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2000	https://www.gauthmath.com/solution/1836493876356145/You-are-given-the-polynomial-x3-6x2-x-30-and-one-of-its-zeros-c-2-Use-the-techni	Solved: You are given the polynomial x^3-6x^2-x+30 and one of its zeros c=-2 Use the techniques in [Math] Drag Image or Click Here to upload Command+to paste Upgrade Sign in Homework Homework Assignment Solver Assignment Calculator Calculator Resources Resources Blog Blog App App Gauth Unlimited answers Gauth AI Pro Start Free Trial Homework Helper Study Resources Math Expression Questions Question You are given the polynomial x^3-6x^2-x+30 and one of its zeros c=-2 Use the techniques in Section 3.2 to fnd the rest of the real zeros and factor the polynomial. Choose the answer that represents the factored polynomial. Show transcript Gauth AI Solution 100%(2 rated) Answer The answer is (x+2)(x-3)(x-5) Explanation Utilize the Factor Theorem: Since $$c = -2$$c=−2 is a zero of the polynomial $$x^{3} - 6x^{2} - x + 30$$x 3−6 x 2−x+30, then $$(x + 2)$$(x+2) is a factor. Perform Polynomial Long Division: Divide the polynomial $$x^{3} - 6x^{2} - x + 30$$x 3−6 x 2−x+30 by $$(x + 2)$$(x+2) to obtain the quotient. \multicolumn 2 r x 2\cline 2−5 x+2\multicolumn 2 r x 3\cline 2−3\multicolumn 2 r 0\multicolumn 2 r\cline 3−4\multicolumn 2 r\multicolumn 2 r\cline 4−5\multicolumn 2 r−8 x x 3+2 x 2−8 x 2−8 x 2 0+15−6 x 2−x−16 x 15 x 15 x 0−x+30+30 0+30 The quotient is $$x^{2} - 8x + 15$$x 2−8 x+15 Factor the Quadratic: Factor the quadratic expression $$x^{2} - 8x + 15$$x 2−8 x+15 to find its roots. $$x^{2} - 8x + 15 = (x - 3)(x - 5)$$x 2−8 x+15=(x−3)(x−5) Complete Factorization: Combine the factors to obtain the complete factorization of the original polynomial. $$x^{3} - 6x^{2} - x + 30 = (x + 2)(x - 3)(x - 5)$$x 3−6 x 2−x+30=(x+2)(x−3)(x−5) Identify the Zeros: The zeros of the polynomial are the values of $$x$$x that make each factor equal to zero. Therefore, the zeros are $$x = -2$$x=−2, $$x = 3$$x=3, and $$x = 5$$x=5 Helpful Not Helpful Explain Simplify this solution Gauth AI Pro Back-to-School 3 Day Free Trial Limited offer! Enjoy unlimited answers for free. Join Gauth PLUS for $0 Previous questionNext question Related For the polynomial given below, you are given one of its zeros. Use the techniques in this section to find the rest of the real zeros. Enter your answers as a comma-separated list. If there are no other zeros, enter the zero given. x3 12x2+48x-64,c=4 Factor the polynomial. Verified Answer In Exercises 31 - 40, you are given a polynomial and one of its zeros. Use the techniques in this section to find the rest of the real zeros and factor the polynomial. 31. x3-6x2+11x-6,c=1 100% (7 rated) Below you are given a polynomial and one of its zeros. Use the techniques in this section to find the rest of the real zeros and factor the polynomial. a. x3-6x2+11x-6,c=1 b, 2x3-3x2-11x+6,c= 1/2 C. 4x4-28x3+61x2-42x+9,c= 1/2 is a zero of multiplicity 2 100% (5 rated) For the polynomial given below, you are given one of its zeros. Use the techniques in this section to find the rest of the real zeros. Enter your answers as a comma-separated list. If there are no other zeros, enter the zero given. x4-49x2,c=0 x= Factor the polynomial. Additional Materials Reading 100% (2 rated) For the polynomial given below, you are given one of its zeros. Use the techniques in this section to find the rest of the real zeros. Enter your answers as a comma-separated list. If there are no other zeros, enter the zero given. 4x4-60x3+253x2-210x+49,c= 1/2 is a zero of multiplicity 2 x=square Factor the polynomial. square 100% (3 rated) For the polynomial given below, you are given one of its zeros. Use the techniques in this section to find the rest of the real zeros. Enter your answers as a comma-separated list. If there are no other zeros, enter the zero given. 4x4-68x3+321x2-272x+64,c= 1/2 is a zero of multiplicity 2 x=square × Factor the polynomial. square × Additional Materials 100% (7 rated) Eur the polynomial given below, you are given one of its zeros. Use the techniques in this section to find the rest of the real zeros. Enter your answers as a comma-seporatnson i the zero given. 4x4-44x3+141x2-110x+25,c= 1/2 is a zero of multiplicity 2 x=\| 5 square Factor the polynomial square x Additional Material= Reading 100% (1 rated) Write the quotient in the form a+bi. 7-i/3-6i 7-i/3-6i =square Simplify your answer. Type your answer in the form a+bi . Use integers or fractions for any numbers in the expressio 100% (4 rated) Solve the following inequality algebraically. 5x-5/x+2 ≤ 4 What is the solution? -2,13 Type your answer in interval notation. Simplify your answer. Use integers or fractions for any numbers in the expression. 100% (4 rated) How may different arrangements are there of the letters in The number of possible arrangements is MISSISSIPPI？ 100% (2 rated) Gauth it, Ace it! contact@gauthmath.com Company About UsExpertsWriting Examples Legal Honor CodePrivacy PolicyTerms of Service Download App
2001	https://www.osmosis.org/learn/Development_of_the_muscular_system	Fall in love with Osmosis at 20% off! Save now until September 30 at 11:59 PM PT. Learn more Skip to the video Development of the muscular system Foundational SciencesEmbryologyOrgan system developmentMusculoskeletal system 12,216views 00:00 / 00:00 Development of the muscular system Organ system development Cardiovascular system Development of the cardiovascular system Fetal circulation Eyes, ears, nose, and throat Development of the ear Development of the eye Development of the face and palate Pharyngeal arches, pouches, and clefts Gastrointestinal system Development of the digestive system and body cavities Development of the gastrointestinal system Development of the teeth Development of the tongue Integumentary system Development of the integumentary system Musculoskeletal system Development of the axial skeleton Development of the limbs Development of the muscular system Nervous system Development of the nervous system Renal system Development of the renal system Reproductive system Development of the reproductive system Respiratory system Development of the respiratory system Assessments Flashcards 0 / 16 complete USMLE® Step 1 questions 0 / 1 complete High Yield Notes 28 pages Flashcards Development of the muscular system 0 of 16 complete Questions USMLE® Step 1 style questions USMLE 0 of 1 complete A clinician is testing the function of myotomes in the upper and lower extremities via isometric resisted muscle testing. This provides the clinician with information regarding the level in the spine where a lesion may be present. Which of the following best describes the embryological origin of a myotome? Transcript Watch video only Content Reviewers Rishi Desai, MD, MPH Contributors Marisa Pedron, Tanner Marshall, MS The muscular system starts taking shape when the embryo is just a flat little pancake made up of two layers: the epiblast on the dorsal, or back, side, and the hypoblast on the ventral, or front, side. A line called the primitive streak appears on the epiblast “back” of this two-layered creature. Cells migrate along the primitive streak during gastrulation, leading to a now three-layered embryo pancake, with each layer containing germ cells that form organs and tissues of the body. The ventral, or bottom, germ layer is called endoderm, the dorsal, or top, germ layer is called ectoderm, and the layer in between these two is called mesoderm. Collectively, these germ cells produce all of the organs and tissues in the body. During week 3, the embryo transitions from a flat organism to a more tubular creature by folding along its longitudinal and lateral axes. At the same time, a solid rod of mesoderm called the notochord forms on the midline of the embryo. Above the notochord, the ectoderm invaginates to form the neural tube, an early precursor of the central nervous system. This is the embryo’s first symmetry axis, and the mesoderm on either side of the neural tube differentiates into three distinct portions: immediately flanking the neural tube there’s the paraxial mesoderm; next, there’s the intermediate mesoderm; and finally, the lateral plate mesoderm. Between the cells of the lateral plate mesoderm, small gaps appear and coalesce to form the intraembryonic coelom, a cavity inside the embryo’s body. This cavity separates the lateral plate mesoderm into two layers: a parietal layer that’s in contact with the ectoderm, and a visceral layer that’s in contact with the endoderm. The paraxial and lateral plate mesoderm will become the skeletal muscles in our body. Before the mesoderm cells develop into skeletal muscle, they first organize into cell blocks called somites. Somites arise in pairs from a combination of paraxial mesoderm cells and mesenchyme, which is a soupy fetal tissue containing pluripotent cells. Around day 20 of development, somites begin to form in the occipital region of the embryo, which is at the base of the head. Somites continue to form cranio-caudally, or from head-to-tail end of the embryo, with about three pairs forming each day. Up to 40 somite pairs form by the end of week 5. Some degenerate, while the rest go on to form bone and muscle structures. Each somite undergoes a split, with cells from the ventral portion forming sclerotome, creating the vertebrae and the ribs. Cells from the dorsomedial lip of the somite (the top right layer of cubes here) mix with some cells from the ventrolateral lip in the opposite corner of the cube (the bottom left) to form a new, mixed tissue called dermomyotome. Dermomyotome cells further differentiate into dermatome and myotome cells, which turn into the dermis layer of the skin and into muscles, respectively. Now, fast forwarding a bit, the muscles of the myotome start to develop. One way to categorize the muscles is according to their innervation. Summary Development of the muscular system starts at around week d of gestation. The muscular system begins with the formation of muscle cells called myoblasts. Myoblasts originate from the mesoderm and fuse together to form long and multinucleated fibers called muscle fibers. Muscle fibers are attached by collagenous connective tissues, and the entire muscle is enclosed in a fibrous capsule. All skeletal and cardiac muscles and most smooth muscles arise from mesoderm cells, except pupillary muscles and the sweat and mammary glands, which arise from ectoderm. We use cookies that are necessary to make our site work. We may also use additional cookies to analyze, improve, and personalize our content and your digital experience. For more information, see ourCookie Policy Cookie Preference Center We use cookies which are necessary to make our site work. We may also use additional cookies to analyse, improve and personalise our content and your digital experience. For more information, see our Cookie Policy and the list of Google Ad-Tech Vendors. You may choose not to allow some types of cookies. However, blocking some types may impact your experience of our site and the services we are able to offer. 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2002	https://mathoverflow.livejournal.com/49513500.html	A formula for minimum number of $k$-element nonnegative-sum subsets of a zero-sum set: Revisiting Ma Main Top Interesting Checklist ☕🧦🍂 Help mathoverflow Follow us: Follow us on Twitter Applications Download Huawei RuStore COMPANY About News Help PRODUCTS "Share" button COMMUNITY Frank CHOOSE LANGUAGE English ▾ English Deutsch Dansk español Français Italiano Русский Українська Беларуская 日本語 Português Esperanto עברית Nederlands Magyar Gaeilge íslenska suomi Ελληνικά Norsk bokmål Svenska polski 简体中文 Latviešu Türkçe Bahasa Melayu हिन्दी Brazilian Portuguese Chinese Traditional Lietuvių Norsk nynorsk Privacy Policy User Agreement Recommendation technologies Help LiveJournal — v.880 Main Top Interesting Checklist ☕🧦🍂 Help Follow us: Follow us on Twitter Applications Download Huawei RuStore COMPANY About News Help PRODUCTS "Share" button COMMUNITY Frank CHOOSE LANGUAGE English ▾ English Deutsch Dansk español Français Italiano Русский Українська Беларуская 日本語 Português Esperanto עברית Nederlands Magyar Gaeilge íslenska suomi Ελληνικά Norsk bokmål Svenska polski 简体中文 Latviešu Türkçe Bahasa Melayu हिन्दी Brazilian Portuguese Chinese Traditional Lietuvių Norsk nynorsk Privacy Policy User Agreement Recommendation technologies Help LiveJournal — v.880 LiveJournal Find more Communities RSS Reader Shop Help Search Log in mathoverflow Log in Join free Join English (en) English (en) Русский (ru) Українська (uk) Français (fr) Português (pt) español (es) Deutsch (de) Italiano (it) Беларуская (be) Log in No account?Create an account [x] Remember me Forgot password Log in Log in QR code No account?Create an account By logging in to LiveJournal using a third-party service you accept LiveJournal's User agreement mathoverflow Subscribe April 9 2025, 08:57 0 00 0 0 0 0 00 0 A formula for minimum number of $k$-element nonnegative-sum subsets of a zero-sum set: Revisiting Ma Define the quantity$A(n,k)$as follows: $A(n,k)$denotes theminimumnumber of$k$-element non-negative subsets of$n$arbitrary real numbers$x_1,\dots,x_n$whose sum is non-negative. More formally,$A(n,k)$is given by $$\min_{x \in \mathbb R^n: \sum_{i\in [n]}x_i \ge 0} \left \| \left {S\subseteq [n]: \sum_{i\in S}x_i \ge 0 \right } \right \|.$$ Now consider the following result and two conjectures: Devisable case: Given$\sum_{i\in[n]}x_i \ge 0$, for$k\|n$:$$A(n,k)=\binom{n-1}{k-1}.$$ MMS Conjecture:Given$\sum_{i\in[n]}x_i \ge 0, $for$n\ge 4k$:$$A(n,k)=\binom{n-1}{k-1}.$$ MM Conjecture:, Given$\sum_{i\in[n]}x_i \ge 0,$for$n <4k$with$k\not \| n$:$$A(n,k) = \binom{n-\lfloor \frac n k \rfloor}{k}.$$ In page 391 of this 1987 paper by Manickam and Miklós, you can find both the Manickam-Miklós-Singhi (MMS) and the Manickam-Miklós (MM) conjectures. The above combinatorial form of the MMS conjecture was phrased in terms of calculating invariants of Johnson Schemes in linear algebra by Manickam and Singhi in 1988 . The MMS conjecture seems to be correct (see these 2014, 2015 and 2017 papers , , , and references therein). In fact, the MMS conjecture has been already proved for$n\ge 10^{46}k,$and$n\ge 8 k^2$. In the literature, I could not find any discussion on the MM conjecture, which only appeared in Manickam and Miklós's paper . Two different proofs for the devisable case are given in and . WLOG, we can consider$\color{blue}{\sum_{i\in[n]}x_i=0}$instead of$\sum_{i\in[n]}x_i \ge 0$. Then, the number of$k$-element subsets of$x_1,\dots,x_n$whose sum is non-negative equals the number of$(n-k)$-element subsets of$x_1,\dots,x_n$whose sum is non-positive. In this case,the MMS and MM conjectures imply that $$ \small {A(n,k) = \min \left ( \binom{n-1}{k-1}, \binom{n-1}{n-k-1} \right )} \; k\|n \; \text{or} \; n-k\|n\tag{1.1} $$$$ \small {A(n,k) = \min \left ( \binom{n-1}{k-1}, \binom{n-\lfloor \frac n k \rfloor}{k},\binom{n-\lfloor \frac n {n-k} \rfloor}{n-k} , \binom{n-1}{n-k-1} \right )} \; \text{otherwise} \; \tag{1.2} $$ In my attempt to answer this question , first, I realized thatthe MM conjecture is not generally correct. Second, I guess that$A(n,k)$has the following general formula: $$ \fbox{$ A(n,k) = \min_{a \in[n-1]} \sum_{0 \le t \le \frac k n a} \binom{a}{t} \binom{n-a}{k-t}.$} \tag{2} $$ The above is derived based on the intuition: $ A(n,k)$is always attained for a sequence$x_1,\dots,x_n \in {-\alpha, \beta }$for some$\alpha, \beta>0$. The structure$x_1,\dots,x_n \in {-\alpha, \beta }$has the minimum flexibility for a non-zero feasible solution of$\sum_{i\in[n]}x_i=0$, included in all other possible structures$$x_1,\dots,x_n \in {-\alpha_p,\dots,-\alpha_1, \beta_1,\dots,\beta_q } \tag{3}$$with$p,q\in \mathbb N$and distinct$\alpha_1,\dots,\alpha_p, \beta_1,\dots,\beta_q>0$. It is why the above conjecture seems to be correct. If the above is the case,$a$of$x_1,\dots,x_n$are$-\alpha$, and$n-a$of them are$\beta$for some$a \in [n-1]$. As$\sum_{i\in[n]}x_i=0$, then$\alpha= \beta \left ( \frac na -1\right)$. Moreover, for some$ t \in {0,\dots,a}$,$t$of the elements of a$k$-element subset are$-\alpha$, and the subset has a non-negative sum if and only if$-t\alpha+ (k-t) \beta \ge 0$, which is equivalent to$t \le \frac{k}{n} a$(independent of$\alpha$and$\beta$). Then, one can derive the formula (2) by counting different ways of selecting$t$out of$a$and$k-t$out of$n-a$choices, and minimizing over all the possible values of$a \in[n-1]$. In the following, you can find the plots of the normalized quantity$\frac{A(n,k)}{\binom{n}{k}}$for$n=6 4, 99, 100, 101$based on the old formula in (1.1) and (1.2) (shown in red), and the new formula in (2) (shown in green): $n=64$ $n=99$ enter image description here $n=100$ $n=101$ You may notice that the new formula (2)nicely handle both cases in the devisable case and MMS conjecturefor$k\|n$and$n\ge 4k$(see where red and green lines overlap; theprooffollows from the results given here ), and it shows thatthe MM conjecture cannot be generally correct(see where red and green lines do not overlap). In fact, the old and new formulas can differ only over the range$n < 4k <3n$, and they are the same for$n \ge 4k$and$4k \ge 3n$($n \ge 4(n-k)$), giving$\binom{n-1}{k-1}$and$\binom{n-1}{n-k-1}$, respectively. In a case that a counterexample is given, the formula in (2) is an upper bound for$A(n,k)$, which is better than the old one given in (1.2) over$n < 4k <3n$. I hope one can prove or disprove (2) by seeing the problem with new eyes though a part of it (the MMS conjecture) has been not fully proved yet after about three decades. Using the result provedhere, one can see thatthe new fromula (2) covers both the devisable case and MMS conjecture, that is, it is equal to$A(n,k)=\binom{n-1}{k-1}$for$k\|n$and$n\g e 4k$. The result also suggests thatthe MMS conjecture holds over a wider range of$n\ge \frac {27}{8} k.$ 0 0Subscribe0 0 0 0 0 00 0 LJ Video No comments yet Post a new comment )_ Your IP address will be recorded For a line break, press Shift + Enter Press Enter to send comment Applications Download Huawei RuStore Follow us: Follow us on Twitter LiveJournal COMPANY About News Help PRODUCTS "Share" button COMMUNITY Frank CHOOSE LANGUAGE English ▾ English Deutsch Dansk español Français Italiano Русский Українська Беларуская 日本語 Português Esperanto עברית Nederlands Magyar Gaeilge íslenska suomi Ελληνικά Norsk bokmål Svenska polski 简体中文 Latviešu Türkçe Bahasa Melayu हिन्दी Brazilian Portuguese Chinese Traditional Lietuvių Norsk nynorsk v.880 Privacy Policy User Agreement Recommendation technologies Help
2003	https://flexbooks.ck12.org/cbook/ck-12-algebra-i-concepts-honors/section/7.8/primary/lesson/applications-of-quadratic-equations-alg-i-hnrs/	Skip to content Elementary Math Grade 1 Grade 2 Grade 3 Grade 4 Grade 5 Math 6 Math 7 Math 8 Algebra I Geometry Algebra II Math 6 Math 7 Math 8 Algebra I Geometry Algebra II Probability & Statistics Trigonometry Math Analysis Precalculus Calculus What's the difference? Grade K to 5 Earth Science Life Science Physical Science Biology Chemistry Physics Advanced Biology Math FlexLets Science FlexLets Writing Spelling Social Studies Economics Geography Government History World History Philosophy Sociology More Astronomy Engineering Health Photography Technology College College Algebra College Precalculus Linear Algebra College Human Biology The Universe Adult Education Basic Education High School Diploma High School Equivalency Career Technical Ed English as 2nd Language Country Bhutan Brasil Chile Georgia India Translations Spanish Korean Deutsch Chinese Greek Polski EXPLORE Flexi A FREE Digital Tutor for Every Student FlexBooks 2.0 Customizable, digital textbooks in a new, interactive platform FlexBooks Customizable, digital textbooks Schools FlexBooks from schools and districts near you Study Guides Quick review with key information for each concept Adaptive Practice Building knowledge at each student’s skill level Simulations Interactive Physics & Chemistry Simulations PLIX Play. Learn. Interact. eXplore. CCSS Math Concepts and FlexBooks aligned to Common Core NGSS Concepts aligned to Next Generation Science Standards Certified Educator Stand out as an educator. Become CK-12 Certified. Webinars Live and archived sessions to learn about CK-12 Other Resources CK-12 Resources Concept Map Testimonials CK-12 Mission Meet the Team CK-12 Helpdesk FlexLets Know the essentials. Pick a Subject Donate Sign Up 7.8 Applications of Quadratic Equations Written by:Brenda Meery \| Kaitlyn Spong Fact-checked by:The CK-12 Editorial Team Last Modified: Sep 01, 2025 Two cars leave an intersection at the same time. One car travels north and the other car travels west. When the car traveling north had gone 24 miles, the distance between the cars was four miles more than three times the distance traveled by the car heading west. Find the distance between the cars at that time. Quadratic Equations Quadratic functions can be used to help solve many different real world problems. Here are two hints for solving quadratic word problems: It is often helpful to start by drawing a picture in order to visualize what you are asked to solve. Once you have solved the problem, it is important to make sure that your answers are realistic given the context of the problem. For example, if you are solving for the age of a person and one of your answers is a negative number, that answer does not make sense in the context of the problem and is not actually a solution. Let's apply quadratic equations to solve the following problems: The number of softball games that must be scheduled in a league with @$\begin{align}n\end{align}@$ teams is given by @$\begin{align}G(n)= \frac{n^2 - n}{2}\end{align}@$. Each team can only play every other team exactly once. A league schedules 21 games. How many softball teams are in the league? You are given the function @$\begin{align}G(n)= \frac{n^2-n}{2}\end{align}@$ and you are asked to find @$\begin{align}n\end{align}@$ when @$\begin{align}G(n)=21\end{align}@$. This means, you have to solve the equation: @$\begin{align}21=\frac{n^2-n}{2}\end{align}@$ Start by setting the equation equal to zero: @$$\begin{align}42&=n^2-n\ n^2-n-42&=0\end{align}@$$ Now solve for @$\begin{align}n\end{align}@$ to find the number of teams @$\begin{align}(n)\end{align}@$ in the league. Start by factoring the left side of the equation and rewriting the equation: @$\begin{align}n^2-n-42=0 \end{align}@$ becomes @$\begin{align}(n-7)(n+6)=0\end{align}@$ There are 7 teams in the softball league. When a home-made rocket is launched from the ground, it goes up and falls in the pattern of a parabola. The height, in feet, of a home-made rocket is given by the equation @$\begin{align}h(t) = 160t - 16t^2\end{align}@$ where @$\begin{align}t\end{align}@$ is the time in seconds. How long will it take for the rocket to return to the ground? The formula for the path of the rocket is @$\begin{align}h(t)=160t-16t^2\end{align}@$. You are asked to find @$\begin{align}t\end{align}@$ when @$\begin{align}h(t)=0\end{align}@$, or when the rocket hits the ground and no longer has height. Start by factoring: @$\begin{align}160t-16t^2=0\end{align}@$ becomes @$\begin{align}16t(10-t)=0\end{align}@$ This means @$\begin{align}16t=0\end{align}@$ (so @$\begin{align}t=0\end{align}@$) or @$\begin{align}10-t=0\end{align}@$ (so @$\begin{align}t=10\end{align}@$). @$\begin{align}t=0\end{align}@$ represents the rocket being on the ground when it starts, so it is not the answer you are looking for. @$\begin{align}t=10\end{align}@$ represents the rocket landing back on the ground. The rocket will hit the ground after 10 seconds. Using the information in the previous problem, what is the height of the rocket after 2 seconds? To solve this problem, you need to replace @$\begin{align}t\end{align}@$ with 2 in the quadratic function. @$$\begin{align}h(t)&=160t-16t^2\ h(2)&=160(2)-16(2)^2\ h(2)&=320-64\ h(2)&=256.\end{align}@$$ Therefore, after 2 seconds, the height of the rocket is 256 feet. Examples Example 1 Earlier, you were told that two cars leave an intersection at the same time, one traveling north and the other traveling west. When the car traveling north had gone 24 miles the distance between the cars was four miles more than three times the distance traveled by the car heading west. What is the distance between the cars at that time? First draw a diagram. Since the cars are traveling north and west from the same starting position, the triangle made to connect the distance between them is a right triangle. Since you have a right triangle, you can use the Pythagorean Theorem to set up an equation relating the lengths of the sides of the triangle. The Pythagorean Theorem is a geometry theorem that says that for all right triangles, @$\begin{align}a^2+b^2=c^2\end{align}@$ where @$\begin{align}a\end{align}@$ and @$\begin{align}b\end{align}@$ are legs of the triangle and @$\begin{align}c\end{align}@$ is the longest side of the triangle, the hypotenuse. The equation for this problem is: @$$\begin{align}x^2+24^2&=(3x+4)^2\ x^2+576&=(3x+4)(3x+4)\ x^2+576&=9x^2+24x+16\end{align}@$$ Now set the equation equal to zero and factor the quadratic expression so that you can use the zero product property. @$$\begin{align}x^2+576&=9x^2+24x+16\ 0&=8x^2+24x-560\ 0&=8(x^2+3x-70)\ 0&=8(x-7)(x+10)\end{align}@$$ So you now know that @$\begin{align}x = 7\end{align}@$. Since the distance between the cars is represented by the expression @$\begin{align}3x + 4\end{align}@$, the actual distance between the two cars after the car going north has traveled 24 miles is: @$$\begin{align}3x+4&=3(7)+4\ &=21+4\ &=25 \ miles\end{align}@$$ Example 2 A rectangle is known to have an area of 520 square inches. The lengths of the sides are shown in the diagram below. Solve for both the length and the width. The rectangle has an area of 520 square inches and you know that the area of a rectangle has the formula: @$\begin{align}A = l \times w\end{align}@$. Therefore: : @$$\begin{align}520&=(x+7)(2x)\ 520&=2x^2+14x\ 0&=2x^2+14x-520\ 0&=2(x^2+7x-260)\ 0&=2(x-13)(x+20)\end{align}@$$ : Therefore the value of @$\begin{align}x\end{align}@$ is 13. This means that the width is @$\begin{align}2x\end{align}@$ or @$\begin{align}2(13) = 26 \ inches\end{align}@$. The length is @$\begin{align}x + 7 = 13 + 7 = 20 \ inches\end{align}@$. Example 3 The height of a ball in feet can be found by the quadratic function @$\begin{align}h(t)=-16t^2+80t+5\end{align}@$ where @$\begin{align}t\end{align}@$ is the time in seconds that the ball is in the air. Determine the time(s) at which the ball is 69 feet high. The equation for the ball being thrown is @$\begin{align}h(t)=-16t^2+80t+5\end{align}@$. If you drew the path of the thrown ball, you would see something like that shown below. : You are asked to find the time(s) when the ball hits a height of 69 feet. In other words, solve for: : @$$\begin{align}69=-16t^2+80t+5\end{align}@$$ : To solve for @$\begin{align}t\end{align}@$, you have to factor the quadratic and then solve for the value(s) of @$\begin{align}t\end{align}@$. : @$$\begin{align}&\quad \qquad 0=-16t^2+80t-64\ &\quad \qquad 0=-16(t^2-5t+4)\ &\quad \qquad 0=-16(t-1)(t-4)\ &\qquad \quad \qquad \swarrow \qquad \qquad \searrow\ & \quad t - 1 = 0 \qquad \qquad t-4 = 0\ &\qquad \ \ t = 1 \quad \qquad \qquad \ \ t = 4\end{align}@$$ : Since both values are positive, you can conclude that there are two times when the ball hits a height of 69 feet. These times are at 1 second and at 4 seconds. Example 4 A manufacturer measures the number of cell phones sold using the binomial @$\begin{align}0.015c + 2.81\end{align}@$. She also measures the wholesale price on these phones using the binomial @$\begin{align}0.011c +3.52\end{align}@$. Calculate her revenue if she sells 100,000 cell phones. The number of cell phones sold is the binomial @$\begin{align}0.015c + 2.81\end{align}@$. The wholesale price on these phones is the binomial @$\begin{align}0.011c +3.52.\end{align}@$ The revenue she takes in is the wholesale price times the number that she sells. Therefore: : @$\begin{align}R(c)=(0.015c+2.81)(0.011c+3.52)\end{align}@$ : First, let’s expand the expression for @$\begin{align}R\end{align}@$ to get the quadratic expression. Therefore: : @$$\begin{align}R(c)&=(0.015c+2.81)(0.011c+3.52)\ R(c)&=0.000165c^2+0.08371c+9.8912\end{align}@$$ : The question then asks if she sold 100,000 cell phones, what would her revenue be. Therefore what is @$\begin{align}R(c)\end{align}@$ when @$\begin{align}c = 100,000\end{align}@$. : @$$\begin{align}R(c)&=0.000165c^2+0.08371c+9.8912\ R(c)&=0.000165(100,000)^2+0.08371(100,000)+9.8912\ R(c)&=1,658,380.89\end{align}@$$ : Therefore she would make $1,658,380.89 in revenue. Review A rectangle is known to have an area of 234 square feet. The length of the rectangle is given by @$\begin{align}x+3\end{align}@$ and the width of the rectangle is given by @$\begin{align}x+8\end{align}@$. What is the value of @$\begin{align}x\end{align}@$? Solve for @$\begin{align}x\end{align}@$ in the rectangle below given that the area is 9 units. Solve for @$\begin{align}x\end{align}@$ in the triangle below given that the area is 10 units. A pool is treated with a chemical to reduce the amount of algae. The amount of algae in the pool @$\begin{align}t\end{align}@$ days after the treatment can be approximated by the function @$\begin{align}A(t)=40t^2-300t+500\end{align}@$. How many days after treatment will the pool have the no algae? How much algae is in the pool before treatments are started? How much less algae is in the pool after 1 day? A football is kicked into the air. The height of the football in meters can be found by the quadratic function @$\begin{align}h(t)=-5t^2+25t\end{align}@$ where @$\begin{align}t\end{align}@$ is the time in seconds since the ball has been kicked. How high is the ball after 3 seconds? At what other time is the ball the same height? When will the ball be 20 meters above the ground? After how many seconds will the ball hit the ground? A ball is thrown into the air. The height of the ball in meters can be found by the quadratic function @$\begin{align}h(t)=-5t^2+30t\end{align}@$ where @$\begin{align}t\end{align}@$ is the time in seconds since the ball has been thrown. How high is the ball after 3 seconds? When will the ball be 25 meters above the ground? After how many seconds will the ball hit the ground? Kim is drafting the windows for a new building. Their shape can be modeled by the function @$\begin{align}h(w)=-w^2+4\end{align}@$, where @$\begin{align}h\end{align}@$ is the height and @$\begin{align}w\end{align}@$ is the width of points on the window frame, measured in meters. Find the width of each window at its base. Find the width of each window when the height is 3 meters. What is the height of the window when the width is 1 meter? Review (Answers) Click HERE to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option. \| Image \| Reference \| Attributions \| --- Student Sign Up Are you a teacher? Having issues? Click here By signing up, I confirm that I have read and agree to the Terms of use and Privacy Policy Already have an account? Save this section to your Library in order to add a Practice or Quiz to it. (Edit Title)35/ 100 This lesson has been added to your library. \|Searching in: \| \| \| Looks like this FlexBook 2.0 has changed since you visited it last time. 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2004	https://openstax.org/books/university-physics-volume-2/pages/6-2-explaining-gausss-law	Skip to Content Go to accessibility page Keyboard shortcuts menu Log in University Physics Volume 2 6.2 Explaining Gauss’s Law University Physics Volume 2 6.2 Explaining Gauss’s Law Search for key terms or text. Learning Objectives By the end of this section, you will be able to: State Gauss’s law Explain the conditions under which Gauss’s law may be used Apply Gauss’s law in appropriate systems We can now determine the electric flux through an arbitrary closed surface due to an arbitrary charge distribution. We found that if a closed surface does not have any charge inside where an electric field line can terminate, then any electric field line entering the surface at one point must necessarily exit at some other point of the surface. Therefore, if a closed surface does not have any charges inside the enclosed volume, then the electric flux through the surface is zero. Now, what happens to the electric flux if there are some charges inside the enclosed volume? Gauss’s law gives a quantitative answer to this question. To get a feel for what to expect, let’s calculate the electric flux through a spherical surface around a positive point charge q, since we already know the electric field in such a situation. Recall that when we place the point charge at the origin of a coordinate system, the electric field at a point P that is at a distance r from the charge at the origin is given by E⃗ P=14πε0qr2rˆ, where rˆ is the radial unit vector from the charge at the origin to the point P. We can use this electric field to find the flux through the spherical surface of radius R, as shown in Figure 6.13. Figure 6.13 A closed spherical surface surrounding a point charge q. Then we apply Φ=∫SE⃗ ⋅nˆdA to this system and substitute known values. On the sphere, nˆ=rˆ and r=R, so for an infinitesimal area dA, dΦ=E→⋅nˆdA=14πε0qR2rˆ⋅rˆdA=14πε0qR2dA. We now find the net flux by integrating this flux over the surface of the sphere: Φ=14πε0qR2∮SdA=14πε0qR2(4πR2)=qε0. where the total surface area of the spherical surface is 4πR2. This gives the flux through the closed spherical surface at radius r as Φ=qε0. 6.4 A remarkable fact about this equation is that the flux is independent of the size of the spherical surface. This can be directly attributed to the fact that the electric field of a point charge decreases as 1/r2 with distance, which just cancels the r2 rate of increase of the surface area. Electric Field Lines Picture An alternative way to see why the flux through a closed spherical surface is independent of the radius of the surface is to look at the electric field lines. Note that every field line from q that pierces the surface at radius R1 also pierces the surface at R2 (Figure 6.14). Figure 6.14 Flux through spherical surfaces of radii R1 and R2 enclosing a charge q are equal, independent of the size of the surface, since all E-field lines that pierce one surface from the inside to outside direction also pierce the other surface in the same direction. Therefore, the net number of electric field lines passing through the two surfaces from the inside to outside direction is equal. This net number of electric field lines, which is obtained by subtracting the number of lines in the direction from outside to inside from the number of lines in the direction from inside to outside gives a visual measure of the electric flux through the surfaces. You can see that if no charges are included within a closed surface, then the electric flux through it must be zero. A typical field line enters the surface at dA1 and leaves at dA2. Every line that enters the surface must also leave that surface. Hence the net “flow” of the field lines into or out of the surface is zero (Figure 6.15(a)). The same thing happens if charges of equal and opposite sign are included inside the closed surface, so that the total charge included is zero (part (b)). A surface that includes the same amount of charge has the same number of field lines crossing it, regardless of the shape or size of the surface, as long as the surface encloses the same amount of charge (part (c)). Figure 6.15 Understanding the flux in terms of field lines. (a) The electric flux through a closed surface due to a charge outside that surface is zero. (b) Charges are enclosed, but because the net charge included is zero, the net flux through the closed surface is also zero. (c) The shape and size of the surfaces that enclose a charge does not matter because all surfaces enclosing the same charge have the same flux. Statement of Gauss’s Law Gauss’s law generalizes this result to the case of any number of charges and any location of the charges in the space inside the closed surface. According to Gauss’s law, the flux of the electric field E→ through any closed surface, also called a Gaussian surface, is equal to the net charge enclosed (qenc) divided by the permittivity of free space (ε0): ΦClosed Surface=qencε0. This equation holds for charges of either sign, because we define the area vector of a closed surface to point outward. If the enclosed charge is negative (see Figure 6.16(b)), then the flux through either SorS' is negative. Figure 6.16 The electric flux through any closed surface surrounding a point charge q is given by Gauss’s law. (a) Enclosed charge is positive. (b) Enclosed charge is negative. The Gaussian surface does not need to correspond to a real, physical object; indeed, it rarely will. It is a mathematical construct that may be of any shape, provided that it is closed. However, since our goal is to integrate the flux over it, we tend to choose shapes that are highly symmetrical. If the charges are discrete point charges, then we just add them. If the charge is described by a continuous distribution, then we need to integrate appropriately to find the total charge that resides inside the enclosed volume. For example, the flux through the Gaussian surface S of Figure 6.17 is Φ=(q1+q2+q5)/ε0. Note that qenc is simply the sum of the point charges. If the charge distribution were continuous, we would need to integrate appropriately to compute the total charge within the Gaussian surface. Figure 6.17 The flux through the Gaussian surface shown, due to the charge distribution, is Φ=(q1+q2+q5)/ε0. Recall that the principle of superposition holds for the electric field. Therefore, the total electric field at any point, including those on the chosen Gaussian surface, is the sum of all the electric fields present at this point. This allows us to write Gauss’s law in terms of the total electric field. Gauss’s Law The flux Φ of the electric field E→ through any closed surface S (a Gaussian surface) is equal to the net charge enclosed (qenc) divided by the permittivity of free space (ε0): Φ=∮SE→⋅nˆdA=qencε0. To use Gauss’s law effectively, you must have a clear understanding of what each term in the equation represents. The field E→ is the total electric field at every point on the Gaussian surface. This total field includes contributions from charges both inside and outside the Gaussian surface. However, qenc is just the charge inside the Gaussian surface. Finally, the Gaussian surface is any closed surface in space. That surface can coincide with the actual surface of a conductor, or it can be an imaginary geometric surface. The only requirement imposed on a Gaussian surface is that it be closed (Figure 6.18). Figure 6.18 A Klein bottle partially filled with a liquid. Could the Klein bottle be used as a Gaussian surface? Example 6.5 Electric Flux through Gaussian Surfaces Calculate the electric flux through each Gaussian surface shown in Figure 6.19. Figure 6.19 Various Gaussian surfaces and charges. Strategy From Gauss’s law, the flux through each surface is given by qenc/ε0, where qenc is the charge enclosed by that surface. Solution For the surfaces and charges shown, we find Φ=2.0μCε0=2.3×105N⋅m2/C. Φ=−2.0μCε0=−2.3×105N⋅m2/C. Φ=2.0μCε0=2.3×105N⋅m2/C. Φ=−4.0μC+6.0μC−1.0μCε0=1.1×105N⋅m2/C. Φ=4.0μC+6.0μC−10.0μCε0=0. Significance In the special case of a closed surface, the flux calculations become a sum of charges. In the next section, this will allow us to work with more complex systems. Check Your Understanding 6.3 Calculate the electric flux through the closed cubical surface for each charge distribution shown in Figure 6.20. Figure 6.20 A cubical Gaussian surface with various charge distributions. Interactive Use this simulation to adjust the magnitude of the charge and the radius of the Gaussian surface around it. See how this affects the total flux and the magnitude of the electric field at the Gaussian surface. Previous Next Order a print copy Citation/Attribution This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission. Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute OpenStax. Attribution information If you are redistributing all or part of this book in a print format, then you must include on every physical page the following attribution: Access for free at If you are redistributing all or part of this book in a digital format, then you must include on every digital page view the following attribution: Access for free at Citation information Use the information below to generate a citation. We recommend using a citation tool such as this one. Authors: Samuel J. Ling, William Moebs, Jeff Sanny Publisher/website: OpenStax Book title: University Physics Volume 2 Publication date: Oct 6, 2016 Location: Houston, Texas Book URL: Section URL: © Jul 8, 2025 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . 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2005	https://math.stackexchange.com/questions/1658476/inequalities-with-floor-function	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 Inequalities with floor function Ask Question Asked Modified 9 years, 7 months ago Viewed 7k times 6 $\begingroup$ I need some help with this exercise, I'm pretty new solving this exercises. $$ \lfloor x \rfloor + \lfloor y \rfloor \le \lfloor x + y \rfloor \le \lfloor x \rfloor + \lfloor y \rfloor + 1$$ I know that I had to use the formal definition of the floor function, which is: $$ \lfloor x \rfloor = \max {{ m \in \Bbb Z \mid m \le x}}$$ ceiling-and-floor-functions Share edited Feb 16, 2016 at 17:32 3SAT 7,63744 gold badges2929 silver badges4747 bronze badges asked Feb 16, 2016 at 17:08 Jesus RamosJesus Ramos 6111 silver badge33 bronze badges $\endgroup$ 1 $\begingroup$ Hi @vonbrand, this a new topic to me, I just start seeing it yesterday. I tried to use the inequality property to prove $ \lfloor x \rfloor + \lfloor y \rfloor \le \lfloor x + y \rfloor \le \lfloor x \rfloor + \lfloor y \rfloor + 1$ first. I use this property: $$ a \le c; b \le d; a + b \le c+d $$ $\endgroup$ Jesus Ramos – Jesus Ramos 2016-02-17 17:33:09 +00:00 Commented Feb 17, 2016 at 17:33 Add a comment \| 4 Answers 4 Reset to default 5 $\begingroup$ Use that $x=\lfloor x\rfloor +\alpha,\,\alpha\in[0,1)$ and $y=\lfloor y\rfloor +\beta,\,\beta\in[0,1)$ $$\lfloor x+y\rfloor =\lfloor \lfloor x\rfloor+\alpha+\lfloor y\rfloor +\beta\rfloor\ge \lfloor x\rfloor+\lfloor y\rfloor $$ because $\lfloor x\rfloor+\alpha+\lfloor y\rfloor +\beta\ge \lfloor x\rfloor+\lfloor y\rfloor$ and therefore $\lfloor x+y\rfloor$ is at least $\lfloor x\rfloor+\lfloor y\rfloor$ For the other inequality: $$\lfloor x\rfloor+\alpha+\lfloor y\rfloor +\beta < \lfloor x\rfloor+\lfloor y\rfloor+2\Rightarrow $$ From here you see that one integer $m\in\mathbb Z$ that satisfies $$()\quad\quad m\leq \lfloor x\rfloor+\alpha+\lfloor y\rfloor +\beta $$ is $m=\lfloor x\rfloor+\lfloor y\rfloor$ and the greatest possible is $\lfloor x\rfloor+\lfloor y\rfloor+1$, because already $\lfloor x\rfloor+\lfloor y\rfloor+2>\lfloor x\rfloor+\lfloor y\rfloor+\alpha+\beta$ -see $()$. Therefore $$\lfloor x+y\rfloor =\lfloor \lfloor x\rfloor+\alpha+\lfloor y\rfloor +\beta\rfloor\leq \lfloor x\rfloor+\lfloor y\rfloor+1$$ Share edited Feb 16, 2016 at 18:06 answered Feb 16, 2016 at 17:28 SvetoslavSvetoslav 5,32522 gold badges1919 silver badges3636 bronze badges $\endgroup$ 0 Add a comment \| 3 $\begingroup$ $$ \lfloor x \rfloor + \lfloor y \rfloor \le \lfloor x + y \rfloor \le \lfloor x \rfloor + \lfloor y \rfloor + 1$$ Note that you can pull integers out from floor\ceil function Hint: for three first cases Case one: $x,y \in \mathbb{Z}$ so $$x+y\le x+y \le x+y+1 $$ Case two $x\in \mathbb{Z}$ but $y\in \mathbb{R}$ and $y\notin \mathbb{Z}$ $$x + \lfloor y \rfloor \le x+\lfloor y \rfloor \le x+\lfloor y \rfloor +1$$ Case three $y\in \mathbb{Z}$ but $x\in \mathbb{R}$ and $x\notin \mathbb{Z}$ $$y+\lfloor x \rfloor \le y +\lfloor x \rfloor \le y+1+\lfloor x \rfloor $$ Now, try to prove case four: $x,y\in \mathbb{R}$ and $x,y\notin \mathbb{Z}$ Share edited Feb 16, 2016 at 18:30 answered Feb 16, 2016 at 17:30 3SAT3SAT 7,63744 gold badges2929 silver badges4747 bronze badges $\endgroup$ 1 $\begingroup$ Thank you for update and take your time to answer my question @Nehorai. $\endgroup$ Jesus Ramos – Jesus Ramos 2016-02-17 17:39:04 +00:00 Commented Feb 17, 2016 at 17:39 Add a comment \| 3 $\begingroup$ I do not know how rigorous or formal you need/want to be but it's straight forward that $[x]$ is the unique integer such that $[x] \le x < [x] + 1$[] Therefore $[x] \le x < [x] + 1$ and $[y] \le y < [y] + 1$ so $[x] + [y] \le x + y < [x] +[y] + 2$. So there are two possible cases: Case 1: $[x] + [y] \le x + y < [x] + [y] + 1$ This means $[x + y ] = [x] + [y]$. So $[x]+[y] = [x+y] < [x] + [y] + 1$. Case 2: $[x] + [y] + 1 \le x + y < [x]+ [y] + 2$ This means $[x + y] = [x] + [y] + 1$. So $[x]+[y] < [x+y] = [x]+[y] + 1$. And that's it. [] Rigor and formality?? By the archemedian principal, for every real $x$, there is a unique integer $n$ such that $n \le x < n+ 1$. $n \in \mathbb Z$ and $n \le x$ so $n \in {m \in \mathbb Z\| m \le x}$. If $m \in \mathbb Z; m > n$ then $m \ge n+1 > x$ so $m \not \in {m \in \mathbb Z\| m \le x}$. So $n = \max {m \in \mathbb Z\| m \le x} = [x]$. Share edited Feb 16, 2016 at 22:50 answered Feb 16, 2016 at 18:13 fleabloodfleablood 132k55 gold badges5252 silver badges142142 bronze badges $\endgroup$ 1 $\begingroup$ Thank you for your time @fleablood. $\endgroup$ Jesus Ramos – Jesus Ramos 2016-02-17 17:37:13 +00:00 Commented Feb 17, 2016 at 17:37 Add a comment \| 2 $\begingroup$ As said by Nehorai, you can pull any integer from the floor function, so it suffices to validate with $x,y\in[0,1)$. Then $$ \lfloor x \rfloor + \lfloor y \rfloor \le \lfloor x + y \rfloor \le \lfloor x \rfloor + \lfloor y \rfloor + 1$$ becomes $$ 0\le \lfloor x + y \rfloor \le 1,$$ which is obviously true as $$0\le x+y<2.$$ Share answered Feb 16, 2016 at 17:52 user65203user65203 $\endgroup$ 0 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 ceiling-and-floor-functions See similar questions with these tags. 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2006	https://www.youtube.com/watch?v=MBRiaZeuS8c	Algebra 2: 1-2 Evaluate Expressions MaeMap 38200 subscribers 7 likes Description 1077 views Posted: 1 Sep 2016 In this lesson video, we will review the order of operation (PEMDAS) which will be used to evaluate variable expressions throughout the remainder of this Algebra 2 unit. Please pay careful attention and take time to memorize the order in which to evaluate! This is the lesson video. 👩‍🏫 To download the lesson worksheet please go to ➡️ Please support me: 💸 ☕ Happy learning! Transcript: Intro in this next video we're going to continue reviewing and we're going to go over evaluating expressions but Before we jump into that let's go over order of operations again so when you hear order Order of Operations of operations think PEMDAS remember parenthesis exponents multiplication and division and then addition and subtraction that is the order that you complete operations let's also review variable and what a variable is so variable is a letter or symbol that represents a certain number we can have X as our variable and X in one instance can equal to but in a later expression or equation X could equal negative 5 remember that the variable is changing in different problems so one problem you can have x equals 2 another problem you can have x equals negative 5 sometimes you'll use wise a b c and so on a Algebraic Expressions mathematical expression is like a statement so if i was to say 3-2 that's an expression an algebraic expression is a mathematical statement that contains a variable so an example of an algebraic expression could be B plus negative 5.3 or you could have x squared minus 2 those are algebraic expressions because they contain a variable so let's go ahead and evaluate an algebraic expression the algebraic expression I'm going to use in this video will be the following this is my algebraic expression 5 plus a squared minus 4 over B and i'm going to give you what your variables are equal to a is equal to 1 and b is equal to negative 2 so what's our first step in evaluating this expression first we're going to want to plug in our variables into the expression itself so we're going to have 5 plus a which a is equal to one so 5 plus 1 squared minus 4 over B which is negative 2 now we can evaluate it using the order of operations so what's first parentheses we have one set of parentheses and that's around the 5+1 so we evaluate what's inside of the parentheses first so 5 plus 1 is 6 so we're left with 6 squared minus 4 over negative 2 next we're going to evaluate the exponents there's one exponent in this problem which is the two so we're going to evaluate 6 squared 6 squared is 36 so now we've evaluated our exponent next is multiplication and division remember the order of multiplication division doesn't matter you just have to make sure you do those before addition and subtraction for us we only had division once in this problem and that's with the fraction sign this is 4 divided by negative 2 so we're going to evaluate that so we're going to get 36 minus 4 divided by negative 2 would be negative 2 however when you subtract a negative you're really adding the positive of that number so this instead of 36 minus negative 2 would be 36 plus 2 so now we've done our multiplication and division so we're left with just addition 36 plus 2 is 38 and we just evaluated our original algebraic expression 5 plus a squared minus 4 over B when I gave us answer of 38 when a was equal to 1 and B was equal to negative 2 so expressions are used with in equations in equations contain an equal sign an equation with a variable is known as an open sentence if that equation represents a relationship such as an area or perimeter of a shape then that equation would be known as a formula but in order to solve equations we must first go over some properties of real numbers which we'll do in the next video and don't forget to like and subscribe to my channel
2007	https://openstax.org/books/elementary-algebra-2e/pages/6-4-special-products	Skip to Content Go to accessibility page Keyboard shortcuts menu Log in Elementary Algebra 2e 6.4 Special Products Elementary Algebra 2e 6.4 Special Products Search for key terms or text. Learning Objectives By the end of this section, you will be able to: Square a binomial using the Binomial Squares Pattern Multiply conjugates using the Product of Conjugates Pattern Recognize and use the appropriate special product pattern Be Prepared 6.8 Before you get started, take this readiness quiz. Simplify: ⓐ 9292 ⓑ (−9)2(−9)2 ⓒ −92.−92. If you missed this problem, review Example 1.50. Square a Binomial Using the Binomial Squares Pattern Mathematicians like to look for patterns that will make their work easier. A good example of this is squaring binomials. While you can always get the product by writing the binomial twice and using the methods of the last section, there is less work to do if you learn to use a pattern. \| \| \| --- \| \| Let's start by looking at (x+9)2(x+9)2. \| \| \| What does this mean? \| (x+9)2(x+9)2 \| \| It means to multiply (x+9)(x+9) by itself. \| (x+9)(x+9)(x+9)(x+9) \| \| Then, using FOIL, we get: \| x2+9x+9x+81x2+9x+9x+81 \| \| Combining like terms gives: \| x2+18x+81x2+18x+81 \| \| \| \| --- \| \| Here's another one: \| (y−7)2(y−7)2 \| \| Multiply (y−7)(y−7) by itself. \| (y−7)(y−7)(y−7)(y−7) \| \| Using FOIL, we get: \| y2−7y−7y+49y2−7y−7y+49 \| \| And combining like terms: \| y2−14y+49y2−14y+49 \| \| \| \| --- \| \| And one more: \| (2x+3)2(2x+3)2 \| \| Multiply. \| (2x+3)(2x+3)(2x+3)(2x+3) \| \| Use FOIL: \| 4x2+6x+6x+94x2+6x+6x+9 \| \| Combine like terms. \| 4x2+12x+94x2+12x+9 \| Look at these results. Do you see any patterns? What about the number of terms? In each example we squared a binomial and the result was a trinomial. (a+b)2=____+____+____ (a+b)2=____+____+____ Now look at the first term in each result. Where did it come from? The first term is the product of the first terms of each binomial. Since the binomials are identical, it is just the square of the first term! (a+b)2=a2+____+____ (a+b)2=a2+____+____ To get the first term of the product, square the first term. Where did the last term come from? Look at the examples and find the pattern. The last term is the product of the last terms, which is the square of the last term. (a+b)2=____+____+b2 (a+b)2=____+____+b2 To get the last term of the product, square the last term. Finally, look at the middle term. Notice it came from adding the “outer” and the “inner” terms—which are both the same! So the middle term is double the product of the two terms of the binomial. (a+b)2=____+2ab+____(a−b)2=____−2ab+____ (a+b)2=____+2ab+____(a−b)2=____−2ab+____ To get the middle term of the product, multiply the terms and double their product. Putting it all together: Binomial Squares Pattern If aandbaandb are real numbers, (a+b)2=a2+2ab+b2(a−b)2=a2−2ab+b2 (a+b)2=a2+2ab+b2(a−b)2=a2−2ab+b2 To square a binomial: square the first term square the last term double their product A number example helps verify the pattern. \| \| \| --- \| \| \| (10+4)2(10+4)2 \| \| Square the first term. \| 102+___+102+___+ \| \| Square the last term. \| 102+___+42102+___+42 \| \| Double their product. \| 102+2·10·4+42102+2⋅10⋅4+42 \| \| Simplify. \| 100+80+16100+80+16 \| \| Simplify. \| 196196 \| To multiply (10+4)2(10+4)2 usually you’d follow the Order of Operations. (10+4)2(14)2196 (10+4)2(14)2196 The pattern works! Example 6.47 Multiply: (x+5)2.(x+5)2. Solution \| \| \| --- \| \| \| \| \| Square the first term. \| \| \| Square the last term. \| \| \| Double the product. \| \| \| Simplify. \| \| Try It 6.93 Multiply: (x+9)2.(x+9)2. Try It 6.94 Multiply: (y+11)2.(y+11)2. Example 6.48 Multiply: (y−3)2.(y−3)2. Solution \| \| \| --- \| \| \| \| \| Square the first term. \| \| \| Square the last term. \| \| \| Double the product. \| \| \| Simplify. \| \| Try It 6.95 Multiply: (x−9)2.(x−9)2. Try It 6.96 Multiply: (p−13)2.(p−13)2. Example 6.49 Multiply: (4x+6)2.(4x+6)2. Solution \| \| \| --- \| \| \| \| \| Use the pattern. \| \| \| Simplify. \| \| Try It 6.97 Multiply: (6x+3)2.(6x+3)2. Try It 6.98 Multiply: (4x+9)2.(4x+9)2. Example 6.50 Multiply: (2x−3y)2.(2x−3y)2. Solution \| \| \| --- \| \| \| \| \| Use the pattern. \| \| \| Simplify. \| \| Try It 6.99 Multiply: (2c−d)2.(2c−d)2. Try It 6.100 Multiply: (4x−5y)2.(4x−5y)2. Example 6.51 Multiply: (4u3+1)2.(4u3+1)2. Solution \| \| \| --- \| \| \| \| \| Use the pattern. \| \| \| Simplify. \| \| Try It 6.101 Multiply: (2x2+1)2.(2x2+1)2. Try It 6.102 Multiply: (3y3+2)2.(3y3+2)2. Multiply Conjugates Using the Product of Conjugates Pattern We just saw a pattern for squaring binomials that we can use to make multiplying some binomials easier. Similarly, there is a pattern for another product of binomials. But before we get to it, we need to introduce some vocabulary. What do you notice about these pairs of binomials? (x−9)(x+9)(y−8)(y+8)(2x−5)(2x+5) (x−9)(x+9)(y−8)(y+8)(2x−5)(2x+5) Look at the first term of each binomial in each pair. Notice the first terms are the same in each pair. Look at the last terms of each binomial in each pair. Notice the last terms are the same in each pair. Notice how each pair has one sum and one difference. A pair of binomials that each have the same first term and the same last term, but one is a sum and one is a difference has a special name. It is called a conjugate pair and is of the form (a−b),(a+b)(a−b),(a+b). Conjugate Pair A conjugate pair is two binomials of the form (a−b),(a+b). (a−b),(a+b). The pair of binomials each have the same first term and the same last term, but one binomial is a sum and the other is a difference. There is a nice pattern for finding the product of conjugates. You could, of course, simply FOIL to get the product, but using the pattern makes your work easier. Let’s look for the pattern by using FOIL to multiply some conjugate pairs. (x−9)(x+9)(y−8)(y+8)(2x−5)(2x+5)x2+9x−9x−81y2+8y−8y−644x2+10x−10x−25x2−81y2−644x2−25 (x−9)(x+9)x2+9x−9x−81x2−81(y−8)(y+8)y2+8y−8y−64y2−64(2x−5)(2x+5)4x2+10x−10x−254x2−25 Each first term is the product of the first terms of the binomials, and since they are identical it is the square of the first term. (a+b)(a−b)=a2−____To get thefirst term, square the first term. (a+b)(a−b)=a2−____To get thefirst term, square the first term. The last term came from multiplying the last terms, the square of the last term. (a+b)(a−b)=a2−b2To get thelast term, square the last term. (a+b)(a−b)=a2−b2To get thelast term, square the last term. What do you observe about the products? The product of the two binomials is also a binomial! Most of the products resulting from FOIL have been trinomials. Why is there no middle term? Notice the two middle terms you get from FOIL combine to 0 in every case, the result of one addition and one subtraction. The product of conjugates is always of the form a2−b2a2−b2. This is called a difference of squares. This leads to the pattern: Product of Conjugates Pattern If aandbaandb are real numbers, The product is called a difference of squares. To multiply conjugates, square the first term, square the last term, and write the product as a difference of squares. Let’s test this pattern with a numerical example. \| \| \| --- \| \| \| (10−2)(10+2)(10−2)(10+2) \| \| It is the product of conjudgates, so the result will be the difference of two squares. \| ____−________−____ \| \| Square the first term. \| 102−____102−____ \| \| Square the last term. \| 102−22102−22 \| \| Simplify. \| 100−4100−4 \| \| Simplify. \| 9696 \| \| What do you get using the order of operations? \| \| \| \| (10−2)(10+2)(8)(12)96(10−2)(10+2)(8)(12)96 \| Notice, the result is the same! Example 6.52 Multiply: (x−8)(x+8).(x−8)(x+8). Solution First, recognize this as a product of conjugates. The binomials have the same first terms, and the same last terms, and one binomial is a sum and the other is a difference. \| \| \| --- \| \| It fits the pattern. \| \| \| Square the first term, x. \| \| \| Square the last term, 8. \| \| \| The product is a difference of squares. \| \| Try It 6.103 Multiply: (x−5)(x+5).(x−5)(x+5). Try It 6.104 Multiply: (w−3)(w+3).(w−3)(w+3). Example 6.53 Multiply: (2x+5)(2x−5).(2x+5)(2x−5). Solution Are the binomials conjugates? \| \| \| --- \| \| It is the product of conjugates. \| \| \| Square the first term, 2x. \| \| \| Square the last term, 5. \| \| \| Simplify. The product is a difference of squares. \| \| Try It 6.105 Multiply: (6x+5)(6x−5).(6x+5)(6x−5). Try It 6.106 Multiply: (2x+7)(2x−7).(2x+7)(2x−7). The binomials in the next example may look backwards – the variable is in the second term. But the two binomials are still conjugates, so we use the same pattern to multiply them. Example 6.54 Find the product: (3+5x)(3−5x).(3+5x)(3−5x). Solution \| \| \| --- \| \| It is the product of conjugates. \| \| \| Use the pattern. \| \| \| Simplify. \| \| Try It 6.107 Multiply: (7+4x)(7−4x).(7+4x)(7−4x). Try It 6.108 Multiply: (9−2y)(9+2y).(9−2y)(9+2y). Now we’ll multiply conjugates that have two variables. Example 6.55 Find the product: (5m−9n)(5m+9n).(5m−9n)(5m+9n). Solution \| \| \| --- \| \| This fits the pattern. \| \| \| Use the pattern. \| \| \| Simplify. \| \| Try It 6.109 Find the product: (4p−7q)(4p+7q).(4p−7q)(4p+7q). Try It 6.110 Find the product: (3x−y)(3x+y).(3x−y)(3x+y). Example 6.56 Find the product: (cd−8)(cd+8).(cd−8)(cd+8). Solution \| \| \| --- \| \| This fits the pattern. \| \| \| Use the pattern. \| \| \| Simplify. \| \| Try It 6.111 Find the product: (xy−6)(xy+6).(xy−6)(xy+6). Try It 6.112 Find the product: (ab−9)(ab+9).(ab−9)(ab+9). Example 6.57 Find the product: (6u2−11v5)(6u2+11v5).(6u2−11v5)(6u2+11v5). Solution \| \| \| --- \| \| This fits the pattern. \| \| \| Use the pattern. \| \| \| Simplify. \| \| Try It 6.113 Find the product: (3x2−4y3)(3x2+4y3).(3x2−4y3)(3x2+4y3). Try It 6.114 Find the product: (2m2−5n3)(2m2+5n3).(2m2−5n3)(2m2+5n3). Recognize and Use the Appropriate Special Product Pattern We just developed special product patterns for Binomial Squares and for the Product of Conjugates. The products look similar, so it is important to recognize when it is appropriate to use each of these patterns and to notice how they differ. Look at the two patterns together and note their similarities and differences. Comparing the Special Product Patterns \| \| \| --- \| \| Binomial Squares \| Product of Conjugates \| \| (a+b)2=a2+2ab+b2(a+b)2=a2+2ab+b2 \| (a−b)(a+b)=a2−b2(a−b)(a+b)=a2−b2 \| \| (a−b)2=a2−2ab+b2(a−b)2=a2−2ab+b2 \| \| \| - Squaring a binomial \| - Multiplying conjugates \| \| - Product is a trinomial \| - Product is a binomial \| \| - Inner and outer terms with FOIL are the same. \| - Inner and outer terms with FOIL are opposites. \| \| - Middle term is double the product of the terms. \| - There is no middle term. \| Example 6.58 Choose the appropriate pattern and use it to find the product: ⓐ (2x−3)(2x+3)(2x−3)(2x+3) ⓑ (8x−5)2(8x−5)2 ⓒ (6m+7)2(6m+7)2 ⓓ (5x−6)(6x+5)(5x−6)(6x+5) Solution ⓐ (2x−3)(2x+3)(2x−3)(2x+3) These are conjugates. They have the same first numbers, and the same last numbers, and one binomial is a sum and the other is a difference. It fits the Product of Conjugates pattern. \| \| \| --- \| \| This fits the pattern. \| \| \| Use the pattern. \| \| \| Simplify. \| \| 2. ⓑ (8x−5)2(8x−5)2 We are asked to square a binomial. It fits the binomial squares pattern. \| \| \| --- \| \| \| \| \| Use the pattern. \| \| \| Simplify. \| \| 3. ⓒ (6m+7)2(6m+7)2 Again, we will square a binomial so we use the binomial squares pattern. \| \| \| --- \| \| \| \| \| Use the pattern. \| \| \| Simplify. \| \| 4. ⓓ (5x−6)(6x+5)(5x−6)(6x+5) This product does not fit the patterns, so we will use FOIL. \| \| \| --- \| \| \| (5x−6)(6x+5)(5x−6)(6x+5) \| \| Use FOIL. \| 30x2+25x−36x−3030x2+25x−36x−30 \| \| Simplify. \| 30x2−11x−3030x2−11x−30 \| Try It 6.115 Choose the appropriate pattern and use it to find the product: ⓐ (9b−2)(2b+9)(9b−2)(2b+9) ⓑ (9p−4)2(9p−4)2 ⓒ (7y+1)2(7y+1)2 ⓓ (4r−3)(4r+3)(4r−3)(4r+3) Try It 6.116 Choose the appropriate pattern and use it to find the product: ⓐ (6x+7)2(6x+7)2 ⓑ (3x−4)(3x+4)(3x−4)(3x+4) ⓒ (2x−5)(5x−2)(2x−5)(5x−2) ⓓ (6n−1)2(6n−1)2 Media Access these online resources for additional instruction and practice with special products: Special Products Section 6.4 Exercises Practice Makes Perfect Square a Binomial Using the Binomial Squares Pattern In the following exercises, square each binomial using the Binomial Squares Pattern. (w+4)2(w+4)2 303. (q+12)2(q+12)2 (y+14)2(y+14)2 305. (x+23)2(x+23)2 (b−7)2(b−7)2 307. (y−6)2(y−6)2 (m−15)2(m−15)2 309. (p−13)2(p−13)2 (3d+1)2(3d+1)2 311. (4a+10)2(4a+10)2 (2q+13)2(2q+13)2 313. (3z+15)2(3z+15)2 (3x−y)2(3x−y)2 315. (2y−3z)2(2y−3z)2 (15x−17y)2(15x−17y)2 317. (18x−19y)2(18x−19y)2 (3x2+2)2(3x2+2)2 319. (5u2+9)2(5u2+9)2 (4y3−2)2(4y3−2)2 321. (8p3−3)2(8p3−3)2 Multiply Conjugates Using the Product of Conjugates Pattern In the following exercises, multiply each pair of conjugates using the Product of Conjugates Pattern. (m−7)(m+7)(m−7)(m+7) 323. (c−5)(c+5)(c−5)(c+5) (x+34)(x−34)(x+34)(x−34) 325. (b+67)(b−67)(b+67)(b−67) (5k+6)(5k−6)(5k+6)(5k−6) 327. (8j+4)(8j−4)(8j+4)(8j−4) (11k+4)(11k−4)(11k+4)(11k−4) 329. (9c+5)(9c−5)(9c+5)(9c−5) (11−b)(11+b)(11−b)(11+b) 331. (13−q)(13+q)(13−q)(13+q) (5−3x)(5+3x)(5−3x)(5+3x) 333. (4−6y)(4+6y)(4−6y)(4+6y) (9c−2d)(9c+2d)(9c−2d)(9c+2d) 335. (7w+10x)(7w−10x)(7w+10x)(7w−10x) (m+23n)(m−23n)(m+23n)(m−23n) 337. (p+45q)(p−45q)(p+45q)(p−45q) (ab−4)(ab+4)(ab−4)(ab+4) 339. (xy−9)(xy+9)(xy−9)(xy+9) (uv−35)(uv+35)(uv−35)(uv+35) 341. (rs−27)(rs+27)(rs−27)(rs+27) (2x2−3y4)(2x2+3y4)(2x2−3y4)(2x2+3y4) 343. (6m3−4n5)(6m3+4n5)(6m3−4n5)(6m3+4n5) (12p3−11q2)(12p3+11q2)(12p3−11q2)(12p3+11q2) 345. (15m2−8n4)(15m2+8n4)(15m2−8n4)(15m2+8n4) Recognize and Use the Appropriate Special Product Pattern In the following exercises, find each product. ⓐ (p−3)(p+3)(p−3)(p+3) ⓑ (t−9)2(t−9)2 ⓒ (m+n)2 ⓓ (2x+y)(x−2y) 347. ⓐ (2r+12)2 ⓑ (3p+8)(3p−8) ⓒ (7a+b)(a−7b) ⓓ (k−6)2 ⓐ (a5−7b)2 ⓑ (x2+8y)(8x−y2) ⓒ (r6+s6)(r6−s6) ⓓ (y4+2z)2 349. ⓐ (x5+y5)(x5−y5) ⓑ (m3−8n)2 ⓒ (9p+8q)2 ⓓ (r2−s3)(r3+s2) Everyday Math Mental math You can use the product of conjugates pattern to multiply numbers without a calculator. Say you need to multiply 47 times 53. Think of 47 as 50−3 and 53 as 50+3. ⓐ Multiply (50−3)(50+3) by using the product of conjugates pattern, (a−b)(a+b)=a2−b2. ⓑ Multiply 47·53 without using a calculator. ⓒ Which way is easier for you? Why? 351. Mental math You can use the binomial squares pattern to multiply numbers without a calculator. Say you need to square 65. Think of 65 as 60+5. ⓐ Multiply (60+5)2 by using the binomial squares pattern, (a+b)2=a2+2ab+b2. ⓑ Square 65 without using a calculator. ⓒ Which way is easier for you? Why? Writing Exercises How do you decide which pattern to use? 353. Why does (a+b)2 result in a trinomial, but (a−b)(a+b) result in a binomial? Marta did the following work on her homework paper: (3−y)232−y29−y2 Explain what is wrong with Marta’s work. 355. Use the order of operations to show that (3+5)2 is 64, and then use that numerical example to explain why (a+b)2≠a2+b2. Self Check ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section. ⓑ On a scale of 1-10, how would you rate your mastery of this section in light of your responses on the checklist? How can you improve this? 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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2008	https://www.cs.toronto.edu/~toni/Courses/Complexity2015/handouts/monotone.pdf	Comput. complex. 8 (1999), 1 – 20 1016-3328/99/010001–20 $ 1.50+0.20/0 c ⃝Birkh¨ auser Verlag, Basel 1999 computational complexity SYMMETRIC APPROXIMATION ARGUMENTS FOR MONOTONE LOWER BOUNDS WITHOUT SUNFLOWERS Christer Berg Staffan Ulfberg Abstract. We propose a symmetric version of Razborov’s method of approximation to prove lower bounds for monotone circuit complex-ity. Traditionally, only DNF formulas have been used as approximators, whereas we use both CNF and DNF formulas. As a consequence we no longer need the Sunﬂower Lemma that has been essential for the method of approximation. The new approximation argument corres-ponds to Haken’s recent method for proving lower bounds for monotone circuit complexity (counting bottlenecks) in a natural way. We provide lower bounds for the BMS problem introduced by Haken, Andreev’s polynomial problem, and for Clique. The exponential bounds obtained are the same as those previously best known for the respective problems. Key words. Circuit complexity; lower bounds; monotone circuits. Subject classiﬁcations. 68Q25. 1. Introduction The diﬃculty of proving super-linear lower bounds for general circuit complex-ity has led to the study of various restrictions of this computational model. One of these is monotone circuits, i.e., only AND gates and OR gates are allowed. Since many important problems in complexity theory are monotone, this com-putational model seems to be a natural one. Examples of monotone graph problems are Clique, Vertex cover, and Hamiltonian path, which are all NP-complete. However, for a long time there had been no stronger lower bounds for monotone than for general circuits; the best one was only 4n (by Tieken-heinrich (1984)). A major breakthrough came in 1985, when Razborov (1985) invented the method of approximation. It allowed him to prove a super-polynomial lower bound, as he showed that Clique requires monotone circuits of size nΩ(log n). Shortly thereafter, Andreev (1985) applied Razborov’s technique 2 Berg & Ulfberg cc 8 (1999) to another function and was thereby able to prove an exponential lower bound. Later, both these results were improved by Alon and Boppana (1987), and in particular, they were the ﬁrst to prove an exponential lower bound for Clique. For a nice exposition of this result, see Section 4 in the survey by Boppana and Sipser (1990). The idea behind the approximation method is the following. The output of each gate in a given circuit is approximated by a function, the approximator for that gate. The goal is to ﬁnd a set of approximator functions with the following properties. For every gate in the circuit, the approximator should introduce but a few errors. Also, the approximator for the output gate should diﬀer from the function computed by the circuit for many input values. If this goal is accomplished, we will have proved that the circuit must be large, since errors in the approximator for the output gate must emerge from the approximation of some gate in the circuit. In 1995, Haken (1995) proposed a new method for proving lower bounds for the size of monotone circuits, which he called “Bottleneck counting”. He demonstrated the method for a graph problem that resembles Clique. Instead of trying to approximate the behavior of the circuit, he deﬁned a function µ which maps input graphs to gates of the circuit. To prove that the circuit contains many gates, he then showed that on the one hand the total number of graphs mapped by µ is large, but on the other hand, only a few graphs are being mapped to each gate. One purpose of this paper is to show that Haken’s approach to lower bounds is in fact an approximation argument in disguise: it can be turned into an approximation argument by using approximators, which for a gate e, introduce errors for exactly those inputs that are mapped to e by Haken’s function µ. The approximation version of Haken’s proof is in Section 5. It is of course nice that Haken’s method of bottleneck counting works in es-sentially the same way as Razborov’s approximation method, but unfortunately this leads to the conclusion that it cannot be used to prove lower bounds better than Ω(n2) for general circuits. The reason for this is, as Razborov showed, that the method of approximation will not work very well for general circuits (1989). Pudl´ ak (1997) extended Razborov’s approximation method to hold for cir-cuits consisting of bounded fan-in gates computing monotone real functions. The corresponding extension of Haken’s method was made by Haken and Cook (1996). Based on Haken’s idea, Jukna (1997) was able to derive a general criterion for lower bounds that holds for both monotone boolean and monotone real circuits. Wigderson independently discovered that Haken’s method can be turned into an approximation argument, and he also noted that this holds cc 8 (1999) Symmetric Approximation Arguments 3 for the extension made by Haken and Cook. The extension to real monotone circuits can be found in Section 4.6 of Berg (1997). Another purpose of the paper is to make the approximation method simpler and more symmetric. Our approximation argument, which is derived from the translation of Haken’s proof, diﬀers from that of Razborov in that we use two approximator functions for each gate. Razborov used only one approximator function which is a monotone depth two circuit. Out of the two possibilities for the form of such circuits, a disjunction of conjunctions or the other way around, Razborov chose the former. In this paper, however, we do not make this choice at all since we use two approximator functions for each gate—one for each of the two possible forms. This seems like a natural thing to do, and one consequence is that we do not have to use the Sunﬂower Lemma by Erd˝ os and Rado (1960). The Sunﬂower Lemma is traditionally used to reduce the size of approximator functions while not reducing their approximation qualities too much. The fact that we do not need this lemma is somewhat surprising since it is a central part in earlier approximation results. Although we are unable to improve the bounds obtained by Alon and Boppana, this approach leads to shorter and, to our thinking, simpler proofs. Note that while most of the known lower bounds obtained using the method of approximation may beneﬁt from being reformulated in this way, it is not clear that all of them do. We do not, for example, know how to present Razborov’s super-polynomial lower bound for perfect matching (which implies that mono-tone circuits are strictly less powerful than non-monotone) using CNF and DNF approximators. 1.1. Recent work. Ideas from the bottleneck counting argument were also used by Amano and Maruoka (1996) to develop approximators similar to the ones presented in this paper. They used the approximators to prove lower bounds for Clique and Andreev’s polynomial problem. Simon and Tsai (1997) showed that the bottleneck counting argument is actually equivalent to the method of approximation. They provided a proof of the fact that errors in the approximation method correspond to mapped inputs in bottleneck counting. 2. Deﬁnitions A monotone circuit Ψ is a boolean circuit with AND gates (∧) and OR gates (∨), but without NOT gates. We let all gates have arbitrary fan-in and fan-out, 4 Berg & Ulfberg cc 8 (1999) and measure the size of a circuit, size(Ψ), by the number of gates contained therein. The monotone functions, for which we provide lower bounds, are all graph problems. An input graph on n vertices is represented in a standard way: as n 2 variables xi,j whose value is 1 if the edge (i, j) exists. The output of a gate e in Ψ when the input x is applied to the circuit is denoted by e(x). Let the output gate be eo so that the circuit Ψ computes the boolean function eo(x). Our proofs are based on the method of approximation which was invented by Razborov (1985). It involves the use of a boolean function to approximate the output of every gate in a given circuit. The approximator functions used by Razborov, Andreev (1985), and Alon and Boppana (1987) are all monotone DNF formulas. The proofs in this paper, however, are inﬂuenced by the work of Haken (1995), and every gate e in a circuit Ψ is approximated by two functions f D and f C e , the approximators for the gate e. The approximator f D has the form C1∨C2∨. . .∨Ct, where Ci is a conjunction of limited size: all conjunctions have less than c distinct literals. The approximator f C e has the form D1∧D2∧. . .∧Ds, where Di is a disjunction containing less than d distinct literals. (Notice that we put no explicit restriction on the numbers s and t.) The following characteristics of the approximator functions f D and f C e are essential to the proof. 1. The approximators f D eo(x) and f C eo(x) fail to correctly represent the output of Ψ for many inputs x. 2. For every gate e, the number of inputs for which the approximators in-troduce errors (measured in a speciﬁc way) is small. For every x for which the approximators for eo fail, the error must have been introduced in at least one of the gates in Ψ. We can therefore draw the conclusion that Ψ contains many gates. For convenience, we consider the inputs to the circuit as being gates them-selves, and for an input variable xi,j, we simply deﬁne f D xi,j = f C xi,j = xi,j. We also deﬁne empty disjunctions to have the value 0, and empty conjunctions to have the value 1. Definition 1. For a gate e and an input x for which e(x) = eo(x), we say that the approximator f D fails for x if f D(x) ̸= e(x), and that the approximator f C e fails for x, if f C e (x) ̸= e(x). cc 8 (1999) Symmetric Approximation Arguments 5 If either f D or f C e fails for x, we say that the approximators for the gate e fail for the input x. A central notion of the proofs is counting the number of errors that are introduced in the gates of Ψ. This is deﬁned as follows. Definition 2. An approximator, f D or f C e , is said to introduce an error for the input x if it fails on the input x, but none of the approximators for the input gates to e fail for the input x. If either f D or f C e introduces an error for the input x, we say that the approximators for the gate e introduce an error for the input x. We now describe how an ∧gate e is approximated, assuming that its input gates are already approximated by f D e1, f D e2, . . . , f D em and f C e1, f C e2, . . . , f C em. The approximator f C e is simply deﬁned as f C e = m ^ i=1 f C ei = s ^ i=1 Di, in which all disjunctions still have length at most d. If none of f C e1, f C e2, . . . , f C em fails for the input x, neither will f C e , so the approximator f C e introduces no errors. The approximator f D should be a disjunction of conjunctions; it is formed by converting f C e into the form t _ i=1 Ci. The standard way of doing this is to form a conjunction Ci for every possible way of picking one literal from each disjunction Di in f C e . All conjunctions that have at least c distinct literals are then discarded, which means that long conjunctions are approximated by the constant 0. In the proofs to follow, we need to establish upper bounds for the number of errors introduced when forming f D from f C e . To get as good bounds as possible, we have to be more careful when forming f D (so that more conjunctions get shorter than c). We will shortly describe the process we actually use in detail. Returning to the deﬁnitions of f D and f C e , we note that f D is produced from f C e , which, in turn, is constructed from the approximators f C ei of the input gates. Thus, when approximating an ∧gate, we do not use the approximators f D ei for the input gates. 6 Berg & Ulfberg cc 8 (1999) The approximator functions for an ∨gate e are formed analogously. We construct f D from the approximators f D ei of the input gates; this introduces no errors. The disjunction of conjunctions is then converted into a conjunction of disjunctions, and disjunctions with at least d distinct literals are discarded (they are in eﬀect approximated by the constant 1). Notice that the way the approximators are constructed, we have f D ≤f C e for all gates e. We now end this section by describing the details involved when converting f C e to f D in an ∧gate (this part is analogous for ∨gates as well). A set of conjunctions is formed which have the property that at least one of them is satisﬁed if and only if all disjunctions in f C e are satisﬁed. The approximator f D is then formed from all resulting conjunctions shorter than c. The rewriting process can be viewed as the construction of a tree: Each edge in the tree is labeled by a literal. For every node v in the tree, we deﬁne a corresponding conjunction that is formed by all literals on the path from the root to v. At the root we create one labeled edge to a new child for each of the literals in the ﬁrst disjunction. We say that we have expanded D1 under the root. Suppose w is a leaf that was created while expanding Di, and that C is the conjunction corresponding to w. Then, D1, . . . , Di are all satisﬁed if C is satisﬁed. We now take care of Di+1. First consider the case when C is satisﬁed. In this case, we know that for all accepting instances of the problem, at least one of the literals in Di+1, say xu,v, must be satisﬁed as well. (This happens if Di+1 contains a literal in common with C, but it could also happen in some other situations if there are restrictions on the possible inputs.) We then make only one new child under w for the literal xu,v and skip the rest of Di+1. This results in fewer leaves in the tree and therefore fewer conjunctions. Otherwise, we expand the disjunction Di+1 under w. The result is that if the conjunction corresponding to one of the new children is satisﬁed, so is Di+1. Conversely, if all of D1, . . . , Di+1 are satisﬁed, there is a node on depth i + 1 in the tree whose corresponding conjunction is satisﬁed. When there are no more leaves for which there are remaining disjunctions to expand, we are done, and we get one conjunction for each leaf in the tree. Leaves whose corresponding conjunctions have at least c distinct literals are then removed, which is why f D may make an error for some inputs. Note that such inputs are always accepted by some conjunction (correspond-ing to a node in the tree) that has exactly c distinct literals, and bounding the number of such conjunctions is the reason for constructing the tree. By count-ing the number of inputs accepted by each conjunction, we can get an upper cc 8 (1999) Symmetric Approximation Arguments 7 bound for the number of errors introduced by f D. 3. The proof method Although they diﬀer in details, the basic strategy for the proofs is the same in the following sections. In this section we therefore describe the strategy by giving generic versions of the theorem, and of the lemmas that are used in the proof of the theorem. The generic theorem and lemmas contain various unspeciﬁed entities, such as numerical functions α, β, and γ. The outline can be turned into an actual proof of an actual theorem by specifying these entities (and checking their required properties). Let MGP(n) be a monotone language on graphs with n input variables (which indicate the presence or absence of edges). For some function h(n), we want to prove the following. Generic Theorem. The monotone circuit complexity for MGP(n) is at least h(n). Generic proof. Suppose we are given a circuit Ψ that correctly decides MGP(n). In order to prove that Ψ must consist of at least h(n) gates, we study two subsets of the possible input graphs: the positive test graphs, which are subsets of the graphs in MGP(n), and the negative test graphs, which are subsets of the graphs not in MGP(n). Let γ1(n) be the number of positive test graphs and γ0(n) the number of negative test graphs. Instead of directly proving that a circuit that decides MGP(n) must be large, we prove that a circuit that separates the positive and negative test graphs from each other must be large; choosing the test graphs in a suitable way is of course essential. We start by showing that the approximators for the output gate of Ψ fail for most inputs. Generic Lemma A. At the output gate eo, the approximators either fail for all negative test graphs, or they fail for at least half of the positive test graphs. Generic proof. Assume that the approximators do not fail for all negative test graphs; otherwise there is nothing to prove. Hence, there exists a negative test graph b, such that f C eo(b) = 0, which implies that f C eo contains a disjunction D. Furthermore, for all positive test graphs g on which f C eo is correct, we must have D(g) = 1. Since D contains less than d distinct literals, we can bound the fraction of positive test graphs for which f C eo = 1 by multiplying the fraction of 8 Berg & Ulfberg cc 8 (1999) positive test graphs for which any speciﬁc edge is present with d. A suitable choice of d proves the lemma. □ The next objective is to bound the number of inputs for which the approx-imators introduce errors at a single gate. Assume e is an ∧gate. In this case f C e introduces no errors at all, and hence, we need only study f D. Also, since f D ≤f C e , no errors are ever introduced for negative test graphs in ∧gates. The next two lemmas introduce the functions α and β as bounds for the number of errors introduced in a single ∧and ∨gate, respectively. Generic Lemma B. At an ∧gate e, the approximator f D introduces an error for at most α(n, c, d) positive test graphs. Generic proof. To prove the lemma, we give an upper bound on the number of positive test graphs g such that C(g) = 1 (g is accepted by C), where C is any conjunction with c distinct literals corresponding to a node in the tree that represents the rewriting process. For each node in the tree that has more than one child, the number of chil-dren is bounded by d, and descending to such a child always adds a distinct literal to the corresponding conjunction. There are therefore at most dc con-junctions with exactly c distinct literals. It remains to ﬁnd out the number of positive test graphs accepted by each such conjunction. This is easily done if the edges in the positive test graphs are present in-dependently from each other (as is the case in the lower bound for Andreev’s polynomial problem in Section 4). If the problems are speciﬁed in terms of vertices (e.g., the BMS problem in Section 5 and Clique in Section 6), we have to be a bit more careful since edges are not independent in the natural test graphs for these problems. Actually, in the proofs for these speciﬁc problems, we do not limit the number of distinct literals in conjunctions, but introduce another more natural measurement of their size. The lemma should now follow from multiplying the number of conjunctions that have c distinct literals by the number of inputs accepted by each such conjunction. □ The following lemma is analogous, and it is proved by bounding the number of errors introduced on negative test graphs when f C e is formed from f D. Generic Lemma C. At an ∨gate e, the approximator f C e introduces an error for at most β(n, c, d) negative test graphs. To ﬁnish the proof of the generic theorem, we need to consider the two cases cc 8 (1999) Symmetric Approximation Arguments 9 from Generic Lemma A. First, if the approximators fail for all negative test graphs, Generic Lemma C implies that size(Ψ) ≥γ0(n)/β(n, c, d). Otherwise, the approximators fail for at least half of the positive test graphs, and then Generic Lemma B implies that size(Ψ) ≥γ1(n)/(2α(n, c, d)). So if h(n) is less than both these values, it bounds MGP(n) as claimed in the Generic Theorem. □ 4. Andreev’s polynomial problem The best lower bound for the size of monotone circuits computing any monotone function was proved by Alon and Boppana (1987). The function for which this bound was obtained was the same that Andreev (1985) used when he was the ﬁrst to prove an exponential lower bound for a monotone problem. We are given a bipartite graph G = (U, V, E) with vertex sets U = GF[q] and V = GF[q], where q is a prime power. The problem POLY(q, s) is to decide whether there exists a polynomial p over GF[q] of degree at most s −1, such that ∀i ∈U: (i, p(i)) ∈E. Theorem 3 (Alon and Boppana (1987)). The monotone circuit com-plexity for POLY(q, s) is qΩ(s) when s ≤1 2 p q/ ln q. Proof. We ﬁrst deﬁne the set of positive test graphs used in the proof. There are qs diﬀerent polynomials of degree at most s −1, each of which corresponds to a positive test graph in a natural way: the polynomial p deﬁnes a test graph with the edges E = {(i, p(i)) \| i ∈U}. The negative test graphs are constructed randomly, by having each possible edge appear with probability 1 −ε for ε = (2s ln q)/q. Notice that this con-struction may result in all possible bipartite graphs on 2q vertices; therefore we need the following lemma. Lemma 4. The probability that a negative test graph is in POLY(q, s) is at most q−s. Proof. The probability that the q speciﬁc edges that correspond to a certain polynomial are present in a randomly chosen negative test graph is (1−ε)q. So the probability that the edges corresponding to at least one of the qs possible polynomials are present in a random graph is at most qs(1 −ε)q ≤qse−εq = qsq−2s = q−s. □ We choose the parameters c = s and d = q2/3/2. 