| 82 \| \| Physical properties \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- \| \| Physical properties provided by CDA \| \| \| \| \| \| \| \| \| \| \| \| \| \| US Customary \| Metric \| \| Melting point - liquidus \| 1750 °F \| 954 °C \| \| Melting point - solidus \| 1680 °F \| 916 °C \| \| Density \| 0.308 lb/in3 at 68 °F \| 8.53 gm/cm3 at 20 °C \| \| Specific gravity \| 8.53 \| 8.53 \| \| Electrical conductivity \| 28% IACS at 68 °F \| 0.162 MegaSiemens/cm at 20 °C \| \| Thermal conductivity \| 70 Btu/sq ft/ft hr/°F at 68 °F \| 121.2 W/m at 20 °C \| \| Coefficient of thermal expansion 68-572 \| 11.1 · 10-6 per °F (68-572 °F) \| 19.2 · 10-6 per °C (20-300 °C) \| \| Specific heat capacity \| 0.09 Btu/lb/°F at 68 °F \| 377.1 J/kg at 20 °C \| \| Modulus of elasticity in tension \| 16000 ksi \| 110317 MPa \| \| Modulus of rigidity \| 6000 ksi \| 41369 MPa \| Fabrication properties \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- \| \| Fabrication properties provided by CDA \| \| \| \| \| \| \| \| \| \| \| \| \| Technique \| Suitability \| \| Soldering \| Excellent \| \| Brazing \| Excellent \| \| Oxyacetylene welding \| Good \| \| Gas shielded arc welding \| Good \| \| Coated metal arc welding \| Not recommended \| \| Spot weld \| Fair \| \| Seam weld \| Not recommended \| \| Butt weld \| Good \| \| Capacity for being cold worked \| Excellent \| \| Capacity for being hot formed \| Fair \| \| Machinability rating \| 30 \| Thermal properties \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- \| \| Thermal properties provided by CDA Temperature is measured in Fahrenheit. \| \| \| \| \| \| \| \| \| \| \| \| \| Treatment \| Minimum \| Maximum \| \| Annealing \| 800 \| 1400 \| \| Hot treatment \| 1350 \| 1550 \| Find a product Search by CDA number # Request a quote C14500 C23000 C24000 C26000 C31400 C31600 C46400 C51000 C52100 C53400 C54400 C62400 C63000 C63020 C64200 C64210 C65100 C67300 C67400 C67410 C67600 C69300 C69400 C69430 C72900 (AMS 4596) C72900 (AMS 4597) C72900 (AMS 4598) C83600 C83800 C84200 C84400 C84800 C85700 C86200 C86300 C86400 C86500 C86700 C87850 C89320 C89325 C89520 C89831 C89833 C89835 C89844 C90300 C90500 C90700 C90800 C90810 C91000 C91100 C91300 C91600 C91700 C92200 C92300 C92500 C92700 C92800 C92900 C93200 C93400 C93500 C93600 C93700 C93800 C93900 C94000 C94100 C94300 C94700 C94700HT C94800 C95200 C95300 C95300HT C95400 C95400HT C95410 C95410HT C95500 C95500HT C95510 C95520HT C95600 C95800 C95900 C96400 C96900HT C97300 C97600 C97800 C99500
10399	https://digitalcommons.harrisburgu.edu/cgi/viewcontent.cgi?article=1000&context=csms_dandt	Harrisburg University of Science and Technology Digital Commons at Harrisburg University Dissertations and Theses Computer and Information Sciences (CSMS) 8-2017 Boundary Value Analysis for Input Variables with Functional Dependency Manmohan Maheshwari Harrisburg University of Science and Technology Follow this and additional works at: Part of the Computer Sciences Commons This Thesis is brought to you for free and open access by the Computer and Information Sciences (CSMS) at Digital Commons at Harrisburg University. It has been accepted for inclusion in Dissertations and Theses by an authorized administrator of Digital Commons at Harrisburg University. For more information, please contact drunyon@harrisburgu.edu, ereed@harrisburgu.edu. Recommended Citation Maheshwari, M. (2017). Boundary Value Analysis for Input Variables with Functional Dependency . Retrieved from 1 Boundary Value Analysis for Input Variables with Functional Dependency Submitted to Harrisburg University for Science and Technology as a Partial fulfillment of the Requiremen ts for the Master of Science degree in Computer Science BY: Manmohan Maheshwa ri manukhator@gmail.com Supervised by Nushwan Al -Nakash , PhD. Assistant Professor of Computer Science Nal -nakash@harrisburgu.edu August 2017 2 Table of Contents Page ABSTRACT ………………………………………………………………… ... …. 4 LIST OF FIGURES …………………………………… ……………………… …. 5 CHAPTER GENERAL INTRODUCT ION …………. ………………………………… …. 7 1.1 Introduction …………. ……………………………………………… 7 1.2 Bounda ry Value Analysis ………….……….……….……… ……… 9 1.3 Functional Dependencies ……… …. ….…… ….…… .….….……… 10 1.4 Automation Testing ………….…………………………….………. 1 1 1.5 Mu tation Testing …………………….…….…….…….…. ….……. 11 1.6 Problem Statement an d Justification …….………………………… 1 3 THEORATICAL BACKGROUND … ………………………………………. 1 8 2.1 Divide and Rule … ….….……. ……. …. …. ….……. ….…… .….…. 1 8 2.1.1 Algorithm …………………………………………………… 1 9 2.2 Th e Function Tree …………………………………………………. 2 4 SYSTEM IMPLEMENTATION …………………………………….………. 2 9 3.1 Introductio n ….……….……….……….….…….………….……… 2 9 3.2 Implementation ………………………… ... ………….……………. 2 9 3.3 Implementing General BVA ………………… .….….……… .……. 2 9 3.3.1 Example of BV A …………………………………………… 30 3.3.2 BVA Analysis ……… .………………….… .…… .…………. 3 2 3.4 Implementing the Funct ion Tree Algorithm ………………………. 3 33 3.5 Implementing the Proposed Approach …… .…… .….… …. ….…… . 36 3.5.1 Algorithm … .….….………… .….……….…….……………. 3 9 EXPERIMENTAL RESUL TS ……… .…………. ….……. ……… .. ..…… …. 4 4 4.1 Introduction ………………….………….………….……………… 4 4 4.2 Analyzing General BVA ………………… .…… .……………… …. 44 4.3 Analyzing the Function Tree Approach …………… .……………. .. 46 4.4 Analyzing the Modified Approach ………………………………. .. 51 4.5 Compari son …………………………………………………….... ... 54 CONCLUSIONS AND FUTURE WORK ……………….… …… …… … 57 5.1 Conclusion ……………………………………………………… … 57 5.2 Future Work ……………………………………………………. ... .. 58 REFERENCES ………………………………….……….……………….…… ... 594 ABSTRACT Software in today’s world is use d more and in different ways as well than ever before. From microwaves and vehicles to space rockets and smart cards. Usually, a software programmer goes through a certain process to establish a software that will follow a given specification. Despite the hard work of the programmer, sometimes they make mistakes or sometimes they forget to include all th e possibilities of the question for which they are writing the program, which is very humanly in nature. And for those mistakes, a testing unit is always th ere. There are numerous techniques of Software Testing, one of which is Boundary Value Analysis. A modified version of Boundary Value Analysis using input parameters with functional dependency is proposed in this work . The idea is derived from the inter de pendency of functions among the input parameters. With this modified algorithm, an automated testing tool is created and implemented. This testing tool shows the advantages of the modified algorithm developed over the Functional Tree Approach and reduces a significant amount of test cases that leads to an exhaustive testing. This modified method will test almost every possible required test case increasing the system’s efficiency. This method will be a very good help for any product based company saving a h uge amount of money and time. Generalized BVA generates 5n number of test cases where n is number of variables while Function Tree method generates the highest of all t hree that is n5^(n -1) and the modified approach generates 7n + k number of test cases where 5 k is the number of mutants killed at each step. So, it shows that the number of test cases in case of modified algorithm is significantly lower than the Function Tree algorithm while almost similar as regular BVA but it covers more functionalities and features Keywords : Software Testing, Boundary Value Analysis, Functional Dependency, Functional Tree, Mutation Testing, Automation Testing .6 LIST OF FIGURES Figure Page 1.1 Boundary Value Analysis Range …. ………………………….… ……… … 9 1.2 Functional Dependencies ………………….…………………….……… … 9 1.3 Defining the Boundary Values ………………………………………… .... 13 1.4 Benefits of Automated Testing .…………….……….…….…….……… .. 14 1.5 Mutation Testing and its Test Su its ….…….……….……….…….……… 15 2.1 System Flow Chart of Di vide and Rule Approach …………………….. … 19 2.2 Boundary Func tion Tree for Xmin ………………………………... ……... 23 2.3 Next -to -Boundary Function Tree for Xmin ……………………... ………. 25 2.4 Middle Function Tree for Xmin …………….….…….….…… ….….….... 26 3.1 Bound ary Value Analysis ………………………………... …….