10 Berg & Ulfberg cc 8 (1999) Lemma 5 (Specialization of Generic Lemma A). At the output gate eo, we either have that the approximator f C eo is identically 1, or that the ap-proximators for eo fail for at least half of the positive test graphs. Proof. If f C eo is identically 1, we are in the ﬁrst case. Otherwise, there is a graph b such that f C eo(b) = 0, and hence there is a disjunction D in f C eo. There are qs−1 positive test graphs that contain every speciﬁc edge, since ﬁxing an edge is equivalent to ﬁxing the value of a polynomial at one point. Thus, there are at most dqs−1 = qs−1/3/2 positive test graphs that contain at least one edge from D. Hence D (and therefore also f C eo) is 1 for at most a fraction qs−1/3/(2qs) = q−1/3/2 of the positive test graphs. □ Lemma 6 (Specialization of Generic Lemma B). At an ∧gate e, f D in-troduces an error for at most dc positive test graphs. Proof. When rewriting f C e to f D, we can form at most dc conjunctions that contain c distinct literals, and for the current function, each removed conjunction introduces an error on at most one positive test graph. This follows since c = s and since a polynomial of degree s −1 is completely speciﬁed by s function points. □ Lemma 7 (Specialization of Generic Lemma C). At an ∨gate e, the probability that f C e introduces an error for a random negative test graph is at most (cε)d. Proof. We bound the probability of an error being introduced for a random negative test graph by multiplying the number of disjunctions with d distinct literals, cd, by the probability that the d literals are 0 in one disjunction, which is εd. □ Notice that, in particular, the upper bound in the lemma holds for the number of errors introduced for negative test graphs that are not in the lan-guage POLY(q, s). To prove the theorem, we have to consider the two cases from Lemma 5. If the approximators fail for at least half of the positive test graphs, we use Lemma 6 which states that the number of errors introduced in a single ∧gate is at most dc, and to get size(Ψ) ≥qs 2dc = 2sqs/3 2 ∈qΩ(s). cc 8 (1999) Symmetric Approximation Arguments 11 If, on the other hand, the approximator f C eo is identically 1, the approximat-ors for eo fail for a random negative test graph with probability 1 −q−s ≥1/2 by Lemma 4. The probability of introducing an error in a single ∨gate is at most (cε)d; thus size(Ψ) ≥ 1 2(cε)d = 1 2 q 2s2 ln q q2/3/2 = 1 22q2/3/2 ∈qΩ(s). □ 5. Broken Mosquito Screens The Broken Mosquito Screens (BMS) problem was introduced by Haken (1995) to illustrate how to count bottlenecks to show that monotone P ̸= NP. In this section we show that the same result can be obtained using our approximator formalism. Haken deﬁnes the BMS problem for graphs with m2−2 vertices as the prob-lem of distinguishing good and bad graphs from each other. Graphs containing m −1 cliques of size m and one clique of size m −2 (but no other edges) are good; graphs containing m−1 independent sets of size m and one independent set of size m −2 but all other edges are bad. Hence, by taking the dual of a good graph (i.e., inverting all the input bits), we get a bad graph. Notice that not all graphs are either good or bad, but the deﬁnition can be extended to include all graphs. Haken shows that a monotone circuit that distinguishes good graphs from bad must be large. We use the following extended deﬁnition of the BMS problem. Definition 8. Instances of BMS(m) are graphs with m2 −2 vertices (m > 2). A graph is in the language BMS(m) if there exists a partition of the vertices into m −1 sets of size m and one of size m −2, so that each of these subsets forms a clique. Using this deﬁnition, it is easy to see that the BMS problem is monotone and in NP. Theorem 9 (Haken (1995)). The monotone circuit complexity for BMS(m) is 2Ω(√m). Proof. We let the set of positive test graphs G for the BMS problem be all graphs with m2 −2 vertices that contain m −1 cliques of size m and one 12 Berg & Ulfberg cc 8 (1999) clique of size m −2, but no other edges. The set of negative test graphs B is all graphs on m2 −2 vertices that contain m−1 independent sets of size m and one independent set of size m −2, and where all other edges are present. Our positive and negative test graphs correspond to Haken’s good and bad graphs respectively. Clearly, all positive test graphs are in the BMS language. Negative test graphs are not in the BMS language since they contain only m −2 cliques of size m. Next we count the number of test graphs. Counting the number of positive test graphs is the same as counting the number of ways that a set of m2 −2 elements can be partitioned into m −1 sets containing m elements and one set containing m −2 elements. We get \|G\| = (m2 −2)! (m!)(m−1)(m −2)!(m −1)! , (1) and by duality \|B\| = \|G\|. For each conjunction and disjunction, let a graph correspond to it in the natural way: for every literal, include the corresponding edge in the graph. Suppose we want to determine whether a positive test graph satisﬁes a con-junction C. This is the case when the graph corresponding to C is a subgraph of the positive test graph. Therefore, we only have to know how the vertices of the graph corresponding to C are divided into connected components to determine what positive test graphs are accepted by C. In this section we do not limit the length of conjunctions and disjunctions by their number of literals, but instead we deﬁne their size by m2 −2 minus the number of connected components in their corresponding graphs, and require that their size be less than c and d respectively. We choose c = d = ⌊√m⌋(since c = d we only use c), i.e., the graphs corresponding to the conjunctions and disjunctions in the approximators have more than m2 −2 −⌊√m⌋connected components. Note that, implicitly, the number of literals in conjunctions and disjunctions is bounded by c2/2 ≤m/2. Lemma 10 (Specialization of Generic Lemma A). At the output gate eo, the approximators either fail for all negative test graphs, or they fail for at least half of the positive test graphs. Proof. If the approximator f C eo for the output gate fails for all the graphs in B, we are in the ﬁrst case. Otherwise, let b ∈B be a graph for which the approximator does not fail at eo. cc 8 (1999) Symmetric Approximation Arguments 13 Since f C eo(b) = 0, there is a disjunction D in f C eo, and as noted above, this disjunction contains less than c2/2 ≤m/2 literals. The fraction of graphs in G that contain a speciﬁc one of these literals is less than 1/m, which can be seen as follows. A graph in G has (m−1) m 2 + m−2 2 edges out of the possible m2−2 2 ones. Since all edges are equally likely, this means that a speciﬁc edge appears in a fraction (m −1) m 2 + m−2 2 m2−2 2 < 1 m of the graphs in G. Thus, the fraction of G that contains at least one of the m/2 literals in D is at most 1/2. So, for at least half of all g ∈G, we have that D(g) = 0, and therefore also that f C eo(g) = 0. □ Next, we establish an upper bound for the number of test graphs for which the approximators introduce an error at a single gate. Lemma 11 (Specialization of Generic Lemma B). At an ∧gate e, the approximator f D introduces an error for at most m4c(m2 −2 −2c)! 2c(m!)(m−1)(m −2)!(m −1)! (2) positive test graphs. Proof. We introduce errors on positive test graphs by removing conjunctions whose corresponding graphs contain at most m2 −2−c connected components. When counting the number of positive test graphs for which an error is intro-duced, we consider diﬀerent orderings of the vertices within the partitions and the order of the partitions as diﬀerent graphs, and in the end we divide by the same denominator as in (1). When building the tree, let w be the node that was created while expanding the disjunction Di, and let C be the conjunction corresponding to w. Consider the case in which Di+1 contains a literal xu,v for which both end-points are in the same connected component in the graph corresponding to C. Then, we create only one single child under w and label the edge xu,v. In this case, we know that xu,v is satisﬁed if C is, so dropping all other children does not introduce errors for any positive test graphs. Clearly, ignoring the creation of some children never introduces errors for negative test graphs. Otherwise, w gets less than c2/2 children, since each disjunction contains less than c2/2 literals. In this case, the number of connected components 14 Berg & Ulfberg cc 8 (1999) decreases by one for the children, so we will make c such expansions before we reach a node where the graph for the corresponding conjunction contains m2−2−c connected components. Therefore, there are at most (c2/2)c ≤mc/2c such conjunctions. We now count the number of inputs accepted by a single conjunction whose corresponding graph H contains m2 −2 −c connected components. Let a denote the number of vertices in H that are part of connected components with more than one vertex. It follows that the number of isolated vertices in H is m2 −2 −a, and that the number of connected components in H with more than one vertex is a −c. To get an upper bound for the number of positive test graphs that are compatible with H, we count as follows. Each connected component in H with more than one vertex must go into one of the m partitions, which makes at most ma−c choices. Then, each vertex in those connected components can be placed in at most m ways each within the partition. Lastly, the remaining m2 −2 −a vertices can be placed freely in (m2 −2 −a)! ways. To sum up, the total number of choices is at most ma−cma(m2 −2 −a)! = m2a−c(m2 −2 −a)!, which is increasing with a. Since the graph H contains a −c connected com-ponents with more than one vertex, the total number a of vertices in such components is at most 2c. This follows since the number of connected com-ponents with more than one vertex is at most c. So an upper bound for the number of choices is m3c(m2 −2 −2c)!. Multiplying the maximum number of conjunctions whose corresponding graphs contain m2 −2 −c connected components by the maximum number of inputs accepted by each, and ﬁnally, dividing by the same denominator as in (1), yields the lemma. □ The proof of the lemma above diﬀers slightly from the corresponding proof by Haken. The reason for this is that we make a construction of the approxim-ator for an ∧gate from the approximators for the input gates, whereas Haken only proves the existence of a set of short conjunctions that describe the gate well enough. Because of the duality of the test graphs and since c = d, the following lemma can be proved analogously to the previous one. cc 8 (1999) Symmetric Approximation Arguments 15 Lemma 12 (Specialization of Generic Lemma C). At an ∨gate e, the number negative test graphs for which the approximator f C e introduces an error is bounded by (2). We now consider the two cases from Lemma 10. Either the approximators for the output gate eo fail for at least half the positive test graphs, or they fail for all negative test graphs. Since \|G\| = \|B\|, and since we have the bound (2) for the maximum number of errors introduced in a single gate, the minimum size of Ψ is 2c(m2 −2)! 2m4c(m2 −2 −2c)!. If we expand this expression, we get size(Ψ) ≥ 2c(m2 −2) · · · (m2 −2 −2c + 1) 2m4c ≥ 2c 2 (m2 −2 −2c + 1)2c (m2)2c = 2⌊√m⌋ 2 m2 −2⌊√m⌋−1 m2 2⌊√m⌋ ∈ 2Ω(√m). □ 6. Clique In this section we prove that deciding whether a graph on n vertices contains any complete subgraph of size m (the problem CLIQUE(m, n)) requires mono-tone circuits of exponential size. Theorem 13 (Alon and Boppana (1987)). The monotone circuit complex-ity for CLIQUE(m, n) is 2Ω(√m) when m ≤n2/3. Proof. Let the positive test graphs be all possible graphs on n vertices with a clique of size m and no other edges. This makes n m positive test graphs. We make one negative test graph for each possible coloring of the n vertices, using m−1 colors by connecting vertices of diﬀerent colors by edges. Note that some colorings result in the same graph, but we treat them as diﬀerent for counting purposes and get (m −1)n negative test graphs. As in the proof of the BMS problem, we deﬁne the size of disjunctions by n minus the number of connected components in their corresponding graphs, and require that their size be less than d. 16 Berg & Ulfberg cc 8 (1999) For the conjunctions, however, we introduce a new notation of size that counts the number of vertices they touch. The number of vertices that a con-junction touches is the number of diﬀerent vertices to which any of the edges connect. A conjunction is required to touch fewer than c vertices. Choose c = ⌊√m⌋, so that conjunctions touch less than ⌊√m⌋vertices. Implicitly, the number of literals in conjunctions is bounded by c2/2 ≤m/2. Let d = ⌊n/(8m)⌋so that the graphs corresponding to disjunctions have more than n −⌊n/(8m)⌋connected components; this implies that they touch less than 2d ≤n/(4m) vertices. Lemma 14 (Specialization of Generic Lemma A). At the output gate eo, the approximators either fail for all negative test graphs, or they fail for at least half of the positive test graphs. Proof. Assume that there is a negative test graph b such that f C eo(b) = 0 (otherwise we are already done). Then there is a disjunction D in f C eo, and we show that this disjunction can be 1 for at most half of the positive test graphs. A necessary condition for D to be 1 on a positive test graph g is that one of the vertices it touches is in the clique of g. Since any given vertex is part of the clique in a fraction m/n of the positive test graphs, a collection of at most n/(8m) vertices has a vertex in common with less than half the positive test graphs. □ Lemma 15 (Specialization of Generic Lemma B). At an ∧gate e, the approximator f D introduces an error for at most n −c m −c n 2m c positive test graphs. Proof. When building the tree, let w be the node that was created while expanding the disjunction Di, and let C be the conjunction corresponding to w. Consider the case in which Di+1 contains a literal xu,v for which both end-points are already touched by C. Then a positive test graph that satisﬁes C also satisﬁes Di+1, so we only create one single child under w containing xu,v. Otherwise, Di+1 contains a mix of literals, for some, none of the two endpoints are touched by C, and for some, one endpoint is touched by C. First, consider the literals that have one endpoint that is already touched by C; they all have another endpoint that is not touched by C. For each such cc 8 (1999) Symmetric Approximation Arguments 17 endpoint, arbitrarily select one of the literals that connects to the endpoint, and form one child for it. The reason that we may disregard some literals from consideration is that two conjunctions that touch the same vertices will accept the same positive test graphs. The total number of children created this way is less than 2d, since Di+1 touches at most 2d vertices. For the literals that connect to two vertices that are not touched by C, we ﬁrst select one of their endpoints in some arbitrary way and create a child for it; however, we do not label the edges to these children. At most 2d such children are created. Then, we create less than 2d children under each of these children for the other endpoint, and the edges down to these children are labeled by literals. The way we have constructed the tree implies that no node has more than 4d = n/(2m) children, and when descending to a child from a node with more than one child, the number of vertices touched by the corresponding conjunc-tion increases by 1. Therefore, there are at most (n/(2m))c conjunctions touch-ing c vertices. The number of positive test graphs accepted by a single conjunction touch-ing c vertices is the number of ways to choose the remaining m −c vertices for the clique out of the total remaining n −c vertices, which is n−c m−c . Finally, the lemma follows from multiplying the bound for the number of conjunctions touching c vertices by the maximum number of inputs they may accept. □ Lemma 16 (Specialization of Generic Lemma C). At an ∨gate e, the approximator f C e introduces an error at most (m/2)d(m −1)n−d negative test graphs. Proof. We start by bounding the number of disjunctions corresponding to nodes in the tree, whose corresponding graphs contain n −d connected components. When building the tree, let w be the node that was created while expanding the conjunction Ci, and let D be the disjunction corresponding to w. Consider the case that Ci+1 contains a literal xu,v for which both endpoints are in the same connected component in the graph corresponding to D. Then a negative test graph that falsiﬁes D also falsiﬁes Ci+1, since u and v must have the same color. It is therefore enough to create only one single child under w and label the edge xu,v. Otherwise, w gets at most m/2 children, since each conjunction contains at most m/2 literals. In this case, the number of connected components decreases 18 Berg & Ulfberg cc 8 (1999) by one for the children, so we will make d such expansions before we reach a node for which the graph for the corresponding disjunction contains n − d connected components. In all, there can hence be at most (m/2)d such disjunctions. It remains to count the number of negative test graphs rejected by a single disjunction, whose corresponding graph H contains n −d connected compon-ents. The negative test graphs that are rejected by such a disjunction are those that use only one color within each connected component, and their number is (m −1)n−d. Multiplying the number of disjunctions whose corresponding graphs contain n−d connected components by the number of negative test graphs rejected by each yields the bound in the lemma. □ We now determine the lower bound for the circuit size. There are two cases to consider: either f D eo fails on at least half the positive test graphs, so that size(Ψ) ≥ n m 2 n−c m−c n 2m c = 1 2 n! (m −c)! (n −c)! m! (2m)c nc ≥ 1 22c(n −c)c mc mc nc ∈ 2Ω(√m), or the approximators fail for all negative test graphs, so that size(Ψ) ≥ (m −1)n (m −1)n−d(m/2)d = 2d m −1 m d ∈ 2Ω(n/m). Since 2Ω(√m) ⊆2Ω(n/m) for m ≤n2/3, the theorem follows. □ Acknowledgment We thank Johan H˚ astad for helpful discussions and comments on earlier ver-sions of this paper. cc 8 (1999) Symmetric Approximation Arguments 19 References Noga Alon and Ravi B. Boppana, The monotone circuit complexity of boolean functions. Combinatorica 7 (1987), 1–22. Kazuyuki Amano and Akira Maruoka, Potential of the approximation method. In Proc. 37th Ann. IEEE Symp. Found. Comput. Sci., 1996, 431–440. Alexander E. Andreev, On a method for obtaining lower bounds for the com-plexity of individual monotone functions. Sov. Math. Dokl. 31 (1985), 530–534. Christer Berg, On Oracles and Circuits—Topics in Computational Complexity. PhD thesis, Royal Institute of Technology, KTH, Stockholm, Sweden, 1997. Ravi B. Boppana and Michael Sipser, The complexity of ﬁnite functions. In Handbook of theoretical computer science, ed. Jan van Leeuwen, vol. A, Al-gorithms and complexity, 757–804. Elsevier/MIT Press, 1990. Paul Erd˝ os and Richard Rado, Intersection theorems for systems of sets. J. London Math. Soc. 35 (1960), 85–90. Armin Haken, Counting bottlenecks to show monotone P ̸= NP. In Proc. 36th Ann. IEEE Symp. Found. Comput. Sci., 1995, 36–40. Armin Haken and Stephen Cook, An exponential lower bound for the size of monotone real circuits. Submitted to J. Comput. System Sci., 1996. Stasys Jukna, Finite limits and monotone computations over the reals. In Twelfth Annual IEEE Conference on Computational Complexity, 1997. Pavel Pudl´ ak, Lower bounds for resolution and cutting planes proofs and mono-tone computations. J. Symbolic Logic, to appear (1997). Alexander A. Razborov, Lower bounds on the monotone complexity of some boolean functions. Sov. Math. Dokl. 31 (1985), 354–357. Alexander A. Razborov, On the method of approximations. In Proc. Twenty-ﬁrst Ann. ACM Symp. Theor. Comput., 1989, 167–176. Janos Simon and Shi-Chun Tsai, A note on the bottleneck counting argument. In Twelfth Annual IEEE Conference on Computational Complexity, 1997. J¨ urgen Tiekenheinrich, A 4n lower bound on the monotone network complexity of a one-output boolean function. Inform. Process. Lett. 18 (1984), 201–202. 20 Berg & Ulfberg cc 8 (1999) Manuscript received 16 July 1996 Christer Berg Royal Institute of Technology S-100 44 Stockholm, SWEDEN berg@nada.kth.se Staffan Ulfberg Royal Institute of Technology S-100 44 Stockholm, SWEDEN staffanu@nada.kth.se
2009	https://radiopaedia.org/articles/kidneys?lang=us	Kidneys Last revised by Tariq Walizai on 11 Oct 2024 Edit article Report problem with article View revision history Citation, DOI, disclosures and article data Citation: Knipe H, Walizai T, Southi J, et al. Kidneys. Reference article, Radiopaedia.org (Accessed on 28 Aug 2025) DOI: Permalink: rID: 25813 Article created: 10 Nov 2013, Henry Knipe Disclosures: At the time the article was created Henry Knipe had no recorded disclosures. View Henry Knipe's current disclosures Last revised: 11 Oct 2024, Tariq Walizai Disclosures: At the time the article was last revised Tariq Walizai had no financial relationships to ineligible companies to disclose. View Tariq Walizai's current disclosures Revisions: 65 times, by 31 contributors - see full revision history and disclosures Systems: Urogenital Sections: Anatomy Tags: renal Synonyms: Kidney Renal anatomy Kidney anatomy kidneys The kidneys are paired retroperitoneal organs that lie at the level of the T12 to L3 vertebral bodies. On this page: Article: Gross anatomy Arterial supply Venous drainage Innervation Variant anatomy Radiographic features Development Related pathology Related articles References Images: Cases and figures Gross anatomy Location The kidneys are located to either side of the vertebral column in the perirenal space of the retroperitoneum, within the posterior abdominal wall. The long axis of the kidney is parallel to the lateral border of the psoas muscle and it lies anterior to the quadratus lumborum muscle. Being parallel to the psoas muscle, the kidneys lie at an oblique angle, with its superior pole more medial and posterior than its inferior pole. Due to the right lobe of the liver, the right kidney usually lies slightly lower than the left kidney 16. Size In adults, the normal kidney is 10-14 cm long in males and 9-13 cm long in females, 3-5 cm wide, 3 cm in antero-posterior thickness and weighs 150-260 g. The left kidney is usually slightly larger than the right. normal kidney size in adults normal kidney size in children Structure The kidney is bean-shaped with a superior and an inferior pole, anterior and posterior surfaces, and lateral and medial borders. The midportion of the kidney is often called the midpole. The kidney has a fibrous capsule, which is surrounded by perirenal fat. The kidney itself can be divided into renal parenchyma, consisting of renal cortex and medulla, and the renal sinus containing renal pelvis, calyces, renal vessels, nerves, lymphatics and perirenal fat. The renal parenchyma has two layers: cortex and medulla. The renal cortex lies peripherally under the capsule while the renal medulla consists of 10-14 renal pyramids, which are separated from each other by an inward extension of the renal cortex called renal columns. Urine is produced in the renal lobes, which consists of the renal pyramid with the associated overlying renal cortex and adjacent renal columns. Each renal lobe drains at a papilla into a minor calyx, four or five of these unite to form a major calyx. Each kidney normally has two or three major calyces, which unite to form the renal pelvis. The renal hilum is the entry to the renal sinus and lies vertically at the anteromedial aspect of the kidney. It contains the renal vessels and nerves, fat and the renal pelvis, which typically emerges posterior to the renal vessels, with the renal vein being anterior to the renal artery. Function filter the blood to remove excess water, minerals, and waste products of protein metabolism, producing urine in the process are involved in blood pressure regulation regulation of body fluid volume, osmolality and pH vitamin D and red blood cell (RBC) production Tests of renal function include: estimated glomerular filtration rate (eGFR) electrolytes blood urea nitrogen (BUN) or urea creatinine levels and creatinine clearance cystatin C levels Relations posterior diaphragm and associated parietal pleura (upper pole of kidney) 11th rib (left kidney only), 12th rib (both kidneys) medial and lateral arcuate ligaments psoas major, quadratus lumborum, transversus abdominis (medial to lateral) subcostal nerve, iliohypogastric nerve anterior right kidney right lobe of the liver hepatorenal recess (Morison's pouch) D2 duodenum (medial margin of the kidney) right colic flexure (just above inferior pole) left kidney stomach omental bursa/lesser sac (upper pole) spleen (lateral margin) tail of the pancreas (hilum) left colic flexure (lateral to lower pole) superior adrenal glands Arterial supply renal arteries (from the abdominal aorta) Venous drainage renal veins (to inferior vena cava) Innervation sympathetic innervation: renal and aorticorenal plexuses, which is derived from lesser and least splanchnic nerves parasympathetic innervation: vagus nerve via the celiac, renal and aorticorenal plexuses Variant anatomy See article: developmental anomalies of the kidney and ureter. Radiographic features Plain radiograph Kidney length should not be less than three vertebral body lengths, and no more than four vertebral body lengths 10. Ultrasound Antenatally, fetal kidneys show varying texture depending on gestational age. It is echogenic in the first trimester, with decreasing echogenicity as the pregnancy progresses. Corticomedullary differentiation can be appreciated after 15 weeks of gestation but clear demarcation between cortex and medulla can be seen at 20 weeks. Renal echogenicity decreases compared to liver and spleen after 17 weeks12. Normal kidney appearance in adult 11: cortex is less echogenic than the liver medullary pyramids are slightly less echogenic than the cortex cortex thickness equals/is more than 6 mm 14 if the pyramids are difficult to differentiate, the parenchymal thickness can be measured instead and should be 15-20 mm 11 central renal sinus, consisting of the calyces, renal pelvis and fat, is more echogenic than the cortex renal pelvis may appear as a central slit of anechoic fluid at the hilum normal ureters are generally not well seen on ultrasound CT On unenhanced CT the renal pyramids can appear hyperdense 7 (see: white pyramid sign). MRI Renal cortex has slightly higher signal than medulla on T1-weighted images. On T2-weighted images, medulla has slightly higher signal than renal cortex 16. Development The collecting system arises from the ureteric bud, which arises from the mesonephric duct in the fourth week of gestation. The renal parenchyma arises from the metanephros, which appears in the fifth week, a derivative of the intermediate mesoderm 15. The ureteric bud penetrates the metanephric mesoderm, which forms as a cup-shaped tissue cap. The ureteric bud dilates and subdivides to form twelve or so generations of tubules with the first generations fusing to form the renal pelvis, major and minor calyces, and renal pyramids with the later generations forming approximately a million renal tubules 15. Under the regulation of complex signaling pathways, the ureteric bud incites the metanephric tissue to form small renal vesicles that eventually form primitive S-nephrons which are invaginated by endothelial cells from nearby angioblasts; before going on to form the definitive nephron 15. Glomerular filtration begins at 9th week of gestation. Nephron development is complete at birth except in premature infants. The kidney matures functionally after birth 15. The kidney develops in the pelvis but assumes its normal cranial abdominal location in adults due to disproportionate growth of the body in the lumbar and sacral regions 8-9. Related pathology A discrepancy of >2 cm between renal lengths should be considered abnormal 10 and may indicate an underlying disease. Common diseases affecting the kidneys include: gamuts delayed nephrogram persistent nephrogram striated nephrogram spotted nephrogram renal emphysema neoplastic benign renal neoplasm malignant renal cell carcinoma metabolic renal hemosiderosis renal cortical nephrocalcinosis infections renal abscess renal antibioma perinephric abscess pyelonephritis xanthogranulomatous pyelonephritis (XGP) emphysematous pyelonephritis others renal pseudotumor renal replacement lipomatosis renal stones hydronephrosis Quiz questions Question 2401 Report problem with question Pyonephrosis usually requires antibiotics and what kind of follow-up? Skip question References Keith L. Moore, Arthur F. Dalley, A. M. R. Agur. Clinically Oriented Anatomy. (2013) ISBN: 9781451119459 - Google Books Drake R, Vogl AW, Mitchell AWM. Gray's Basic Anatomy: with STUDENT CONSULT Online Access (Grays Anatomy for Students). Churchill Livingstone. ISBN:B0092DEQ1U. Read it at Google Books - Find it at Amazon McMINN. Lasts Anatomy Regional and Applied. CHURCHILL LIVINGSTONE. (2003) ISBN:B0084AQDG8. Read it at Google Books - Find it at Amazon Emilio Quaia. Radiological Imaging of the Kidney. (2010) ISBN: 9783540875963 - Google Books Frederic H. Martini, Judi L. Nath, Edwin F. Bartholomew. Fundamentals of Anatomy and Physiology. (2011) ISBN: 9780321719799 - Google Books Ronald W. Dudek. High-Yield Kidney. (2006) ISBN: 9780781755696 - Google Books Arjun Kalyanpur. Step by Step in Emergency Radiology: Helical CT in Acute Abdomen. (2006) ISBN: 9788180618031 - Google Books Dan Longo, Anthony Fauci, Dennis Kasper et al. Harrison's Principles of Internal Medicine, 18th Edition. (2011) ISBN: 9780071748896 - Google Books Thomas W. Sadler. Langman's Medical Embryology. (2011) ISBN: 9781451113426 - Google Books Ronald J. Zagoria. Genitourinary Radiology. (2004) ISBN: 9780323018425 - Google Books Marchal G, Verbeken E, Oyen R, Moerman F, Baert AL, Lauweryns J; Ultrasound of the normal kidney: a sonographic, anatomic and histologic correlation. Ultrasound Med Biol. 1986 Dec;12(12):999-1009. PMID: 3547990 Hindryckx A & De Catte L. Prenatal Diagnosis of Congenital Renal and Urinary Tract Malformations. Facts Views Vis Obgyn. 2011;3(3):165-74. PMC3991456 - Pubmed Christopher J. Lote. Principles of Renal Physiology. (2012) ISBN: 9781461437857 - Google Books El-Reshaid W & Abdul-Fattah H. Sonographic Assessment of Renal Size in Healthy Adults. Med Princ Pract. 2014;23(5):432-6. doi:10.1159/000364876 - Pubmed Zweyer M. Embryology of the Kidney. Med Radiol. 2010;:3-15. doi:10.1007/978-3-540-87597-0_1 Stephanie Ryan, Michelle McNicholas, Stephen J. Eustace. Anatomy for Diagnostic Imaging. (2011) ISBN: 9780702029714 - Google Books Incoming Links Articles: Solid and hollow abdominal viscera Urogenital curriculum Pancreas transplant Nephroptosis Kidney sweat sign Renal emphysema Organomegaly Late arterial phase Persistent fetal lobulation of the kidneys Normal radiological reference values Extramedullary haematopoiesis Renal pelvis Transversalis fascia Retrorenal spleen Splenic vein Subcostal artery Urinary system Left colic artery Renal vascular pedicle injury Ascending colon Load more articles Cases: Normal CT KUB Horseshoe kidney with lithiasis Perirenal space (Gray's illustrations) Renal vessels (Gray's illustrations) Acquired cystic kidney disease Normal catheter renal angiogram Focused assessment with sonography for trauma (negative eFAST) Hair on end sign Adult polycystic kidney disease Ureterolithiasis Page kidney Normal upper abdominal ultrasound Ureterolithiasis Renal cell carcinoma Ureterolithiasis Renal cell carcinoma Ureterolithiasis Gastric antropyloric adenocarcinoma Ureterolithiasis Renal sinus lymphangioma Load more cases Multiple choice questions: Question 2401 Question 2400 Question 2399 Question 49 Related articles: Anatomy: Abdominopelvic Anatomy: Abdominopelvic skeleton of the abdomen and pelvis[+][+] lumbar spine bony pelvis innominate bones pubis pubic symphysis obturator foramen ischium greater sciatic notch lesser sciatic notch ilium sacrum sacral hiatus sacrotuberous ligament sacrospinous ligament greater sciatic foramen lesser sciatic foramen coccyx anterior angulation of the coccyx muscles of the abdomen and pelvis[+][+] diaphragm sternocostal triangle anterior abdominal wall Scarpa's fascia muscles external oblique muscle inguinal ligament lacunar ligament internal oblique muscle conjoint tendon transversus abdominis muscle rectus abdominis muscle rectus sheath linea alba umbilicus arcuate line semilunar line transversalis fascia pyramidalis muscle surface anatomy posterior abdominal wall psoas major muscle psoas minor muscle quadratus lumborum muscle iliacus muscle pelvic floor levator ani muscle coccygeus muscle piriformis muscle obturator internus muscle spaces of the abdomen and pelvis[+][+] anterior abdominal wall sternocostal triangle posterior abdominal wall superior lumbar triangle inferior lumbar triangle iliopsoas compartment abdominal cavity peritoneum peritoneal ligaments left triangular ligament right triangular ligament falciform ligament umbilical vein ligamentum teres veins of Sappey omentum hepatogastric ligament gastrosplenic ligament splenorenal ligament mesentery small bowel mesentery root of the mesentery mesoappendix transverse mesocolon sigmoid mesocolon peritoneal spaces supramesocolic space right supramesocolic space right subphrenic space right subhepatic space anterior right subhepatic space posterior right subhepatic space (Morison pouch) lesser sac epiploic foramen (of Winslow) left supramesocolic space (left perihepatic space) left subhepatic space anterior left subhepatic space posterior left subhepatic space left subphrenic space anterior left subphrenic space posterior left subphrenic (perisplenic) space inframesocolic space right inframesocolic space left inframesocolic space right paracolic gutter left paracolic gutter inguinal canal (mnemonic) Hesselbach triangle umbilical folds median umbilical fold medial umbilical folds lateral umbilical folds retroperitoneum perirenal fascia anterior pararenal space perirenal space perinephric bridging septa (of Kunin) posterior pararenal space properitoneal fat great vessel space lateroconal fascia pelvic cavity pelvic peritoneal space rectouterine pouch (pouch of Douglas) rectovesical pouch retropubic space (of Retzius) paravesical space lateral fossa supravesical fossa presacral space canal of Nuck pudendal canal obturator canal perineum urogenital triangle superficial perineal pouch perineal membrane deep perineal pouch anal triangle transverse perineal muscles deep transverse perineal muscles superficial transverse perineal muscles ischioanal fossa urogenital diaphragm abdominal and pelvic viscera gastrointestinal tract[+][+] esophagus phrenic ampulla gastro-esophageal junction Z line stomach rugal folds small intestine duodenum Brunner gland duodenojejunal flexure jejunum ileum Meckel diverticulum ileocecal valve jejunum vs ileum valvulae conniventes large intestine cecum appendix duplex appendix colon ascending colon right colic flexure transverse colon left colic flexure descending colon loop-to-loop colon sigmoid colon rectum mesorectal fascia appendix epiploica haustral folds taeniae coli anal canal anal sphincters spleen[+][+] variants splenunculus wandering spleen asplenia polysplenia splenogonadal fusion retrorenal spleen hepatobiliary system[+][+] liver beaver tail liver ductus venosus ligamentum venosum porta hepatis Riedel lobe segmental liver anatomy (mnemonic) supradiaphragmatic liver biliary tree common hepatic duct cystic duct common bile duct ampulla of Vater sphincter of Oddi subvesical bile ducts gallbladder accessory gallbladder gallbladder duplication gallbladder triplication gallbladder agenesis Rokitansky-Aschoff sinuses Phrygian cap septate gallbladder endocrine system[+][+] adrenal gland adrenal vessels adrenal arteries adrenal veins chromaffin cells variants horseshoe adrenal gland pancreas pancreatic ducts pancreatic duct diameter variants pancreas divisum meandering main pancreatic duct ansa pancreatica anomalous pancreaticobiliary junction variants ectopic pancreatic tissue annular pancreas organs of Zuckerkandl urinary system kidney renal pelvis extrarenal pelvis renal sinus avascular plane of Brodel variants number[+][+] renal agenesis supernumerary kidney fusion[+][+] horseshoe kidney sigmoid kidney crossed fused renal ectopia junctional parenchymal defect pancake kidney location[+][+] ectopic kidney pelvic kidney fused pelvic kidney intrathoracic kidney crossed renal ectopia abnormal renal rotation nephroptosis shape[+][+] persistent fetal lobulation hypertrophied column of Bertin renal hilar lip dromedary hump faceless kidney ureter[+][+] variants ectopic ureter duplex collecting system Weigert-Meyer law bifid ureter ureteral duplication retrocaval ureter ureterocele urinary bladder[+][+] detrusor bladder neuroanatomy (micturition) urethra[+][+] fossa navicularis male urethra verumontanum prostatic utricle musculus compressor nuda female urethra paraurethral glands paraurethral ducts embryology[+][+] metanephric blastema ureteric bud Wolffian duct male reproductive system[+][+] prostate seminal vesicles ejaculatory duct bulbourethral glands urethral glands of Littré spermatic cord (mnemonic) ductus deferens cremaster muscle scrotum (mnemonic) dartos muscle epididymis testes descent tunica vaginalis tunica albuginea testicular appendages appendix of the testis appendix of the epididymis polyorchidism bilobed testis penis female reproductive system[+][+] vulva mons pubis labia majora labia minora clitoris bulbs of the vestibule vestibule of the vulva vaginal opening hymen Bartholin glands vagina fornix uterus cervix parametrium chorion amnion endometrium placenta basal plate chorionic plate variation in morphology retroplacental complex umbilical cord Wharton jelly adnexa Fallopian tubes ovaries ovarian follicle dominant ovarian follicle primary follicle secondary follicle mature vesicular follicle corpus luteum corpus albicans cumulus oophorus broad ligament (mnemonic) variant anatomy retroverted uterus congenital uterovaginal anomalies transverse vaginal septum Mullerian duct anomalies uterine duplication anomalies uterus didelphys bicornuate uterus septate uterus uterine agenesis unicornuate uterus arcuate uterus T-shaped uterus embryology Mullerian duct blood supply of the abdomen and pelvis[+][+] arteries abdominal aorta inferior phrenic artery superior suprarenal artery celiac artery common hepatic artery (variant anatomy) hepatic artery proper left hepatic artery middle hepatic artery falciform artery right hepatic artery cystic artery gastroduodenal artery supraduodenal artery right gastroepiploic artery superior pancreaticoduodenal artery right gastric artery left gastric artery splenic artery left gastroepiploic artery short gastric arteries greater pancreatic artery dorsal pancreatic artery transverse pancreatic artery superior mesenteric artery inferior pancreaticoduodenal artery jejunal and ileal branches ileocolic artery appendicular artery accessory appendicular artery colic branch right colic artery middle colic artery middle suprarenal artery renal artery (variant anatomy) inferior suprarenal artery gonadal artery (ovarian artery \| testicular artery) inferior mesenteric artery left colic artery sigmoid arteries superior rectal artery lumbar arteries median sacral artery common iliac artery external iliac artery inferior epigastric artery cremasteric artery deep circumflex artery internal iliac artery (mnemonic) anterior division umbilical artery superior vesical artery deferential artery obturator artery vaginal artery inferior vesical artery uterine artery middle rectal artery internal pudendal artery inferior rectal artery perineal artery posterior scrotal artery transverse perineal artery artery to the bulb deep artery of the penis/clitoris dorsal artery of the penis/clitoris inferior gluteal artery posterior division (mnemonic) iliolumbar artery lateral sacral artery superior gluteal artery variant anatomy persistent sciatic artery corona mortis portal venous system portal vein splenic vein inferior mesenteric vein superior mesenteric vein left gastric vein (coronary vein) veins inferior vena cava hepatic veins accessory right inferior hepatic vein renal vein ascending lumbar communicant vein gonadal vein pampiniform plexus suprarenal vein variant anatomy retro-aortic left renal vein double retroaortic left renal vein circumaortic left renal vein pancreaticoduodenal veins common iliac vein internal iliac vein internal pudendal vein obturator vein prostatic venous plexus vesical venous plexus uterine venous plexus vaginal venous plexus external iliac vein variant caval anatomy absent infrarenal inferior vena cava azygos continuation circumcaval ureter circumaortic venous collar Eustachian valve left-sided IVC IVC duplication anastomoses arterioarterial anastomoses arc of Barkow arc of Buhler arc of Riolan marginal artery of Drummond portal-systemic venous collateral pathways watershed areas Griffiths point Sudeck point lymphatics[+][+] lymph node groups para-aortic lymph nodes preaortic lymph nodes portal and portocaval lymph nodes