………… 30 3.2 Geometry V iew of the Implementation ……………... …………………… 34 3.3 Flow Chart of the new Modified Method ………... ………………………. 36 3.4 Block Dia gram of the modified Approach ….. …………………………… 3 97 CHAPTER – 1 1.1 Introduction Despite the hard work of the programmer, sometimes they make mistakes or sometimes they forget to include all the possibilities of the question for which they are writing the program, which is very humanly in nature. And for those mistakes, a testing unit is always there. The job of this unit is to figure out what mistake has the programmer made or what did he/she forget to include. In some of the big organizations, they have huge teams of testing for their products because they have an important consideration in mind of th e consequences of a software error. Most of the software usually need to stick to a single rule, i.e. to make sure that what is expected, it does. To use all the available resources in better sense, computers should also be helpful in “the art of software testing” to an improved extent, than is currently the case today. One of the issues today is if humans can make errors in coding then they can also make errors in software testing. The solution of this is not to remove human beings from the process of so ftware testing but to use today’s software development art form and make computers also participate. This thesis will present research aimed at classifying, examining, and improving the basic concept of boundary value analysis through automated software te sting. To describe it further , we should first talk about software testing, boundary value analysis, functional dependencies and then possibility of creating a method that will solve the problem. 8 Software testing is defined as a formal process in which a s oftware unit, several integrated software units or an entire package are examined by running the programs on a computer. All the associated tests are performed according to approved test procedures on approved test cases (Galin, 2004). Software testing, is to test or check whether the software is executing perfectly and the necessary requirements are fulfilled. It is a human tendency to make errors and for elimination of these, tests are done on the product being developed to find out the problem in the sof tware, that is why software testing is necessary. Software Testing is very expensive method in terms of time and money but it is the only way to find out bugs in the system. Due to time and budget constraints, it is impossible for us to perform exhausting testing for every set of test data, especially when there are enormous pools of input combinations. It requires an easy way or special techniques which will select test cases intelligently from the pool of cases, so that every test scenario is covered. So , optimal testing is necessary to save time and money. The hard part in this procedure of testing is to generate the test cases. A test case is a condition put on each input parameter and those sets of conditions will give the tester an output which will h elp in the testing objective. A good testing technique is the one which covers every aspect of the requirement and the objective of testing. In the Late 90’s test cases were derived manually but for some products this procedure takes time much more than th e time of required for the development. So, the method of automated testing is introduced to test the product 9 automatically using program. Detecting and removing errors in the earlier phases will even reduce the cost of whole development. 1.2 Boundary Value A nalysis There are numerous techniques of Software Testing, one of which is Boundary Value Analysis. Regarding boundary value analysis, NIST defines it as , a selection technique in which test data are chosen to lie along ‘boundaries’ of the input domain [or output range] classes, data structures, procedure parameters. The basic idea of boundary value analysis can be judged from the word boundary, as we know most of things in this world have boundaries. Such bounded values are used in this procedure. This process tests the product being developed on the boundary values of each input parameter and then generate the desired test cases. The basic idea of boundary value analysis(BVA) is to generate the number of test cases using the parameter values at their boundary points such as minimum (min) and maximum (max), the values next to the boundary points such as just above the minimum (min +) and just below the maximum (max −) and the median i.e. the nominal value (nom). In a program, the input parameters consist of th e upper and lower bounds, so the test cases are obtained following the boundary value analysis, by holding the values of all but one parameter at their nominal values and letting that parameter assume its extreme values . 10 Fig 1 .1 : Boundary Value Analysis Range 1.3 Functional Dependencies There are various examples of programs on which boundary value analysis works, one of which is Next Date function. Though it seems like a perfect method for software testing but it has some limitations too. It does n ot work on input parameters with functional dependencies e.g. X, Y and Z are three input parameters, now Y = f(X) and Z = g (X, Y), thus, Y is a function of X and Z is a function of X and Y, so the above method may not be applied. These input parameters wi th functional dependencies were studied previously. One of the previous methods in uses the boundary value analysis with divide and rule approach. The other method in uses the function tree for generalization of boundary value analysis to input par ameters that are functionally dependent. Fig 1. 2 : Functional Dependencies 11 1.4 Automation Testing The aspect of automation in software testing is focused on keeping human involvement to a minimum. Software Test automation uses specialized tools for the e xecution of tests of the product and compares the actual results against the expected result. But the question here is, why we need automation testing. Well, automation testing is important because of many reasons. One such reason is that it increases the speed of the testing process by executing test cases on its own. As we know, manual testing is both time and cost consuming and can be boring sometimes. An example for difficulty in manually test is testing for multi lingual sites. For that, automation tes ting is required. One of the advantages of automation is that we can run the testing overnight even unattended as it does not require human interaction. So again, through this thesis I am trying to make the technique of boundary value analysis more efficie nt and easy while making the process automated. It is a very difficult job to make an automation tool as every piece of program or every product requires a different automation tool so that it can create only the test cases required for the product. This t o go in the aims of the thesis. 1.5 Mutation Testing The method of Mutation testing is out there for ages and testers have known this for many years. However, few of them are using it for different -different reasons. Most of the procedural steps used are auto mated, for example, mutant software 12 creation and the white box testing. In this kind of testing, modifying a program code, re -running a suit of correct tests, and deliberately altering are included against the mutated program. The tester can easily assess the quality of a test suite using Mutation testing. So, by mutating different elements in the software in the product’s source code and after that checking if the test code can find the mutated err ors and detecting those errors. Mutation testing is a misno mer. Mutation testing tool is a super powerful tool which can be used for checking coverage or detecting testing inadequacies on testing software. It can choose correct mutant programs and going through a comprehensive testing of these correctly chosen pro grams. It is more of an analytical method than testing technique. For reliable and better results, one should find better methods that can reduce the number of tests and find efficient number of mutants. More efficient numbers of mutants lead to longer tes ting time duration. 1.6 Problem Statement and Justification To develop the problem statement, one should know what is the question you are trying to answer and why. In this work , I am trying to question whether the testing techniques nowadays are enough to an swer or test all the possible outcomes that a program has. Can I make an automation tool that will help me solve this problem of efficient testing? With the functional dependencies of the attributes, is it possible to create a testing technique that will b e far much better than normal Boundary Value Analysis? 13 Now after defining the problem, we should discuss why this problem should be solved and who all are affected with this problem solving is discussed. In this thesis, the problem is, can I make an autom ation tool that will help me solve this problem of efficient testing? And with the functional dependencies of the attributes, is it possible to create a testing technique that will be far much better than normal Boundary Value Analysis? It is very importan t that testers of any field should be able to write test cases based on Boundary value analysis because the hardest part in the procedure of testing is to generate enough test cases which can test all the possibilities of the question for which they are wr iting the program. Testing components will always be a challenging area for research. During the design, the component properties are going to be adopted. And one of the motive is to validate and test against these adopted component properties. We can gain higher productivity with generic components. To test a generic component which contains several generic parameters is very hard to test. Therefore, all possible applications through customization should be tested. There are various problems regarding test ing in this world. For boundary value analysis, an efficient number of test cases should be generated which will check for all the wrong and right possible outcomes. But with inputs with functional dependencies, sometimes test cases fail to check all the o utcomes efficiently. So, in this thesis, a modified approach is created which will check for test cases for inputs with dependencies. This modified approach will be created using two already present techniques known as “Divide -and -Conquer -Approach” and “Th e Function 14 Tree”. Then an automation tool will be created which will first use mutation testing to check and then automatically create all the test cases for those input variables. Boundary value analysis is another type of Black box testing and is also a part of stress and negative testing. By the name, boundary value analysis indicates limitation on something. So, when the inputs are supplied within the boundary values, it is positive testing and inputs are considered as negative testing beyond the bounda ry values. For example, if an application accepts VLAN Ids ranging from 0 –255, then 0, 255 will be the boundary values and any input going below 0 or above 255 will be invalid and will constitute negative testing. Fig 1. 3: Defining the boundary of the values This technique of Boundary Value Analysis is used for figuring out the bugs that usually occurs at the boundaries rather than ones that exist in the center. This technique is used to reduce the number of test cases to minimal, while assuring that 15 the test cases selected are effective test cases which would help in testing whole scenarios. To continuously maintain and improve the efficiency and quality of the software system’s development is a very hard challenge for any company or organization. In many software projects, because of time or cost constraints organizations neglect the testing phase. Sometimes, this might lead to customer dissatisfaction, followed by a lack of product quality and ultimately to increased overall quality costs. Despite t he hard work of the programmer and tester, an automation tool will reduce the man power in some extent and help the industry save money. Poor test strategy, delay in testing, underestimated effort of test case generation and subsequent test maintenance mig ht lead to additional cost of the software project. Automated tests are fast and can run frequently within different tests due to reused modules, and for the software products with a long maintenance life, it is cost -effective. While testing in an agile environment, changing software requirements of the system is necessary. As the software project progresses, test cases should be modified for a period in both manual and automated testing. It is important that people should know that using test automation, complete coverage of all the tests is unrealistic. Test automation allows performing different types of testing efficiently and effectively. 16 Fig 1. 4 : Benefits of Automated Testing The main reason of using the automation testing is that “Automated re gression tests which ensure the continuous system stability and functionality after changes to the software were made lead to shorter development cycles combined with better quality software and thus the benefits of automated testing quickly outgain the in itial costs” . Fig 1. 5 : Mutation Testing and its Test Suits Manual testing can be mundane and exasperating while test automation can remove all the frustrations of a tester and allows test execution without human 17 interaction guaranteeing accura cy and testers can now concentrate on more difficult test scenarios. Sometimes it happens that your tests can not find a bug, will you believe that there are no bugs? For that, one of the tests that allow you to assess the tests is Mutation testing. Unlike the other testing methods which will give you a proper testing result, in mutation testing there is no Pass or Fail result of disposition even if the output fails in the test cases to meet the standard output. Instead, to improve the lack of whatever is c ausing it, adjustments are made, then the test is done again until we attain a satisfactory score. 18 CHAPTER – 2 In the previous chapter, a description to the effect of solving the problem and who all will the effort affect. There have already been some ground -breaking discoveries in this matter such as techniques like “Divide -and -Conquer -Approach” which deals with inter -dependent parameters and “The Function Tree” which uses a technique in which there are various levels of testing the functional dep endencies. 2.1 DIVIDE AND RULE APPROACH This algorithm named divide and rule creates various independent sets of parameters by breaking the dependencies among those parameters. This method will make sure that there is no interference within the boundary v alues or with the boundary values of each other by the parameters in the new independent sets. And this is the reason, why this technique is named Divide and -Rule approach . Using this method, different independent parameter sets can be created from the dependent parameter sets and after that on these independent parameter sets, the traditional method of boundary value analysis is performed. This technique usually deals with the inter -dependent parameters which means that the boundary values of those par ameters are dependent on the value of some other parameter. So, such type of dependency in parameters is known as boundary 19 dependency. In this approach, there are 3 types of parameters: Dependent Parameters, Independent Parameters, and Boundary -determining Parameters. In the next section, written is an algorithm which helps in creating the test cases on boundary value analysis using divide -and rule approach. 