gastric lymph node stations peripancreatic lymph nodes common iliac lymph nodes external iliac lymph nodes internal iliac lymph nodes cisterna chyli innervation of the abdomen and pelvis[+][+] thoracic splanchnic nerves lumbar plexus subcostal nerve Iliohypogastric nerve ilioinguinal nerve genitofemoral nerve lateral femoral cutaneous nerve femoral nerve obturator nerve lumbosacral trunk sacral plexus lumbosacral trunk sciatic nerve superior gluteal nerve inferior gluteal nerve nerve to piriformis perforating cutaneous nerve posterior femoral cutaneous nerve parasympathetic pelvic splanchnic nerves pudendal nerve perineal branch inferior rectal nerve dorsal nerve of the penis or clitoris nerve to quadratus femoris and inferior gemellus muscles nerve to internal obturator and superior gemellus muscles autonomic ganglia and plexuses phrenic plexus celiac plexus hepatic plexus aorticorenal plexus renal plexus superior mesenteric plexus inferior mesenteric plexus superior hypogastric plexus inferior hypogastric plexus ganglion impar Promoted articles (advertising) ADVERTISEMENT: Supporters see fewer/no ads Cases and figures Figure 1: normal kidney Case 1: normal on renal MRA Figure 2: normal development Case 2: normal on CT Figure 3 Figure 4a: renal vessels Figure 4b: renal vessels Figure 5: perirenal space Figure 6: perirenal space © 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2010	https://www.sciencedirect.com/topics/immunology-and-microbiology/pericardium	Skip to Main content Sign in Chapters and Articles You might find these chapters and articles relevant to this topic. Epicardial Ventricular Tachycardia Anatomical Considerations The pericardium is a double-walled, flask-shaped sac that contains the heart and the roots of the great vessels, superior vena cava (SVC), and pulmonary veins (PVs). By separating the heart from its surroundings—the descending aorta, lungs, diaphragm, esophagus, trachea, and tracheobronchial lymph nodes—the pericardial space allows complete freedom of cardiac motion within this sac.10 The pericardium consists of two sacs intimately connected with one another: an outer fibrous envelope (the fibrous pericardium) and an inner serous sac (the serous pericardium). The fibrous pericardium consists of fibrous tissue and forms a flask-shaped bag, the neck of which is closed by its fusion with the external coats of the great vessels, while its base is attached by loose fibroareolar tissue to the central tendon and to the muscular fibers of the left side of the diaphragm. The fibrous pericardium is also attached to the posterior sternal surface by superior and inferior sternopericardial ligaments.10 These attachments are essential to maintain the normal cardiac position in relation to the surrounding structures, to restrict the volume of the thin-walled cardiac chambers (right atrium and ventricle), and also to serve as direct protection against injuries. The vessels receiving fibrous prolongations from the fibrous pericardium are the aorta, the SVC, the right and left pulmonary arteries, and the four PVs. The inferior vena cava (IVC) enters the pericardium through the central tendon of the diaphragm, and receives no covering from the fibrous layer. The serous pericardium is a delicate membrane that lies within the fibrous pericardium and lines its walls; it is composed of two layers: the parietal pericardium and the visceral pericardium. The parietal pericardium is fused to and inseparable from the fibrous pericardium. On the other hand, the visceral pericardium, which is composed of a single layer of mesothelial cells, is part of the epicardium (i.e., the layer immediately outside of the myocardium) and covers the heart and the great vessels except for a small area on the posterior wall of the atria. The visceral layer extends to the beginning of the great vessels, and is reflected from the heart onto the parietal layer of the serous pericardium along the great vessels in tubelike extensions. This happens at two areas: where the aorta and pulmonary trunk leave the heart and where the SVC, IVC, and PVs enter the heart.10 The serous pericardium is also metabolically active. At the pericardial reflections and at the posterior wall between the great vessels, the pericardial space is divided up into a contiguous network of recesses and sinuses. There are three sinuses in the pericardial space: the superior sinus, the transverse sinus, and the oblique sinus. The superior sinus (superior aortic recess) lies anterior to the upper ascending aorta and main pulmonary artery. The transverse sinus is limited by a pericardial reflection between the superior PVs and contains the right pulmonary artery. The oblique sinus is confined by the pericardial reflections around the PVs and the IVC. The postcaval recess lies behind the SVC, the right pulmonary artery, and the right superior PV. The right and left PV recesses extend between their respective superior and inferior PVs.10 The pericardial cavity or sac is a continuous virtual space that lies between the parietal and visceral layers of serous pericardium. The heart invaginates the wall of the serous sac from above and behind, and practically obliterates its cavity, the space being merely a potential one. The sac normally contains 20 to 40 mL of clear fluid that occupies the virtual space between the two layers. Because all pericardial reflections are located basally in relation to the great vessels, the entire epicardial surface is accessible from the pericardial space, except for the atrial and ventricular septa, which are not in direct contact with the pericardium. Unlike the endovascular approach, the pericardial space is notable for the absence of obstacles and the relative ease with which catheter manipulation can be performed.10 By the same token, achieving firm, stable contact with the catheter tip at the target site may be difficult. View chapterExplore book Read full chapter URL: Book2012, Clinical Arrhythmology and Electrophysiology: A Companion to Braunwald's Heart Disease (Second Edition)Ziad F. Issa MD, ... Douglas P. Zipes MD Chapter Epicardial Ventricular Tachycardia 2019, Clinical Arrhythmology and Electrophysiology (Third Edition)Ziad F. Issa MD, ... Douglas P. Zipes MD Anatomical Considerations The pericardium is a double-walled sac that contains the heart and the roots of the great arteries, the superior vena cava (SVC), and pulmonary veins (PVs). By separating the heart from its surroundings—the descending aorta, lungs, diaphragm, esophagus, trachea, and tracheobronchial lymph nodes—the pericardial space allows complete freedom of cardiac motion within this sac.35,36 The pericardium consists of two sacs intimately connected with one another: an outer fibrous envelope (the fibrous pericardium) and an inner serous sac (the serous pericardium). The fibrous pericardium (0.8 to 2.5 mm in thickness) consists of fibrous tissue and forms a flask-shaped bag, the neck of which is closed by its fusion with the adventitia of the great vessels, while its base is attached by loose fibro-areolar tissue to the central tendon and to the muscular fibers of the left side of the diaphragm. The fibrous pericardium is also attached to the posterior sternal surface by the superior and inferior sterno-pericardial ligaments. These attachments are essential to maintain the normal cardiac position in relation to the surrounding structures, to restrict the volume of the thin-walled cardiac chambers (right atrium [RA] and ventricle), and to serve as direct protection against injuries. The fibrous pericardium extends up to 5 to 6 cm along the great vessels, including the aorta, the SVC, the right and left pulmonary arteries, and the four PVs. The inferior vena cava (IVC) enters the pericardium through the central tendon of the diaphragm and receives no covering from the fibrous pericardium.37 Nerve fibers from the parietal layer of the pericardium transmitted by the phrenic nerve are sensitive to pain (e.g., during pericarditis). In contrast, the visceral pericardial layer on the cardiac surface is insensitive to pain.38 The serous pericardium is a delicate membrane that lies within the fibrous pericardium and lines its walls; it is composed of two layers: the parietal pericardium and the visceral pericardium. The parietal pericardium is fused to and inseparable from the fibrous pericardium. On the other hand, the visceral pericardium, which is composed of a single layer of mesothelial cells, is part of the epicardium (i.e., the layer immediately outside the myocardium) and covers the heart and the great vessels except for a small area on the posterior wall of the atria. The visceral layer extends to the beginning of the great vessels, and is reflected from the heart onto the parietal layer of the serous pericardium along the great vessels in tube-like extensions. This happens at two areas: where the aorta and pulmonary trunk leave the heart and where the SVC, IVC, and PVs enter the heart. At the pericardial reflections and at the posterior wall between the great vessels, the pericardial space is divided up into a contiguous network of recesses and sinuses (eFig. 27.2). There are three sinuses in the pericardial space: superior, transverse, and oblique. The two pericardial space sinuses that can be accessed in electrophysiological (EP) procedures are the transverse and oblique sinuses. The superior sinus (superior aortic recess), which lies anterior to the upper ascending aorta and main pulmonary artery, is irrelevant to the EP procedure.19 The oblique sinus is a cul-de-sac located posterior to the atria (particularly the left atrium [LA] in the region between the four PVs) and anterior to the esophagus and descending aorta. The oblique sinus is an inverted U-shaped pericardial reflection bordered by the right PVs and IVC on the right side, the left PVs on the left side, and the right and left PVs superiorly (Fig. 27.2). The right and left PV recesses extend from the oblique sinus between the upper and lower PVs on each side.19,35–37 The tunnel-shaped transverse sinus lies posterior to the ascending aorta and pulmonary trunk, anterior to the SVC and superior PVs (eFig. 27.3). The roof of the transverse sinus is formed in parts by the aortic arch, the floor of the right pulmonary artery, and a part of the main pulmonary artery. Inferiorly, the transverse sinus is bound by the LA roof and by the pericardial reflection between the right and left superior PVs (which separates the transverse sinus from the oblique sinus below). The transverse sinus contains the right pulmonary artery and the inferior aortic recess (between the ascending aorta and the LA).37 Catheter exploration of the sinus allows access to the anterior portion of the LA, the area of Bachman's bundle, and via the inferior aortic recess, the noncoronary and right coronary aortic sinuses of Valsalva.19,39 The transverse sinus can be accessed by a catheter passed posteriorly around the lateral wall of the LV and LA, and then under the pulmonary arteries.35,36 The pericardial cavity or sac is a continuous virtual space that lies between the parietal and visceral layers of serous pericardium. The heart invaginates the wall of the serous sac from above and behind, and practically obliterates its cavity, the space being merely a potential one. The sac normally contains 20 to 40 mL of clear fluid that occupies the virtual space between the two layers. Although pericardial reflections can become anatomical obstacles for catheter navigation, all those reflections are located basally in relation to the great vessels; thus most of the epicardial surface is freely accessible from the pericardial space, except for the atrial and ventricular septa (which are not in direct contact with the pericardium) and the atrioventricular (AV) grooves and interventricular sulci (which are separated from the visceral pericardium by epicardial fat). Unlike the endovascular approach, the pericardial space is notable for the absence of obstacles and the relative ease with which catheter manipulation can be performed. By the same token, achieving firm, stable contact with the catheter tip at the target site can be difficult.19 View chapterExplore book Read full chapter URL: Book2019, Clinical Arrhythmology and Electrophysiology (Third Edition)Ziad F. Issa MD, ... Douglas P. Zipes MD Chapter Chiralic contractile field (C-CF) 2010, Muscles and MeridiansPhillip Beach DO DAc OSNZ The pericardium The pulsating heart is encased by a structure called the pericardium. It is a thin sac that surrounds the heart. The pericardium has two layers, one of which is intimately connected to the heart, whilst the outer layer is attached to the sternum, the proximal ends of the major heart vessels, the top of the diaphragm, and the mediastinum. Between the layers is a microscopic film of serous fluid that acts to reduce friction between the pulsating heart and the chest cavity. The pericardium is a tough structure that anchors the heart to stop it writhing like a high-pressure garden hose in your chest, and it will resist the swelling of the heart during high blood pressure events. Pericarditis is an inflammation between the layers that may arise slowly or suddenly, usually felt as chest pain described as sharp and stabbing. The pain may radiate to the back, scapula, neck, or arm. Pericarditis is often positional, where lying on one's back increases the pain and shortness of breath whereas bending forward usually relieves it. The pleurae A closed invaginated sac surrounds each lung. The visceral pleura (the surface attached to the lung) slide on the parietal pleura (the surface attached to the chest wall, diaphragm, and the mediastinal contents) via a serous pleural fluid. The pleurae allow the lungs to move within the chest with little friction, and surface tension within the pleural cavity holds the lung closely to the internal chest wall. Hence large excursions of the rib cage will change the shape of the lungs. When the pleural cavity is ruptured or inflamed (called pleurisy or pleuritis) the symptoms usually include severe chest pain that is often worse when breathing in, an inability to take a deep breath, a shortness of breath, and if the cause is infective, fever/chills will accompany this distress. The peritoneum The peritoneum is similar in form and function to the other remnants of the coelom. It is a bilaminar serous membrane that contains a slippery fluid that encloses most of the abdominal contents. Visceral organs are tethered to the wall of the torso by mesenteries that are derived from the peritoneum. Blood vessels and nerves enter the viscera via these mesenteries. Each organ has a range of normal physiological movement that can be contemplated only with reference to the other viscera that closely constrain it. Peritonitis, i.e. inflammation of the peritoneum, may affect the whole peritoneum or it may be isolated to an area as a pus-filled abscess. Rupture or obstruction anywhere along the GIT is the most common pathway for the entry of an infectious agent. Peritonitis causes acute severe pain in all or part of the abdomen; the abdominal wall muscles contract to a board-like rigidity, the abdomen becomes swollen and bloated, there is nausea, vomiting and sweating, the skin is pale and clammy, and peristalsis stops. It is a serious condition that requires antibiotics and often surgical intervention. View chapterExplore book Read full chapter URL: Book2010, Muscles and MeridiansPhillip Beach DO DAc OSNZ Chapter Anatomic Considerations and Examination of Cardiovascular Specimens (Excluding Devices) 2016, Cardiovascular Pathology (Fourth Edition)J.J. Maleszewski, ... J.P. Veinot Pericardium and Pericardial Reflections The pericardium is a roughly conical structure that both covers and surrounds the heart as the serous visceral and fibrous parietal pericardia, respectively (Figure 1.1a). Between these two layers, in a space called the pericardial sac, is contained 15-50 mL of straw-colored, serous fluid . The parietal pericardium attaches along the great vessels such that the ascending aorta, main pulmonary artery, terminal superior vena cava, and terminal pulmonary veins are intrapericardial (Figure 1.1b). The pericardium is attached to the central tendon of the diaphragm and the adjacent diaphragmatic muscle at its base. The parietal pericardium consists of an outer fibrous layer with an inner layer of mesothelial cells. Its outer surface also contains variable amounts of adipose tissue that may contribute to the cardiac silhouette radiologically. The dense collagen limits the acute distensibility of the parietal pericardium such that as little as 200 mL of rapidly accumulated fluid can compromise diastolic filling (cardiac tamponade) (see Chapter 13). The visceral pericardium (also called the epicardium) consists of a lining of mesothelial cells with variable deposits of subjacent adipose tissue. Typically, the visceral pericardium is transparent with the subjacent structures readily visible; however, focal pale white thickening can be encountered in many structurally normal specimens. Often referred to as “soldier's plaques,” the incidence of these epicardial collagen plaques rises with age and may represent healed pericarditis (Figure 1.2) . The fatty deposits are most pronounced in the region of the atrioventricular, interventricular, and interatrial grooves. Prominent fat tags also cover the origins of the coronary arteries between the aorta and atrial appendages. Embryologically, rapid bulboventricular growth pushes the developing heart into the pericardial space analogous to a fluid-filled balloon with a fist being pressed into it. The parietal and visceral pericardia meet along the great vessels at the pericardial reflection. Likewise, the pericardium reflects around the region of the great veins, forming the oblique pericardial sinus immediately posterior to the left atrium. A tunnel-shaped reflection is also formed by the great arteries and the atrial walls, called the transverse pericardial sinus. The appreciation of these sinuses is increasingly important with catheter-based intrapericardial procedures being developed. View chapterExplore book Read full chapter URL: Book2016, Cardiovascular Pathology (Fourth Edition)J.J. Maleszewski, ... J.P. Veinot Chapter Safety Assessment including Current and Emerging Issues in Toxicologic Pathology 2013, Haschek and Rousseaux's Handbook of Toxicologic Pathology (Third Edition)Eric D. Lombardini, ... Mark A. Melanson Pericardium The pericardium is the primary heart tissue affected, presenting frequently as an acute pericarditis with associated pericardial effusion or fibrosis leading to restrictive cardiac function. While the pathogenesis is poorly understood, it is theorized to be the result of insult to the microvasculature as discussed earlier in the chapter. The acute lesion can lead to chronic constrictive pericarditis, myocarditis, conduction abnormalities, coronary arteritis, and valvular dysfunction. Chronic restrictive pericarditis was noted in up to 5% of human patients who underwent irradiation for lung and breast cancers, but increased to 10% in those treated for lymphoma. The disease typically presents with pericardial effusion and cardiac tamponade. View chapterExplore book Read full chapter URL: Book2013, Haschek and Rousseaux's Handbook of Toxicologic Pathology (Third Edition)Eric D. Lombardini, ... Mark A. Melanson Chapter The Pericardium 2019, Cardiovascular Magnetic Resonance (Third Edition)Jay S. Leb, Susie N. Hong Conclusion The pericardium is a complex and important structure in the evaluation of the cardiovascular system. Pericardial pathology can result in a spectrum of clinical presentations, ranging from little or no symptoms to severe hemodynamic compromise and collapse. Advanced noninvasive imaging modalities such as CMR have broadened our understanding of pericardial diseases and their impact on the cardiovascular system. When combined with a high degree of clinical suspicion, CMR can serve as an important tool to diagnose and guide clinical management and therapeutic strategies. View chapterExplore book Read full chapter URL: Book2019, Cardiovascular Magnetic Resonance (Third Edition)Jay S. Leb, Susie N. Hong Chapter Pericarditis, Pericardial Constriction, and Pericardial Tamponade 2018, Cardiology Secrets (Fifth Edition)Rahul Thomas, Brian D. Hoit 1 The pericardium is not necessary for life. What does it do? Why is it important? The pericardium serves many important but subtle functions. It limits distention and facilitates interaction of the cardiac chambers, influences ventricular filling, prevents excessive torsion and displacement of the heart, minimizes friction with surrounding structures, prevents the spread of infection from contiguous structures, and equalizes gravitational, hydrostatic, and inertial forces over the surface of the heart. The pericardium also has immunologic, vasomotor, fibrinolytic, and metabolic activities. Therapeutically the pericardial space can be used for drug delivery. 2 What diseases affect the pericardium? The pericardium is affected by virtually every category of disease (Box 67.1), including idiopathic, infectious, neoplastic, immune/inflammatory, metabolic, iatrogenic, traumatic, and congenital. 3 What is pericarditis? What are the clinical manifestations? What are the causes? Acute pericarditis is a syndrome of pericardial inflammation characterized by typical chest pain (sharp, retrosternal pain that radiates to the trapezius ridge, often aggravated by lying down and relieved by sitting up), a pathognomonic pericardial friction rub (characterized as superficial, scratchy, crunchy, and evanescent), and specific electrocardiographic changes (diffuse ST-T wave changes [Fig. 67.1] with characteristic evolutionary changes and PR segment depression). These manifestations vary in terms of presentation (chest pain >85% to 90%, pericardial friction rub ≤33%, electrocardiogram [ECG] changes 60%). The 2015 European Society of Cardiology (ESC) Guidelines suggest that at least two of four of the following criteria are needed to make the diagnosis: pericardial pain, pericardial rub, ECG changes, and pericardial effusion. Additional supporting findings include elevation in inflammatory markers (erythrocyte sedimentation rate [ESR], C-reactive protein [CRP], white blood cell [WBC]) count and evidence of pericardial inflammation by computed tomography (CT) or magnetic resonance imaging (MRI). Causes of pericarditis include infection (viral, bacterial, fungal, mycobacterial, human immunodeficiency virus [HIV] associated), neoplasm (usually metastatic from lung or breast; melanoma, lymphoma, or acute leukemia), myocardial infarction, injury (post pericardiotomy, traumatic), radiation, myxedema, and connective tissue disease. In the developed world the most common cause remains viral etiologies, whereas tuberculosis is the most frequent cause in developing countries. 4 Should patients presenting with acute pericarditis be hospitalized? Why? Hospitalization is warranted for high-risk patients with an initial episode of acute pericarditis to determine a cause and to observe for the development of cardiac tamponade; close, early follow-up is critically important for the remainder of patients not hospitalized. Features indicative of high-risk pericarditis include fever greater than 38°C, subacute onset, an immunosuppressed state, trauma, oral anticoagulant therapy, myopericarditis, a moderate or large pericardial effusion, cardiac tamponade, and failure of initial outpatient medical therapy. 5 What is the treatment for acute pericarditis? Acute pericarditis usually responds to oral nonsteroidal anti-inflammatory drugs (NSAIDs), such as aspirin (650-1000 mg every 4-6 hours) or ibuprofen (600-800 mg every 6-8 hours). Colchicine (1 mg/day) may be used to supplement the NSAIDs because it may reduce symptoms and decrease the rate of recurrences. Chest pain is usually alleviated in 1 to 2 days, and the friction rub and ST-segment elevation resolve shortly thereafter. Most mild cases of idiopathic and viral pericarditis are adequately treated within a week or two of treatment, but the duration of therapy is variable and patients should be treated until inflammation or an effusion, if present, has resolved. The intensity of therapy is dictated by the distress of the patient, and narcotics may be required for severe pain. Corticosteroids should be avoided unless there is a specific indication (such as connective tissue disease or uremic pericarditis) because they enhance viral multiplication and may result in recurrences when the dosage is tapered. Based on the colchicine for acute pericarditis (COPE) trial the ESC Guidelines recommend a weight-adjusted colchicine dose of 0.5 mg once daily for patients less than 70 kg or 0.5 mg BID for those ≥70 kg for at least 3 months to improve response to medical therapy and prevent recurrences. 6 What is recurrent pericarditis? How is it treated? Recurrences of pericarditis (with or without pericardial effusion) occur in up to one-third of patients, usually within 18 months of the acute attack and may follow a course of many years. Although they may be spontaneous, occurring at varying intervals after discontinuation of drug, they are more commonly associated with either discontinuation or tapering of anti-inflammatory drugs. A poor initial response to therapy with NSAIDs and the use of corticosteroids predict recurrences. Two randomized placebo-controlled trials of colchicine for recurrent pericarditis (CORE, CORP) reported marked and significant reductions in symptom persistence at 72 hours and recurrence at 18 months when colchicine was added to conventional therapy. Although painful recurrences of pericarditis may require corticosteroids (preferably at low to moderate doses with slow tapering), once administered, dependency and the development of steroid-induced abnormalities are potential perils. Tapering of therapy should be done gradually, with tapering of a single class of drug at a time before colchicine is discontinued. For patients who require high-dose, long-term steroids or who do not respond to anti-inflammatory therapies, several drugs (azathioprine, intravenous immunoglobulin (IVIG), and anakinra, a recombinant interleukin IL- 1) have been used; however, strong evidence-based data supporting these therapies are lacking. Pericardiectomy should be considered only when repeated attempts at medical treatment have clearly failed. 7 What is postcardiac injury syndrome? Postcardiac injury syndrome (PCIS) refers to pericarditis or pericardial effusion that results from injury of the pericardium. The principal conditions considered under these headings include postmyocardial infarction syndrome, postpericardiotomy syndrome, and traumatic (blunt, sharp, or iatrogenic) pericarditis. Clinical features include the following: • : Prior injury of the pericardium, myocardium, or both • : A latent period between the injury and development of pericarditis or pericardial effusion • : A tendency for recurrence • : Responsiveness to NSAIDs and corticosteroids • : Fever, leukocytosis, and elevated ESR (and other markers of inflammation) • : Pericardial and sometimes pleural effusion, with or without a pulmonary infiltrate • : Alterations in the populations of lymphocytes in peripheral blood When the PCIS occurs after an acute myocardial infarction, it is also known as Dressler syndrome, which is now much less common than in the past. In the randomized multicenter Colchicine for the Prevention of Post-pericardiotomy Syndrome (COPPS) study, prophylactically administered colchicine reduced the incidence of postpericardiotomy syndrome after cardiac surgery. According to the 2015 ESC Guidelines on Pericardial Disease, colchicine should be considered (class IIa) after cardiac surgery, using weight-based doses, for a 1-month duration as prevention. 8 What are the pericardial compressive syndromes? What are their variants? The complications of acute pericarditis include cardiac tamponade, constrictive pericarditis, and effusive-constrictive pericarditis. Cardiac tamponade is characterized by the accumulation of pericardial fluid under pressure and may be acute, subacute, low pressure (occult), or regional. Constrictive pericarditis is the result of thickening, calcification, and loss of elasticity of the pericardial sac. Pericardial constriction is typically chronic but may be subacute, transient, and occult. Effusive-constrictive pericarditis is characterized by constrictive physiology with a coexisting pericardial effusion, usually with tamponade. Elevation of the right atrial and pulmonary wedge pressures persists after drainage of the pericardial fluid. 9 What are the similarities between tamponade and constrictive pericarditis? Characteristic of both tamponade and constrictive pericarditis is greatly enhanced ventricular interaction (interdependence), in which the hemodynamics of the left and right heart chambers are directly influenced by each other to a much greater degree than normal. Other similarities include diastolic dysfunction and preserved ventricular ejection fraction; increased respiratory variation of ventricular inflow and outflow; equally elevated central venous, pulmonary venous, and ventricular diastolic pressures; and mild pulmonary hypertension. 10 What are the differences between tamponade and constrictive pericarditis? In tamponade the pericardial space is open and transmits the respiratory variation in thoracic pressure to the heart, whereas in constrictive pericarditis the cavity is obliterated and the pericardium does not transmit these pressure changes. The dissociation of intrathoracic and intracardiac pressures (along with ventricular interaction) is the basis for the physical, hemodynamic, and echocardiographic findings of constriction. In tamponade, systemic venous return increases with inspiration, enlarging the right side of the heart and encroaching on the left, whereas in constrictive pericarditis, systemic venous return does not increase with inspiration. The mechanism of diminished left ventricular and increased right ventricular volume in constrictive pericarditis is impaired left ventricular filling because of a lesser pressure gradient from the pulmonary veins. In tamponade, early ventricular filling is impaired, whereas it is enhanced in constriction. 11 What are the physical findings of tamponade? Cardiac tamponade is a hemodynamic condition characterized by equal elevation of atrial and pericardial pressures, an exaggerated inspiratory decrease in arterial systolic pressure (pulsus paradoxus), and arterial hypotension. The physical findings are dictated by both the severity of cardiac tamponade and the time course of its development. Inspection of the jugular venous pulse waveform reveals elevated venous pressure with a loss of the Y descent (because of the decrease in intrapericardial pressure that occurs during ventricular ejection, the systolic atrial filling wave and the X descent are maintained). Pulsus paradoxus is an inspiratory decline of systolic arterial pressure exceeding 10 mm Hg, which is measured by subtracting the pressure at which Korotkoff sounds are heard only during expiration from the pressure at which sounds are heard throughout the respiratory cycle. Tachycardia and tachypnea are usually present. 12 What are the physical findings of constrictive pericarditis? Constrictive pericarditis resembles the congestive states caused by myocardial disease and chronic liver disease. Physical findings include ascites, hepatosplenomegaly, edema, and, in long-standing cases, severe wasting. The venous pressure is elevated and displays deep Y and often deep X descents. The venous pressure fails to decrease with inspiration (Kussmaul sign). A pericardial knock that is similar in timing to the third heart sound is pathognomonic but occurs infrequently. Except in severe cases, the arterial blood pressure is normal. 13 What is the role of echocardiography in tamponade? Although tamponade is a clinical diagnosis, echocardiography plays major roles in the identification of pericardial effusion and in the assessment of its hemodynamic significance (Fig. 67.2). The use of echocardiography for the evaluation of all patients with suspected pericardial disease was given a class I recommendation by a 2003 task force of the American College of Cardiology (ACC), the American Heart Association (AHA), and the American Society of Echocardiography (ASE). This recommendation was supported by a 2013 ASE Expert Consensus Statement and the 2015 ESC Guidelines. Except in hyperacute cases, a moderate to large effusion is usually present, and swinging of the heart within the effusion may be seen. Reciprocal changes in left and right ventricular volumes occur with respiration. Echocardiographic findings suggesting hemodynamic compromise (atrial and ventricular diastolic collapses) are the result of transiently reversed right atrial and right ventricular diastolic transmural pressures and typically occur before hemodynamic embarrassment. The respiratory variation of mitral and tricuspid flow velocities is greatly increased and out of phase, reflecting the increased ventricular interaction. Less than a 50% inspiratory reduction in the diameter of a dilated inferior vena cava reflects a marked elevation in central venous pressure, and abnormal right-sided venous flows (systolic predominance and expiratory diastolic reversal) are diagnostic. In patients who do not have tamponade on first assessment, repeat echocardiography during clinical follow-up was given a class IIa recommendation by the 2003 ACC/AHA/ASE task force. 14 What is the role of echocardiography in constrictive pericarditis? Echocardiography is an essential adjunctive procedure in patients with suspected pericardial constriction. The use of echocardiography for the evaluation of all patients with suspected pericardial disease is a class I ACC/AHA/ASE task force recommendation. Echocardiography findings to be sought include increased pericardial thickness (best with transesophageal echocardiography), abrupt inspiratory posterior motion of the ventricular septum in early diastole, plethora of the inferior vena cava and hepatic veins, enlarged atria, and an abnormal contour between the posterior left ventricular the left atrial posterior walls. Although no sign or combination of signs on M-mode is diagnostic of constrictive pericarditis, a normal study virtually rules out the diagnosis. Doppler is particularly useful, showing a high E velocity of right and left ventricular inflow and rapid deceleration, a normal or increased tissue Doppler E′, and a 25% to 40% fall in transmitral flow and marked increase of tricuspid velocity in the first beat after inspiration. Increased respiratory variation of mitral inflow may be missing in patients with markedly elevated left atrial pressure but may be brought out by preload reduction (e.g., head-up tilt). Hepatic vein flow reversals increase with expiration, reflecting the ventricular interaction and the dissociation of intracardiac and intrathoracic pressures, and pulmonary venous flow shows marked respiratory variation. 15 Are other imaging modalities useful in pericardial disease? Other imaging techniques, such as CT and cardiovascular magnetic resonance (CMR), are not necessary if two-dimensional and Doppler echocardiography are available. However, pericardial effusion may be detected, quantified, and characterized by CT and CMR. CT scanning of the heart is extremely useful in the diagnosis of constrictive pericarditis; findings include increased pericardial thickness (>4 mm) and calcification. CMR provides direct visualization of the normal pericardium, which is composed of fibrous tissue and has a low MRI signal intensity. CMR is claimed by some to be the diagnostic procedure of choice for the detection of constrictive pericarditis (Fig. 67.3). Late gadolinium enhancement of the pericardium may predict reversibility of transitory constrictive pericarditis (see later) following treatment with anti-inflammatory agents. The 2015 ESC Guidelines recommend CT and/or CMR as second-level testing in the diagnostic workup of pericarditis. 16 What is the role for medical therapy in constrictive pericarditis? Constrictive pericarditis is a surgical disease, except in cases of very early constriction or in severe, advanced disease. Medical therapy of constrictive pericarditis plays a small but important role. Medical therapy of specific etiologies (e.g., tuberculosis pericarditis) may significantly reduce progression to constriction. Medical therapy with anti-inflammatory drugs may resolve constriction (transitory constriction). Finally, medical therapy may be supportive therapy to control symptoms in patients not candidates for pericardiectomy due to their surgical risk. Diuretics and digoxin (in the presence of atrial fibrillation) are useful in these patients. Preoperative, before pericardiectomy, diuretics should be used sparingly with the goal of reducing, not eliminating, elevated jugular pressure, edema, and ascites. Postoperatively, diuretics should be given if spontaneous diuresis does not occur; the central venous pressure may take weeks to months to return to normal after pericardiectomy. In some patients, constrictive pericarditis resolves either spontaneously or in response to various combinations of NSAIDs, steroids, and antibiotics (transitory constriction). Therefore, before pericardiectomy is recommended, conservative management for 2 to 3 months in hemodynamically stable patients with subacute constrictive pericarditis is recommended. View chapterExplore book Read full chapter URL: Book2018, Cardiology Secrets (Fifth Edition)Rahul Thomas, Brian D. Hoit Chapter The Cardiovascular System 2009, Equine Anesthesia (Second Edition)Colin C. Schwarzwald, ... William W. Muir Pericardium The pericardial space is formed by the reflection of the two pericardial membranes, the parietal pericardium and the visceral pericardium (epicardium) and normally contains a small amount of fluid. The pericardium, although not an essential structure, serves to restrain and protect the heart, acts as a barrier against contiguous infection, balances right and left ventricular output through diastolic and systolic interactions, limits acute chamber dilation, and exerts lubricant effects that minimize friction between cardiac chambers and surrounding structures.8 Pericardial constraint limits chamber dilation at high filling volumes or in the presence of pathologic conditions such as pericardial effusion or constriction, particularly of the thinner-walled right atrium and ventricle; augments ventricular interdependence; and limits diastolic ventricular filling.8 View chapterExplore book Read full chapter URL: Book2009, Equine Anesthesia (Second Edition)Colin C. Schwarzwald, ... William W. Muir Chapter History and background of traditional Chinese medicine in neurological conditions 2010, Acupuncture in Neurological ConditionsVal Hopwood PhD FCSP Dip Ac Nanjing, Clare Donnellan MSc MCSP Dip Shiatsu MRSS Pericardium (Xin Bao) The Pericardium is closely related to the Heart; traditionally it is thought to shield the Heart from the invasion of External Pathogenic factors. It is also known as the Heart Protector. The ancient manuscripts, most particularly the Spiritual Pivot , do not grant the Pericardium true Zang Fu status, describing the Heart as the master of the five Zang and the six Fu. The Heart is considered to be the dwelling of the Shen and no Pathogen can be allowed past the barrier of the Pericardium in case the Heart is damaged and the Shen departs and death occurs. The Pericardium displays some of the characteristics of Xin Heart but is far less important, in that it only assists with the government of Blood and housing the Mind. However, as stated earlier, Worsley considered the Pericardium more of a function than an entity and the points on the channel are often used to treat emotional problems, having a perceived cheering effect . They are also frequently used for their sedative effect. The meridian is also used in treatment of the Heart but is considered to be a less intense form of therapy than the use of Heart points. In effect the Pericardium is considered as the active mechanism of the Heart, the physical pumping activity, whereas the Heart itself is more involved with containing the Spirit and maintaining full consciousness. In spite of this lesser importance in Zang Fu terms, the meridian is a very useful one with many internal connections and wide-ranging physiological effects. View chapterExplore book Read full chapter URL: Book2010, Acupuncture in Neurological ConditionsVal Hopwood PhD FCSP Dip Ac Nanjing, Clare Donnellan MSc MCSP Dip Shiatsu MRSS Chapter Uncommon Cardiac Diseases 2018, Kaplan's Essentials of Cardiac Anesthesia (Second Edition)Jonathan F. Fox MD, ... William C. OliverJr. MD Pericardial Heart Disease The pericardium is a two-layer sac that encloses the heart and great vessels. The inner layer is a serous membrane (visceral pericardium) covering the surface of the heart. The outer layer is a fibrinous sac (parietal pericardium), which is attached to the great vessels, diaphragm, and sternum. The parietal pericardium is a stiff collagenous membrane that is resistant to acute expansion. The space between the two layers is the pericardial space, normally containing up to 50 mL of clear fluid that is an ultrafiltrate of plasma. It can gradually dilate to accept large volumes of fluid if slowly accumulated; however, rapid fluid accumulation leads to cardiac tamponade. The two layers of the pericardium are joined at the level of the great vessels and at the central tendon of the diaphragm caudally, and a serous layer extends past these junctions to line the inside of the fibrinous sac (parietal pericardium). The lateral course of the phrenic nerve on either side of the heart is an important anatomic relationship because this nerve is encapsulated in the pericardium and thus can easily be damaged during pericardiectomy. The pericardium is not essential for life, and pericardiectomy causes no apparent disability, but it has many subtle functions that are advantageous. Foremost, it acts to minimize torsion of the heart and reduces the friction from surrounding organs. Acute Pericarditis Acute pericarditis is common, but the actual incidence is unknown because it often goes unrecognized. It is generally self-limited, with symptoms lasting 6 weeks. Acute pericarditis has many causes (Box 18.6), the most common of which is viral (30–50%). Anesthesiologists encounter patients with acute pericarditis in the context of malignancy, MI, postcardiotomy syndrome, uremia, or infection when surgery is required, because symptoms are incapacitating and medical therapy has failed. Constrictive Pericarditis CP is a dense fusion of the parietal and visceral pericardium that limits diastolic filling of the heart, irrespective of the cause. The changes in the pericardium can be due to scarring, induced by a single episode of acute pericarditis, or caused by a prolonged exposure to a recurrent or chronic inflammatory process. Table 18.5 lists some of the causes of chronic CP. Up to 18% of pericardiectomies are attributed to previous cardiac surgeries, which may explain the increase in the number of cases of CP over the last 15 years. Surgical Considerations for Pericardial Disease Pericardiectomy is performed for recurrent pericardial effusion and CP refractory to conservative therapies. Pericardial dissection for effusive pericarditis is straightforward; however, pericardiectomy for CP is often a surgical challenge, with an operative mortality of 5.9% to 11.9%. The occurrence of tricuspid regurgitation may occur, similar to CP with signs of right-sided heart failure and volume overload. The presence of tricuspid regurgitation may identify a subset of patients with CP who have more advanced disease. Persistent low CO immediately after pericardiectomy is a major cause of morbidity and mortality, occurring in 14% to 28% of patients in the immediate postoperative period. These findings are in contrast to the accepted belief that the pericardium is the problem in these patients and the myocardium is normal. Although patients with cardiac tamponade usually improve clinically once the pericardium is opened, improvement is not always apparent immediately after pericardiectomy for patients with CP. Noticeable improvement in cardiac function may take weeks; however, 90% of patients will experience relief of symptoms postoperatively. Anesthetic Considerations Anesthetic goals for managing patients with CP for pericardiectomy include minimizing bradycardia, myocardial depression, and decreases in preload and afterload. PAC monitoring is often used because of the risk of low CO syndrome postoperatively. Low CO syndrome develops in a subset of patients with CP, irrespective of the approach or extent of pericardiectomy. Low CO, hypotension, and arrhythmias (atrial and ventricular) are common during chest dissection. Because of limited and relatively fixed ventricular diastolic filling, CO becomes rate-dependent. If myocardial function or heart rate is depressed, then β-agonists or pacing will improve CO. Catastrophic hemorrhage can occur suddenly if the atrium or ventricle is perforated, necessitating adequate central venous access. Damage to coronary arteries may also occur during dissection; so careful monitoring of the ECG for signs of ischemia is prudent. Cardiac Tamponade Tamponade exists when fluid accumulation in the pericardial space dramatically increases intrapericardial pressure and limits filling of the heart. The rate of pericardial fluid accumulation, rather than the absolute fluid volume, is the determinant of tamponade sequelae. The classic Beck triad of acute tamponade, consisting of: (1) decreasing arterial pressure; (2) increasing venous pressure; and (3) a small, quiet heart, is only observed in 10% to 40% of patients. Pulsus paradoxus (Fig. 18.14) may also be observed, which is a fall in systolic blood pressure of more than 10 mm Hg during inspiration, caused by a reduced left ventricular stroke volume that is generated by increased filling of the right-sided heart during inspiration. Pulsus paradoxus is not sensitive or specific for tamponade, because it may be present in those with obstructive pulmonary disease, right ventricular infarction, or CP. Hemodynamic monitoring may aid in the diagnosis of cardiac tamponade. Ultimately, the right atrial pressure, pulmonary artery diastolic pressure, and pulmonary capillary wedge pressure equilibrate. Equilibration of these pressures (within 5 mm Hg of each other) merits immediate action to rule out acute tamponade. Echocardiography is the current method of choice and the most reliable noninvasive method to detect pericardial effusion and exclude tamponade. Echocardiographic features of tamponade include an exaggerated motion of the heart within the pericardial sac in conjunction with atrial and ventricular collapse. Specific two-dimensional echocardiographic findings that support cardiac tamponade include diastolic collapse of the RV, inversion of the RA during diastole, abnormal ventricular septal motion, and variation of ventricular size with the respiratory cycle. Diastolic collapse of the right-sided chambers occurs because of pericardial pressure exceeding intracardiac pressure during diastole. Right atrial collapse is a specific finding during echocardiographic examination if it is present for more than one-third of the cardiac cycle. Pericardiocentesis is indicated for life-threatening cardiac tamponade in conjunction with fluid resuscitation to maintain adequate filling pressures. Hemodynamic improvement should ensue after pericardiocentesis. Although pericardiocentesis relieves the symptoms of tamponade, definitive therapy directed at the underlying cause should be pursued. Major complications of pericardiocentesis include coronary artery laceration, cardiac puncture, and pneumothorax. Tamponade attributable to hemorrhage in the patient after cardiac surgery requires immediate mediastinal exploration to determine the source of bleeding and to stabilize hemodynamics. Anesthetic Considerations Severe hypotension or cardiac arrest has followed the induction of general anesthesia in patients with tamponade. The causes include myocardial depression, sympatholysis, decreased venous return, and changes in heart rate that often accompany anesthetic medications and positive pressure ventilation. Resuscitation requires immediate drainage of pericardial fluid. Pericardiotomy via a subxiphoid incision with only local anesthetic infiltration or light sedation is an option. If intrapericardial injury is confirmed, then general anesthesia can be induced after decompression of the pericardial space. Ketamine (0.5 mg/kg) and 100% oxygen have been used with local anesthetic infiltration of the preexisting sternotomy to drain severe pericardial tamponade. Spontaneous respirations, instead of positive-pressure ventilation, will support CO more effectively until tamponade is relieved. Correction of metabolic derangements is mandatory. Volume expansion is likely warranted in hypotensive patients. Similar to CP, patients with tamponade have a relatively low and fixed stroke volume and thus rely on heart rate and adequate filling to maintain CO. Catecholamine infusions or pacing may be used to avoid bradycardia. View chapterExplore book Read full chapter URL: Book2018, Kaplan's Essentials of Cardiac Anesthesia (Second Edition)Jonathan F. Fox MD, ... William C. OliverJr. MD Related terms: Pulmonary Artery Constrictive Pericarditis Pericarditis Pleurae Hemodynamic Qi Surface Property Click Liquid Diaphragm View all Topics
2011	https://cplusplus.com/reference/ostream/basic_ostream/operator-free/	C++ Reference C library: Containers: Input/Output: Multi-threading: Other: class templates classes manipulators basic_ostream member classes member functions non-member overloads protected members std::operator<< (basic_ostream) \| \| \| --- \| \| single character (1) \| template<class charT, class traits>basic_ostream<charT,traits>& operator<< (basic_ostream<charT,traits>& os, charT c);template<class charT, class traits>basic_ostream<charT,traits>& operator<< (basic_ostream<charT,traits>& os, char c);template<class traits>basic_ostream<char,traits>& operator<< (basic_ostream<char,traits>& os, char c);template<class traits>basic_ostream<char,traits>& operator<< (basic_ostream<char,traits>& os, signed char c);template<class traits>basic_ostream<char,traits>& operator<< (basic_ostream<char,traits>& os, unsigned char c); \| \| character sequence (2) \| template<class charT, class traits>basic_ostream<charT,traits>& operator<< (basic_ostream<charT,traits>& os, const charT s);template<class charT, class traits>basic_ostream<charT,traits>& operator<< (basic_ostream<charT,traits>& os, const char s);template<class traits>basic_ostream<char,traits>& operator<< (basic_ostream<charT,traits>& os, const char s);template<class traits>basic_ostream<char,traits>& operator<< (basic_ostream<charT,traits>& os, const signed char s);template<class traits>basic_ostream<char,traits>& operator<< (basic_ostream<charT,traits>& os, const unsigned char s); \| \| \| \| --- \| \| single character (1) \| template<class charT, class traits>basic_ostream<charT,traits>& operator<< (basic_ostream<charT,traits>& os, charT c);template<class charT, class traits>basic_ostream<charT,traits>& operator<< (basic_ostream<charT,traits>& os, char c);template<class traits>basic_ostream<char,traits>& operator<< (basic_ostream<char,traits>& os, char c);template<class traits>basic_ostream<char,traits>& operator<< (basic_ostream<char,traits>& os, signed char c);template<class traits>basic_ostream<char,traits>& operator<< (basic_ostream<char,traits>& os, unsigned char c); \| \| character sequence (2) \| template<class charT, class traits>basic_ostream<charT,traits>& operator<< (basic_ostream<charT,traits>& os, const charT s);template<class charT, class traits>basic_ostream<charT,traits>& operator<< (basic_ostream<charT,traits>& os, const char s);template<class traits>basic_ostream<char,traits>& operator<< (basic_ostream<charT,traits>& os, const char s);template<class traits>basic_ostream<char,traits>& operator<< (basic_ostream<charT,traits>& os, const signed char s);template<class traits>basic_ostream<char,traits>& operator<< (basic_ostream<charT,traits>& os, const unsigned char s); \| \| rvalue insertion (3) \| template<class charT, class traits, class T>basic_ostream<charT,traits>& operator<< (basic_ostream<charT,traits>&& os, const T& val); \| << os<<val os.width() Parameters char os.widen(c) char os.widen Return Value \| flag \| error \| --- \| \| eofbit \| failbit \| The function failed while formatting the output (it may also be set if the construction of sentry failed). \| \| badbit \| Either the insertion on the stream failed, or some other error happened (such as when this function catches an exception thrown by an internal operation). When set, the integrity of the stream may have been affected. \| Example \| \| \| --- \| \| 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 \| // example on insertion #include <iostream> // std::cout int main () { const char str[] = "Example string"; char ch = 'X'; std::cout << str << '\n'; // Insert string std::cout.width(5); std::cout.fill(''); std::cout << ch << '\n'; // Insert single character return 0; } \| 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 `// example on insertion include // std::cout int main () { const char str[] = "Example string"; char ch = 'X'; std::cout << str << '\n'; // Insert string std::cout.width(5); std::cout.fill(''); std::cout << ch << '\n'; // Insert single character return 0; }` \| \| \| Example string X \| Data races Exception safety See also
2012	https://blog.csdn.net/qixiang_chen/article/details/112028613	等差数列前n项和推导过程_等差数列前n项和公式二次函数-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·新人礼包消息历史创作中心创作等差数列前n项和推导过程原创已于 2022-12-13 16:51:18 修改·4.1k 阅读 · 2 · 0· CC 4.0 BY-SA版权版权声明：本文为博主原创文章，遵循CC 4.0 BY-SA版权协议，转载请附上原文出处链接和本声明。文章标签： #数学于 2020-12-31 16:46:10 首次发布初中数学解题方法汇总专栏收录该内容 16 篇文章订阅专栏由于未提供博客具体内容，无法生成包含关键信息的摘要。确定要放弃本次机会？福利倒计时 : : 立减 ¥ 普通VIP年卡可用立即使用华科易迅关注关注 2点赞踩 0 收藏觉得还不错? 一键收藏 0评论分享复制链接分享到 QQ 分享到新浪微博扫一扫举报举报专栏目录算法：求等差数列的n 项和 weixin_41964616的博客 07-05 423 @华为算法题等形式读取数据 ) python实现 while True: try: a = int(input()) v = 2 s = 0 for i in range(a): s += v 等差数列 java用等差公式写_ 等差数列求和公式及推导方法 weixin_33693600的博客 02-25 2040 等差数列是指从第二项起，每一项与它的前一项的差等于同一个常数的一种数列，常用A、P表示。这个常数叫做等差数列的公差。前 n 项和公式为：Sn=a1n+[n(n-1)d]/2或Sn=[n(a1+an)]/2。等差数列公式 1.定义式2.通项公式 3.求和公式 4.前 n 项和公式等差数列推论(1)从通项公式可以看出，a(n)是n的一次函数(d≠0)或常数函数(d=0)，(n，an)排在一条直线上，由前 n 项... 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求和公式 - 等差数列解析 9-17 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……(2n-1)。等差数列{an}的通项公式为:an=a1+(n-1)d。前 n 项和公式为:Sn=na1+n(n-1)d/2或Sn=n(a1+an)/2。注意: 管理类联考——数学——汇总篇——知识点突破——代数——等差数列 9-11 该博客主要围绕等差数列展开,介绍了其分类方式,罗列了定义、通项、前 n 项和等公式,讲解了考点、考试解读和命题方向,还阐述了通项公式、性质、前 n 项和的相关内容,以及基本问题、奇数项和之比、Sn最值问题的求解方法。读书笔记等差数列可以分为: 1.等差数列基本问题:求和、求项数、求某项、轮换对称性。 2. 两... 2.3-等差数列前项 n 和的性质及推导--2.3(第二课时).pdf 04-21 人教版--高中数学必修5--2.3 等差数列前项 n 和的常用的性质以及这些性质的推导证明，适用于老师复习和学生复习等差数列的前 n 项和 12-13 等差数列的前 n 项和等差数列解析 9-18 (2)式知,Sn是n的二次函数(d≠0)或一次函数(d=0,a1≠0),且常数项为0。在等差数列中,等差中项: , 且任意两项 am,an的关系为: an=am+(n-m)d 它可以看作等差数列广义的通项公式。从等差数列的定义、通项公式,前 n 项和公式还可推出: a1+an=a2+an-1=a3+an-2=…=ak+an-k+1,k∈{1,2,…... 数学术语大全 9-21 argument of a function 函数的自变量 arithmetic 算术 arithmetic mean 算术平均;等差中顶;算术中顶 arithmetic progression 算术级数;等差级数 arithmetic sequence 等差序列 arithmetic series 等差级数 arm 边 array 数组; 数组 arrow 前号 ascending order 递升序 ascending powers of X X 的升幂 assertion 断语; ... 等差数列前 n 项和公式证明那么巧合的专栏 02-25 1万+ 等差数列前 n 项和公式证明等差数列第一项为 a1公差为 d则前 n 项的和为：Sn=[(a1+an)∗n]/2Sn = [ (a1 + an)n ] / 2证明过程如下： Sn = a1 + (a1 + d) + ... + (a1 + (n-1)d) Sn = an + (an - d) + ... + (an - (n-1)d) 上边两式相加得到 2Sn = (a1 + an)n 即 Sn = 等差数列前 n 项和网站开发初中级代码参考-转载 11-08 1242 #include<iostream.h> void main() { int s,i,n,sum; cout<<"输入初始值s："<<"\t"; cin>>s; cout<<"输入增量i:"<<"\t"; cin>>i; 二分查找、三分查找求极点、二分求等比数列【模板】找出最大等比数列 _和... 9-16 二分求等比数列 LL Power(LL p,LL n) { LL ret=1;while(n>0) {if(n&1) ret=retp%9901; p=pp%9901; n>>=1; }returnret; } LLSum(LL p,LL n)//递归二分求(1 + p^1 + p^2 + … + p^n)%mod{if(n==0)return1;if(n&1)return(Sum(p,n/2)(1+Power(p,n/2+1)))%... 辽宁省专升本考数学样例真题及解析_专升本数学辽宁真题 9-26 文章讨论了如何通过分析二次函数在特定区间上的性质找到参数a的值,以及如何利用等差数列前 n 项和公式解决给定S10 和 S100求S110的问题。题目:已知函数 f(x) = x^2 + ax + 3 在区间 [-1, 2] 上的最小值为-3,求 a 的值。解析:首先,二次函数 f(x) = x^2 + ax + 3 的对称轴为 x = -a/2... 高二数学等差数列前 n 项和 PPT课件.pptx 10-10 等差数列前 n 项和的求解是等差数列理论的重要组成部分，它在实际问题解决中有着广泛的应用。首先，我们要理解什么是等差数列。等差数列是一个序列，其中任意两项之间的差是一个常数，这个常数被称为公差。例如，... 高中数学数列等差数列的前 n 项和北师大必修PPT课件.pptx 10-10 此外，等差数列的前 n 项和公式体现了数学中的对称美，以及从具体实例到数学模型的构建过程，有助于培养学生的数学兴趣和自信心。在教学反思环节，教师应关注学生是否真正掌握了等差数列的前 n 项和公式，是否能够灵活... ...这些看看有没有掌握_ 二次函数顶点式、等差数列通项公式 9-17 三角函数也是重点,正弦函数 sinα,余弦函数 cosα,正切函数 tanα,它们之间存在众多的关系式,如 sin²α + cos²α = 1。数列方面,等差数列通项公式为 an = a1 + (n - 1)d,前 n 项和公式为 Sn = n(a1 + an) / 2 ;等比数列通项公式为 an = a1×q^(n - 1),前 n 项和公式为 Sn ... 求数列中连续数值组成的等差数列最大长度_ 二次函数与等差数列也可以扯... 9-15 等差数列前 n 项之和 Sn是项数n的二次函数: 它没有常数项(图像过原点),开口朝向可上可下,视d的正负而定,以d>0为例: 若n的取值为连续实数,则图像为完整的抛物线。但在等差数列中,n的取值为正整数,故图像为一连串的点(图中红点)。例2:已知等差数列{an}的前 n 项之和为Sn。若a1<0,S8=S14,则{Sn}有最... 等差数列前 n 项和教案.doc 10-21 教学内容主要围绕等差数列前 n 项和的推导过程和简单应用展开。这一部分的教学是在学生已经掌握了等差数列通项公式的基础上进行的，因此学生应具备一定的函数和数列基础知识，包括等差数列的定义和性质，以及高斯求和... 等差数列前 n 项和公开课.ppt 11-07 通过本节公开课，学生不仅能够复习和巩固等差数列的基础知识，还能深入理解等差数列前 n 项和的推导过程及其几何意义，进一步提升数学思维和问题解决能力。通过课堂练习和小测验，学生可以检验自己的学习效果，并在... 推导等差数列前 n 项积 weixin_42613018的博客 02-09 2161 等差数列前 n 项积可以使用递推公式推导：设等差数列前 n 项积为S_n，公差为d，首项为a_1 则：S_n = a_1 (a_1 + d) (a_1 + 2d) …… (a_1 + (n-1)d) 由此可得：S_n = a_1^n ((a_1 + d) / a_1) ((a_1 + 2d) / (a_1 + d)) …… ((a_1 + (n-1)d) / (a_1 +... 等差数列 2，5，8，11，14。。。。输入:正整数N>0 输出:求等差数列前 N 项和黄昏的晨曦 08-17 3427 #include using namespace std; int main() { int a1=2,d=3,n,sum; while(cin>>n) { sum=na1+n(n-1)/2d; //等差数列求和公式 cout<<sum<<endl; ... 语言求等差数列前 n 项和【尧哥说数列2】_等差数列 weixin_42372573的博客 01-12 758 教学理论上的目标：1.理解等差数列的概念；2.掌握等差数列的通项公式与前 n 项和公式；3.能在具体的问题情境中识别数列的等差关系，并能用有关知识解决相应的问题；4.了解等差数列与一次函数、二次函数的关系．尧哥的教学目标：熟练掌握万能法知识点一等差数列的定义如果一个数列从第2 项起，每一项与它的前一项的差等于同一个常数，那么这个数列就叫做等差数列，这个常数叫做等差数列的公差，公差通常用字... Java编程等差数列琉和四月的博客 07-09 6716 sum=2+5+8+11+14+⋯，输入正整数nn，求sumsum的前 nn 项和。公式描述：公式中首项为a1，末项为an，项数为n，公差为d，前 n 项和为Sn。第一种是直接写出公式的方法 public static void main(String[] args) { Scanner sc=new Scanner(System.in); System.out.println("请输入数")... Java：1006: 求等差数列的和 bairimeng16的博客 05-25 735 1006: 求等差数列的和时间限制:1Sec内存限制:30 MB 提交:55615解决:35498 [状态] [讨论版] [提交] [命题人:admin] 题目描述给出三个整数，分别表示等差数列的第一项、最后一项和公差，求该数列的和。输入输入三个整数，之间用空格隔开。第1个数作为首项，第2个数作为末项，第3个数作为公差。输出输出占一行，包含一个整数，为该等差数列的和。样例输入Copy 2 11 3 样例输出Copy 26 提示本题在... 2.3 等差数列的前 n 项和 kingBook928的博客 05-06 615 一般地，我们称 a 1+a 2+a 3+…+a n a 1+a 2+a 3+…+a n 为数列{a n}{a n}的前 n 项和，用 S n S n 表示，即 S n=a 1+a 2+a 3+…+a n.S n=a 1+a 2+a 3+…+a n. 由高斯算法的启示，对于公差为 d d 的等差数列，我们用两种方式表示 S n S n： [S_n=a_1+(a_1+d)+(a_1+2d)+\ldots+[a_1+(n-1)d],\ \ ... 等比数列前 n 项和推导 qixiang_chen的博客 12-31 2099 等比数列前 n 项和推导 c语言求分数序列前 N 项和热门推荐 inooll的博客 03-08 1万+ 本题要求编写程序，计算序列 2/1+3/2+5/3+8/5+… 的前 N 项之和。注意该序列从第 2 项起，每一项的分子是前一项分子与分母的和，分母是前一项的分子。输入格式: 输入在一行中给出一个正整数 N。输出格式: 在一行中输出部分和的值，精确到小数点后两位。题目保证计算结果不超过双精度范围。尝试 #include int main() { int n; ... 差分法求高阶等差数列的通项公式 Happig的博客 09-21 4871 高阶等差数列对于一个给定的数列，将连续两项之间的差bn=an+1−anb_n=a_{n+1}-a_nbn=an+1−an得到一个新的数列，那么bnb_nbn称为原数列的一阶等差数列，若cn=bn+1−bnc_n=b_{n+1}-b_ncn=bn+1−bn，那么cnc_ncn称为原数列的二阶等差数列，以此类推… 高阶等差数列都有一个多项式的通项公式。差分法给定序列aaa，依次求出该序列的kkk阶等差序列，直到某个序列全为000为止，按照下列排列规则排列在纸上 Cn+11&nb 等差数列前 n 项和公式最新发布 03-19 ### 等差数列前 n 项和的数学定义等差数列是一种特殊的数列，其特点是每一项与其前一项之差是一个固定常数，称为公差 d d。设首项为 a 1 a 1，则第 k k 项可以表示为： a k=a 1+(k−1)da k=a 1+(k−1)d 对于等差数列的前 n n 项和 S n S n 的定义如下： S n=a 1+a 2+a 3+⋯+a nS n=a 1+a 2+a 3+⋯+a n 通过代入通项公式可得： S n=∑k=1 n a k=∑k=1 n(a 1+(k−1)d)S n=∑k=1 n a k=∑k=1 n(a 1+(k−1)d) 展开求和表达式并简化得到最终公式： S n=n 2⋅(2 a 1+(n−1)d)S n=n 2⋅(2 a 1+(n−1)d) 或者更常见的形式为： S n=n 2⋅(a 1+a n)S n=n 2⋅(a 1+a n) 其中 a n a n 是该序列中的最后一项。 ### 等差数列前 n 项和的推导过程为了验证上述公式，可以通过高斯法来直观理解这一结论。假设我们有以下等差数列及其反向排列：正序：a 1,a 2,...,a n a 1,a 2,...,a n 倒序：a n,a n−1,...,a 1 a n,a n−1,...,a 1 当我们将这两组相加时，每一对对应位置上的元素总和均为相同的定值： (a 1+a n),(a 2+a n−1),...,(a n+a 1)6,(a 2+a n−1),...,(a n+a 1) 由于共有 n n 对这样的组合，因此整个序列的两倍和等于这些配对的总数乘以单个配对的数值： 2 S n=n⋅(a 1+a n)2 S n=n⋅(a 1+a n) 由此得出标准公式： S n=n 2⋅(a 1+a n)S n=n 2⋅(a 1+a n) 如果替换掉 a n a n 使用通项公式，则会获得另一种表述方式： S n=n 2⋅(2 a 1+(n−1)d)S n=n 2⋅(2 a 1+(n−1)d) 以下是实现计算的一个简单 Python 函数示例： python def arithmetic_sum(n, first_term, common_difference): """ 计算 _等差数列_ _前_ n _项_ _和_ 参数: n : int _前_ 几 _项_ 的数量 first_term : float 首 _项_ common_difference : float 公差返回: sum_of_series : float 序列 _前_ n _项_ _和_ """ return n / 2 (2 first_term + (n - 1) common_difference) 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华科易迅博客等级码龄7年 497 原创463 点赞 934 收藏 305 粉丝关注私信 🔥2核2G ARM架构云服务器每月750小时免费试用🆓 灵活弹性又可靠💪早用早省～广告热门文章 SpringBoot四大组件 30699 VUE集成ElementUI开发 18766 管理后台界面基本框架设计 18731 Java常用的设计模式 18015 证明三角形tanA+tanB+tanC=tanAtanBtanC 15413 分类专栏人工智能22篇高考备考4篇中考备考14篇 SQL(Mysql-Oracle)5篇华科易迅信息化系统4篇初中物理6篇高中数学24篇初中数学123篇高中数学解题方法汇总9篇初中数学解题方法汇总16篇笔记3篇初中英语4篇 Mint-ui2篇 ShardingSphere1篇微信小程序2篇 Android4篇 SpringBoot29篇 Java17篇 Freemarker15篇 Redis11篇 Solr8篇 RabbitMQ11篇 Dubbo4篇 Nginx9篇 SQL(Mysql,Oracle)3篇华科易迅创客平台项目21篇 Git SpringCloud10篇 NodeJS3篇 Shiro2篇 VUE18篇 Docker1篇 Angular2篇 Fdfs分布文件系统3篇 Test11篇 Mybatis2篇 PostgreSQL1篇面试知识集锦4篇 Elasticsearch1篇 Java基础27篇 Activiti 展开全部收起上一篇：绝对值和最小逆向法则下一篇：等比数列前n项和推导最新评论 SpringBoot集成JasperReport 生产队队长:jasperreports 6.0.0版本无法下载 Ubuntu 20安装微信客户端华科易迅:操作系统版本是ubuntu-20.04.5-desktop-amd64，试过其他安装方式都未成功，只要这种方式可以安装并使用电脑版微信，只是消息不弹屏的缺陷。 Ubuntu 20安装微信客户端 Silent976:没，现在只能用网页的微信文件传输助手传文件了 Ubuntu 20安装微信客户端 Silent976:下载的版本太低了，打开软件要求更新 VUE入门案例 CSDN-Ada助手:哇, 你的文章质量真不错，值得学习！不过这么高质量的文章, 还值得进一步提升, 以下的改进点你可以参考下: (1)使用更多的站内链接；(2)增加除了各种控件外，文章正文的字数；(3)使用标准目录。大家在看 JavaScript身份证号校验算法 XXE - 实体注入（xml外部实体注入）论文精读（六）：微服务系统服务依赖发现技术综述 5100 二.栈和队列【每日一问】为什么PCB布线要避免直角？ 322 最新文章使用Dlib实现人脸识别人工智能学习83-Yolo训练类人工智能学习81-Yolo预测类 2025 07月 10篇 06月 35篇 05月 29篇 2024年 1篇 2023年 6篇 2022年 7篇 2021年 14篇 2020年 215篇 2019年 78篇 2018年 102篇 🔥2核2G ARM架构云服务器每月750小时免费试用🆓 灵活弹性又可靠💪早用早省～广告上一篇：绝对值和最小逆向法则下一篇：等比数列前n项和推导分类专栏人工智能22篇高考备考4篇中考备考14篇 SQL(Mysql-Oracle)5篇华科易迅信息化系统4篇初中物理6篇高中数学24篇初中数学123篇高中数学解题方法汇总9篇初中数学解题方法汇总16篇笔记3篇初中英语4篇 Mint-ui2篇 ShardingSphere1篇微信小程序2篇 Android4篇 SpringBoot29篇 Java17篇 Freemarker15篇 Redis11篇 Solr8篇 RabbitMQ11篇 Dubbo4篇 Nginx9篇 SQL(Mysql,Oracle)3篇华科易迅创客平台项目21篇 Git SpringCloud10篇 NodeJS3篇 Shiro2篇 VUE18篇 Docker1篇 Angular2篇 Fdfs分布文件系统3篇 Test11篇 Mybatis2篇 PostgreSQL1篇面试知识集锦4篇 Elasticsearch1篇 Java基础27篇 Activiti 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
2013	https://artofproblemsolving.com/wiki/index.php/2025_AIME_I_Problems/Problem_2?srsltid=AfmBOoq2LaPgTePTyMrsqi4ABjI0p41TnWHsy7XclsZMSWZQM7-cbIZY	Art of Problem Solving 2025 AIME I Problems/Problem 2 - 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 2025 AIME I Problems/Problem 2 Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search 2025 AIME I Problems/Problem 2 Contents 1 Problem 2 Video solution by grogg007 3 Solution 1 4 Solution 2 5 Solution 3 6 Solution 4 (Vectors) 7 Video Solution 1 by SpreadTheMathLove 8 Video Solution(Fast! Easy!) 9 Video Solution by Mathletes Corner 10 Video Solution by yjtest 11 See also Problem On points , , , and lie in that order on side with , , and . Points , , , and lie in that order on side with , , and . Let be the reflection of through , and let be the reflection of through . Quadrilateral has area . Find the area of heptagon . Video solution by grogg007 Solution 1 Note that the triangles outside have the same height as the unshaded triangles in . Since they have the same bases, the area of the heptagon is the same as the area of triangle . Therefore, we need to calculate the area of . Denote the length of as and the altitude of to as . Since , and the altitude of is . The area . The area of is equal to . ~alwaysgonnagiveyouup Solution 2 Because of reflections, and various triangles having the same bases, we can conclude that . Through the given lengths of on the left and on the right, we conclude that the lines through are parallel, and the sides are in a ratio. Because these lines are parallel, we can see that , are similar, and from our earlier ratio, we can give the triangles side ratios of , or area ratios of . Quadrilateral corresponds to the , which corresponds to the ratio . Dividing by , we get , and finally multiplying gives us our answer of ~shreyan.chethan, cleaned up by cweu001 Solution 3 By area lemma, we can see that the areas of the shaded areas are equivalent to the areas of the unshaded areas. Thus, we see that the desired area is equivalent to the area of the triangle . Since , we have , meaning . Thus, since , we can calculate . ~cweu001, cleaned up by shreyan.chethan Solution 4 (Vectors) Let △A B C be given with points D,E on segment A B such that A,D,E,B lie in that order, and A D=4, D E=16, and E B=8. Similarly, points F,G lie on segment A C such that A,F,G,C lie in that order with A F=13, F G=52, and G C=26. Note that the segment A B is partitioned into three parts with lengths 4:16:8, which simplifies to the ratio 1:4:2. Similarly, segment A C is partitioned into parts 13:52:26, also reducing to the ratio 1:4:2. This implies that the points D,E divide A B and points F,G divide A C in the same proportions. Because the divisions correspond proportionally on A B and A C, the line segments D E and F G are parallel to B C. In particular, triangles A D F, A E G, and A B C are similar by the Angle-Angle similarity criterion. Let us introduce vector notation for convenience. Represent points A,B,C by vectors A→,B→,C→ respectively. Then the points on A B satisfy: D→=A→+1 7(B→−A→), E→=A→+5 7(B→−A→), D=A+1 7(B−A), E=A+5 7(B−A), and the points on A C satisfy: F→=A→+1 7(C→−A→), G→=A→+5 7(C→−A→). F=A+1 7(C−A), G=A+5 7(C−A). The point M is the reflection of D about F, so M→=2 F→−D→, M=2 F−D, and N is the reflection of G about E, so N→=2 E→−G→. N=2 E−G. The polygon A F N B C E M has vertices at A→,F→,N→,B→,C→,E→,M→. Because M and N are reflections of points D and G about points F and E respectively, the triangles △D F M and △E G N are congruent to △A D F and △A E G respectively. Thus, the area of polygon A F N B C E M can be decomposed as Since △A D F, △A E G, and △A B C are similar with similarity ratios in lengths of 1:5:7, their areas are in the ratio Given the quadrilateral D E G F is the region between △A E G and △A D F, its area is where k=[A D F]. From the problem, [D E G F]=288, so Hence, Since the heptagon's area is [A B C]+[A D F]+[A E G]=[A B C]+k+25 k=[A B C]+26 k, but recalling that △D F M and △E G N are precisely the reflected copies of △A D F and △A E G that replace these smaller triangles inside A B C, the total area of the heptagon A F N B C E M is exactly the area of △A B C: 588. ~Pinotation Video Solution 1 by SpreadTheMathLove Video Solution(Fast! Easy!) ~MC Video Solution by Mathletes Corner ~GP102 Video Solution by yjtest See also 2025 AIME I (Problems • Answer Key • Resources) Preceded by Problem 1Followed by Problem 3 1•2•3•4•5•6•7•8•9•10•11•12•13•14•15 All AIME Problems and Solutions These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Retrieved from " Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
2014	https://www.scoilnet.ie/uploads/resources/28771/28507.pdf	4.4.5 - De Moivre’s Theorem II - Proof 4.4 - Algebra - Complex Numbers Leaving Certiﬁcate Mathematics Higher Level ONLY 4.4 - Algebra - Complex Numbers 4.4.5 - De Moivre’s Theorem II - Proof Higher Level ONLY 1 / 4 De Moivre’s Theorem Theorem: (cos θ + i sin θ)n = cos nθ + i sin nθ, for all n. 4.4 - Algebra - Complex Numbers 4.4.5 - De Moivre’s Theorem II - Proof Higher Level ONLY 2 / 4 Theorem: (cos θ + i sin θ)n = cos nθ + i sin nθ, 4.4 - Algebra - Complex Numbers 4.4.5 - De Moivre’s Theorem II - Proof Higher Level ONLY 3 / 4 Theorem: (cos θ + i sin θ)n = cos nθ + i sin nθ, for n ∈N. 4.4 - Algebra - Complex Numbers 4.4.5 - De Moivre’s Theorem II - Proof Higher Level ONLY 3 / 4 Theorem: (cos θ + i sin θ)n = cos nθ + i sin nθ, for n ∈N. Proof: 4.4 - Algebra - Complex Numbers 4.4.5 - De Moivre’s Theorem II - Proof Higher Level ONLY 3 / 4 Theorem: (cos θ + i sin θ)n = cos nθ + i sin nθ, for n ∈N. Proof: Use induction. 4.4 - Algebra - Complex Numbers 4.4.5 - De Moivre’s Theorem II - Proof Higher Level ONLY 3 / 4 Theorem: (cos θ + i sin θ)n = cos nθ + i sin nθ, for n ∈N. Proof: Use induction. Step 1: Test for n = 1. 4.4 - Algebra - Complex Numbers 4.4.5 - De Moivre’s Theorem II - Proof Higher Level ONLY 3 / 4 Theorem: (cos θ + i sin θ)n = cos nθ + i sin nθ, for n ∈N. Proof: Use induction. Step 1: Test for n = 1. LHS = (cos θ + i sin θ)1 4.4 - Algebra - Complex Numbers 4.4.5 - De Moivre’s Theorem II - Proof Higher Level ONLY 3 / 4 Theorem: (cos θ + i sin θ)n = cos nθ + i sin nθ, for n ∈N. Proof: Use induction. Step 1: Test for n = 1. LHS = (cos θ + i sin θ)1 = cos θ + i sin θ 4.4 - Algebra - Complex Numbers 4.4.5 - De Moivre’s Theorem II - Proof Higher Level ONLY 3 / 4 Theorem: (cos θ + i sin θ)n = cos nθ + i sin nθ, for n ∈N. Proof: Use induction. Step 1: Test for n = 1. LHS = (cos θ + i sin θ)1 = cos θ + i sin θ RHS = cos(1θ) + i sin(1θ) 4.4 - Algebra - Complex Numbers 4.4.5 - De Moivre’s Theorem II - Proof Higher Level ONLY 3 / 4 Theorem: (cos θ + i sin θ)n = cos nθ + i sin nθ, for n ∈N. Proof: Use induction. Step 1: Test for n = 1. LHS = (cos θ + i sin θ)1 = cos θ + i sin θ RHS = cos(1θ) + i sin(1θ) = cos θ + i sin θ 4.4 - Algebra - Complex Numbers 4.4.5 - De Moivre’s Theorem II - Proof Higher Level ONLY 3 / 4 Theorem: (cos θ + i sin θ)n = cos nθ + i sin nθ, for n ∈N. Proof: Use induction. Step 1: Test for n = 1. LHS = (cos θ + i sin θ)1 = cos θ + i sin θ RHS = cos(1θ) + i sin(1θ) = cos θ + i sin θ LHS = RHS 4.4 - Algebra - Complex Numbers 4.4.5 - De Moivre’s Theorem II - Proof Higher Level ONLY 3 / 4 Theorem: (cos θ + i sin θ)n = cos nθ + i sin nθ, for n ∈N. Proof: Use induction. Step 1: Test for n = 1. LHS = (cos θ + i sin θ)1 = cos θ + i sin θ RHS = cos(1θ) + i sin(1θ) = cos θ + i sin θ LHS = RHS ∴ Equation is true for n = 1. 4.4 - Algebra - Complex Numbers 4.4.5 - De Moivre’s Theorem II - Proof Higher Level ONLY 3 / 4 Theorem: (cos θ + i sin θ)n = cos nθ + i sin nθ, for n ∈N. Step 2: Assume true for n = k. 4.4 - Algebra - Complex Numbers 4.4.5 - De Moivre’s Theorem II - Proof Higher Level ONLY 4 / 4 Theorem: (cos θ + i sin θ)n = cos nθ + i sin nθ, for n ∈N. Step 2: Assume true for n = k. i.e. (cos θ + i sin θ)k = 4.4 - Algebra - Complex Numbers 4.4.5 - De Moivre’s Theorem II - Proof Higher Level ONLY 4 / 4 Theorem: (cos θ + i sin θ)n = cos nθ + i sin nθ, for n ∈N. Step 2: Assume true for n = k. i.e. (cos θ + i sin θ)k = cos kθ + i sin kθ 4.4 - Algebra - Complex Numbers 4.4.5 - De Moivre’s Theorem II - Proof Higher Level ONLY 4 / 4 Theorem: (cos θ + i sin θ)n = cos nθ + i sin nθ, for n ∈N. Step 2: Assume true for n = k. i.e. (cos θ + i sin θ)k = cos kθ + i sin kθ Step 3: Prove true for n = k + 1, assuming true for n = k. 4.4 - Algebra - Complex Numbers 4.4.5 - De Moivre’s Theorem II - Proof Higher Level ONLY 4 / 4 Theorem: (cos θ + i sin θ)n = cos nθ + i sin nθ, for n ∈N. Step 2: Assume true for n = k. i.e. (cos θ + i sin θ)k = cos kθ + i sin kθ Step 3: Prove true for n = k + 1, assuming true for n = k. (cos θ + i sin θ)k+1 4.4 - Algebra - Complex Numbers 4.4.5 - De Moivre’s Theorem II - Proof Higher Level ONLY 4 / 4 Theorem: (cos θ + i sin θ)n = cos nθ + i sin nθ, for n ∈N. Step 2: Assume true for n = k. i.e. (cos θ + i sin θ)k = cos kθ + i sin kθ Step 3: Prove true for n = k + 1, assuming true for n = k. (cos θ + i sin θ)k+1 = (cos θ + i sin θ)k(cos θ + i sin θ) 4.4 - Algebra - Complex Numbers 4.4.5 - De Moivre’s Theorem II - Proof Higher Level ONLY 4 / 4 Theorem: (cos θ + i sin θ)n = cos nθ + i sin nθ, for n ∈N. Step 2: Assume true for n = k. i.e. (cos θ + i sin θ)k = cos kθ + i sin kθ Step 3: Prove true for n = k + 1, assuming true for n = k. (cos θ + i sin θ)k+1 = (cos θ + i sin θ)k(cos θ + i sin θ) = (cos kθ + i sin kθ)(cos θ + i sin θ) . . . assumed 4.4 - Algebra - Complex Numbers 4.4.5 - De Moivre’s Theorem II - Proof Higher Level ONLY 4 / 4 Theorem: (cos θ + i sin θ)n = cos nθ + i sin nθ, for n ∈N. Step 2: Assume true for n = k. i.e. (cos θ + i sin θ)k = cos kθ + i sin kθ Step 3: Prove true for n = k + 1, assuming true for n = k. (cos θ + i sin θ)k+1 = (cos θ + i sin θ)k(cos θ + i sin θ) = (cos kθ + i sin kθ)(cos θ + i sin θ) . . . assumed = cos(kθ + θ) + i sin(kθ + θ) 4.4 - Algebra - Complex Numbers 4.4.5 - De Moivre’s Theorem II - Proof Higher Level ONLY 4 / 4 Theorem: (cos θ + i sin θ)n = cos nθ + i sin nθ, for n ∈N. Step 2: Assume true for n = k. i.e. (cos θ + i sin θ)k = cos kθ + i sin kθ Step 3: Prove true for n = k + 1, assuming true for n = k. (cos θ + i sin θ)k+1 = (cos θ + i sin θ)k(cos θ + i sin θ) = (cos kθ + i sin kθ)(cos θ + i sin θ) . . . assumed = cos(kθ + θ) + i sin(kθ + θ) = cos [(k + 1)θ] + i sin [(k + 1)θ] 4.4 - Algebra - Complex Numbers 4.4.5 - De Moivre’s Theorem II - Proof Higher Level ONLY 4 / 4 Theorem: (cos θ + i sin θ)n = cos nθ + i sin nθ, for n ∈N. Step 2: Assume true for n = k. i.e. (cos θ + i sin θ)k = cos kθ + i sin kθ Step 3: Prove true for n = k + 1, assuming true for n = k. (cos θ + i sin θ)k+1 = (cos θ + i sin θ)k(cos θ + i sin θ) = (cos kθ + i sin kθ)(cos θ + i sin θ) . . . assumed = cos(kθ + θ) + i sin(kθ + θ) = cos [(k + 1)θ] + i sin [(k + 1)θ] Step 4: Conclusion. 4.4 - Algebra - Complex Numbers 4.4.5 - De Moivre’s Theorem II - Proof Higher Level ONLY 4 / 4 Theorem: (cos θ + i sin θ)n = cos nθ + i sin nθ, for n ∈N. Step 2: Assume true for n = k. i.e. (cos θ + i sin θ)k = cos kθ + i sin kθ Step 3: Prove true for n = k + 1, assuming true for n = k. (cos θ + i sin θ)k+1 = (cos θ + i sin θ)k(cos θ + i sin θ) = (cos kθ + i sin kθ)(cos θ + i sin θ) . . . assumed = cos(kθ + θ) + i sin(kθ + θ) = cos [(k + 1)θ] + i sin [(k + 1)θ] Step 4: Conclusion. True for n = k + 1, 4.4 - Algebra - Complex Numbers 4.4.5 - De Moivre’s Theorem II - Proof Higher Level ONLY 4 / 4 Theorem: (cos θ + i sin θ)n = cos nθ + i sin nθ, for n ∈N. Step 2: Assume true for n = k. i.e. (cos θ + i sin θ)k = cos kθ + i sin kθ Step 3: Prove true for n = k + 1, assuming true for n = k. (cos θ + i sin θ)k+1 = (cos θ + i sin θ)k(cos θ + i sin θ) = (cos kθ + i sin kθ)(cos θ + i sin θ) . . . assumed = cos(kθ + θ) + i sin(kθ + θ) = cos [(k + 1)θ] + i sin [(k + 1)θ] Step 4: Conclusion. True for n = k + 1, assuming true for n = k. 4.4 - Algebra - Complex Numbers 4.4.5 - De Moivre’s Theorem II - Proof Higher Level ONLY 4 / 4 Theorem: (cos θ + i sin θ)n = cos nθ + i sin nθ, for n ∈N. Step 2: Assume true for n = k. i.e. (cos θ + i sin θ)k = cos kθ + i sin kθ Step 3: Prove true for n = k + 1, assuming true for n = k. (cos θ + i sin θ)k+1 = (cos θ + i sin θ)k(cos θ + i sin θ) = (cos kθ + i sin kθ)(cos θ + i sin θ) . . . assumed = cos(kθ + θ) + i sin(kθ + θ) = cos [(k + 1)θ] + i sin [(k + 1)θ] Step 4: Conclusion. True for n = k + 1, assuming true for n = k. But true for n = 1. 4.4 - Algebra - Complex Numbers 4.4.5 - De Moivre’s Theorem II - Proof Higher Level ONLY 4 / 4 Theorem: (cos θ + i sin θ)n = cos nθ + i sin nθ, for n ∈N. Step 2: Assume true for n = k. i.e. (cos θ + i sin θ)k = cos kθ + i sin kθ Step 3: Prove true for n = k + 1, assuming true for n = k. (cos θ + i sin θ)k+1 = (cos θ + i sin θ)k(cos θ + i sin θ) = (cos kθ + i sin kθ)(cos θ + i sin θ) . . . assumed = cos(kθ + θ) + i sin(kθ + θ) = cos [(k + 1)θ] + i sin [(k + 1)θ] Step 4: Conclusion. True for n = k + 1, assuming true for n = k. But true for n = 1. ∴ True for n = 2, 4.4 - Algebra - Complex Numbers 4.4.5 - De Moivre’s Theorem II - Proof Higher Level ONLY 4 / 4 Theorem: (cos θ + i sin θ)n = cos nθ + i sin nθ, for n ∈N. Step 2: Assume true for n = k. i.e. (cos θ + i sin θ)k = cos kθ + i sin kθ Step 3: Prove true for n = k + 1, assuming true for n = k. (cos θ + i sin θ)k+1 = (cos θ + i sin θ)k(cos θ + i sin θ) = (cos kθ + i sin kθ)(cos θ + i sin θ) . . . assumed = cos(kθ + θ) + i sin(kθ + θ) = cos [(k + 1)θ] + i sin [(k + 1)θ] Step 4: Conclusion. True for n = k + 1, assuming true for n = k. But true for n = 1. ∴ True for n = 2, 3, 4.4 - Algebra - Complex Numbers 4.4.5 - De Moivre’s Theorem II - Proof Higher Level ONLY 4 / 4 Theorem: (cos θ + i sin θ)n = cos nθ + i sin nθ, for n ∈N. Step 2: Assume true for n = k. i.e. (cos θ + i sin θ)k = cos kθ + i sin kθ Step 3: Prove true for n = k + 1, assuming true for n = k. (cos θ + i sin θ)k+1 = (cos θ + i sin θ)k(cos θ + i sin θ) = (cos kθ + i sin kθ)(cos θ + i sin θ) . . . assumed = cos(kθ + θ) + i sin(kθ + θ) = cos [(k + 1)θ] + i sin [(k + 1)θ] Step 4: Conclusion. True for n = k + 1, assuming true for n = k. But true for n = 1. ∴ True for n = 2, 3, . . . and all n ∈N 4.4 - Algebra - Complex Numbers 4.4.5 - De Moivre’s Theorem II - Proof Higher Level ONLY 4 / 4 Theorem: (cos θ + i sin θ)n = cos nθ + i sin nθ, for n ∈N. Step 2: Assume true for n = k. i.e. (cos θ + i sin θ)k = cos kθ + i sin kθ Step 3: Prove true for n = k + 1, assuming true for n = k. (cos θ + i sin θ)k+1 = (cos θ + i sin θ)k(cos θ + i sin θ) = (cos kθ + i sin kθ)(cos θ + i sin θ) . . . assumed = cos(kθ + θ) + i sin(kθ + θ) = cos [(k + 1)θ] + i sin [(k + 1)θ] Step 4: Conclusion. True for n = k + 1, assuming true for n = k. But true for n = 1. ∴ True for n = 2, 3, . . . and all n ∈N by induction. 4.4 - Algebra - Complex Numbers 4.4.5 - De Moivre’s Theorem II - Proof Higher Level ONLY 4 / 4
2015	https://gap-packages.github.io/numericalsgps/doc/chap6.html	Goto Chapter: Top 1 2 3 4 5 6 7 8 9 10 11 12 13 14 A B C Bib Ind [Top of Book] [Contents] [Previous Chapter] [Next Chapter] [MathJax on] [Style] 6 Irreducible numerical semigroups 6.1 Irreducible numerical semigroups 6.1-1 IsIrreducible 6.1-2 IsSymmetric 6.1-3 IsPseudoSymmetric 6.1-4 AnIrreducibleNumericalSemigroupWithFrobeniusNumber 6.1-5 IrreducibleNumericalSemigroupsWithFrobeniusNumber 6.1-6 IrreducibleNumericalSemigroupsWithFrobeniusNumberAndMultiplicity 6.1-7 DecomposeIntoIrreducibles 6.2 Complete intersection numerical semigroups 6.2-1 AsGluingOfNumericalSemigroups 6.2-2 IsCompleteIntersection 6.2-3 CompleteIntersectionNumericalSemigroupsWithFrobeniusNumber 6.2-4 IsFree 6.2-5 FreeNumericalSemigroupsWithFrobeniusNumber 6.2-6 IsTelescopic 6.2-7 TelescopicNumericalSemigroupsWithFrobeniusNumber 6.2-8 IsUniversallyFree 6.2-9 IsNumericalSemigroupAssociatedIrreduciblePlanarCurveSingularity 6.2-10 NumericalSemigroupsPlanarSingularityWithFrobeniusNumber 6.2-11 IsAperySetGammaRectangular 6.2-12 IsAperySetBetaRectangular 6.2-13 IsAperySetAlphaRectangular 6.3 Almost-symmetric numerical semigroups 6.3-1 AlmostSymmetricNumericalSemigroupsFromIrreducible 6.3-2 AlmostSymmetricNumericalSemigroupsFromIrreducibleAndGivenType 6.3-3 IsAlmostSymmetric 6.3-4 AlmostSymmetricNumericalSemigroupsWithFrobeniusNumber 6.3-5 AlmostSymmetricNumericalSemigroupsWithFrobeniusNumberAndType 6.4 Several approaches generalizing the concept of symmetry 6.4-1 IsGeneralizedGorenstein 6.4-2 IsNearlyGorenstein 6.4-3 NearlyGorensteinVectors 6.4-4 IsGeneralizedAlmostSymmetric 6 Irreducible numerical semigroups An irreducible numerical semigroup is a semigroup that cannot be expressed as the intersection of numerical semigroups properly containing it. It is not difficult to prove that a semigroup is irreducible if and only if it is maximal (with respect to set inclusion) in the set of all numerical semigroups having its same Frobenius number (see [RB03]). Hence, according to [FGR87] (respectively [BDF97]), symmetric (respectively pseudo-symmetric) numerical semigroups are those irreducible numerical semigroups with odd (respectively even) Frobenius number. In [RGGJ03] it is shown that a nontrivial numerical semigroup is irreducible if and only if it has only one special gap. We use this characterization. In old versions of the package, we first constructed an irreducible numerical semigroup with the given Frobenius number (as explained in [RG04]), and then we constructed the rest from it. The present version uses a faster procedure presented in [BR13]. Every numerical semigroup can be expressed as an intersection of irreducible numerical semigroups. If S can be expressed as S=S_1∩ ⋯∩ S_n, with S_i irreducible numerical semigroups, and no factor can be removed, then we say that this decomposition is minimal. Minimal decompositions can be computed by using Algorithm 26 in [RGGJ03]. 6.1 Irreducible numerical semigroups In this section we provide membership tests to the two families that conform the set of irreducible numerical semigroups. We also give a procedure to compute the set of all irreducible numerical semigroups with fixed Frobenius number (or equivalently genus, since for irreducible numerical semigroups once the Frobenius number is fixed, so is the genus). Also we give a function to compute the decomposition of a numerical semigroup as an intersection of irreducible numerical semigroups. 6.1-1 IsIrreducible \| \| \| --- \| \| ‣ IsIrreducible( s ) \| ( property ) \| \| \| \| --- \| \| ‣ IsIrreducibleNumericalSemigroup( s ) \| ( property ) \| s is a numerical semigroup. The output is true if s is irreducible, false otherwise. This filter implies IsAlmostSymmetricNumericalSemigroup (6.3-3) and IsAcuteNumericalSemigroup (3.1-30). ``` gap> IsIrreducible(NumericalSemigroup(4,6,9)); true gap> IsIrreducibleNumericalSemigroup(NumericalSemigroup(4,6,7,9)); false ``` 6.1-2 IsSymmetric \| \| \| --- \| \| ‣ IsSymmetric( s ) \| ( attribute ) \| \| \| \| --- \| \| ‣ IsSymmetricNumericalSemigroup( s ) \| ( attribute ) \| s is a numerical semigroup. The output is true if s is symmetric, false otherwise. This filter implies IsIrreducibleNumericalSemigroup (6.1-1). ``` gap> IsSymmetric(NumericalSemigroup(10,23)); true gap> IsSymmetricNumericalSemigroup(NumericalSemigroup(10,11,23)); false ``` 6.1-3 IsPseudoSymmetric \| \| \| --- \| \| ‣ IsPseudoSymmetric( s ) \| ( property ) \| \| \| \| --- \| \| ‣ IsPseudoSymmetricNumericalSemigroup( s ) \| ( property ) \| s is a numerical semigroup. The output is true if s is pseudo-symmetric, false otherwise. This filter implies IsIrreducibleNumericalSemigroup (6.1-1). ``` gap> IsPseudoSymmetric(NumericalSemigroup(6,7,8,9,11)); true gap> IsPseudoSymmetricNumericalSemigroup(NumericalSemigroup(4,6,9)); false ``` 6.1-4 AnIrreducibleNumericalSemigroupWithFrobeniusNumber \| \| \| --- \| \| ‣ AnIrreducibleNumericalSemigroupWithFrobeniusNumber( f ) \| ( function ) \| f is an integer. When f=0 or f≤ -2, the output is fail. Otherwise, the output is an irreducible numerical semigroup with Frobenius number f. From the way the procedure is implemented, the resulting semigroup has at most four generators (see [RG04]). ``` gap> s := AnIrreducibleNumericalSemigroupWithFrobeniusNumber(28); gap> MinimalGenerators(s); [ 3, 17, 31 ] gap> FrobeniusNumber(s); 28 ``` 6.1-5 IrreducibleNumericalSemigroupsWithFrobeniusNumber \| \| \| --- \| \| ‣ IrreducibleNumericalSemigroupsWithFrobeniusNumber( f ) \| ( function ) \| f is an integer. The output is the set of all irreducible numerical semigroups with Frobenius number f. The algorithm is inspired in [BR13]. ``` gap> Length(IrreducibleNumericalSemigroupsWithFrobeniusNumber(19)); 20 ``` 6.1-6 IrreducibleNumericalSemigroupsWithFrobeniusNumberAndMultiplicity \| \| \| --- \| \| ‣ IrreducibleNumericalSemigroupsWithFrobeniusNumberAndMultiplicity( f, m ) \| ( function ) \| f and m are integers. The output is the set of all irreducible numerical semigroups with Frobenius number f and multiplicity m. The implementation appears in [BOR21]. ``` gap> Length(IrreducibleNumericalSemigroupsWithFrobeniusNumberAndMultiplicity(31,11)); 16 ``` 6.1-7 DecomposeIntoIrreducibles \| \| \| --- \| \| ‣ DecomposeIntoIrreducibles( s ) \| ( function ) \| s is a numerical semigroup. The output is a set of irreducible numerical semigroups containing it. These elements appear in a minimal decomposition of s as intersection into irreducibles. ``` gap> DecomposeIntoIrreducibles(NumericalSemigroup(5,6,8)); [ , ] ``` 6.2 Complete intersection numerical semigroups The cardinality of a minimal presentation of a numerical semigroup is always greater than or equal to its embedding dimension minus one. Complete intersection numerical semigroups are numerical semigroups reaching this bound, and they are irreducible. It can be shown that every complete intersection (other that N) is a complete intersection if and only if it is the gluing of two complete intersections. When in this gluing, one of the copies is isomorphic to N, then we obtain a free semigroup in the sense of [BC77]. Two special kinds of free semigroups are telescopic semigroups ([KP95]) and those associated to an irreducible planar curve ([Zar86]). We use the algorithms presented in [AG13] to find the set of all complete intersections (also free, telescopic and associated to irreducible planar curves) numerical semigroups with given Frobenius number. 