2.1.1 Algorithm This algorithm is based on the idea that a multiple set of independent parameters can be generated from a single set of dependent parameters. Using the simple or traditional boundary value analysis will not generate various test cases that are required for testing all possibilities. Therefore, a generic algorithm is required so that we will b e able to generate the independent sets of parameters with any kind of dependency. This method can be used for generating test cases for programs such as Next -Date function [2,12]. This program is based on 6 different types of modules and the program resul ts in six screens with each representing a module of the tool and the six modules are: getVariables, getRelations, getNumOfSets, getBounds, getSetRelations, and generateBVATestCases. So, these six modules are shown in the algorithm as well as in the flow c hart representation of this technique as shown in Fig 2.1. User will enter some input and that input will be accepted by each module and then each module will forward its output to the next module. There exists a button for each and every module that is th e “Submit Query” button, which will allow the user submission of the input. In the following, the design for each module is presented individually. 20 Fig 2.1 : System Flow Chart of Divide and Rule Approach The above representation is explained as : Module1 (getVariables): In this module, the user is prompted to enter all the variable names. 10 text boxes are displayed in the screen of this module and all the users can enter the names up to 10 variables. Due to some design purposes, there is a limitat ion on the number of variables. Module2 (getRelations): In this module, relationships among the variables are defined by the user. Based on the relationships entered by the user, the tool determines the boundary -determining, boundary -dependent, and indepen dent 21 variables. For n variables, the screen displays n2−1 checkboxes in n rows and n columns. To specify the relationship among variables, the proper checkboxes are checked. For example, to specify that variable i depends on variable j, the checkbox in ith row and jth column is checked. Module3 (getNumOfSets): Values for all boundary dependent and boundary - determining variables can be divided into two or more sets. This module prompts the user to enter how many sets can be formed for each boundary -dependent and boundary -determining variable. These sets are refe rred to as boundary -dependent sets and boundary -determining sets. For k boundary -dependent variables and j boundary determining variables, the screen displays k + j textboxes. Module4 (getBounds): This module prompts the user to specify the {min , min+ , nom , max− , max} values for every dependent variable, all boundary -dependent sets and all boundary -determining sets. For p independent variables, 5 p text boxes are displayed. For q boundary -dependent variables with s1,...,s q sets each, tq = 5( s1 + ··· + sq) te xt boxes are displayed. For r boundary -dependent variables with s1,...,s r sets each, tr = 5( s1 × ··· × sr) text boxes are displayed. Module5 (getSetRelations): This module generates all possible combinations of boundary -determining sets for each boundary -dependent variable. It then prompts 22 the user to associate each combination of the boundary determining sets with a boundary -dependent set. The screen displays all possible boundary -determining set combinations for each boundary -dependent variable. For every combination, it displays n radio buttons, where n is the total number of sets for that boundary dependent variable. Module6 (generateBVATestCases): This module generates all possible combinations of all the boundary determining sets. It then associates ev ery combination with a boundary -dependent set of every boundary dependent variable, based on the associations specified by the user in Module 5. It then associates all the combinations with all the independent variables. These boundary -determining set comb inations associated with boundary -dependent, and independent variables form the BVA sets. The module then generates the test cases by performing boundary value analysis for every BVA set. The following algorithm is written in 10 steps: Step 1: Divide the parameters in three sets: D: dependent parameters, B: boundary -determining parameters, I: independent parameters. Step 2: For every parameter d D, create a set of its determining parameters. Each element of this set shall also be an element of set B. 23 Step 3: For every parameter d D, create separate sets of all possible boundary value ranges for d. Mark these sets d1 to dx. These sets must contain these bounds (min, min+, nom, max−, max) based on its boundary -determining parameters. Step 4: For e very parameter b B, create sets of values that will affect its dependent parameter on the boundary values. Mark these sets b1 to by. The relation between sets (d1 to dx) and sets (b1 to by) are specified in Step 6. Step 5: For every parameter d D, gen erate all possible combinations of its sets. For e.g., if a and b are 2 elements of this set and a has 2 determining sets (a1 and a2) and b has 3 determining sets (b1, b2 and b3), then possible combinations will be: a1b1, a1b2, a1b3, a2b1, a2b2, a2b3. Ste p 6: Assign a boundary -value set to every possible combination of determining sets. The relation between these sets is that if variables a and b assume the values from a combination apbq, then d can assume a value from corresponding dr only. Step 7: For b oundary -determining sets, generate all possible combinations of all boundary -determining parameters. Step 8: For every combination, there is in the previous step, assign the boundary values to the sets for each of the dependent parameters using Step6. St ep 9: For every combination, there is, assign all the independent parameters which will create an independent set of combinations. Step 10: For every independent parameter set that is produced, perform traditional boundary value analysis to generate test c ases. 24 2.2 THE FUNCTION TREE The function tree approach is used to generate test cases for the input parameters with functional dependencies using boundary value analysis. This method is not for any specific ideology or specific problem but this is very generi c solution to the black box testing technique which can use 2, 3 or even many more input parameters dependent on one another. Following are the 3 different levels for implementing this technique. Fig 2.2 : Boundary Function Tree for Xmin Level 1: Assume that the lower and upper bounds for X are given respectively Xmin and Xmax. Now, the function tree is drawn for each of the two bounds as shown in Fig 2.2. Boundary Test, in this level all the parameters take their boundary values for the implementat ion and those values include min and max. The function tree for this level is always a binary tree and every path in that tree is a test case. The function tree here is a binary tree which is usually generated by the upper and lower bounds function for X and Y. Each path on the function tree corresponds to that one test case. There are four testing cases from left to right as: 25 (Xmin , fL (Xmin) , gL (Xmin , fL (Xmin)) (Xmin , fL (Xmin) , gU (Xmin , fL (Xmin)) (Xmin , fU (Xmin) , gL (Xmin , fU (Xmin)) (Xmin , fU (Xmin ), gU (Xmin , fL (Xmin)) There are some test cases that may be same and some may not satisfy the constraint function. The function tree for Xmax and function tree for Xmin can be drew similarly except that all the Xmin is replaced by Xmax and that to everyw here in the tree. So, the four testing cases that can be generated from the function tree for Xmax are: (Xmax , fL (Xmax) , gL (Xmax , fL (Xmax)) (Xmax , fL (Xmax) , gU (Xmax , fL (Xmax)) (Xmax , fU (Xmax) , gL (Xmax , fU (Xmax)) (Xmax ,fU (Xmax) , gU (Xmax ,fL (Xmax)) Boundary tests can be formed by remove the duplicated and those that do not satisfy the condition h (X, Y, Z ) ≥ 0 or h (X, Y, Z ) = 0. Level 2: Next -To -Boundary Test, in this level of test at least 1 of the parameters has the next to boundary values that in cludes max -1 or min+1 and all the other parameters can have any value. So, in total there are 4 values assigned to any parameter which includes min, max, min+1, max -1. 