6.2-1 AsGluingOfNumericalSemigroups \| \| \| --- \| \| ‣ AsGluingOfNumericalSemigroups( s ) \| ( function ) \| s is a numerical semigroup. Returns all partitions {A_1,A_2} of the minimal generating set of s such that s is a gluing of ⟨ A_1⟩ and ⟨ A_2⟩ by gcd(A_1)gcd(A_2). ``` gap> s := NumericalSemigroup( 10, 15, 16 ); gap> AsGluingOfNumericalSemigroups(s); [ [ [ 10, 15 ], [ 16 ] ], [ [ 10, 16 ], [ 15 ] ] ] gap> s := NumericalSemigroup( 18, 24, 34, 46, 51, 61, 74, 8 ); gap> AsGluingOfNumericalSemigroups(s); [ ] ``` 6.2-2 IsCompleteIntersection \| \| \| --- \| \| ‣ IsCompleteIntersection( s ) \| ( property ) \| \| \| \| --- \| \| ‣ IsACompleteIntersectionNumericalSemigroup( s ) \| ( property ) \| s is a numerical semigroup. The output is true if the numerical semigroup is a complete intersection, that is, the cardinality of a (any) minimal presentation equals its embedding dimension minus one. This filter implies IsSymmetricNumericalSemigroup (6.1-2) and IsCyclotomicNumericalSemigroup (10.1-8). ``` gap> s := NumericalSemigroup( 10, 15, 16 ); gap> IsCompleteIntersection(s); true gap> s := NumericalSemigroup( 18, 24, 34, 46, 51, 61, 74, 8 ); gap> IsACompleteIntersectionNumericalSemigroup(s); false ``` 6.2-3 CompleteIntersectionNumericalSemigroupsWithFrobeniusNumber \| \| \| --- \| \| ‣ CompleteIntersectionNumericalSemigroupsWithFrobeniusNumber( f ) \| ( function ) \| f is an integer. The output is the set of all complete intersection numerical semigroups with Frobenius number f. ``` gap> Length(CompleteIntersectionNumericalSemigroupsWithFrobeniusNumber(57)); 34 ``` 6.2-4 IsFree \| \| \| --- \| \| ‣ IsFree( s ) \| ( property ) \| \| \| \| --- \| \| ‣ IsFreeNumericalSemigroup( s ) \| ( property ) \| s is a numerical semigroup. The output is true if the numerical semigroup is free in the sense of [BC77]: it is either N or the gluing of a copy of N with a free numerical semigroup. This filter implies IsACompleteIntersectionNumericalSemigroup (6.2-2). ``` gap> IsFree(NumericalSemigroup(10,15,16)); true gap> IsFreeNumericalSemigroup(NumericalSemigroup(3,5,7)); false ``` 6.2-5 FreeNumericalSemigroupsWithFrobeniusNumber \| \| \| --- \| \| ‣ FreeNumericalSemigroupsWithFrobeniusNumber( f ) \| ( function ) \| f is an integer. The output is the set of all free numerical semigroups with Frobenius number f. ``` gap> Length(FreeNumericalSemigroupsWithFrobeniusNumber(57)); 33 ``` 6.2-6 IsTelescopic \| \| \| --- \| \| ‣ IsTelescopic( s ) \| ( property ) \| \| \| \| --- \| \| ‣ IsTelescopicNumericalSemigroup( s ) \| ( property ) \| s is a numerical semigroup. The output is true if the numerical semigroup is telescopic in the sense of [KP95]: it is either N or the gluing of ⟨ n_e⟩ and s'=⟨ n_1/d,..., n_e-1/d⟩, and s' is again a telescopic numerical semigroup, where n_1 < ⋯ < n_e are the minimal generators of s. This filter implies IsAperySetBetaRectangular (6.2-12) and IsFree (6.2-4). ``` gap> IsTelescopic(NumericalSemigroup(4,11,14)); false gap> IsTelescopicNumericalSemigroup(NumericalSemigroup(4,11,14)); false gap> IsFree(NumericalSemigroup(4,11,14)); true ``` 6.2-7 TelescopicNumericalSemigroupsWithFrobeniusNumber \| \| \| --- \| \| ‣ TelescopicNumericalSemigroupsWithFrobeniusNumber( f ) \| ( function ) \| f is an integer. The output is the set of all telescopic numerical semigroups with Frobenius number f. ``` gap> Length(TelescopicNumericalSemigroupsWithFrobeniusNumber(57)); 20 ``` 6.2-8 IsUniversallyFree \| \| \| --- \| \| ‣ IsUniversallyFree( s ) \| ( property ) \| \| \| \| --- \| \| ‣ IsUniversallyFreeNumericalSemigroup( s ) \| ( property ) \| s is a numerical semigroup. The output is true if the numerical semigroup is free for all the arrangements of its minimal generators. This filter implies IsTelescopic (6.2-6). ``` gap> s:=NumericalSemigroup(10,15,18);; gap> IsUniversallyFree(s); true gap> s:=NumericalSemigroup(4,6,9);; gap> IsFree(s); true gap> IsUniversallyFree(s); false ``` 6.2-9 IsNumericalSemigroupAssociatedIrreduciblePlanarCurveSingularity \| \| \| --- \| \| ‣ IsNumericalSemigroupAssociatedIrreduciblePlanarCurveSingularity( s ) \| ( property ) \| s is a numerical semigroup. The output is true if the numerical semigroup is associated to an irreducible planar curve singularity ([Zar86]). These semigroups are telescopic. This filter implies IsAperySetAlphaRectangular (6.2-13) and IsTelescopicNumericalSemigroup (6.2-6). ``` gap> ns := NumericalSemigroup(4,11,14);; gap> IsNumericalSemigroupAssociatedIrreduciblePlanarCurveSingularity(ns); false gap> ns := NumericalSemigroup(4,11,19);; gap> IsNumericalSemigroupAssociatedIrreduciblePlanarCurveSingularity(ns); true ``` 6.2-10 NumericalSemigroupsPlanarSingularityWithFrobeniusNumber \| \| \| --- \| \| ‣ NumericalSemigroupsPlanarSingularityWithFrobeniusNumber( f ) \| ( function ) \| f is an integer. The output is the set of all numerical semigroups associated to irreducible planar curves singularities with Frobenius number f. ``` gap> Length(NumericalSemigroupsPlanarSingularityWithFrobeniusNumber(57)); 7 ``` 6.2-11 IsAperySetGammaRectangular \| \| \| --- \| \| ‣ IsAperySetGammaRectangular( S ) \| ( function ) \| S is a numerical semigroup. Test for the γ-rectangularity of the Apéry Set of a numerical semigroup. This test is the implementation of the algorithm given in [DMS14]. Numerical Semigroups with this property are free and thus complete intersections. This filter implies IsFreeNumericalSemigroup (6.2-4). ``` gap> s:=NumericalSemigroup(30, 35, 42, 47, 148, 153, 157, 169, 181, 193);; gap> IsAperySetGammaRectangular(s); false gap> s:=NumericalSemigroup(4,6,11);; gap> IsAperySetGammaRectangular(s); true ``` 6.2-12 IsAperySetBetaRectangular \| \| \| --- \| \| ‣ IsAperySetBetaRectangular( S ) \| ( function ) \| S is a numerical semigroup. Test for the β-rectangularity of the Apéry Set of a numerical semigroup. This test is the implementation of the algorithm given in [DMS14]; β-rectangularity implies γ-rectangularity. This filter implies IsAperySetGammaRectangular (6.2-11). ``` gap> s:=NumericalSemigroup(30, 35, 42, 47, 148, 153, 157, 169, 181, 193);; gap> IsAperySetBetaRectangular(s); false gap> s:=NumericalSemigroup(4,6,11);; gap> IsAperySetBetaRectangular(s); true ``` 6.2-13 IsAperySetAlphaRectangular \| \| \| --- \| \| ‣ IsAperySetAlphaRectangular( S ) \| ( function ) \| S is a numerical semigroup. Test for the α-rectangularity of the Apéry Set of a numerical semigroup. This test is the implementation of the algorithm given in [DMS14]; α-rectangularity implies β-rectangularity. This filter implies IsAperySetBetaRectangular (6.2-12). ``` gap> s:=NumericalSemigroup(30, 35, 42, 47, 148, 153, 157, 169, 181, 193);; gap> IsAperySetAlphaRectangular(s); false gap> s:=NumericalSemigroup(4,6,11);; gap> IsAperySetAlphaRectangular(s); true ``` 6.3 Almost-symmetric numerical semigroups A numerical semigroup is almost-symmetric ([BF97]) if its genus is the arithmetic mean of its Frobenius number and type. We use a procedure presented in [RG14] to determine the set of all almost-symmetric numerical semigroups with given Frobenius number. In order to do this, we first calculate the set of all almost-symmetric numerical semigroups that can be constructed from an irreducible numerical semigroup. 6.3-1 AlmostSymmetricNumericalSemigroupsFromIrreducible \| \| \| --- \| \| ‣ AlmostSymmetricNumericalSemigroupsFromIrreducible( s ) \| ( function ) \| s is an irreducible numerical semigroup. The output is the set of almost-symmetric numerical semigroups that can be constructed from s by removing some of its generators (as explained in [RG14]). ``` gap> ns := NumericalSemigroup(5,8,9,11);; gap> AlmostSymmetricNumericalSemigroupsFromIrreducible(ns); [ , , ] gap> List(last,MinimalGeneratingSystemOfNumericalSemigroup); [ [ 5, 8, 9, 11 ], [ 5, 8, 11, 14, 17 ], [ 5, 9, 11, 13, 17 ] ] ``` 6.3-2 AlmostSymmetricNumericalSemigroupsFromIrreducibleAndGivenType \| \| \| --- \| \| ‣ AlmostSymmetricNumericalSemigroupsFromIrreducibleAndGivenType( s, t ) \| ( function ) \| s is an irreducible numerical semigroup and t is a positive integer. The output is the set of almost-symmetric numerical semigroups with type t that can be constructed from s by removing some of its generators (as explained in [BOR18]). ``` gap> ns := NumericalSemigroup(5,8,9,11);; gap> AlmostSymmetricNumericalSemigroupsFromIrreducibleAndGivenType(ns,4); [ , ] gap> List(last,MinimalGenerators); [ [ 5, 8, 11, 14, 17 ], [ 5, 9, 11, 13, 17 ] ] ``` 6.3-3 IsAlmostSymmetric \| \| \| --- \| \| ‣ IsAlmostSymmetric( s ) \| ( function ) \| \| \| \| --- \| \| ‣ IsAlmostSymmetricNumericalSemigroup( s ) \| ( function ) \| s is a numerical semigroup. The output is true if the numerical semigroup is almost symmetric. ``` gap> IsAlmostSymmetric(NumericalSemigroup(5,8,11,14,17)); true gap> IsAlmostSymmetricNumericalSemigroup(NumericalSemigroup(5,8,11,14,17)); true ``` 6.3-4 AlmostSymmetricNumericalSemigroupsWithFrobeniusNumber \| \| \| --- \| \| ‣ AlmostSymmetricNumericalSemigroupsWithFrobeniusNumber( f[, ts] ) \| ( function ) \| f is an integer, and so is ts. The output is the set of all almost symmetric numerical semigroups with Frobenius number f, and type greater than or equal to ts. If ts is not specified, then it is considered to be equal to one, and so the output is the set of all almost symmetric numerical semigroups with Frobenius number f. ``` gap> Length(AlmostSymmetricNumericalSemigroupsWithFrobeniusNumber(12)); 15 gap> Length(IrreducibleNumericalSemigroupsWithFrobeniusNumber(12)); 2 gap> List(AlmostSymmetricNumericalSemigroupsWithFrobeniusNumber(12,4),Type); [ 12, 10, 8, 8, 6, 6, 6, 6, 4, 4, 4, 4, 4 ] ``` 6.3-5 AlmostSymmetricNumericalSemigroupsWithFrobeniusNumberAndType \| \| \| --- \| \| ‣ AlmostSymmetricNumericalSemigroupsWithFrobeniusNumberAndType( f, t ) \| ( function ) \| f is an integer and so is t. The output is the set of all almost symmetric numerical semigroups with Frobenius number f and type t. ``` gap> Length(AlmostSymmetricNumericalSemigroupsWithFrobeniusNumberAndType(12,4)); 5 ``` 6.4 Several approaches generalizing the concept of symmetry Let S be a numerical semigroup and let R be its semigroup ring K. We say that S has the generalized Gorenstein property if its semigroup ring has this property. For the definition and characterization of generalized Gorenstein rings please see [tttt17]. A numerical semigroup is said to be nearly Gorenstein if its maximal ideal is contained in its trace ideal [HHS19]. Every almost symmetric numerical semigroup is nearly Gorenstein. A numerical semigroup S with canonical ideal K is a generalized almost symmetric numerical semigroup if either 2K=K (symmetric) or 2K∖ K={F(S)-x_1,dots, F(S)-x_r,F(S)} for some x_1,dots,x_r ∈ M∖ 2M (minimal generators) and x_i-x_jnot\in (S-M)∖ S (not pseudo-Frobenius numbers), see [DS21]. As expected, every almost symmetric numerical semigroup is a generalized almost symmetric numerical semigroup. 6.4-1 IsGeneralizedGorenstein \| \| \| --- \| \| ‣ IsGeneralizedGorenstein( s ) \| ( property ) \| s is a numerical semigroup. The output is true if the semigroup ring K is generalized Gorenstein using the characterization by Goto-Kumashiro [MK17]. ``` gap> s:=NumericalSemigroup(3,7,8);; gap> IsAlmostSymmetric(s); false gap> IsGeneralizedGorenstein(s); true ``` 6.4-2 IsNearlyGorenstein \| \| \| --- \| \| ‣ IsNearlyGorenstein( s ) \| ( property ) \| s is a numerical semigroup. The output is true if the semigroup is nearly Gorenstein, and false otherwise. ``` gap> s:=NumericalSemigroup(10,11,12,25);; gap> IsAlmostSymmetric(s); false gap> IsNearlyGorenstein(s); true gap> s:=NumericalSemigroup(3,7,8);; gap> IsNearlyGorenstein(s); false ``` 6.4-3 NearlyGorensteinVectors \| \| \| --- \| \| ‣ NearlyGorensteinVectors( s ) \| ( operation ) \| s is a numerical semigroup. The output is a lists of lists (making the cartesian product of them yields all possible NG-vectors). If n_i is the ith generator of s, in the ith position of the list it returns all pseudo-Frobenius numbers f of s such that n_i+f-f' is in s for all f a pseudo-Frobenius number of s, [MS21]. ``` gap> s:=NumericalSemigroup(10,11,12,25);; gap> IsAlmostSymmetric(s); false gap> IsNearlyGorenstein(s); true gap> s:=NumericalSemigroup(3,7,8);; gap> IsNearlyGorenstein(s); false ``` 6.4-4 IsGeneralizedAlmostSymmetric \| \| \| --- \| \| ‣ IsGeneralizedAlmostSymmetric( s ) \| ( property ) \| s is a numerical semigroup. Determines whether or not s is a generalized almost symmetric numerical semigroup. ``` gap> s:=NumericalSemigroup(9, 24, 39, 43, 77);; gap> IsGeneralizedAlmostSymmetric(s); true gap> IsAlmostSymmetric(s); false ``` [Top of Book] [Contents] [Previous Chapter] [Next Chapter] Goto Chapter: Top 1 2 3 4 5 6 7 8 9 10 11 12 13 14 A B C Bib Ind generated by GAPDoc2HTML
2016	https://math.stackexchange.com/questions/3310421/reverse-triangle-inequality-proof-verification	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 Reverse triangle inequality proof verification Ask Question Asked Modified 6 years, 2 months ago Viewed 1k times 2 $\begingroup$ Whilst trying to come up with a proof for the reverse triangle inequality, I came up with this. I know that I overcomplicated things, but still wanted to know whether what I did was correct. What I need to prove is that $\forall x, y \in \mathbb{R}: \|x-y\|\geq \|\|xy\|\|$. I tackled this problem with a proof by contradiction. So, assume that $\forall x, y\in \mathbb{R}: \|x-y\| < \|\|xy\|\|$. Then I squared both sides in order to get rid of the absolute values. Because both the left and right hand side of the inequality are positive, the inequality stays the same. $(x-y)^2 < (\|xy\|)^2$ $x^2 - 2xy + y^2 < x^2 - 2\|xy\| + y^2$ Simplifying this gives $-2xy < -2\|xy\|$ Division by a negative number, so inequality changes $xy > \|xy\|$ Which can never be true because $\forall a \in \mathbb{R}: \|a\| \geq a$, thus we have a contradiction and so $\forall x, y \in \mathbb{R}: \|x-y\|\geq \|\|xy\|\|$ must be true. I'm fairly new to writing my own proofs, so any help would be much appreciated! real-analysis proof-verification Share asked Aug 1, 2019 at 13:13 mrMoonpenguinmrMoonpenguin 7566 bronze badges $\endgroup$ 1 2 $\begingroup$ Your proof looks good to me, and I do not think it's overly complicated. I would even say it's more elegant than the brute force one from the Wikipedia page :) $\endgroup$ janosch – janosch 2019-08-01 13:25:58 +00:00 Commented Aug 1, 2019 at 13:25 Add a comment \| 1 Answer 1 Reset to default 2 $\begingroup$ I think your proof is correct. There is an easier proof though. Notice that $\|x\| = \|x - y + y\| \leq \|x - y\| + \|y\|$ by the triangle equality. This expression rearranges to $\|xy\| \leq \|x - y\|$. Similarly, $\|y\| = \|y - x + x\| \leq \|y - x\| + \|x\| = \|x - y\| + \|x\|$, which rearranges to $\|yx\| \leq \|x - y\|$. Multiplying by -1 yields $\|xy\| \geq -\|x - y\|$. Combining this inequality with the previous one yields $-\|x - y\| \leq \|xy\| \leq \|x - y\|$, or $\|\|xy\|\| \leq \|x - y\|$. Share answered Aug 1, 2019 at 13:37 Amy NgoAmy Ngo 89044 silver badges77 bronze badges $\endgroup$ Add a comment \| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions real-analysis proof-verification See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Related 4 Generalization of Bernoulli's Inequality. 1 Proof of the Reverse Triangle Inequality 1 (Proof verification) Proving the equivalence between two statements about a limit. 1 Proof verification: Determine whether $f(x)$ is one-to-one or not Proof Verification: Infinitely many Primes Using Euclid's Algorithm 4 Is there a reverse triangle inequality equivalent for $L^p$ spaces with $0 2 Alternative proofs to inequality problems 1 Function limits proof problem. Hot Network Questions What happens if you miss cruise ship deadline at private island? Is existence always locational? Real structure on a complex torus I have a lot of PTO to take, which will make the deadline impossible Verify a Chinese ID Number Why is the definite article used in “Mi deporte favorito es el fútbol”? Copy command with cs names Calculating the node voltage Checking model assumptions at cluster level vs global level? Clinical-tone story about Earth making people violent Is encrypting the login keyring necessary if you have full disk encryption? Is it ok to place components "inside" the PCB Identifying a movie where a man relives the same day Numbers Interpreted in Smallest Valid Base Spectral Leakage & Phase Discontinuites How do trees drop their leaves? How can blood fuel space travel? What is the meaning of 率 in this report? A time-travel short fiction where a graphologist falls in love with a girl for having read letters she has not yet written… to another man Is it safe to route top layer traces under header pins, SMD IC? What can be said? Analog story - nuclear bombs used to neutralize global warming Where is the first repetition in the cumulative hierarchy up to elementary equivalence? Fix integral lower bound kerning in textstyle or smaller with unicode-math more hot questions Question feed
2017	https://www.ncbi.nlm.nih.gov/books/NBK513264/	An official website of the United States government The .gov means it's official. Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you're on a federal government site. The site is secure. The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely. Log in Account Logged in as:username Dashboard Publications Account settings Log out Access keys NCBI Homepage MyNCBI Homepage Main Content Main Navigation Browse Titles Advanced Help Disclaimer NCBI Bookshelf. A service of the National Library of Medicine, National Institutes of Health. StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan-. StatPearls [Internet]. Show details Treasure Island (FL): StatPearls Publishing; 2025 Jan-. Hepatocellular Adenoma Aparna P. Shreenath; Lafaine M. Grant; Arslan Kahloon. Author Information and Affiliations Authors Aparna P. Shreenath1; Lafaine M. Grant2; Arslan Kahloon3. Affiliations 1 University of Tennessee 2 University of Texas Southwestern Medical Center 3 University of Tennessee, Chattanooga Last Update: May 6, 2024. Continuing Education Activity Hepatocellular adenoma, also called "hepatic adenoma," is a rare, benign epithelial liver tumor frequently associated with oral contraceptive pill intake. Conditions like anabolic steroid abuse, Fanconi anemia, and aplastic anemia have also been implicated in the development of this lesion. Hepatic adenomas are benign but carry a higher hemorrhage and malignant transformation risk. Elective resection is recommended in all men with adenomas of any size and women with tumors larger than 5 cm. Rupture may complicate untreated lesions, manifesting with hemorrhage, severe abdominal pain, and hypotension, and may lead to hypovolemic shock and death. Women on oral contraceptives are most susceptible to liver adenoma rupture, which requires immediate recognition due to its potentially catastrophic consequences. This activity for healthcare professionals is designed to improve learners' competence in evaluating and managing hepatocellular adenoma. Participants gain profound insights into the condition's pathophysiology, symptomatology, and evidence-based diagnostic and treatment practices. Learners become prepared to collaborate within an interprofessional team caring for patients with or at risk for hepatic adenoma. Objectives: Identify the historical and physical examination features suggestive of a hepatocellular adenoma. Create a clinically guided diagnostic strategy for a person with suspected hepatic adenoma. Implement current evidence-based treatment options for an individual diagnosed with hepatocellular adenoma. Collaborate with the interprofessional team to educate, treat, and monitor individuals with or at risk for hepatocellular adenoma. Access free multiple choice questions on this topic. Introduction Hepatocellular adenoma (HCA), otherwise known as hepatic adenoma, is a rare, benign epithelial hepatic neoplasm often linked to exogenous estrogen intake, usually in the form of oral contraceptive pills. The lesion is also associated with steroid abuse, Fanconi anemia, aplastic anemia, metabolic syndrome, and glycogen storage disease (GSD). HCAs are benign but are at high risk of hemorrhage and malignant transformation. Although typically solitary, multiple adenomas can occur. Hepatic adenomatosis refers to the presence of 10 or more tumors. Advances in radiologic diagnosis and subtype classifications based on molecular behavior have emerged, which provide a more systematic approach to treating patients with hepatic adenomas. Elective resection is recommended in men with adenomas regardless of size and in women with adenomas greater than 5 cm. Liver Anatomy and Histology The liver is located inferior to the diaphragm and occupies the abdominal right upper quadrant (RUQ). This organ has a complex anatomical configuration crucial for its multifunctional roles. The liver is encased in the visceral peritoneum and extends from the midclavicular line to the right costal margin at the level of the 5th intercostal space. The organ's superoposterior aspect houses the bare area where the diaphragm and inferior vena cava (IVC) converge. The Couinaud classification subdivides the liver into 8 functionally independent segments based on vascularization, bile duct distribution, and lymphatic drainage. These wedge-shaped segments have apices directed toward the hepatic hilum and receive a single branch each of the bile duct, portal vein, and hepatic artery. Hepatic veins run between adjacent segments, ultimately draining into the IVC, while the middle hepatic vein delineates the liver's right and left lobes. The right hepatic vein further partitions the right lobe into posterior and anterior segments, while the falciform ligament separates the left lobe into medial and lateral segments. The portal vein horizontally divides segments into superior and inferior sections. Segment I is the caudate lobe, distinguished from other lobes by its unique characteristics, such as a dual blood supply and direct drainage into the IVC. Segments II, III, and IV constitute the left hepatic lobe sections. Segments V, VI, VII, and VIII constitute the right lobe hepatic sections. The liver's blood supply, primarily from the portal vein, enhances liver imaging during the portal venous phase—a property crucial for diagnostic purposes. Tumors supplied by the hepatic artery exhibit enhanced imaging during arterial phases, a principle utilized in therapeutic interventions like transarterial chemoembolization. Hepatic veins appear as anechoic tubes on ultrasound and drain into the IVC. The portal triad—consisting of portal veins, hepatic arteries, and bile ducts—manifests as echogenic foci within the liver parenchyma. The hepatic lobule serves as the primary functional unit, comprising hexagonal arrays of hepatocytes surrounding central veins. Hepatocytes organize into cords within lobules, enveloping sinusoids housing Kupffer and stellate cells. The lobule's organization into portal triads ensures efficient blood flow across zones delineated by their proximity to portal tracts or central veins. This alternative organization, known as the portal acinus, delineates functional zones. Zone 1 surrounds portal tracts involved in oxidative metabolism. Zone 3 envelopes central veins primarily engaged in drug biotransformation. Zone 2 exhibits mixed functionality. Etiology Chen and colleagues enhanced our understanding of hepatocellular adenoma formation when their 2002 study identified the role of β-catenin in the Wnt signaling pathway. Subsequently, Bioulac-Sage and associates developed a phenotypic-genotypic classification system for hepatocellular adenomas based on molecular behavior patterns.Other groups have since validated this classification scheme, which categorizes hepatocellular adenomas into 4 main groups: Hepatocyte Nuclear Factor -1α Inactivated Mutations Patients with hepatocyte nuclear factor-1α (HNF-1α) inactivated mutations comprise 35% to 40% of cases. The condition involves biallelic mutations of the T-cell factor-1 gene that encodes the transcription factor HNF-1α. This protein is involved in hepatocyte differentiation, liver development, and glucose and lipid metabolism. HCA lesions in this group are noticeably more steatotic on biopsy. Women are mostly affected. The mutation is likewise often associated with mature-onset diabetes of the young (MODY3). About 90% of the mutations are somatic, although MODY3 association can involve the germline and commonly manifests with adenomatosis. Lesion size smaller than 5 cm reduces complication risk. β-Catenin Activated Mutations Individuals with β-catenin activated mutationsconstitute 15% to 20% of liver adenoma cases. The condition is frequently linked to androgen exposure, glycogenesis, and familial adenomatous polyposis. Mild hepatic cytological or architectural abnormalities lead to an acinar, pseudoglandular pattern. The mutation is rarely detected in liver steatosis and inflammatory cases. Adenomas in this group have an increased malignant transformation risk. The lesions typically exhibit positivity for glutamine synthetase and abnormal β-catenin expression on immunohistochemical staining. Inflammatory Hepatocellular Adenomas Inflammatory HCA (IHCA) is seen in 40% to 50% of patients with HCA (see Image. Inflammatory Hepatocellular Adenoma Histopathology). Risk factors include female sex, high body mass index (BMI), excessive alcohol consumption, and systemic inflammatory syndrome. The lesions lack both HNF-1α and β-catenin mutations. IL-6 inflammatory pathway activation is implicated in IHCA development, manifesting with dystrophic vessels and telangiectasia. Specifically, the arteries have thick walls, and the sinusoids are dilated, a condition called "peliosis." Adenomas in this category were formerly called “telangiectatic focal nodular hyperplasia.” The tumors exhibit positivity for serum amyloid A (SAA) and C-reactive protein (CRP) and demonstrate inflammatory infiltrates and ductular reactions. Unclassified Type Patients with unclassified HCA comprise 10% of cases. The lesions are negative for CRP, SAA, β-catenin, and glutamine synthetase but exhibit typical liver fatty acid-binding protein (LFABP) staining. Epidemiology HCA's annual incidence among patients taking oral contraceptive pills is 30 to 40 cases per million, compared to 1 case per million in people who do not take these medications. A higher risk is seen after more than 2 years of use. A study found a 25-fold relative risk increase in women using oral contraceptive pills for over 109 months compared to those using them for less than 12 months. Discontinuation of oral contraceptives often results in spontaneous tumor regression, supporting the link to sex hormones. HCA is more common in women than men, with a ratio of 4 to 1, although this ratio is changing due to sports-related anabolic drug use. Other populations at risk for developing HCA include individuals with GSD types I and III, iron-overload conditions like β-thalassemia and hemochromatosis, and states of endogenous sex hormonal imbalance such as Klinefelter and polycystic ovarian syndromes. Except for polycystic ovarian syndrome, these other conditions predominantly affect men and are often diagnosed during childhood. Pathophysiology Genetic mutations offer new insights into the behavior of these tumors. However, traditional knowledge highlights the strong association between sex hormones, notably oral contraceptive pills and anabolic steroids, and HCA development. Other drugs that have been implicated include clomiphene, recombinant human growth hormones, and barbiturates. Histopathology Gross examination of HCAs reveals yellow or light brown well-circumscribed lesions with a soft consistency. The lesions are usually solitary, with sizes ranging from 2 to 15 cm. HCAs tend to be larger in women on oral contraceptives. Most are located in the right lobe and are subscapular. Microscopic examination shows a lack of malignancy features and abundant glycogen or fat. Liver architecture is absent, with central veins, bile ducts, and portal tracts often not visualized. However, HCAs are well-vascularized with small, thin-walled arterioles. HCAs are typically solitary tumors with monoclonal cell lines. These tumors lack a true capsule, though hepatocyte sheets in the outer boundaries form their "pseudo-capsule" and are surrounded by otherwise normal liver tissue. The lack of biliary ducts distinguishes the lesion from normal liver tissue (see Image. Hepatocellular Adenoma Histopathology). History and Physical History About half of patients with HCAs are asymptomatic, with the tumor discovered incidentally on imaging. The condition presents variably in symptomatic individuals, ranging from mild, ill-defined epigastric pain with bloating to severe, acute RUQ pain. Adenoma rupture may be accompanied by dizziness, intractable vomiting, and lethargy. The pain may radiate to the right flank in some cases, resembling cholecystitis or a urinary tract pathology. Rupture occurs in up to 27% of HCAs, resulting in a 5% to 10% mortality rate. Individuals with suspected HCA should be asked about the symptoms, especially abdominal pain and its characteristics. Risk factors must be assessed, including prolonged sex hormone use and underlying conditions like GSD and obesity. The family history may reveal liver diseases or malignancies and other genetic conditions predisposing to HCA evolution. Physical Examination A complete physical examination helps determine the diagnosis and the patient's hemodynamic status. Individuals with abdominal pain due to gastrointestinal pathology often lie still, in contrast to patients with renal colic. Tachycardia and hypotension are possible indicators of shock. Fever may manifest due to inflammation and is a physical feature shared with many other conditions, including hepatitis. Some patients may have dry oral mucosa and poor skin turgor from dehydration. Pallor and poor pulses may be appreciated due to hemorrhage. Normal chest auscultation findings may rule out a cardiac or pulmonary cause of referred abdominal pain unless such illnesses coexist. The abdominal examination may reveal hepatomegaly with tenderness. Abdominal rigidity may be a sign of peritoneal irritation from hemoperitoneum due to tumor rupture. Evaluation Laboratory tests are generally unhelpful in diagnosing hepatocellular adenomas, though they can help rule out other disorders and determine hematologic and metabolic status. α-fetoprotein (AFP) levels are typically normal but can rise if malignant transformation to hepatocellular carcinoma (HCC) occurs. Hepatitis tests should be conducted to rule out these infections. Alkaline phosphatase and γ-glutamyl transferase levels may increase two- to threefold, especially in IHCAs. White blood cell count, fibrinogen, and CRP may also be elevated. Core needle biopsy has limited diagnostic value in HCA. However, immunohistochemical markers may be helpful in expert centers. Biopsy should be limited to rare cases where imaging is equivocal, and the diagnosis may alter management strategy. Ultrasonography often fails to distinguish between benign and malignant tumors. Doppler ultrasonography may demonstrate arterial hypervascularity with vessels running along the lesion's border in a "basket" pattern. Some tumors may appear hyperechoic due to the hepatocytes' steatotic content. Dynamic magnetic resonance imaging (MRI) with a hepatocyte-specific contrast agent like gadobenate dimeglumine is the modality of choice for diagnosing HCAs. This method can distinguish between hepatocellular adenomas and other benign and malignant liver tumors (see Image. Hepatocellular Adenoma on Magnetic Resonance Imaging). The lesion may exhibit a clearly defined central margin with nearly parallel vessels entering from the periphery, giving the appearance of a spoked wheel. In some cases, dynamic MRI may demonstrate a tortuosity of peripheral vessels with central necrosis. A dynamic computed tomography (CT) scan is also potentially useful in evaluating HCAs. Dynamic CT may show peripheral enhancement of the tumors during the early phase, transitioning to centripetal flow in the later portal venous phase. Treatment / Management HCAs less than 5 cm in size and linked to oral contraceptive pills are initially approached conservatively. Oral contraceptive withdrawal with regular imaging surveillance has demonstrated significant tumor regression in numerous cases. Resuming oral contraceptive intake requires careful radiologic monitoring. The optimal follow-up duration remains unestablished, with some authors suggesting surveillance until menopause. Refractory tumors often correlate with obesity. Most hepatic adenomas remain stable during pregnancy. Tertiary centers often monitor adenomas smaller than 5 cm serially in pregnant women every 3 months and during the postpartum period. Patients with small adenomas are not discouraged from getting pregnant. Surgical resection is recommended for all male patients regardless of the tumor size and for women with tumors larger than 5 cm. The technique does not require a wide margin or regional lymphadenectomy. Emergent surgery for a ruptured hepatic adenoma with intraperitoneal bleeding has a mortality rate of 5% to 10%. In contrast, elective resection's mortality rate is less than 1%. Transarterial embolization (TAE) is recommended for hepatic adenomas complicated by hemorrhage. Patients with intratumoral hemorrhage rarely present with hemodynamic instability. In such cases, TAE may be followed by elective surgical resection of the adenoma. TAE is indicated within 2 to 3 days of tumoral hemorrhage. Radiofrequency ablation is appropriate only for a very select patient population. The modality is not appropriate for surgical candidates and individuals with hormone-sensitive tumors, underlying liver disease, or a desire for pregnancy. Radiofrequency ablation should be used to treat patients with adenomas less than 4 cm in size. Obesity, nonalcoholic steatohepatitis, and metabolic syndromes increase the risk of developing hepatic adenomatosis. Complications like hemorrhage or malignant transformation correlate less with adenomatosis than the tumors' molecular signatures. Moreover, most liver adenomatoses are typically associated with HNF-1α mutations, which have a low risk for malignant transformations. Thus, a liver transplant is generally not indicated for nonresectable hepatic adenomas. Exceptions involve men with intrahepatic portosystemic venous shunts with nonresectable hepatic adenomas. Genetic counseling is recommended for patients with liver adenomatosis, especially if associated with familial adenomatous polyposis or MODY3. About 4% to 5% of hepatic adenomas are at risk for malignant transformation to HCC, with the risk increased by Fanconi anemia and androgen treatments. The Wnt and β-catenin pathways are particularly associated with malignant transformation. Surgical resection is typically recommended for this subtype. Differential Diagnosis The differential diagnosis of liver lesions suspicious for HCA includes hemangioma, focal nodular hyperplasia (FNH), HCC, and metastatic tumors. Dynamic contrast-enhanced MRI can differentiate between HCA and hemangiomas, which generally exhibit peripheral enhancement and a centripetal fill-in pattern. Gadolinium-based contrast MRI often shows FNH taking up more contrast in the hepatobiliary phase than HCA, along with a central scar and lobulated appearance. The clinical scenario is crucial in distinguishing between HCA and HCC or metastatic disease. Individuals with HCC typically have underlying chronic liver disease with or without cirrhosis. People with metastatic disease typically have a known extrahepatic primary malignancy and associated risk factors. Prognosis The prognosis for patients with hepatocellular adenoma remains poorly established. Oral contraceptive discontinuation may result in the regression or resolution of some lesions. As previously stated, about 27% of HCAs may rupture, which has a 5% to 10% mortality risk. The risk of HCA's malignant transformation is 4.2%. The risk of malignancy generally persists even after discontinuing oral contraceptives. Complete HCA resolution is unlikely in most patients. About 25% of women will continue to have RUQ pain, with hemorrhage potentially occurring in up to 30% to 45% of women. Bleeding may occur within the lesion or peritoneum. The larger the lesion, the higher the risk of hemorrhage. Pregnancy has been associated with HCA enlargement and increased rupture risk. Complications HCA's complications include tumor rupture and hemorrhage, which may result in hemorrhagic shock and death. Tumors larger than 5 cm have an increased rupture risk. The malignant transformation of HCAs is greater in men than women with tumors larger than 5 cm. Consultations HCA management requires interprofessional care coordination involving gastroenterologists, radiologists, and surgeons. The interprofessional approach enhances diagnostic accuracy and therapeutic or surveillance efficacy. Deterrence and Patient Education Primary prevention of HCAs involves minimizing exposure to known risk factors, such as oral contraceptives and anabolic steroids. Education about the risks associated with these medications and their potential to induce HCA development is crucial. Making healthy lifestyle choices, including weight management and avoiding excessive alcohol consumption, may also help decrease the risk of developing HCAs. Meanwhile, patients with the condition or genetic risks for its development must be counseled about potential complications, eg, rupture and HCC, and their likelihood if no treatment is taken. Female patients should be advised that pregnancy is not contraindicated, especially with lesions of 5 cm or less. However, pregnant patients with HCA must receive interprofessional care and meticulous tumor monitoring. If surgery is warranted during pregnancy, resections should ideally occur during the 2nd trimester when risks to both mother and fetus are at their lowest. Patients must be fully informed of their condition and any action's options, risks, and benefits. Pearls and Other Issues HCAs are rare liver tumors often linked to oral contraceptive intake in reproductive-age women. Obesity, metabolic dysfunction associated with steatohepatitis, and metabolic syndromes are associated with hepatocellular adenomatosis. The most dreaded complications of this condition include rupture and HCC. The phenotypic-genotypic classification system for this condition is validated and valuable in understanding molecular behavior patterns. The HNF-1α subtype, associated with hepatocyte differentiation, is prevalent in women and individuals with MODY3. β-catenin-activated mutations, found in 15% to 20% of HCAs, are often linked to male hormone exposure and carry a greater risk of malignant transformation. Surgical resection is recommended for this subtype. IHCAs, prevalent in women with higher BMI and alcohol intake, stain positively for SAA and CRP. Unclassified types lack the features exhibited by other types and comprise 10% of HCA cases. Laboratory diagnosis is often unhelpful diagnostically. However, MRI can help reliably distinguish HCAs from other liver conditions. Imaging complements clinical examination during the diagnostic process. Discontinuation of oral contraceptives can induce regression in adenomas less than 5 cm, warranting continued imaging surveillance. Surgery is generally recommended for tumors greater than 5 cm in size but does not require wide margins or regional lymphadenectomy. Enhancing Healthcare Team Outcomes Prevention rather than treatment is the main focus of HCA management. Patient education involves the primary care physician, pharmacist, and nurse, contributing to better outcomes through interprofessional management. Once detected, the gastroenterologist and surgeon's roles—primarily surveillance and possibly treatment—become invaluable in minimizing morbidity and mortality. Pregnant women must be under the care of a high-risk obstetric care specialist, who may coordinate with the interprofessional team. All women prescribed oral contraceptives should be warned about the risk of developing this lesion, with lower doses potentially reducing the risk. Patients without other contraception options should be educated on HCA symptoms and when to seek help. Pregnancy increases HCA's hemorrhage risk and maternal mortality. Women with large or symptomatic tumors who desire pregnancy should be counseled about considering HCA removal before conceiving. Ongoing imaging surveillance, possibly annually, is recommended, especially for lesions larger than 5 cm. Review Questions Access free multiple choice questions on this topic. Click here for a simplified version. Comment on this article. Figure Inflammatory Hepatocellular Adenoma Histopathology. A woman born in 1947 experienced massive bleeding with no history of oral contraceptive use and a BMI of 21.5. She underwent a right hepatectomy in 2000, revealing massive liver necrosis and (more...) Figure Hepatocellular Adenoma Histopathology. High-magnification micrograph of a hepatic adenoma showing sheets of hepatocytes with vacuolated cytoplasm, arranged on a regular reticulin scaffold and measuring less than or equal to 3 cells thick. Portal (more...) Figure Hepatocellular Adenoma on Magnetic Resonance Imaging. This image shows multiple hepatocellular adenomas (yellow arrows). The imaging characteristics of these tumors include signals ranging from hyperintense to mildly hypointense compared (more...) References 1. : Tsilimigras DI, Rahnemai-Azar AA, Ntanasis-Stathopoulos I, Gavriatopoulou M, Moris D, Spartalis E, Cloyd JM, Weber SM, Pawlik TM. Current Approaches in the Management of Hepatic Adenomas. J Gastrointest Surg. 2019 Jan;23(1):199-209. [PubMed: 30109469] 2. : Aziz H, Underwood PW, Gosse MD, Afyouni S, Kamel I, Pawlik TM. Hepatic adenoma: evolution of a more individualized treatment approach. J Gastrointest Surg. 2024 Jun;28(6):975-982. [PubMed: 38521190] 3. : Vernon H, Wehrle CJ, Alia VSK, Kasi A. 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Update on the new classification of hepatic adenomas: clinical, molecular, and pathologic characteristics. Arch Pathol Lab Med. 2014 Aug;138(8):1090-7. [PubMed: 25076298] 14. : Mittal S, Gopal P, Khatri G, Singal AG. Evaluation and Management of Hepatocellular Adenomas. Clin Liver Dis (Hoboken). 2021 Feb;17(2):57-60. [PMC free article: PMC7916436] [PubMed: 33680436] 15. : Kis LE, Centeno BA, Anaya DA, Kis B. Hepatic adenoma rupture following portal vein embolization. Radiol Case Rep. 2020 Jun;15(6):664-667. [PMC free article: PMC7136602] [PubMed: 32280397] 16. : Yoon PD, Chen AZL, Tovmassian D, Pleass H. Spontaneous hepatic haemorrhage secondary to ruptured hepatocellular adenoma in a young male patient. BMJ Case Rep. 2020 Aug 25;13(8) [PMC free article: PMC7449593] [PubMed: 32843462] 17. : He MN, Lv K, Jiang YX, Jiang TA. Application of superb microvascular imaging in focal liver lesions. World J Gastroenterol. 2017 Nov 21;23(43):7765-7775. [PMC free article: PMC5703936] [PubMed: 29209117] 18. : Thyagarajan MS, Sharif K. Space Occupying Lesions in the Liver. Indian J Pediatr. 2016 Nov;83(11):1291-1302. [PubMed: 27783314] 19. : Chiorean L, Cui XW, Tannapfel A, Franke D, Stenzel M, Kosiak W, Schreiber-Dietrich D, Jüngert J, Chang JM, Dietrich CF. Benign liver tumors in pediatric patients - Review with emphasis on imaging features. World J Gastroenterol. 2015 Jul 28;21(28):8541-61. [PMC free article: PMC4515836] [PubMed: 26229397] 20. : Klompenhouwer AJ, de Man RA, Dioguardi Burgio M, Vilgrain V, Zucman-Rossi J, Ijzermans JNM. New insights in the management of Hepatocellular Adenoma. Liver Int. 2020 Jul;40(7):1529-1537. [PMC free article: PMC7383747] [PubMed: 32464711] 21. : Patacsil SJ, Noor M, Leyva A. A Review of Benign Hepatic Tumors and Their Imaging Characteristics. Cureus. 2020 Jan 29;12(1):e6813. 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2018	https://link.springer.com/chapter/10.1007/978-1-4757-2600-8_5	Advertisement Quadratic Programming with Box Constraints Part of the book series: Nonconvex Optimization and Its Applications ((NOIA,volume 18)) 1104 Accesses 46 Citations Abstract The global minimization of quadratic problems with box constraints naturally arises in many applications and as a subproblem of more complex optimization problems. In this paper we briefly describe the main results on global optimality conditions. Moreover, some of the most interesting computational approaches for the problem will be summarized. Download to read the full chapter text Chapter PDF Similar content being viewed by others Adaptive Global Algorithm for Solving Box-Constrained Non-convex Quadratic Minimization Problems On global minimizers of quadratic functions with cubic regularization On global integer extrema of real-valued box-constrained multivariate quadratic functions Explore related subjects Abbreviations box constrained quadratic problem Karush, Kuhn, Tucker stationarity conditions. References H.H. Benson. Concave Minimization: Theory, Applications and Algorithms. Handbook of Global Optimization (Eds: R. Horst and P.M. Pardalos), Kluwer, 43–142, (1995). Google Scholar I.M. Bomze. Copositivity Conditions for Global Optimality in Indefinite Quadratic Programming Problems. Czechoslovak J. for OR 1:7–19, (1992). MATH Google Scholar I.M. Bomze and G. Danninger. A Global Optimization Algorithm for Concave Quadratic Problems. Siam J. Optim., 3:826–842, (1993). Article MathSciNet MATH Google Scholar I.M. 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Article MathSciNet MATH Google Scholar Download references Author information Authors and Affiliations Facoltà di Economia, Istituto Universitario Navale, Via A. de Gasperi, 5, 80136, Napoli, Italy Pasquale L. De Angelis Dept. of Industrial and System Engineering, University of Florida, Gainesville, FL, 32611, USA Panos M. Pardalos Facoltà di Agraria, Istituto di Idraulica, University of Naples “Federico II” & CPS, Via Cintia, ed. T, 80126, Napoli, Italy Gerardo Toraldo Search author on:PubMed Google Scholar Search author on:PubMed Google Scholar Search author on:PubMed Google Scholar Editor information Editors and Affiliations University of Vienna, Vienna, Austria Immanuel M. Bomze József Attila University, Szeged, Hungary Tibor Csendes University of Trier, Trier, Germany Reiner Horst University of Florida, Gainesville, Florida, USA Panos M. Pardalos Rights and permissions Reprints and permissions Copyright information © 1997 Springer Science+Business Media Dordrecht About this chapter Cite this chapter De Angelis, P.L., Pardalos, P.M., Toraldo, G. (1997). Quadratic Programming with Box Constraints. In: Bomze, I.M., Csendes, T., Horst, R., Pardalos, P.M. (eds) Developments in Global Optimization. Nonconvex Optimization and Its Applications, vol 18. Springer, Boston, MA. Download citation DOI: Publisher Name: Springer, Boston, MA Print ISBN: 978-1-4419-4768-0 Online ISBN: 978-1-4757-2600-8 eBook Packages: Springer Book Archive Share this chapter Anyone you share the following link with will be able to read this content: Sorry, a shareable link is not currently available for this article. 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2020	https://fluidpower.pro/temperature-viscosity-chart/	Temperature-Viscosity Chart – FluidPower.Pro Fluid Power Pro - LinkedIn Become an author \| Login Articles Formulas Unit Converter Calculators Links About Temperature-Viscosity Chart Dzyanis September 8, 201920Calculatorstemperature, viscosity, Viscosity Index 86010 Views The interpolation calculator provided below is used to determine: the kinematic viscosity values, depends from temperature, calculated by the formulas specified in ASTM D341 – (‘Standard Practice for Viscosity-Temperature Charts for Liquid Petroleum Products’); the Viscosity Index (VI) by the formulas specified in ASTM D2270 (Standard Practice for Calculating Viscosity Index from Kinematic Viscosity at 40 and 100°C 1) or in ISO 2909 (Standard Practice for Calculating Viscosity Index from Kinematic Viscosity at 40 and 100°C 1) or in ГОСТ 25371-2018 (Нефтепродукты. Расчёт индекса вязкости по кинематической вязкости) Annotation. Any mineral oil is supplied with the specification of kinematic viscosity values at two different temperatures, usually at 40°C and at 100°C (or at 100°F and at 210°F). These data is enough to determine kinematic viscosity at any other values of temperature and as well Viscosity Index. You can calculate pre-filled values of oil classified by ISO 3448 with grades VG22, VG32, VG46 or VG68 (at viscosity Index = 100) or enter specific values for any other oils. Warning! The calculator below is for informative purposes only. The use of any of the calculators’ results is at the user’s sole risk. How to use calculator Just put your values (either imperial or metrical) in the appropriate fields below and press ENTER. All other values will be re-calculated automatically. The graphic curve will be updated as well. \| Operation data \| Metrical \| Imperial \| --- \| Select hydraulic oil: \| \| \| Temperature #1 \| °C \| °F \| \| Kinematic viscosity at Temperature #1 \| cSt \| SUS \| \| Temperature #2 \| °C \| °F \| \| Kinematic viscosity at Temperature #2 \| cSt \| SUS \| \| Temperature to calculate \| °C \| °F \| \| Results: \| \| \| Kinematic viscosity at calc. temperature \| cSt \| SUS \| \| Viscosity Index \| by ISO 2909:2002 Procedures A and B by ASTM D2270-04 Procedures A and B by ГОСТ 25371-2018 Методы А и Б \| \| Temperature, C \| Viscosity, cSt \| --- \| \| -30 \| 18,046.788 \| \| -29 \| 15,568.815 \| \| -28 \| 13,468.949 \| \| -27 \| 11,684.393 \| \| -26 \| 10,163.55 \| \| -25 \| 8,863.9 \| \| -24 \| 7,750.298 \| \| -23 \| 6,793.613 \| \| -22 \| 5,969.629 \| \| -21 \| 5,258.163 \| \| -20 \| 4,642.347 \| \| -19 \| 4,108.044 \| \| -18 \| 3,643.38 \| \| -17 \| 3,238.353 \| \| -16 \| 2,884.516 \| \| -15 \| 2,574.723 \| \| -14 \| 2,302.908 \| \| -13 \| 2,063.916 \| \| -12 \| 1,853.35 \| \| -11 \| 1,667.457 \| \| -10 \| 1,503.023 \| \| -9 \| 1,357.289 \| \| -8 \| 1,227.886 \| \| -7 \| 1,112.77 \| \| -6 \| 1,010.18 \| \| -5 \| 918.591 \| \| -4 \| 836.68 \| \| -3 \| 763.301 \| \| -2 \| 697.455 \| \| -1 \| 638.273 \| \| 0 \| 584.996 \| \| 1 \| 536.958 \| \| 2 \| 493.579 \| \| 3 \| 454.348 \| \| 4 \| 418.815 \| \| 5 \| 386.586 \| \| 6 \| 357.313 \| \| 7 \| 330.686 \| \| 8 \| 306.434 \| \| 9 \| 284.316 \| \| 10 \| 264.118 \| \| 11 \| 245.649 \| \| 12 \| 228.74 \| \| 13 \| 213.24 \| \| 14 \| 199.014 \| \| 15 \| 185.943 \| \| 16 \| 173.919 \| \| 17 \| 162.844 \| \| 18 \| 152.633 \| \| 19 \| 143.209 \| \| 20 \| 134.5 \| \| 21 \| 126.444 \| \| 22 \| 118.985 \| \| 23 \| 112.072 \| \| 24 \| 105.657 \| \| 25 \| 99.699 \| \| 26 \| 94.161 \| \| 27 \| 89.007 \| \| 28 \| 84.207 \| \| 29 \| 79.732 \| \| 30 \| 75.557 \| \| 31 \| 71.658 \| \| 32 \| 68.013 \| \| 33 \| 64.604 \| \| 34 \| 61.412 \| \| 35 \| 58.421 \| \| 36 \| 55.616 \| \| 37 \| 52.983 \| \| 38 \| 50.511 \| \| 39 \| 48.186 \| \| 40 \| 46 \| \| 41 \| 43.942 \| \| 42 \| 42.003 \| \| 43 \| 40.175 \| \| 44 \| 38.451 \| \| 45 \| 36.823 \| \| 46 \| 35.285 \| \| 47 \| 33.832 \| \| 48 \| 32.457 \| \| 49 \| 31.156 \| \| 50 \| 29.923 \| \| 51 \| 28.755 \| \| 52 \| 27.647 \| \| 53 \| 26.596 \| \| 54 \| 25.598 \| \| 55 \| 24.65 \| \| 56 \| 23.75 \| \| 57 \| 22.893 \| \| 58 \| 22.077 \| \| 59 \| 21.301 \| \| 60 \| 20.562 \| \| 61 \| 19.857 \| \| 62 \| 19.185 \| \| 63 \| 18.544 \| \| 64 \| 17.932 \| \| 65 \| 17.347 \| \| 66 \| 16.789 \| \| 67 \| 16.255 \| \| 68 \| 15.744 \| \| 69 \| 15.256 \| \| 70 \| 14.788 \| \| 71 \| 14.34 \| \| 72 \| 13.911 \| \| 73 \| 13.5 \| \| 74 \| 13.106 \| \| 75 \| 12.727 \| \| 76 \| 12.364 \| \| 77 \| 12.016 \| \| 78 \| 11.681 \| \| 79 \| 11.359 \| \| 80 \| 11.05 \| \| 81 \| 10.752 \| \| 82 \| 10.466 \| \| 83 \| 10.191 \| \| 84 \| 9.926 \| \| 85 \| 9.67 \| \| 86 \| 9.424 \| \| 87 \| 9.187 \| \| 88 \| 8.959 \| \| 89 \| 8.738 \| \| 90 \| 8.526 \| \| 91 \| 8.321 \| \| 92 \| 8.123 \| \| 93 \| 7.931 \| \| 94 \| 7.747 \| \| 95 \| 7.568 \| \| 96 \| 7.395 \| \| 97 \| 7.229 \| \| 98 \| 7.067 \| \| 99 \| 6.911 \| \| 100 \| 6.76 \| \| 101 \| 6.614 \| \| 102 \| 6.472 \| \| 103 \| 6.335 \| \| 104 \| 6.202 \| \| 105 \| 6.073 \| \| 106 \| 5.949 \| \| 107 \| 5.828 \| \| 108 \| 5.71 \| \| 109 \| 5.596 \| \| 110 \| 5.486 \| \| 111 \| 5.379 \| \| 112 \| 5.274 \| \| 113 \| 5.173 \| \| 114 \| 5.075 \| \| 115 \| 4.98 \| \| 116 \| 4.887 \| \| 117 \| 4.797 \| \| 118 \| 4.709 \| \| 119 \| 4.624 \| \| 120 \| 4.541 \| Temperature-Viscosity Chart If you find some mismatch or an error or just have any questions please do not hesitate to leave your comments below. ← Previous post Next post → 20 Comments ALIAugust 18, 2020 at 5:36 am OUR FUEL VISCOSITY IS 39.9MM AT 50 DEG C I WANT VISCOSITY OF 18MM2 AT BOILER WHAT WILL BE TEMPERATURE TO MAINTAIN THAT VISCOSITY. Reply↓ 1. Dzyanis(Post author)October 21, 2020 at 8:09 am Hello Ali. I assume when you say 39.9MM and 18MM2 you mean viscosity 39.9mm^2/s and 18mm^2/s what are equivalents to 39.9 cSt and 18.0 cSt accordingly. Unfortunately, it means you have oil with the custom grade and just one parameter of viscosity at temp is not enough to make this calculation in accordance with the standard. Please check your oil specification – there should be two lines of viscosity at temp. Put them in the table above and find the value of temp at any value of viscosity. Reply↓ SHOctober 23, 2020 at 9:39 am Amazing tool! Extremely helpful. Thank you very much for making it available to everybody! Reply↓ 1. Dzyanis(Post author)October 23, 2020 at 11:12 am Thank you! Reply↓ IkhsanNovember 4, 2020 at 6:25 pm Amazing tool! thank you very much for your effort. is this calculator applicable to another oil type (engine, gear, etc)? Reply↓ 1. Dzyanis(Post author)November 5, 2020 at 8:19 am Thanks for your feedback! This chart uses equations as per ASTM D341 standard to calculate relationship temp-visc for any liquid petroleum products and lubricants. Reply↓ asfDecember 4, 2020 at 11:22 am hey , given viscosity of 1000 cSt at -20 deg C , and ASTM slope of 0.6 , how to find VI? Reply↓ 5. Carlos J AguirreMarch 20, 2021 at 1:32 pm GREAT TOOL GOO JOB!!! Reply↓ 6. Pankaj PatelJune 22, 2021 at 3:18 pm I am using a SAE 75-90 wt synthetic gear oil. I am needing to find out viscosity as per temperature – viscosity chart in reyn. Can you help me how I can do it. Reply↓ 1. Dzyanis(Post author)July 13, 2021 at 12:00 pm Hello! Reyn is a British unit of dynamic viscosity. This chart is for kinematic viscosity extrapolation. Reply↓ ThomasJuly 13, 2021 at 5:54 am Hi, Great work! The ASTM D341 states that it’s possible to use 3 points for the extrapolation. Do you know how this could be achieved? Thanks! Reply↓ 1. Dzyanis(Post author)July 13, 2021 at 11:56 am hello Thomas. Thanks for you comment. Standard D341-03, paragraph 1.1: “1.1 The kinematic viscosity-temperature charts (see Figs. 1 and 2) covered by this standard are a convenient means to ascertain the kinematic viscosity of a petroleum oil or liquid hydrocarbon at any temperature within a limited range, pro- vided that the kinematic viscosities at two temperatures are known.” So, like you see this standard for the extrapolation using 2 points. Sorry, I didn’t find any states about 3 points in this standard, did you? And… what is the sense to make a calculator for 3 points if all oil manufacturers provide temp/visc parameters for their oil only for 2 points? Reply↓ Mohd SaquibOctober 27, 2021 at 5:29 am My lubricant is SAE 20W-40 at 60 °C, what would be its absolute viscosity in mPa. s Reply↓ 9. haniFebruary 1, 2022 at 5:29 am For ISO VG 100 oil, how to find constants A&B using Walther-ASTM equation to determine the effect of temperature on viscosity of the oil as per the parameters given below. Given parameters:- a. Temperature range (273 – 523 K) b. Kinematic viscosity: 99.4 cSt at 40 C and 11.12 cSt at 100 C. and Show the relationship between viscosity (vertical) and temperature (horizontal) of ISO VG 100 oil using a graph and discuss the results as the temperature varies. please any one answer this question? Reply↓ 10. GeorgeMarch 4, 2022 at 4:34 pm Would it be possible to list the A & B Constants for the oils in the chart? I’m trying to do an Excel calculation with a curve fit based on the data points in the tool but it is not accurate enough. I’d like to use the equations directly but I don’t know where to get the A and B constants. Reply↓ 11. maryJuly 11, 2022 at 2:24 am Hello, Would you be able to send me the excel file so i can create graphs using andrade equation? How is it calculated? Reply↓ 1. Dzyanis Sukhanitski(Post author)July 11, 2022 at 6:49 am I do not have an Excel file, the kinematic viscosity values are calculated by the formulas specified in ASTM D341 – (‘Standard Practice for Viscosity-Temperature Charts for Liquid Petroleum Products’). Reply↓ MitchJanuary 13, 2023 at 2:57 pm Hi Dzyanis, I really love this tool and it’s helped me to understand both VI and the relationship between viscosity and temperature in a way that other explanations have not. I think this is great for those of us that lack your engineering knowledge and experience. I have one suggestion. It would be really great, if you could input TWO different oil products, and have them both charted on the same graph with the the x/y scaling fixed to whichever is the larger data range. With two lines showing overlapped, you could make very convincing visual comparisons between TWO different oil products. Reply↓ 1. Dzyanis(Post author)August 22, 2023 at 9:28 am Good suggestion, I will work on it. Also, I’m thinking to add autofill into input fields from the URL to get the required diagram by link. Reply↓ Metin SimsekJune 11, 2025 at 8:09 am Hello Great help, very useful. Would it be possible to adjust this for HVİ oils where the curve will go near-flat? Regards. Metin Reply↓ Leave a Comment Cancel reply Your email address will not be published.Required fields are marked [x] Save my name, email, and website in this browser for the next time I comment. This site is protected by reCAPTCHA and the Google Privacy Policy and Terms of Service apply. The reCAPTCHA verification period has expired. Please reload the page. Δ This site uses Akismet to reduce spam. Learn how your comment data is processed. FluidPower.Pro is a free resource where any Fluid Power specialists can share their knowledge, interesting info about new products or just make a discussion on any topics related to the hydraulic or pneumatic. CONTACT US LinkedIn: Follow Us! 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2021	https://stackoverflow.com/questions/3809044/how-many-values-can-be-represented-with-n-bits	Stack Overflow About For Teams Stack Overflow for Teams Where developers & technologists share private knowledge with coworkers Advertising Reach devs & technologists worldwide about your product, service or employer brand Knowledge Solutions Data licensing offering for businesses to build and improve AI tools and models Labs The future of collective knowledge sharing About the company Visit the blog Collectives¢ on Stack Overflow Find centralized, trusted content and collaborate around the technologies you use most. Learn more about Collectives Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams How many values can be represented with n bits? Ask Question Asked Modified 6 years, 4 months ago Viewed 263k times 46 For example, if n=9, then how many different values can be represented in 9 binary digits (bits)? My thinking is that if I set each of those 9 bits to 1, I will make the highest number possible that those 9 digits are able to represent. Therefore, the highest value is 1 1111 1111 which equals 511 in decimal. I conclude that, therefore, 9 digits of binary can represent 511 different values. Is my thought process correct? If not, could someone kindly explain what I'm missing? How can I generalize it to n bits? binary bit Share Improve this question edited Nov 1, 2012 at 18:03 NullUserException 85.7k3131 gold badges212212 silver badges239239 bronze badges asked Sep 28, 2010 at 1:26 SeanSean 5,42355 gold badges2525 silver badges2626 bronze badges 4 13 You forgot 000000000. David R Tribble – David R Tribble 2010-09-28 02:05:19 +00:00 Commented Sep 28, 2010 at 2:05 9 +1 for a very well-asked homework question. If all homework questions on Stack Overflow were like this, it would be a much better place. James McNellis – James McNellis 2010-09-28 02:26:41 +00:00 Commented Sep 28, 2010 at 2:26 2 Why it closed :( It is not related to "specific moment in time" ..... Kanagavelu Sugumar – Kanagavelu Sugumar 2015-02-17 06:53:46 +00:00 Commented Feb 17, 2015 at 6:53 e.g. 2 pow 3 represents two combinations (0,1) for three place holders like 000, 001, 010, 100, 101, 110, 111. If you have 8 (octal 0-7) combination for five places then total possible combinations will be 8 pow 5. Kanagavelu Sugumar – Kanagavelu Sugumar 2015-02-17 07:01:17 +00:00 Commented Feb 17, 2015 at 7:01 Add a comment \| 7 Answers 7 Reset to default 63 29 = 512 values, because that's how many combinations of zeroes and ones you can have. What those values represent however will depend on the system you are using. If it's an unsigned integer, you will have: 000000000 = 0 (min) 000000001 = 1 ... 111111110 = 510 111111111 = 511 (max) In two's complement, which is commonly used to represent integers in binary, you'll have: 000000000 = 0 000000001 = 1 ... 011111110 = 254 011111111 = 255 (max) 100000000 = -256 (min) <- yay integer overflow 100000001 = -255 ... 111111110 = -2 111111111 = -1 In general, with k bits you can represent 2k values. Their range will depend on the system you are using: Unsigned: 0 to 2k-1 Signed: -2k-1 to 2k-1-1 Share Improve this answer edited Sep 28, 2010 at 2:58 answered Sep 28, 2010 at 1:29 NullUserExceptionNullUserException 85.7k3131 gold badges212212 silver badges239239 bronze badges 2 Comments Sean Sean You raise a very interesting point. I hadn't thought of signed and unsigned integers. Nathan Nathan but in any case the number of different values is always 2^k 11 What you're missing: Zero is a value Share Improve this answer answered Sep 28, 2010 at 1:27 SamStephensSamStephens 5,87166 gold badges3939 silver badges4646 bronze badges Comments 3 A better way to solve it is to start small. Let's start with 1 bit. Which can either be 1 or 0. That's 2 values, or 10 in binary. Now 2 bits, which can either be 00, 01, 10 or 11 That's 4 values, or 100 in binary... See the pattern? Share Improve this answer answered Sep 28, 2010 at 1:33 Randy the DevRandy the Dev 26.9k66 gold badges4646 silver badges5454 bronze badges Comments 1 Okay, since it already "leaked": You're missing zero, so the correct answer is 512 (511 is the greatest one, but it's 0 to 511, not 1 to 511). By the way, an good followup exercise would be to generalize this: How many different values can be represented in n binary digits (bits)? Share Improve this answer answered Sep 28, 2010 at 1:29 schnaaderschnaader 49.9k1010 gold badges108108 silver badges139139 bronze badges 1 Comment Sean Sean Thank you for explaining that I was missing the 0 value. The way you reworded the question will definitely help me in the future. 1 Without wanting to give you the answer here is the logic. You have 2 possible values in each digit. you have 9 of them. like in base 10 where you have 10 different values by digit say you have 2 of them (which makes from 0 to 99) : 0 to 99 makes 100 numbers. if you do the calcul you have an exponential function base^numberOfDigits: 10^2 = 100 ; 2^9 = 512 Share Improve this answer answered Sep 28, 2010 at 1:30 community wiki pastjean Comments 1 There's an easier way to think about this. Start with 1 bit. This can obviously represent 2 values (0 or 1). What happens when we add a bit? We can now represent twice as many values: the values we could represent before with a 0 appended and the values we could represent before with a 1 appended. So the the number of values we can represent with n bits is just 2^n (2 to the power n) Share Improve this answer answered Sep 28, 2010 at 1:32 davedave 13.4k44 gold badges2323 silver badges1313 bronze badges Comments 1 The thing you are missing is which encoding scheme is being used. There are different ways to encode binary numbers. Look into signed number representations. For 9 bits, the ranges and the amount of numbers that can be represented will differ depending on the system used. Share Improve this answer edited Sep 28, 2010 at 1:57 answered Sep 28, 2010 at 1:43 James KastrantasJames Kastrantas 7611 silver badge33 bronze badges 5 Comments Lie Ryan Lie Ryan The range of numbers that is represented will differ depending on the encoding, but the number of distinct values that can be represented stays the same no matter the encoding used James Kastrantas James Kastrantas I just worked out 3-bit fixed precision on paper to check myself. The unsigned and 2's compliment representations can represent 8 values. I count 7 values for signed magnitude and 1's compliment because those systems have representations for positive and negative zero. Are positive and negative zero also counted? Lie Ryan Lie Ryan for 1's compliment, positive and negative zeros are considered as two distinct values. I'm not sure what you meant by "signed magnitude", can you elaborate what you meant by that? James McNellis James McNellis @Lie: One could argue that +0 and -0 are two different datums that represent the same value. Using that terminology, the number of datums in a sequence of k bits is always 2^k, but the number of values those datums may represent is less than or equal to 2^k. James Kastrantas James Kastrantas @Lie Ryan: en.wikipedia.org/wiki/… I didn't know (or forgot...) that +0 and -0 are both counted. I'll remember that in the future. Thanks. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions binary bit See similar questions with these tags. 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2022	https://www.mathsisfun.com/calculus/derivative-vs-integral.html	Graphical Intro toDerivatives and Integrals Derivatives and Integrals have a two-way relationship! Let's start by looking at sums and slopes: Example: walking in a straight line Walk slow, the distance increases slowly Walk fast, the distance increases fast Stand still and the distance won't change A distance increase of 4 km in 1 hour gives a speed of 4 km per hour Or, walking at 4 km per hour for 1 hour increases the distance by 4 km Speed is the rate of change of distance Change in distance is the sum of the speed over time It will make more sense when you play with it below: change the distance line, or the speed line, to see its affect on the other: images/deriv-integ.js?mode=1&topic=walking Play with that a little and get comfortable with the two-way relationship. Try zero speed, or negative speed. The slope of the distance line gives us the speed line, like this: The "area" under the speed line gives us the increase in distance, like this: Many things have that same two-way relationship: Wealth and income Volume and flow rate Energy and power lots more! Here is the same app as above, but you can choose different topics: images/deriv-integ.js?mode=multi Integrals and Derivatives also have that two-way relationship! Try it below, but first note: Δx (the gap between x values) only gives an approximate answer dx (when Δx approaches zero) gives the actual derivative and integral images/deriv-integ.js?mode=fn Note: this is a computer model and actually uses a very small Δx to simulate dx, and can make erors. For true derivatives refer to Derivative Rules, and for integrals refer to Introduction to Integration Derivative Rules Introduction to Integration Integral Approximations Calculus Index Copyright © 2023 Rod Pierce
2023	https://en.wikipedia.org…gleAnimated2.gif	Jump to content Search File:PascalTriangleAnimated2.gif File Talk Read View on Commons Edit local description View history Tools Actions Read View on Commons Edit local description View history General What links here Related changes Upload file Permanent link Page information Get shortened URL Download QR code Print/export Download as PDF Printable version In other projects Appearance From Wikipedia, the free encyclopedia File File history File usage Global file usage No higher resolution available. PascalTriangleAnimated2.gif (260 × 240 pixels, file size: 27 KB, MIME type: image/gif, looped, 84 frames, 12 s) \| \| \| This is a file from the Wikimedia Commons. Information from its description page there is shown below.Commons is a freely licensed media file repository. You can help. \| \| \| \| This image was selected as picture of the month on the Mathematics Portal for October 2008. \| Summary \| \| \| --- \| \| DescriptionPascalTriangleAnimated2.gif \| English: Pascal's triangle is a geometric arrangement of integers representing the binomial coefficients in a polynominal equation of the format (x + y)n. The formation also demonstrates many other mathematical properties, such as listing the entire set of the natural numbers in the first diagonal rows. This phenomenon is named after Blaise Pascal in the western world, however was studied in detail before his time in many Asian countries. It is also called the Halayudha's triangle, in honor of the Sanskrit prosody scholar who described it. (See: Alexander Zawaira and Gavin Hitchcock (2008), A Primer for Mathematics Competitions, Oxford University Press, ISBN 978-0-19-156170-2, page 237) It is alternately referred to as "Khayyam's triangle" after the Persian Omar Khayyám. Each number is the sum of the two directly above it. This animation shows this relation in the construction of the first five rows, however the pattern applies for an infinite range. This version has the 1 cells already filled in, and includes actual animation to better demonstrate the construction. Čeština: Animace zobrazuje vytváření Pascalova trojúhelníku. \| \| Date \| \| \| Source \| Own work \| \| Author \| Hersfold \| \| Other versions \| \| Licensing \| \| \| Public domainPublic domainfalsefalse \| \| \| \| --- \| \| \| I, the copyright holder of this work, release this work into the public domain. This applies worldwide.In some countries this may not be legally possible; if so:I grant anyone the right to use this work for any purpose, without any conditions, unless such conditions are required by law. \| Captions Pascal's triangle Pascalův trojúhelník треугольник Паскаля - это геометрическое расположение целых чисел, представляющих биномиальные коэффициенты в полиномиальном уравнении формата (x + y)n. формирование также демонстрирует множество других математических свойств, таких как перечисление Items portrayed in this file depicts Pascal's triangle copyright status copyrighted, dedicated to the public domain by copyright holder copyrighted copyright license released into the public domain by the copyright holder inception 18 April 2008 source of file original creation by uploader media type File history Click on a date/time to view the file as it appeared at that time. \| Date/Time \| Thumbnail \| Dimensions \| User \| Comment \| --- --- \| current \| 18:59, 21 May 2023 \| 260 × 240 (27 KB) \| Phreneticc \| Reduced file weight with FileOptimizer in lossless compression mode. \| \| 19:45, 2 May 2012 \| 260 × 240 (28 KB) \| Dratini0 \| Optimized and fixed with Gimp, because it did not work for me in LibreOffice and Gimp. I might have made a mistake. \| \| 20:49, 18 April 2008 \| 260 × 240 (41 KB) \| Hersfold \| {{Information \|Description=Pascal's triangle is a geometric arrangement of integers representing the binomial coefficients in a polynominal equation of the format (x + y)n. The \| File usage The following 9 pages use this file: Blaise Pascal Mathematics education in the United States Pascal's triangle Sanskrit prosody Wikipedia:Featured picture candidates/April-2008 Wikipedia:Featured picture candidates/Pascal's triangle Wikipedia:Graphics Lab/Illustration workshop/Archive/Nov 2012 Portal:Mathematics/Featured picture/2008 10 Portal:Mathematics/Featured picture archive Global file usage The following other wikis use this file: Usage on als.wikipedia.org Pascalsches Dreieck Usage on ar.wikipedia.org بوابة:رياضيات/صورة مختارة مثلث باسكال بوابة:رياضيات/صورة مختارة/أرشيف بوابة:رياضيات/صورة مختارة/3 Usage on ary.wikipedia.org قيسارية:لماط Usage on arz.wikipedia.org باسكال Usage on az.wikipedia.org Tusi-Paskal üçbucağı Usage on ba.wikipedia.org Блез Паскаль Usage on be.wikipedia.org Гісторыя тэорыі імавернасцей Usage on bn.wikipedia.org প্যাসকেলের ত্রিভুজ Usage on bs.wikipedia.org Blaise Pascal Pascalov trougao Usage on ca.wikipedia.org Triangle de Tartaglia Usage on ckb.wikipedia.org سێگۆشەی پاسکاڵ Usage on cs.wikipedia.org Pascalův trojúhelník Usage on cv.wikipedia.org Паскаль, Блез Паскаль виçкĕтеслĕхĕ Usage on da.wikipedia.org Blaise Pascal Pascals trekant Usage on de.wikipedia.org Blaise Pascal Pascalsches Dreieck Binomische Formeln Benutzer:Das Volk/Spielzeugladen Wikipedia:Auskunft/Archiv/2012/Woche 42 Benutzer:Googolplexian1221/Binomische Formeln Usage on de.wikibooks.org Mathe für Nicht-Freaks: Binomialkoeffizient: Rechenregeln Benutzer:Dirk Hünniger/mnf Mathematrix: AT BRP/ Theorie/ Mittleres Niveau 3 Mathematrix: MA TER/ Theorie/ Arbeiten mit Termen Mathematrix: AT PSA/ Theorie/ Expertenniveau 2 Mathematrix: AT BRP/ Theorie nach Thema/ Arbeiten mit Termen Mathematrix: AT PSA/ Theorie nach Thema/ Arbeiten mit Termen Mathematrix: AT AHS/ Theorie nach Thema/ Arbeiten mit Termen Mathematrix: AT AHS/ Theorie/ Klasse 4 Mathematrix: BY GYM/ Theorie/ Klasse 8 Mathematrix: BY GYM/ Theorie nach Thema/ Arbeiten mit Termen Mathematrix: AT PSA/ Theorie/ X Ing Mathematik: Permutationen, Kombinationen, binomischer Lehrsatz Usage on diq.wikipedia.org Blaise Pascal Usage on en.wikiquote.org Tao Te Ching Usage on eo.wikipedia.org Triangulo de Pascal Usage on es.wikipedia.org Blaise Pascal Triángulo de Pascal Usuario:Envilchez/Taller Usage on et.wikipedia.org Tuumamagnetresonantsspektroskoopia Usage on eu.wikipedia.org Pascalen hirukia Usage on fa.wikipedia.org بلز پاسکال Usage on fi.wikipedia.org Pascalin kolmio Usage on fr.wikipedia.org Triangle de Pascal View more global usage of this file. 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2024	https://math.libretexts.org/Bookshelves/Applied_Mathematics/Contemporary_Mathematics_(OpenStax)/10%3A__Geometry/10.04%3A_Polygons_Perimeter_and_Circumference	Skip to main content 10.4: Polygons, Perimeter, and Circumference Last updated : Jan 2, 2025 Save as PDF 10.3.0: Exercises 10.4.0: Exercises Page ID : 129643 OpenStax OpenStax ( \newcommand{\kernel}{\mathrm{null}\,}) Learning Objectives Identify polygons by their sides. Identify polygons by their characteristics. Calculate the perimeter of a polygon. Calculate the sum of the measures of a polygon’s interior angles. Calculate the sum of the measures of a polygon’s exterior angles. Calculate the circumference of a circle. Solve application problems involving perimeter and circumference. In our homes, on the road, everywhere we go, polygonal shapes are so common that we cannot count the many uses. Traffic signs, furniture, lighting, clocks, books, computers, phones, and so on, the list is endless. Many applications of polygonal shapes are for practical use, because the shapes chosen are the best for the purpose. Modern geometric patterns in fabric design have become more popular with time, and they are used for the beauty they lend to the material, the window coverings, the dresses, or the upholstery. This art is not done for any practical reason, but only for the interest these shapes can create, for the pure aesthetics of design. When designing fabrics, one has to consider the perimeter of the shapes, the triangles, the hexagons, and all polygons used in the pattern, including the circumference of any circular shapes. Additionally, it is the relationship of one object to another and experimenting with different shapes, changing perimeters, or changing angle measurements that we find the best overall design for the intended use of the fabric. In this section, we will explore these properties of polygons, the perimeter, the calculation of interior and exterior angles of polygons, and the circumference of a circle. Identifying Polygons A polygon is a closed, two-dimensional shape classified by the number of straight-line sides. See Figure 10.82 for some examples. We show only up to eight-sided polygons, but there are many, many more. If all the sides of a polygon have equal lengths and all the angles are equal, they are called regular polygons. However, any shape with sides that are line segments can classify as a polygon. For example, the first two shapes, shown in Figure {PageIndex3, are both pentagons because they each have five sides and five vertices. The third shape Figure {PageIndex31 is a hexagon because it has six sides and six vertices. We should note here that the hexagon in Figure {PageIndex3 is a concave hexagon, as opposed to the first two shapes, which are convex pentagons. Technically, what makes a polygon concave is having an interior angle that measures greater than 180∘. They are hollowed out, or cave in, so to speak. Convex refers to the opposite effect where the shape is rounded out or pushed out. While there are variations of all polygons, quadrilaterals contain an additional set of figures classified by angles and whether there are one or more pairs of parallel sides. See Figure {PageIndex4. Example {PageIndex1: Identifying Polygons Identify each polygon. Answer : 1. This shape has six sides. Therefore, it is a hexagon. 2. This shape has four sides, so it is a quadrilateral. It has two pairs of parallel sides making it a parallelogram. 3. This shape has eight sides making it an octagon. 4. This is an equilateral triangle, as all three sides are equal. 5. This is a rhombus; all four sides are equal. 6. This is a regular octagon, eight sides of equal length and equal angles. Your Turn {PageIndex1 Identify the shape. Example {PageIndex2: Determining Multiple Polygons What polygons make up Figure 10.85? Answer : Shapes 1 and 5 are hexagons; shapes 2, 3, 4, 6, 7, 9, 10, 12, 13, 14, 15, and 16 are triangles; shapes 8 and 17 are parallelograms; and shape 11 is a trapezoid. Your Turn {PageIndex2 What polygons make up the figure shown? Figure {PageIndex6 Perimeter Perimeter refers to the outside measurements of some area or region given in linear units. For example, to find out how much fencing you would need to enclose your backyard, you will need the perimeter. The general definition of perimeter is the sum of the lengths of the sides of an enclosed region. For some geometric shapes, such as rectangles and circles, we have formulas. For other shapes, it is a matter of just adding up the side lengths. A rectangle is defined as part of the group known as quadrilaterals, or shapes with four sides. A rectangle has two sets of parallel sides with four angles. To find the perimeter of a rectangle, we use the following formula: FORMULA The formula for the perimeter P of a rectangle is P=2L+2W, twice the length L plus twice the width W. For example, to find the length of a rectangle that has a perimeter of 24 inches and a width of 4 inches, we use the formula. Thus, 2424−8168=2l+2(4)=2l+8=2l=2l=l The length is 8 units. The perimeter of a regular polygon with n sides is given as P=n⋅s. For example, the perimeter of an equilateral triangle, a triangle with three equal sides, and a side length of 7 cm is P=3(7)=21 cm. Example {PageIndex3: Finding the Perimeter of a Pentagon Find the perimeter of a regular pentagon with a side length of 7 cm (Figure {PageIndex7). Figure {PageIndex7 Answer : A regular pentagon has five equal sides. Therefore, the perimeter is equal to P=5(7)=35 cm. Your Turn {PageIndex3 Find the perimeter of a square table that measures 30 inches across from one side to its opposite side. Example {PageIndex4: Finding the Perimeter of an Octagon Find the perimeter of a regular octagon with a side length of 14 cm (Figure {PageIndex8). Figure {PageIndex8 Answer : A regular octagon has eight sides of equal length. Therefore, the perimeter of a regular octagon with a side length of 14 cm is P=8(14)=112 cm. Your Turn {PageIndex1 Find the perimeter of a regular heptagon with a side length of 3.2 as shown in the figure. Figure {PageIndex9 Sum of Interior and Exterior Angles To find the sum of the measurements of interior angles of a regular polygon, we have the following formula. FORMULA The sum of the interior angles of a polygon with n sides is given by S=(n−2)180∘. For example, if we want to find the sum of the interior angles in a parallelogram, we have S=(4−2)180∘=2(180)=360∘ Similarly, to find the sum of the interior angles inside a regular heptagon, we have S=(7−2)180∘=5(180)=900∘. To find the measure of each interior angle of a regular polygon with n sides, we have the following formula. FORMULA The measure of each interior angle of a regular polygon with n sides is given by a=(n−2)180∘n For example, find the measure of an interior angle of a regular heptagon, as shown in Figure {PageIndex10. We have a=(7−2)180∘7=128.57∘ Figure {PageIndex10: Interior Angles Example {PageIndex5: Calculating the Sum of Interior Angles Find the measure of an interior angle in a regular octagon using the formula, and then find the sum of all the interior angles using the sum formula. Answer : An octagon has eight sides, so n=8. Step 1: Using the formula a=(n−2)180∘8 : a=(8−2)180∘8=(6)180∘8=135∘. So, the measure of each interior angle in a regular octagon is 135∘. Step 2: The sum of the angles inside an octagon, so using the formula: S=(n−2)180∘=(8−2)180∘=6(180)=1,080∘. Step 3: We can test this, as we already know the measure of each angle is 135∘. Thus, 8(135∘)=1,080∘. Your Turn {PageIndex5 Find the measure of each interior angle of a regular pentagon and then find the sum of the interior angles. Example {PageIndex6: Calculating Interior Angles Use algebra to calculate the measure of each interior angle of the five-sided polygon (Figure {PageIndex11). Figure {PageIndex11 Answer : Step 1: Let us find out what the total of the sum of the interior angles should be. Use the formula for the sum of the angles in a polygon with n sides: S=(n−2)180∘. So, S=(5−2)180∘=540∘. Step 2: We add up all the angles and solve for x : 5(x+7)+120+(6x+25)+5(2x+5)+5(3x−5)5x+6x+10x+15x+18036xx=540=540=360=10 Step 3: We can then find the measure of each interior angle: m∡A=5(10+7)=85∘m∡B=120∘m∡C=6(10)+25=85∘m∡D=5(2∗10+5)=125∘m∡E=5(3∗10−5)=125∘ Your Turn {PageIndex6 Find the sum of the measures of the interior angles and then find the measure of each interior angle in the figure shown. Figure {PageIndex12 An exterior angle of a regular polygon is an angle formed by extending a side length beyond the closed figure. The measure of an exterior angle of a regular polygon with n sides is found using the following formula: FORMULA To find the measure of an exterior angle of a regular polygon with n sides we use the formula b=360∘n. In Figure {PageIndex13, we have a regular hexagon ABCDEF. By extending the lines of each side, an angle is formed on the exterior of the hexagon at each vertex. The measure of each exterior angle is found using the formula, b=360∘6=60∘. Figure {PageIndex13: Exterior Angles Now, an important point is that the sum of the exterior angles of a regular polygon with n sides equals 360∘. This implies that when we multiply the measure of one exterior angle by the number of sides of the regular polygon, we should get 360∘. For the example in Figure {PageIndex13, we multiply the measure of each exterior angle, 60∘, by the number of sides, six. Thus, the sum of the exterior angles is 6(60∘)=360∘. Example {PageIndex7: Calculating the Sum of Exterior Angles Find the sum of the measure of the exterior angles of the pentagon (Figure {PageIndex14). Figure {PageIndex14 Answer : Each individual angle measures 3605=72∘. Then, the sum of the exterior angles is 5(72∘)=360∘. Your Turn {PageIndex7 Find the sum of the measures of the exterior angles in the figure shown. Figure {PageIndex15 Circles and Circumference The perimeter of a circle is called the circumference. To find the circumference, we use the formula C=πd where d is the diameter, the distance across the center, or C=2πr where r is the radius. FORMULA The circumference of a circle is found using the formula C=πd, where d is the diameter of the circle, or C=2πr, where r is the radius. The radius is 1/2 of the diameter of a circle. The symbol π=3.141592654… is the ratio of the circumference to the diameter. Because this ratio is constant, our formula is accurate for any size circle. See Figure {PageIndex16. Figure {PageIndex16: Circle Diameter and Radius Let the radius be equal to 3.5 inches. Then, the circumference is C=2π(3.5)=21.99in. Example {PageIndex8: Finding Circumference with Diameter Find the circumference of a circle with diameter 10 cm. Answer : If the diameter is 10 cm, the circumference is C=10π=31.42 cm. Your Turn {PageIndex8 Find the circumference of a circle with a radius of 2.25 cm. Example {PageIndex9: Finding Circumference with Radius Find the radius of a circle with a circumference of 12 in. Answer : If the circumference is 12 in, then the radius is 12=2πr122π=r=1.91in. Your Turn {PageIndex9 Find the radius of a circle with a circumference of 15.71cm. Example {PageIndex10: Calculating Circumference for the Real World You decide to make a trim for the window in Figure {PageIndex17. How many feet of trim do you need to buy? Figure {PageIndex17 Answer : The trim will cover the 6 feet along the bottom and the two 12-ft sides plus the half circle on top. The circumference of a semicircle is ½ the circumference of a circle. The diameter of the semicircle is 6 ft. Then, the circumference of the semicircle would be 12πd=12π(6)=3πft=9.4ft.12πd=12π(6)=3πft=9.4ft. Therefore, the total perimeter of the window is 6+12+12+9.4=39.4ft. You need to buy 39.4 ft of trim. Your Turn {PageIndex10 To make a trim for the window in the figure shown, you need the perimeter. How many feet of trim do you need to buy? Figure {PageIndex18 People in Mathematics: Archimedes The overwhelming consensus is that Archimedes (287–212 BCE) was the greatest mathematician of classical antiquity, if not of all time. A Greek scientist, inventor, philosopher, astronomer, physicist, and mathematician, Archimedes flourished in Syracuse, Sicily. He is credited with the invention of various types of pulley systems and screw pumps based on the center of gravity. He advanced numerous mathematical concepts, including theorems for finding surface area and volume. Archimedes anticipated modern calculus and developed the idea of the “infinitely small” and the method of exhaustion. The method of exhaustion is a technique for finding the area of a shape inscribed within a sequence of polygons. The areas of the polygons converge to the area of the inscribed shape. This technique evolved to the concept of limits, which we use today. Figure {PageIndex19: Archimedes (credit: “Archimedes” Engraving from the book Les vrais pourtraits et vies des hommes illustres grecz, latins et payens (1586)/Wikimedia Commons, Public Domain) One of the more interesting achievements of Archimedes is the way he estimated the number pi, the ratio of the circumference of a circle to its diameter. He was the first to find a valid approximation. He started with a circle having a diameter of 1 inch. His method involved drawing a polygon inscribed inside this circle and a polygon circumscribed around this circle. He knew that the perimeter of the inscribed polygon was smaller than the circumference of the circle, and the perimeter of the circumscribed polygon was larger than the circumference of the circle. This is shown in the drawing of an eight-sided polygon. He increased the number of sides of the polygon each time as he got closer to the real value of pi. The following table is an example of how he did this. \| Sides \| Inscribed Perimeter \| Circumscribed Perimeter \| \| 4 \| 2.8284 \| 4.00 \| \| 8 \| 3.0615 \| 3.3137 \| \| 16 \| 3.1214 \| 3.1826 \| \| 32 \| 3.1365 \| 3.1517 \| \| 64 \| 3.1403 \| 3.1441 \| Archimedes settled on an approximation of π≈3.1416 after an iteration of 96 sides. Because pi is an irrational number, it cannot be written exactly. However, the capability of the supercomputer can compute pi to billions of decimal digits. As of 2002, the most precise approximation of pi includes 1.2 trillion decimal digits. Who Knew?: The Platonic Solids The Platonic solids (Figure 10.100) have been known since antiquity. A polyhedron is a three-dimensional object constructed with congruent regular polygonal faces. Named for the philosopher, Plato believed that each one of the solids is associated with one of the four elements: Fire is associated with the tetrahedron or pyramid, earth with the cube, air with the octahedron, and water with the icosahedron. Of the fifth Platonic solid, the dodecahedron, Plato said, “… God used it for arranging the constellations on the whole heaven.” Figure {PageIndex20: Platonic Solids Plato believed that the combination of these five polyhedra formed all matter in the universe. Later, Euclid proved that exactly five regular polyhedra exist and devoted the last book of the Elements to this theory. These ideas were resuscitated by Johannes Kepler about 2,000 years later. Kepler used the solids to explain the geometry of the universe. The beauty and symmetry of the Platonic solids have inspired architects and artists from antiquity to the present. Exercise 10.4.1 In the following exercises, identify the regular polygons. Exercise 10.4.2 Find the perimeter of the regular pentagon with side length of 6 cm. Exercise 10.4.3 Find the sum of the interior angles of a regular hexagon. Exercise 10.4.4 Find the measure of each interior angle of a regular hexagon. Exercise 10.4.5 Find the measurements of each angle in the given figure. Exercise 10.4.6 Find the circumference of the circle with radius equal to 3 cm. 10.3.0: Exercises 10.4.0: Exercises