26 Fig 2.3 : Next -to -boundary Function Tree for Xmin Next -to -Boundary tests can be fo rmed by selecting the one’s that has at least one of parameters is not boundary value and remove the duplicated and those that do not satisfy the condition h (X, Y, Z ) ≥ 0 or h (X, Y, Z ) = 0. The Next -to -boundary Function Tree for Xmax can easily be obtained by replacing Xmin with Xmax . Next -to -boundary Function Trees for Xmin + and Xmax − would be different with Figure 2. For example, to obtain the tree for Xmin +, we fi rst replacing Xmin in Figure 2 by Xmin +, then we need to connect the paths from fL(•) to gL(• ,f L(•)) and from fL(•) to gU(• ,f L(•)) respectively. Other the right most side, we need to connect the paths from fU(•) to gL(• ,f U(•)) and from fU(•) to gU(• ,f U(•) ) respectively. This is because these four test cases were not included in the cases generated in Step 1. Similar changes need to be made to get the Next -to -boundary Function Trees for Xmax −. 27 Fig 2.4 : Middle Function Tree for Xmin Level 3: Normal Test, in this level of test there is a middle value included with the 4 values of Level 2 that is nom. At least one of these parameters has the middle value. This level is almost like tradit ional boundary value analysis except that in traditional boundary value analysis, all except 1 has the middle value. Level 3 tests or Middle tests are same as the cases in the traditional boundary value analysis, thus 2 of the 3 parameters taking middle va lues. The third parameters go through its 5 elements sequence: min , min+ , nom , max− and max. Similar as in step 1 and 2, we use the Middle Function Tree to generate test cases in level 3. The middle function tree is the most complete tree. Every element in the X’s 5 elements sequence Xmin , X min+ , X nom , X max− , and Xmax has a middle fu nction tree. So, using these 3 levels, an algorithm is designed which will automatically generate test cases. There are two real time systems working on this algorithm. 28 One of which is Net Provision Activator (NPA), a product of Syndesis Limit where this algorithm is being used for generation of their test cases. And the second one resides in the Atmospheric Physics Lab at Trent University is a Celestial Tracking Software (CTS). It is used to conduct spectral analysis of light from different - different sou rces such as the Sun, hardware such as spectroscope and mirrors used in the analysis are being controlled by (CTS). So, shown above are the previously defined work that has been already done but, the Function tree approach induces a lot of test cases of or der n5^(n -1). So, it is difficult to cope with such a large data and to check on these values which are more than 18k if n=6 and this results in exhaustive testing. While Divide -and -Rule approach is efficient but this method is limited to some extent e.g. it can solve Next Date problem and those like this problem. That is why a new method is required so that it can generalize the problem solving for input variables with functional dependencies. 29 CHAPTER – 3 3.1. Introduction Software testing is one of th e most widely and vividly used software analysis techniques in today’s world, one of which is Boundary Value Analysis ( BVA ). A modified version of BVA using input parameters with functional dependency is proposed in this chapter . The idea is derived from t he inter dependency of functions among the input parameters. With this modified algorithm, an automated testing tool is created and implemented. 3.2. Implementation In this chapter, three different models of BVA with respect to their functional dependencies ar e discussed and implemented. A comparison amongst them is made to find out the most suitable method as far as execution time. 3.3. Implementing General BVA One of the three methods that is compared in this thesis is the general BVA. Boundary value analysis can be referred to as a technique or a method of a black box testing in which all the values as input at the boundaries of the domain of the input are tested. However, it has been widely recognized that the values at the 30 extreme ends of the input, and may be just on the outside of, domains may tend to create significant amount of errors in system functionality. This technique helps in boundary value testing which generally is done between both the valid and the invali d boundary partitions. With the help of thi s method or technique, all the boundary v alues are tested by creating the number of those test cases for a particular input field. This method does generate a very small number of test cases but it surely does not cover all the necessary cases as well. In general, this method generates 7n number of test cases, with n being number of variables. 3.3.1 Example of BVA Consider an organization that should test a software program which takes the values of the integers ranging between -100 to +100. Now in this case, total of three different sets of the valid equivalent partitions should be taken, which must be – the negative range which is from -100 to -1, the zero (0), and the positive range which is from 1 to 100. All these ranges have a maxi mum an d a minimum boundar y value. A low er value of -100 and -1 as the upper value will be the n egative range and the p ositive range will contain 1 as the lower value and the upper value as 100 . Though, w hile testing these values, we should consider the fact that some of the values will overlap once the boundary values for each partition are selected . So, during the testing of these conditions, the overlapping values will appear when these boundaries being checked. 31 These values that have been overlapping must be removed so that the elimination of these redundant test cases can be done with ease . Now , after doing all that, the input box test cases that accepts the range of the integers between -100 and +100 through BVA are: i All the test cases which have the same data as the input bo un daries of input domain: in this case, -100 and +100. ii All the test cases that have values just below the extreme edges of input domain: in this case, -101 and 99 iii All the test cases having values which are just above the extreme edges of input domain: in thi s case, -99 and 101 With the BVA technique, it is quite an easy task of test ing such a small piece of data instead of doing a test on the whole data set. That is the reason , in major testing fields such as quality management services and software testing , this method is often adopted instead of other testing methods . Fig 3.1: Boundary Value Analysis 32 3.3.2 BVA Analysis For the generalized BVA method, the following are the advantages and the disadvantages: Advantages are: This technique is very good at exposing the potential user interface or user input problems It has v ery clear guidelines on determining test cases It is the best approach in all the cases in which the software functionality is based on numerous variables which generally represents the physical quantities. It generates v ery small set of test cases Disadvantages are: It d oes not test all the possible inputs It does no t fit well with respect to the Boolean Variables It generally works well only with those independent variables that usually depict qu antity It d oes not test the dependencies between combinations of inputs Now, as this work is based on solving the problem of functional dependencies in the input variables, there is no scope for this generalized approach of boundary value analysis since t his method does not deal with the dependencies .33 3.4 Implementing Function Tree Algorithm As discussed in Chapter 2, Function Tree Algorithm has in total three levels. Each level i s an upgrade of the other. So, it’s decided only to implement the last level i.e . Level 3 . The reason for this choice is that this level is almost like the traditional boundary value analysis except that in traditional boundary value analysis, al l except 1 has the middle value but in Level 3 of Function Tree Algorithm has the middle v alue in at -least one of these parameters. Now, let assume that these three input parameters are X, Y and Z. Also, the following assumptions are made relating to the functions to be constrained: (1) Xmin will be the lower bound and X max will be the upper b ound for X. For Y, the upper bound and the lower bound will be the functions of X. So, Ymax = fU(X) ……………………………... equation 3.1 Ymin = fL(X) ……………………………… equation 3.2 (2) For the third parameter, the upper bound and the lower bound will be functions of X a nd Y. So, Zmin = gL(X, Y) …………………………... equation 3.3 Zmax = gU(X, Y) ………………………….. equation 3.4 (3) All the three parameters: X, Y and Z must satisfy the following constraint function: h(X, Y, Z) ≥ 0 or h(X, Y, Z) = 0 ……………. equation 3.5 34 Now, it is known in general that the test cases which might involve the boundary values are usually more important than those cases which involve “middle” values. Though, it does not necessary mean that the “middle” value is required for testing, especially when factors such as cost and time are permitted. The test cases in this method are almost like the cases in the general boundary value analysis, taking two of the three parameters as middle values. The third parameters must go throug h the defined five elements sequence wh ich is : min, min+, nom, max− and max. The most complete tree is the middle function tree . Every element has a middle function tree in the five elements sequence of the X’s that is: Xmin , Xmin+ , Xnom , Xmax− , and Xmax . Using this method, a total of five test cases can be generated from the left most path. So, i n total, there are 25 test cases that are being generated from the Level 3 tree shown in Chapter 2 . Though some of these test cases might not satisfy the Function Tree definition of the test cases . So, we will have to use constraint functions to remove all the se unwanted or repetitive cases that do not satisfy the method’s conditions. However, i f a test case satisfies the Function Tree definition of the test cases , it will be added to the test case list. The Level 3 approach of the function tree method was originally inspired from the general geometry view point. Fig 3.2 shows a representation of the boundary val ue selection for each parameter in a geometry point of view. 35 Fig 3.2: Geometry view of the i mplementation An algorithm was devised for the automation testing tool . Note that due to complexity constraint, the algorithm caters for only two cases. For two parameters (X, Y): • Generate x’s 5 -element sequence x -lower, x -lower+ 1, x -middle, x -upper - 1, x-upper, where x -middle = (x -lower + x -upper) / 2. • Loop through x in x -sequence to find the y’s bounds. • Generate y’s 5 -element sequence y -lower, y -lower+ 1, y -middle, y -upper - 1, y-upper. • Loop through y in y -sequence. For each pair (x, y) if at least one element is a middle element then add it to the test cases list. For three parameters (X, Y, Z): • Generate x’s 5 -element sequence x -lower, x -lower+ 1, x -middle, x -upper - 1, x-upper. 36 • Loop through x in x -sequence to find the y’s bounds. Generate y’s 5 -element sequence y -lower, y -lower + 1, y -middle, y -upper - 1, y -upper. • Loop through y in y -sequence. For each pair (x, y) if it satisfies the constraints is In -Z-Constraints (x, y) find z’s lower and upper bounds. • Generate z’s 5 -element sequence z -lower, z -lower+ 1, z -middle, z -upper - 1, z-upper. • For each triple (x, y, z) if at least one element is a middle element then add it to the test cases list. Though, only algorithms for two cases are written i.e. two and three parameters but t he function tree approach show n in this work is general and one can apply this approach to any constraints even with more than three parameters. Using this function tree algorithm, an automation testing tool was designed and implemented to decrease the time of testing . 3.5 Implementing th e Proposed Approach In the proposed approach, a new modified, based on boundary value analysis (BVA) algorithm works on the input parameters with functional dependencies being generated even with the invalid cases alongside valid cases to give more precise testing objectives. Figure 3.5 represents the proposed modified method which works on the algorithm. Then, the variables defined will be designated its value. As you know, each variable will be given a boundary value t hat is, min as in minimum value of 37 the range and max as in maximum value of the range. This summarizes the block number 2 of the block diagram. Fig 3.4 : Block Diagram of the modified approach The following will explain the algorithm below in conjunction with fig.3.5 above, which represents the proposed modified method. For figure 3.5, starting with the stage number 1 and Step 1 in the algorithm by askin g for variables details. First, it asks for the number of variables and their names. Then, we must define the number of dependent and independent variables. Though in this case, it is taken for 38 granted that the first variable taken is the only independent variable and the other variables are dependent variables. After designating the variable values, the algorit hm moves to stage 3 in which various functions are created known as variable functions that are basically functions to represent the dependencies of the dependent variable. The values for the coefficients and the constant term are then entered for the func tion. For example, suppose x is the independent variable and y is the dependent variable. Then the function will look like: Y = f(X) = aX 4 + bX 3 + cX 2 + dX + e ………………… ……………. Equation 3.6 Depending on the function type that whether the function is linear or non -linear and if non -linear then what degree. These are various factors that concludes the creation of the function. According to equation 3.6, if the function is linear, then value of a, b, and c will be 0. Otherwise, the values of a, b, and c depends o n the degree of the function and dependency. Now, after creating these variable functions, they are divided into monomials by making them dependent on each variable one at a time. And i f you are dividing a polynomial function by a monomial, it means that y ou are split ting the problem into different pieces . For instance, suppose if Y = 3X 4 + 2X 2 +5, then by dividing this polynomial function by X 2 and if we assume X 2 as Z, we can split the problem and make this polynomial function a monomial. After dealing wi th the variable function, we will now generate random mutants in the function. For example, consider the following C++ code fragment :39 This is a generic example to prove the case It is not from the code if (a && b) { c = 1; } else { c = 0; } The condition mutation operator would replace && with \|\| and produce the following mutant: if (a \|\| b) { c = 1; } else { c = 0; } This is how the mutants are generated. After we finish installing a mutant at every step of the monomial, several test cases are generated. Now these test cases are generated according to the traditional boundary value analysis methods or techniques. 40 Now, with these mutants generated at every step, we will compare the values of the functions as well as the monomials also at every step respect to the mutation and the test case. If the value at the variable function is equal to the value at the monomials, then we should repeat the step of mutation and test case generation again because it means mutant is still there and noth ing happened to it. Otherwise, we will follow the next step because mutant is killed. As soon as the mutant is killed, the random value used to generate the test cases and the mutants is chosen and returned to be the kth test case where k being the number of mutants killed. So, this is how we create mutants and kill them to generate test cases. This concludes the step 6 of the algorithm. Now, all the test cases are generated which are required or necessary to generate by an easy and efficient automation te sting method using the mutation testing. 3.5.1 Modified Algorithm : 1) First 7n test cases are computed by applying the traditional approach on the matrix??? created by Level 3 algorithm. 2) The dependent variable functions or polynomials are generated. 3) These functions are divided into monomials by making them dependent on each variable one at a time. 4) Generate test cases by randomly creating the mutants at every step. 5) If the computed value of the F(Polynomial) is equal to the computed monomial value then repeat Step 4 and Step 5 else Mutant is killed. 41 6) Return the random value as soon as the mutant is killed being the 5n +kth test case, with K being the nu mber of mutants killed up to this point. Now, in the algorithm given in chapter 2 i.e. the Function tree, there are five values for every parameter min, min+1, nom, max -1 and max. However, a modification is made by the inclusion of two parameters i.e. min -1 and max+1 thereby giving the advantage of testing on the invalid cases. In level 3 of the Function tree i.e. the Normal Test in which at -least one of the parameters should be nom, others now can have seven different values which will make the testing objective more precise. The seven values considered as parameters are min -1, min, min+1, nom, max -1, max and max+1. For instance, X, Y and Z are the three parameters with Y = f (X) and Z = f (X, Y). There are seven boundary values for every parameter. Now assign all but one para meter the middle value, i.e. nom, and that parameter can have any value of those seven. So, X can have any of the seven values then Y and Z can only have the nom value out of those seven. So, the total combinations for X = 7 and again same goes with Y and Z. So, the total number of possible combinations for three input parameters with functional dependency comes out to be 73 i.e. 21. These test cases include invalid cases for testing too. So, in general total number of possible test cases for n input para meters with functional dependencies are: Test Cases = (n7). 42 Fig 3.3: Flowchart of the new modified method Next Step in the algorithm is to divide the polynomial function into subsequent monomial function for each dependent variable. The test cases are generated by randomly creating the mutants at every step. If the computed value of F(Polynomial) is not equal to computed monomial mutant value, then generate this Note: K-NumbersofMutantsKilled N–NumberofVariables 43 random number as the kth test case else repeat the previous step 50 times, which is an arbitrary number, till the condition is satisfied. For instance, if Z= f (X, Y) such that Z=2X+3Y. Now Z is divided into two functions, 2X and 3Y. Basically the coefficient of the other parameter is reduced to This will create mutants in the program causing a deliberate fault in the algorithm to check whether the test cases are still working or not. In this example if X=2 and Y=3 is given then Z will be either 4 or 9 but it should give the output 13. So, the test cases should not work due to the mutant present in it. So, the algorithm randomly creating and killing the mutant before some unknown error can create a mutant by mistake. This will reduce the unwanted errors. 44 CHAPTER – 4 4.1. Introduction In the previous chapter, different algorithms, and different types of techniques on how software technology in terms of testing is improving largely. In this chapter, the d ifference amongst all the implementations in technical terms as well as their result are discussed namely Traditional approach, Function Tree approach, and the proposed approach the results of which are analyzed, and compared. Note that chapter 4 test resu lts are based on 30 different cases, values, and different number of variables, however, only few cases are shown. 4.2. Analyzing General Boundary Value Analysis The following are made for the implementation of the generalized BVA. • Enter the number of variabl es to be tested. • Enter the name of those variables. • Enter the Min and the Max value of those variables. Also, for the above implementation, the following assumptions are made: • It does not check for the values outside the range of Min -Max. • There are no depe ndent variables. • It largely depends on the Mid -value or the average value. 45 Case 4.2. 1: Fig 4.1: Result of Case 4.2. 1 Number of Variables: 3 Name of the variables: a, b, c Min & Max value of a: 10, 100 Min & Max value of b: 5, 50 Min & Max value of c: -100, 100 Total Number of test cases generated: 15 Figure 4.2, below, shows the actual number of test cases versus number of variables that are made throughout this implementation. 46 Fig 4.2: Number of Test Cases per Number of Variables in General BVA In general, this technique generates 5n number of test cases where n is the number of variables. Five different cases being min, min+1, mid, max -1 and max. The dawn side of this method is that it does not solve the problem of dependency while the task for p resented work is based on solving the BVA for input variables with functional dependency. 4.3. Analyzing Function Tree Approach As compared with g eneral BVA above, t his method will deal with functional dependency and it is based upon the following; • To check de pendency, it is required to enter at -least 2 variables. • The first variable to enter is always independent as the dependent variable needs a variable to depend upon as well. 10 15 20 25 0 5 10 15 20 25 30 2345 Number Of Test Cases Number Of Variables Number of Test Cases per Number of Variables 47 • The highest degree considered in the equation of the dependent variable is three be cause of the complexity issue. Also, the following are the steps for method implementation: • Enter the number of variables to be tested. • Enter the name of the variables. • Enter the Min and Max of the first variable [Note: 1 st variable is assumed to be indep endent] • Enter the coefficients based on the nature of variable’s degree. For computational complexities, the highest degree considered is three. If the variable is dependent on the other variable linearly, then the coefficients of other degrees in the equa tions are made zero. The equations that are being created are there to measure the quantity of dependency or may be to tell the degree of dependency of one variable to the other. Case 4.3. 1: Number of Variables: 2 Name of the variables: a, b Min & Max val ue of a: 14, 67 Equation for F(b): Constant Term: 56 Coefficient of a 1: 6 Coefficient of a 2: -7 Coefficient of a 3: 0 48 Total Number of test cases generated: 10 Fig 4.3: Results of Case 4.3.1 Case 4.3.2: Number of Variables: 3 Name of the variables: a, b, c Min & Max value of a: -14, 89 Equation for F(b): 3+4a -8a 2+12a 3 Equation for F(c): 7+18a -3a 2+b+4b 249 Fig 4.4: Inputs for Test Case 4.3.2 Fig 4.5: Results of Case 4.3.2 50 Figure 4.5 below, shows the actual number of test cases versus number of variables that are made throughout this implementation. Fig 4.6: Number of Test Cases per Number of Variables in the Function Tree Approach As shown in fig. 4.5, in the case of 2 variables, total number of test case generated are 10. Similarly, in the case of 3 variables, 75 test cases are being generated and similarly in the case of 4 variables, 500 test cases are being generated and so on. In general, this technique generates n5 (n -1) number of test cases where n is the number of variables. Five different case s being min, min+1, mid, max -1 and max. This approach theoretically gives significantly better results in terms of efficiency but produce more test cases. 10 75 500 3125 0 500 1000 1500 2000 2500 3000 3500 2345 Number Of Test Cases Number Of Variables Number of Test Cases per Number of Variables 51 4.4. Analyzing the Modified Approach In the previous section, the Function Tree approach was analyzed t hrough which it was found that the approach provides overwhelming test cases to deal with. In this section, the proposed method is to deal with the problem of functional dependency efficiently with an average number of test cases. The inputs used to implem ent this modified approach is almost like the Function Tree approach. • Enter the number of variables to be tested. • Enter the name of the variables. • Enter the Min and Max of the first variable • Enter the coefficients based on the nature of variable’s degree . Case 4.4.1: Number of Variables: 3 Name of the variables: a, b, c Min & Max value of a: 1, 10 Equation for F(b): 10+5a+3a 2-a3 Equation for F(c): 7+3a+2a 2+ a 3+8b+3b 2-b3 Fig 4.7 shows that total number of test cases for “Case 4.4.1” in which only three v ariables are used generates 21 regular test cases and 2 additional test cases. 52 Fig 4.7: Results of Case 4.4.1 Case 4.4.2: Number of Variables: 4 Name of the variables: a, b, c, d Min value of a: -1 Max value of a: 10 Equation for F(b): 10+5a+3a 2+2a 3 Equ ation for F(c): 15+3a+7a 2-a3+4 b+2b 2-b3 Function for F(d): 1+a+a 2+a 3+b+b 2+b 3+c+c 2+c 353 Fig 4.8: Results of Case 4.4.2 Total number of test cases generated in this case are 28 + 3 = 31. In general, this technique generates (7n) + k number of test cases wh ere n is the number of variables and k is the number of mutants killed. Seven different cases being min -1, min, min+1, mid, max -1, max, and max+1. This approach theoretically gives significantly better results in terms of efficiency compared to the regular BVA technique b ut provides average amo unt of 54 test cases. In some scenarios, this method is much better than the traditional BVA method as well the Function Tree method. 4.5. Comparison Amongst General BVA, Function Tree, and Modified Approach Fig 4.9 shows th e outcome achieved through this work. The modified method shows good improvement in lowering the number of test cases, with much more improvements as the number of input variables increases. Fig 4.9: General BVA vs Function Tree vs Modified Approach The proposed method will evaluate almost every possible required test case increasing the system’s efficiency. General BVA Function Tree Modified Approach 0 500 1000 1500 2000 2500 3000 2345 10 15 20 25 10 75 500 3125 14 21 28 35 NUMBER OF TEST CASES NUMBER OF VARIABLES General BVA vs Function Tree vs Modified Approach General BVA Function Tree Modified Approach 55 Generalized BVA Function Tree Modified BVA Number of Test Cases 5n n5^(n -1) (7n) + k Mutation Score Least Lesser than Modified BVA More th an both Generalized BVA and FTA Equivalent Mutants > 0 Works Works Doesn’t work Number of Variables Works with finite number of variables Difficult to implement if n > 4 Works with finite number of variables Range Check Between min and max Between min and max Checks for entire range Table 1: Generalized BVA vs Function Tree Algorithm vs Modified Algorithm The table 1 shows a comparison amongst the generalized BVA method, Function Tree approach, and the modified method. In table 1, the red marking of the modified approach shows how it is better in some ways than the other two methods compared with. In the modified BVA, mutation te sting is also included due to which mutation score is counted and the table is created according to the hypothetical inclusion of the mutation in other two techniques. The elements of the Mutation Score (M S) are; D = Dead Mutants; N = Number of Mutants; 56 E = Equivalent Mutants; Where; MS= (D / (N -E)) -----------------------------Eq. 4.1 Mutation Score will be significantly more than the other two approaches because the implementation of this modified method is done considering there are no Equivalent Mutan ts . If there are equivalent mutants, the modified BVA will not recognize it and will fail to work on it while other will work in the case of Equivalent Mutants. Now, moving on to the numbers of test cases and variables. Generalized BVA generates 5n numbe r of test cases while Function Tree method generates the highest of all three i. e. n5^(n -1) and the modified approach generates 7n + k number of test cases where n is number of variables and k being the number of mutants killed at each step. Both general ized BVA and the proposed technique (modified BVA) works well with finite number of variables which can be 5, 10, 20 etc. But in the case of Function tree algorithm, it only works till 4 or 5 variables, since the number of test cases will become so large t hat it will be impossible to test and will therefore encounter complexity issues. 57 CHAPTER 5 5.1. Conclusion There are several techniques of Software Testing in this world but people do not know which technique to use and when. So, this work is a help t o understand testing a little better. O ne of which is Boundary Value Analysis. A modified version of Boundary Value Analysis using input parameters with functional dependency is proposed in this work . The idea is derived from the inter dependency of functi ons among the input parameters. With this modified algorithm, an automated testing tool is created and implemented. Generalized BVA generates 5n number of test cases where n is number of variables while Function Tree method generates the highest of all t hree that is n5^(n -1) and the modified approach generates 7n + k number of test cases where k is the number of mutants killed at each step. So, it shows that the number of test cases in case of modified algorithm is significantly lower than the Function Tree algorithm while almost similar as regular BVA but it covers more functionalities and features as shown in Chapter 4. This testing tool shows the advantages of the modified algorithm developed over the Functional Tree Approach and reduces a significant amount of test cases that leads to an exhaustive testing. This modified method will test almost every possible required test case increasing the system’s efficiency compared to regular 58 BVA in terms of speed and number of test cases . This method will be a very good help for any pro duct based company savin g some amount of money and time. 5.2. Future Work In this work , the Function tree approach induces a lo t of test cases of order n5^(n - 1). So , it is difficult to cope with such a large data and to check on th ese values which are more than 18k if n=6 and this results in exhaustive testing. Thus, the proposed algorithm reduces the test cases significantly to 7n + k; k being the killed mutants. Tho ugh the algorithm requires at -least one independent variable so problems like triangle inequality can’t be handled. So , for the future algorithm can be further modified to incorporate the case when all variables are interdependent. 59 REFERENCES T.Y.Chen, M.Y.Cheng, P.L.Poon, T.H.Tse, and Y.T.Yu, “A study of input domain partitioning”, Proceedings of the 20th IASTED International Multi - Conference on Applied Informatics (AI 2002), ACTA Press, Calgary, Canada, pp. 176 -181, 2002. Karan Vij an d Wenying Feng, “Boundary Value Analysis using Divide - and – rule Approach”, Fifth International Conference on Information and Technology, 2008, IEEE, pg. 70 -75. WENYING FENG," A Generalization of Boundary Value Analysis for Input Parameters with Func tional Dependency",9th IEEE/ACIS International Conference on Computer and Information Science , 2010, IEEE, pg. 776-782. Guru99 2017, Boundary Value Analysis & Equivalence Partitioning with Examples , accessed 14 March 2017, < - partitioning -boundary -value -analysis.html >. Stack Exchange, Functional dependency and normalization , accessed 19 March 2017, < -dependency -and - normalization >. J. C. Cherniavsky, “Validation, Verification, and Testing of Computer Software ”, National Institute of Standards and Tech nology (NIST), February 1981. NIST 500 -75. [7 ] Ali Parsai, Alessandro Murgia, Quinten David Soetens, and Serge Demeyer , “M utation Testing as a Safety Netfor Test Code Refactoring ”, Research Gate, 2015 60 [8 ] Ranorex GmbH , Test Automation and its Benefits , acc essed 15 March 2017, -test -automation.html . [9 ] QATestLab, Automated Testing , accessed 15 March 2017, -are -professionals -in/automated -testing/ . [10 ] N. Debnath and J. R. Haakenson, “Automated testing design and description for certain functions”, Proceedings of the Sixth IEEE International Confer ence on Electro/Information Technology, (IEEE -EIT 2006), Michigan State University, East Lansing, MI, USA, May 7 -10, 2006. [11 ] W. Feng and Z. Zhang, “Sequence algorithms for boundary value analysis with constrained input parameters”, Proceedings of the IS CA 14th International Conference on Intelligent and Adaptive Systems and Software Engineering (IASSE -2005), Toronto, Canada, pp. 255 -260, 2005. [12 ] N. Debnath, W. Feng, M. Burgin, J. Wilson, and J. Cropper: “An auto tester for the modified Next Date funct ion reusing a test case generator”, Proceedings of the IEEE International Conference on Information Reuse and Integration (IEEE IRI -2004), Las Vegas, USA, pp.121126, 2004. R. Pressman, Software Engineering - A Practitioner’s approach, 5th Edition, The McGraw Hill Companies, Inc., New York, 2001.

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