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Note: One of the inputs must be alphabetic (For Example, x). Please enter exactly one variable (x) and three numbers Proportion Calculator: Use this proportion calculator to find missing values in proportional relationships. It uses two methods such as cross-multiplication and proportion formula to represent the equality between two ratios. What Is A Proportion? “When two ratios are set equal to each other, then it is called a proportion” The symbol of proportion is ‘::’ and ‘=’. Two variables are said to be directly proportional if one variable is equal to the product of the other variable and a constant. If the product of the two variables is a constant, then they are said inversely proportional ( x·y is a constant). What Is The Proportion Formula? The formula for proportion is stated as: a:b::c:d=(\frac{a}{b}=\frac{c}{d}) How To Solve Proportions? There are two main and easiest ways to solve proportions: We have used both in the given example for better understanding: Example: The two ratios are 8:?::6:4, solve for the unknown variable x. Solution: Using Cross Multiplication: Step #1: Construct a Proportion (\frac{8}{x}=\frac{6}{4}) Step 2: Apply Cross Multiplication (\ x\ \times\ 6=\ 8\ \times\ 4) (\ 6x= 32) (\ x=\frac{32}{6}) (\ x=\frac{16}{3}= 5.33…) Using The Proportion Formula: a:b::c:d=(\frac{a}{b}=\frac{c}{d}) (\frac{8}{x}=\frac{6}{4}) (\frac{8}{x}=\ 1.5) (\ x =\frac{8}{1.5} = 5.33..) (\ x = 5.33..) In this example, we have found the proportion manually. But for more complex problems, you can use the above proportion calculator. Other Languages: Kalkulator Proporcji, Kalkulator Proporsi, Proportions Rechner, 比例計算, Calculo De Proporção, Calculadora De Proporciones, Calcolo Proporzioni, Калькулятор Пропорций, Mittasuhteet Laskin. 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2026	https://par.nsf.gov/servlets/purl/10313443	An Accurate Computational Model for the Hydration Extent of 1 Atmospherically Relevant Carbonyls on Aqueous Atmospheric 2 Particles 3 4 Matthew J. Elrod , Jane A. Sedlak , and He Ren 5 Department of Chemistry and Biochemistry, Oberlin College, Oberlin, Ohio, 44074 USA 6 Email: mjelrod@oberlin.ed u7 8 Abstract 9 The carbonyl hydration equilibria of several atmospherically relevant carbonyl compounds were 10 studied using nuclear magnetic resonance (NMR) spectroscopy and computationally investigated 11 using the MG2MS elect ronic structure method . The results were used to build an empirically 12 adjusted computational model that resulted in the calculation of free energies of hydration that 13 were accurate to within 1 kcal mol -1. The new model was use d to predict the hydration extent of 14 other potentially atmospherically relevant carbonyl compounds for which no experimental data 15 exists. Because the hydration extent of a carbonyl compound dramatically affects its effective 16 volatility and Henry’s law solu bility, the more accurate estimates for the extent of carbonyl 17 hydration predicted by th is new model will help improve our understanding of which carbonyl 18 compounds are major constituents of aqueous atmospheric particles. 19 Keywords: atmosphere, aerosol, carbonyl, hydration, equilibrium 20 21 Introduction 22 Secondary organic aerosol (SOA) 1 forms in the atmosphere via gas -particle partitioning . SOA 23 chemical composition can also be affected by in -particle chemical processing of condensed 24 components . T he particular chemical makeup of SOA likely plays a role in the effect that it has 25 on human respiratory and cardiovascular disease, 2, 3 visibility loss 4 and climate modification. 5 In 26 orde r to help predict the occurrence of different atmospheric constituents on atmospheric 27 particles such as SOA and clouds , attention has been focused on parametrizing these constituents 28 with respect to their volatility and Henry’s law solubility , which is rel evant to the ir interaction 29 with aqueous atmospheric particles .6 Although many carbonyl compounds are directly emitted 30 into the atmosphere by natural and anthropogenic sources, 7 the oxidizing nature of the 31 atmosphere is also capable of installing carbonyl functional groups into almost any organic 32 compound , including the dominant non -methane hydrocarbon, isoprene. 8 Given their ubiquitous 33 presence in the atmosphere , it is important to understand the physical and chemical properties of 34 the se carbonyl compounds present in the atmosphere under various atmospheric conditions, 35 including their interactions with aqueous atmospheric particles .36 Carbonyl compounds , like propi onaldehyde , are capable of undergoing nucleophilic addition of 37 water across the carbonyl double bond to form hydrated geminal diol species (Figure 1) .9 38 39 Figure 1: Hydration equilibrium for propionaldehyde , a known atmospheric constituent .40 41 This hydration process dramatically affects both the volatility and Henry’s law solubility of 42 propionaldehyde in aqueous solutions. The EPA’s chemical and physical property calculator, 43 EPISUITE, predicts that the hydrated form of propionaldehyde is almost 300 times less volatile 44 and 20 times more water soluble than the unhydrated form 10 owing to the strong effect of the two 45 hydrogen bonding groups in the hydrated form. This clearly illustrates that the ability of a 46 partic ular carbonyl compound to undergo hydration significantly increa ses the probability that it 47 will be incorporated into aqueous atmospheric particles. However, the propensity of a particular 48 compound to undergo hydration can vary enormously – while formaldehyde exists almost 49 entirely in the hydrated form under aqueous c onditions, 11 acetone remains almost entirely 50 unhydrated. 12 51 There have been several previous experimental and computational studies of the hydration of 52 potentially atmospherically relevant carbonyl compounds. 13 -20 However, these studies have 53 focused on parameterizing the properties of a subset of chemical species for a particular 54 atmospheric system, as opposed to developing qualitati ve general structure -property 55 relationships or quantitatively predictive models for carbonyl hydration. We conducted 56 temperature dependent measurements of the hydration equilibrium constant for several simple 57 atmospherically relevant carbonyl compounds in order to establish the van’t Hoff 58 thermodynamic parameters using nuclear magnetic resonance (NMR). We then use d the se 59 results , and other experimental measurements of the free energy of hydration , to empirically 60 calibrate MG2MS ab initio electronic structure calculations , and create a quantitative model that 61 accurately predicts free energies of hydration . We also use d MG2MS results to assess the 62 accuracy of group contribution thermodynamic methods for carbonyl hydration reactions. 63 Finally, t he calibrated computed hydration properties were used to establish structure -free energy 64 of hydration relationships and predict the hydration extent of a number of potentially 65 atmospherically relevant carbonyl compounds for which experimental data is lacking. 66 67 Methods and Materials 68 69 Chemicals . The following commercially available chemicals were used from MilliporeSigma 70 (acetaldehyde, 99.5%; propionaldehyde, 97%, pyruvic acid, 98%, methylglyoxal solution, 40% 71 in H 2O; acetone, 99.5%, hydroxyacetone, 90%; 2,3 -butanedione, 97%), TCI America 72 (dihydroxyacetone, 97% ) and Cambridge Isotope Laboratories, Inc. (deuterium oxide, 99. 5%) .73 The potassium salts of 1 -sulfatoacetone and 3 -sulfato -1-hydroxyacetone were each prepared and 74 characterized according to the procedure outlined by Hettiyadura et al. .21 75 Hydration Equilibrium Constant Measurements . All systems were investigated in bulk 76 solutions with total volumes of 750 μL (the amount required for NMR analysis) with either 10 77 μL or 50 μL (or 10 mg of the appropriate potassium salt of the sulfato compounds) of the 78 carbonyl compound added to an appropriate volume of D 2O. The systems were observed to 79 achieve equilibrium within the five minute preparation and NMR measurement timeframe. This 80 fast establishment of equilibrium is consistent with previous measurements of short hydration 81 lifetimes ( hydration < 30 s) for formaldehyde, 22 glycoaldehyde, 23 glyoxylic acid, 22 and pyruvic 82 acid. 24 NMR spectra were collected on a Bruker 400 MHz instrument using built -in pulse 83 sequences. 1H s pectra were calibrated to the HDO peak at 4.79 ppm . The relative concentrations 84 of the various species were determined by integration of one or more unique peaks in the 1H85 spectra. The overall precision of the relative concentration ratios was estimated to be better than 86 10 % (with much better precisions fo r systems with K H near unity) . van’t Hoff data was collected 87 by performing temperature dependent measurements of the concentration ratios. The built -in 88 Bruker temperature control apparatus was used to vary the temperatures from 298 K to 318 K. A 89 relativ ely narrow temperature range was used to prevent loss of volatile components and thermal 90 degradation of unstable compounds. 91 Computation s. The van’t Hoff parameters for the hydration equilibria of potentially 92 atmospherically relevant carbonyl compounds wer e calculated according to the following 93 procedure. Geometries (determined at the B3LYP/6 -31G(d,p) level) and energies of the relevant 94 species were calculated using a modified version of the G2MS compound method (MG2MS) 25 a95 variation on G2 theory. 26 The Polarizable Continuum Model (PCM) method 27 was used to 96 account for the effects of water solvation on the reactant and product properties. All calculations 97 were carried out with the Gaussian 09 computational suite. 28 Each stationary point was 98 confirmed as a potential energy minimum by inspection of the calculated frequencies. The 99 overall energy expression for the MG2MS scheme is defined in equation 1: 100 Eelec (0 K) = E CCSD(T)/6 -31G(d) + E MP2/6 -311+G(2df,2p) – EMP2/6 -31G (d) + HLC (1) 101 where HLC is an empirically -defined correction term with HLC = An α + Bn β where nα and nβ are 102 the number of α - and β -electrons, respectively, and the constants A and B are 6.06 and 0.19 mH, 103 respectively (all species investigated were closed shell; therefore n  = n  .). The enthalpy is 104 calculated from equation 2: 105 H (298 K) = E elec (0 K) + E zpe + Ethermal (2)106 where Ezpe is the zero point energy and Ethermal includes the corrections to necessary to adjust the 107 internal energy to 298 K and convert to enthalpy. The free energy is calculated from equation 3 :108 G (298 K) = H(298 K) – 298 K  S(298 K) (3) 109 where S(298 K) is the entropy calculated at 298 K. 110 Our previous MG2MS results for atmospherically relevant systems (including radicals and ions) 111 indicate that the MG2MS calculated enthalpies of reaction are typically accurate to within 2 kcal 112 mol -1 for systems similar to those under study here. 29 113 The MG2MS free energy results were observed to be systematically in error for compounds for 114 which experimental values are available . To correct for this error, two methods were used to 115 empirically adjust the MG2MS results in order to achieve more accurate predictions for 116 compounds that have not been experimentally studied. The first method involved using the 117 experimental van’t Hoff parameters to determine a MG2MS c orrection factor for the enthalpy 118 and entropy of hydration individually. The second method involved determining a MG2MS 119 correction factor for the free energy of hydration at 298 K , which was possible b ecause many 120 more experimental 298 K free energy of hydration values are availabl e. The advantage of the 121 first method is that it should lead to more accurate predictions for the free energy of hydration at 122 temperatures other than 298 K, while the advantage of the second method is that it should be 123 better calibrated to experimental values and lead to more accurate predictions for the free energy 124 of hydration at 298 K. 125 126 Results and Discussion 127 NMR Assignments. As an example of the typical NMR spectra collecte d for the hydration of 128 carbonyl compound s, the 1H NMR spectrum and assignments for the species present in the 129 propionaldehyde hydration sy stem are given in Figure 2. Because the hydration of the carbonyl 130 carbon atom causes significant changes in the electron density experienced by the nearby 131 hydrogen atoms, large chemical shifts changes are associated with the hydration of carbonyls and 132 the identification and quantification of all species is straightforward. 133 134 Figure 2: 1H NMR spectrum for the propionaldehyde hydration system at 298 K. 135 Experimental Thermodynamics Parameters. The hydration equilibrium constants were 136 calculated directly from the NMR -determined concentration ratio of the hydrated compound to 137 the carbonyl compound. For example, for the propionaldehyde hydration system, K H is defined 138 as 139 3 2 2H3 2 [CH CH CH(OH) ] K [CH CH C( O)H] = = (4) 140 The standard free energies of hydration were directly determined from the hydration equilibrium 141 constants measured at T = 298 K :142 or HG RTln K  = − (5)143 The van’t Hoff parameters were determined via linear regression analysis from measurem ents of 144 the temperature dependence of the equilibrium constant :145 o or rH H 1 Sln K R T R − =  + (6)146 Figure 3 shows the van’t Hoff plot resulting from experimental measurements for 147 propionaldehyde. In this ideal case where K H is near unity, rHo and rSo were determined with 148 only 3% (1 ) error. Table 1 contains the van’t Hoff parameters (and one standard deviation 149 uncertainties) for all systems experimentally studied in the present work. The expected 150 uncertainty for a rGo value near zero using the previously stated 10% concentration 151 uncertainties is about 0.1 kcal mol -1, which is consistent with the lowest regression analysis -152 determined rGo uncertainties cited in Table 1. For the rGo values , the present results are in 153 very good agreement with most previous literature reports , while statistically significant 154 differences for some of the rHo and rSo values are observed. The good rGo agreement and 155 subpar rHo and rSo agreement suggests that some of the earlier studies might have been 156 impacted by inaccuracies in the temperature dependent measurements and/or analysis. For 157 example, the previous rHo and rSo parameters for acetaldehyde and acetone were determined 158 from measurements at only three different temperatures. 12 159 160 161 Figure 3: van’t Hoff plot for the propionaldehyde hydration system . The corresponding 162 thermodynamic values (and associated uncertainties) are given in Table 1. 163 system experimental, this work experimental, literature MG2MS, this work r rSo rGo r rSo rGo r rSo rGo acetaldehyde -6.44 1 ± 0.093 -21.0 0 ± 0.30 -0.1 9 ± 0.13 -5.3 12 -17 12 -0.2 12 -6.0 -39 5.4 pro pionaldehyde -5.80 ± 0.19 -19.0 1 ± 0.62 -0.1 4 ± 0.26 -0.1 18 -6.0 -39 5.5 pyruvic acid -6.02 1 ± 0.038 -18.8 4 ± 0.12 -0.407 ± 0.052 -8.2 30 -27 30 -0.2 30 -6.5 -40 5.5 methylglyoxal -7.6 2 ± 0.62 -26 .2 ± 2.0 0.19 ± 0.87 0.2 31 -6.9 -41 5.4 acetone -2.6 4 ± 0.37 -21 .3 ± 1.2 3.7 1 ± 0.51 -3.1 12 -23 12 3.7 12 -0.4 -42 12.2 hyd roxyacetone -6.1 ± 0.34 -28 .2 ± 1.1 2. 27 ± 0.47 2.3 32 -5.1 -41 7.2 dihydroxyacetone -6.1 ± 0.70 -19 .3 ± 2.3 -0. 35 ± 0.98 -7.8 -42 4.7 2,3 -butanedione -5. 78 ± 0.84 -16.9 ± 2.8 -0.7 ± 1.2 -0.4 33 -4.7 -39 7.0 164 Table 1: van’t Hoff parameters (and one standard deviation uncertainties) for experimentally studied systems (units: rHo and rGo 165 (kcal mol -1) and rSo (cal mol -1 K-1)). 166 167 168 169 Computational Thermodynamic Model Performance . Table 1 also contains the MG2MS 170 computational thermodynamics results for the set of presently experimentally studied carbonyl 171 compounds, while Table S1 contains the MG2MS computational thermodynamics results for a 172 wider s et of potentially atmospherically relevant compounds. The results in Table 1 indicate 173 that , as expected, the MG2MS rHo values are within the expected 2 kcal mol -1 of the 174 experimental values, with no obvious systematic error. However , large systematic error is 175 observed for the rSo values . This is likely because t he MG2MS method leads to rSo values that 176 are about a factor of two too negative compared to the experi mental results , which then leads to 177 G values that are about five kcal mol -1 too positive at 298 K. This computational systematic 178 error has also been observed in previous ab initio electronic structure studies of the hydration of 179 glyoxal .13 , 19 Kua et al . identify problems with the treatment of translation and rotation, 180 differing solubilities of the re action components, and the problem of inseparability of the thermal 181 corrections to free energy and effect of the solvent as potential reasons for the systematic error .13 182 A method in which the entropy contribution to free energy is estimated as -0.5TS ( i.e., half of its 183 computed contribution) was previously shown to be effecti ve in compensating for the systematic 184 error. 13 While this method impro ved the agreement of the present computational MG2MS values 185 with experimental values, we found that, somewhat surprisingly, a single system -independent 186 correction factor was even more effective . In particular , we used two different models in which 187 the MG2 MS results are empirically calibrated to achieve the best agreement with available 188 experimental results , as discussed above in the Methods and Materials section . The first model 189 has separate correction factors for the rHo and rSo terms which are determined from the 190 experimental van’t Hoff results . For this first model, the rHo MG2MS results are systematically 191 too positive by 0.4 kcal mol -1, while the rSo MG2MS results are systematically too negative by 192 19 cal mol -1 K-1 when compared to the a vailable experimental results (Table 1). The MG2MS 193 results are adjusted by these constant values to achieve new, more accurate “van’t Hoff adjusted” 194 thermodynamic values (Table S1 ). The second model has a single correction factor for rGo is 195 determined f rom a larger set of experimental rGo values. For this second model , the rGo 196 MG2MS results are systematically too positive by 5.4 kcal mol -1 (Table S1 ). Therefore, the 197 MG2MS results are adjusted by this constant value to achieve new, more accurate “ rGexp 198 adjusted” rGo values (Table S1 ). When comparing the predictions from the “van’t Hoff 199 adjusted” rGo values to all available experimental rGo values, the average err or is 1.0 kcal mol - 200 1 , with a standard deviation of errors of 0.7 kcal mol -1. When comparing the predictions from the 201 “rGexp adjusted” rGo values to all available experimental rGo values, the average error is 0.7 202 kcal mol -1, with a standard deviation of errors of 0.8 kcal mol -1. Therefore the “ rGexp adjusted” 203 model leads to slightly more accurate rGo values, but there is a slightly larger fraction of outlier 204 values as well. However , both models lead to similar predicted values which are significantly 205 more accurate than the unadjusted MG2MS values , and more accurate than purely computational 206 models which used much more computational ly intensive high levels of theory. 34 Since 207 equilibrium constants are more useful for the purpose of assessing the potential atmospheric 208 relevance of carbonyl hydration process, Table 2 contains the experimental, van’t Hoff adjusted, 209 difference b etween van’t Hoff adjusted and experimental and rGexp adjusted, difference 210 between rGexp adjusted and experimental log 10 KH values for a number of C 1-C3 atmospherically 211 relevant carbonyls. Table 3 contains the same information for a number of C 4-C7212 atm ospherically relevant carbonyls. All of the values in Tables 2 and 3 were calculated from the 213 thermodynamic data reported in Table S1. 214 experimental van’t Hoff adjusted van’t Hoff (adj -exp) rGezp adjusted rGezp (adj -exp) C1 compounds formaldehyde 3.4 11 3.4 0.0 2.9 -0.5 sulfato formaldehyde ion -4.7 -5.2 C2 compounds acetaldehyde 0.1 a 0.5 0.4 0.0 -0.1 glycoaldehyde 1.0 23 1.6 0.6 1.1 0.1 2-sulfato acetaldehyde ion 1.8 1.2 glyoxal 3.3 2.8 monohydrated glyoxal 3.0 2.5 glyoxylic acid 2.5 22 4.5 2.0 4.0 1.5 C3 compounds propionaldehyde 0.0 a 0.4 0.3 -0.1 -0.2 acetone -2.7 a -4.5 -1.8 -5.0 -2.3 hydroxyacetone -1.7 a -0.8 0.9 -1.3 0.4 fluoroacetone -0.8 33 -0.7 0.1 -1.2 -0.4 chloroacetone -1.0 33 0.0 1.0 -0.5 0.5 sulfatoacetone ion -1.1 a -0.2 0.9 -0.7 0.4 nitratoacetone 0.1 -0.4 1,1 -dichloroacetone 0.4 33 1.0 0.6 0.5 0.1 dihydroxyacetone 0.3 a 1.1 0.8 0.5 0.2 1,3 -dichloroacetone 0.6 33 1.5 0.9 1.0 0.4 3-sulfato -1- hydroxyacetone ion 0.0 a 0.5 0.5 0.0 0.0 lactaldehyde 1.4 35 1.7 0.3 1.2 -0.2 methylglyoxal (1) 2.8 2.3 methylglyoxal (2) 1.2 0.7 monohydrated (1) methylglyoxal -0.2 31 0.6 0.8 0.0 0.2 pyruvic acid 0.1 a 0.6 0.5 0.1 0.0 215 Table 2 : log 10 KH values for C 1-C3 compounds . athis work .216 217 218 219 experimental van’t Hoff adjusted van’t Hoff (adj -exp) rGezp adjusted rGezp (adj -exp) C4 compounds butanal -0.1 12 0.5 0.6 -0.1 0. 0 2-butanone -3.3 -3.8 2,3 -butanedione 0.3 33 -0.6 -0.9 -1.2 -1.5 1-sulfato -2,3 - butanedione ion -0.2 -0.7 methyl vinyl ketone < -2.3 18 -2.1 -2.6 3,4 -dihydroxy -2-butan - 2-one (DHBO) -1.7 -2.3 methacrolein < -2.3 18 0.0 -0.5 2,3 -dihydroxy -2- methylpropanal (DHMP) 0.0 -0.5 4-hydroxy -2-butanone -2.6 -3.1 4-nitrato -2-butanone -1.9 -2.5 4-sulfato -2-butanone ion -2.7 -3.2 C5 compounds 2-pentanone -3.0 -3.6 3-pentanone -4.6 -5.1 2-hydroxy -2-methyl -4- sulfato -3-butanone -0.1 -0.6 3-(hydroxymethyl) -2- methyloxirane -2- carbaldehyde (1 - IEPOXO) 2.0 1.4 3-(hydroxymethyl) -3- methyloxirane -2- carbaldehyde (4 - IEPOXO) 3.1 2.6 2,3,4 -trihydroxy -2- methylbutanal (2 -MT -1- aldehyde) 0.6 0.1 2,3,4 -trihydroxy -3- methylbutanal (2 -MT -4- aldehyde) -0.5 -1.1 C7 compound benzaldehyde -1.9 -2.4 220 Table 3: log 10 KH values for C 4-C7 compounds. athis work. 221 Structure -Free Energy Relationships. The log 10 KH values (Tabl es 2 and 3 ) allow for the 222 opportunity to assess whether there are significant structure -free energy relationships that can be 223 used to qualitatively predict the hydration propensity of potentiall y atmospherically relevant 224 carbonyl compounds. The results clearly indicate that the aldehydes generally have much larger 225 KH values than do the ketones, a distinction that is true of most carbonyl compounds. 9 As most 226 clearly illustrated by acetone and its derivatives in Table 2 , the present results indicate that 227 neighboring electronegative functional groups such as other carbonyl, hydroxyl, and halogen 228 groups serve to increase the extent of hydration, most likely by drawing electron density away 229 from the carbonyl carbon atom whi ch makes for a more electropositive target for nucleophilic 230 attack by water. This effect is quantitatively significant because acetone has a K H value so 231 small (log 10 KH = -2.7) that it is difficult to experimentally measure, while several of its 232 electron egative functional group -derivatized forms have K H values greater than unity (log 10 KH >233 0) . Neighboring carbonyl groups have the largest effect on K H values, with glyoxal and 234 glyoxylic acid having much larger K H values than the base compound acetaldehyde , and 235 methylglyoxal and pyruvic acid having much larger K H values than the base compounds 236 propionaldehyde and acetone. With the exception of formaldehyde, neighboring sulfate and 237 nitrate functional groups have a stronger effect than hydroxyl groups in the ir raising of K H, thus 238 suggesting that carbonyl -containing organosulfates and organonitrates are more likely to be 239 significantly hydrated under atmospheric conditions than multifunctional hydroxy and 240 monofunctional carbonyl compounds .241 Comparison to Group Contribution Methods. In addition to ab initio electronic structure 242 methods, the Joback group contribution method 36 has been used to estimate the thermodynamic 243 values of carbonyl hydration reactions. 20 This method uses the molecular structure and 244 functional group identities of a variety of chemical species to parametrize the contribution of 245 each structural aspect to a particular experimentally measur ed thermodynamic value. This 246 approach has the benefit of a very simple and fast computational algorithm, but its accuracy is 247 constrained by the type of molecules used to “train” the group contribution parameters. For 248 example, this method was shown to per form extremely well , accuracies of better than 1 kcal mol - 249 1 , for the prediction of the gas phase free energies of formation of a number of atmospherically 250 relevant carbonyl compounds , some of which were themselves used in the Joback training set. 20 251 Because the present study focused on the aqueous phase thermodynamic parameters , there is not 252 a simple way to compare the present thermodynamic parameters to the gas phase ones predicted 253 by the Joback method . However, we have carried out separate calculations on the gas phase 254 hydration reaction of acetaldehyde in order to allow a direct comparison to the results obtained 255 by the Joback method. While ab initio methods are often characterized by significant system atic 256 errors, as discussed above for the calculation of rGo values for the carbonyl hydration reactions, 257 the use of isodesmic (bond -conserving) reactions allows for the cancellation of most systematic 258 error and can lead to highly accurate ab initio G valu es. For example, an isodesmic reaction 259 approach was used to calculate the free energy of formation of carbon dioxide, via the 260 disproportionation reaction of formaldehyde to form methane and carbon dioxide, with less than 261 1 kcal mol -1 error using a level of theory significantly lower than the present MG2MS 262 approach. 37 Since the standard gas phase enthalpy of formation of ethylene glycol is a well -263 established experimental property and its standard entropy of formation can be calculated from 264 similarly well -kno wn experimental values, 38 an isodesmic isomerization reaction between 265 ethylene glycol and the acetaldehyde hydration prod uct was used to establish a value for the gas 266 phase free energy of formation of the acetaldehyde hydration product, 267 o o or f 3 2 f 2 2 G G (CH CH(OH) ) G (CH (OH)CH (OH))  =  −  (7) 268 such that 269 o o of 3 2 r f 2 2 G (CH CH(OH) ) G G (CH (OH)CH (OH))  =  +  (8) 270 The MG2MS value for rGo for the isodesmic gas phase ethylene glycol to acetaldehyde 271 hydration product isomerization reaction was found to be -11.4 kcal mol -1. Using the 272 experimental fGo value for gaseous ethylene glycol 38 of -73.1 kcal mol -1, the fGo value for the 273 gaseous acetaldehyde hydration product is determined to be -84.5 kcal mol -1, with an uncertainty 274 on the order of 1 kcal mol -1, as discussed above. Th e Joback method, using the Yaws 36 275 parameterization, predicts a value fGo value for the acetaldehyde hydration product of -74.1 276 kcal mol -1; this large error likely results from the lack of geminal diol species thermodynamic 277 data in the Joback training set. Therefore, while group contribut ion algorithms such as the 278 Joback method are considerably easier to apply than the present empirically adjusted MG2MS 279 method, those methods can be highly inaccurate , particularly when relatively unusual structural 280 components such as geminal diols are prese nt in the reaction system. Because the group 281 contribution term for geminal diols is therefore likely in error in the Joback method and would 282 lead to erroneous results for the hydration of any carbonyl , this method is not recommended for 283 the prediction of the hydration extent of atmospherically relevant carbonyls. 284 Atmospheric Implications. As outlined in the Introduction section, the ability to accurately 285 predict the hydration extent of atmospherically important carbonyl compounds on aqueous 286 atmospheric particles is a primary motivation of the present stud y. Here we have shown that it is 287 possible to generate accurate hydration extent predictions for a number of the most 288 atmospherically abundant carbonyl compounds (Tables 2 and 3) ,7 as well as a number of 289 important SOA components that are thought to derive from isoprene photochemistry and SOA 290 processing (Table 3). 39 -41 Atmospheric proce ssing of organic compounds tends to add 291 electronegative oxygen -containing functional groups . Our model suggests that these new 292 multifunctional carbonyl compounds are more likely to be hydrated under atmospheric 293 conditions than simple aldehydes and ketones. More specifically, our results support th ose of 294 previous studies which have shown that glyoxal 13 -15 and methylglyoxal 16 , 17 undergo significant 295 hydration under atmospheric conditions. Interestingly, experimental measurements illustrate that 296 the terminal carbonyl group of methylglyoxal undergoes hydration , followed by hydration of the 297 internal carbonyl group, but that a compound with a hydrated internal carbonyl group and an 298 unhydrated terminal carbonyl group was not observed .31 Th is result indicates that either our 299 prediction that KH for the internal carbonyl group of methylglyoxal is also greater than unity is 300 incorrect, or that kinetic effects prohibit the experimental observation of this species on the 301 experimental time scales .302 The predictions of this model suggest that some of the carbonyl organosulfates detected in SOA 303 in Atlanta, GA, 41 such as 1 -sulfato acetone ion and 2-sulfato acet aldehyde ion are present in their 304 hydrated form , while other organosulfates, such as 4 -sulfat o-2-butanone, will not be significantly 305 hydrated under atmospheric conditions . Our predictions also suggest that the aldehydic isoprene 306 oxidation intermediates, IEPOXO, 39 , 42 will undergo significant hydration on aqueous 307 atmospheric particles, which would lea d to greater partitioning to the particle phase , which in 308 turn would allow for more efficient processing by nucleophilic reactions involving IEPOXO’s 309 epoxide functional group. However , if the epoxide group of IEPOX is hydrolyzed first, the 310 hydration of th e carbonyl group is predicted to be less favorable, which indicates that epoxide 311 functional groups act to increase the hydration propensity of carbonyl groups even more than 312 hydroxyl groups. The high -NO x isoprene oxidation product nitratoacetone 43 is also pr edicted to 313 be significantly hydrated, which would also increase its partitioning to the particle phase and 314 potentially promote such processes as the hydrolysis of the nitrate group and/or substitution of 315 the nitrate group by sulfate. 44 , 45 316 This model has shown that many atmospher ically important carbonyl compounds are predicted 317 to undergo significant hydration on aqueous atmospheric particles which confers lower volatility 318 and higher Henry’s law solubility on the hydrated compound as compared to the unhydrated 319 compound s, which has important ramifications for gas -particle partitioning and subsequent 320 particle phase reactivity. 321 Supporting Information 322 The Supporting Information is available free of charge on the ACS Publications website at DOI: 323 thermodynamic parameters .324 Acknowledgement s325 This material is based upon work supported by the National Scie nce Foundation under Grant 326 No s. 1559319 and 1841019. 327 328 References and Notes 329 Kanakidou, M.; Seinfeld, J. 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2027	https://lg.m.wikipedia.o…e_Properties.png	File:Circle Properties.png - Wikipedia [x] Olupapupa olusooka Okulondoba empapula Kumpi ne Yingiramu seetingi ez'enjawulo Donate Now If Wikipedia is useful to you, please give today. Okutangaaza ku Wikipedia Okutangaaza ku kkomo ery'obuvunaaniro bwaffe obw'omu mateeka Noonya File:Circle Properties.png Olulimi Goberera olupapula luno View on Wikimedia Commons Add local description Fayiro Ebyafaayo ebya fayiro eno Emiko ekifaananyi kino mwe kikozeseddwa Global file usage Ebikwata ku kifaananyi Tewali kisingako wano. Circle_Properties.png(pikseli 300 ku 300 , bunene bwa fayiro: KB 39, kika kya MIME: image/png) Fayiro eno eva ku Wikimedia Commons era eyinza okukozesebwa pulojekiti endala. Ennyonnyola ku lupapula lwayo [ file description page] eyo eragiddwa wansi. Mu bufunze Ennyinyonnyola Circle Properties.pngEnglish: This diagram consists of the components of a circle (Circumference, Diameter, Radius), of which, the number pi may be derived as the ratio of the circumference to the diameter. pi = C / D = 3.1415926535... Ennaku z'omwezi 21 Gwakkuminogumu(Museenene) 2022 Ensibuko Omulimu gwo kennyini OmuwandiisiZacguymarino Licensing I, the copyright holder of this work, hereby publish it under the following license: This file is licensed under the Creative CommonsAttribution-Share Alike 4.0 International license. You are free: to share – to copy, distribute and transmit the work to remix – to adapt the work Mu bukwakkulizo buno wammanga: attribution – You must give appropriate credit, provide a link to the license, and indicate if changes were made. You may do so in any reasonable manner, but not in any way that suggests the licensor endorses you or your use. share alike – If you remix, transform, or build upon the material, you must distribute your contributions under the same or compatible license as the original. CC BY-SA 4.0 Creative Commons Attribution-Share Alike 4.0 true true Captions Luganda Add a one-line explanation of what this file represents Lungereza Simple diagram of the components of a circle Items portrayed in this file depictsLungereza circleLungereza obwetoloovu π diameterLungereza radiusLungereza creatorLungereza some value author name stringLungereza: Zacguymarino Wikimedia usernameLungereza: Zacguymarino URL: copyright statusLungereza copyrightedLungereza copyright licenseLungereza Creative Commons Attribution-ShareAlike 4.0 InternationalLungereza source of fileLungereza original creation by uploaderLungereza inceptionLungereza 21 Gwakkuminogumu(Museenene) 2022 media typeLungereza image/png checksumLungereza 9f2d65a04ebd2057d9e5022c9f0f11ee2b523af5 determination method or standardLungereza: SHA-1 data sizeLungereza 40,329 byte heightLungereza 300 pixel widthLungereza 300 pixel Ebyafaayo ebya fayiro eno Bw'onyiga ku nnaku n'essaawa, ojjakulaba fayiro nga bwe yali efaanana ku kiseera ekyo. \| \| Ennaku n'obudde \| Kulingiza \| Obuwanvu n'obugazi bwakyo \| Eyakiteekawo \| Okulw'ogerako \| --- --- --- \| \| oluwandika oluliwo kakano \| 17:10, 21 Gwakkuminogumu(Museenene) 2022 \| \| 300 × 300 (KB 39) \| Zacguymarino \| Uploaded own work with UploadWizard \| Emiko ekifaananyi kino mwe kikozeseddwa Empapula 1 ezikuggusa ku fayiro eno ze: Entoloovu = Enkula ennetoloovu Global file usage Wiki endala zino wammanga zikozesa fayiro eno: Enkozesa ku fr.wiktionary.org Wiktionnaire:Entrée étrangère du jour gaandíñja Modèle:Entrée étrangère du jour/2025/04/02 Enkozesa ku ms.wikipedia.org Pi Enkozesa ku yo.wikipedia.org Roboto Ebikwata ku kifaananyi Ekyakozesebwa okutegeka fayiro eno esobole okugenda ku kompyuta, okugeza kamera oba sikaana, kyagiwandikamu ebigikwatako. Ebyawandikibwa ebyo biyinz'okuba nga tebikyamalayo fayiro ebigikwatako bw'ebanga okuva lweyasooka kutegebwa yakolebwako n'ate. \| Unique ID of original document \| xmp.did:5a4fa38e-476b-4e00-8406-68784ab6525e \| \| Software used \| GIMP 2.10 \| \| Horizontal resolution \| 28.35 dpc \| \| Vertical resolution \| 28.35 dpc \| \| File change date and time \| 14:52, 19 Gwamusanvu(Kasambula) 2022 \| Bino bigyidwa ku " Languages This page is not available in other languages. Enkola yaffe ku kukuuma ebikufako Okutangaaza ku Wikipedia Okutangaaza ku kkomo ery'obuvunaaniro bwaffe obw'omu mateeka Enneeyisa Abakola software wa Wiki Ebibalo Ekiwandiiko kya Cookie Ebigobererwa Ekibanja eky'oku mugguukiriro
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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 Intersection point between circle and line (Polar coordinates) Ask Question Asked 8 years, 8 months ago Modified7 years, 4 months ago Viewed 2k times This question shows research effort; it is useful and clear 0 Save this question. Show activity on this post. I'm wondering if there is a way of finding the intersection point between a line and a circle written in polar coordinates. matlab % Line x_line = 10 + r cos(th); y_line = 10 + r sin(th); %Circle circle_x = circle_r cos(alpha); circle_y = circle_r sin(alpha); So far I've tried using the intersect(y_line, circle_y) function without any success. I'm relatively new to MATLAB so bear with me. matlab line geometry intersection Share Share a link to this question Copy linkCC BY-SA 4.0 Improve this question Follow Follow this question to receive notifications edited May 29, 2018 at 18:47 Jonathan Hall 80.3k 19 19 gold badges 161 161 silver badges 206 206 bronze badges asked Jan 31, 2017 at 11:34 bullbobullbo 131 1 1 silver badge 10 10 bronze badges Add a comment\| 2 Answers 2 Sorted by: Reset to default This answer is useful 1 Save this answer. Show activity on this post. I have generalised the below so that other values than a=10 can be used... ```matlab a = 10; % constant line offset th = 0; % constant angle of line % rl = ?? - variable to find % Coordinates of line: % [xl, yl] = [ a + rl cos(th), a + rl sin(th) ]; rc = 1; % constant radius of circle % alpha = ?? - variable to find % Coordinates of circle: % [xc, yc] = [ rc cos(alpha), rc sin(alpha) ]; ``` We want the intersection, so xl = xc, yl = yc matlab % a + rl cos(th) = rc cos(alpha) % a + rl sin(th) = rc sin(alpha) Square both sides of both equations and sum them. Simplifying sin(a)^2 + cos(a)^2 = 1. Expanding brackets and simplifying further gives matlab % rl^2 + 2 a rl (cos(th) + sin(th)) + 2 a - rc^2 = 0 Now you can use the quadratic formula to get the value of rl. Test discriminant: ```matlab dsc = (2 a (cos(th) + sin(th)) )^2 - 4 (2 a - rc^2); rl = []; if dsc < 0 % no intersection elseif dsc == 0 % one intersection at rl = - cos(th) - sin(th); else % two intersection points rl = -cos(th) - sin(th) + [ sqrt(dsc)/2, -sqrt(dsc)/2]; end % Get alpha from an earlier equation alpha = acos( ( a + rl . cos(th) ) ./ rc ); ``` Now you have 0, 1 or 2 points of intersection of the line with the circle, from certain known and unknown values about each line. Essentially this is just simultaneous equations, see the start of this article for a basis of the maths Share Share a link to this answer Copy linkCC BY-SA 3.0 Improve this answer Follow Follow this answer to receive notifications edited Jan 31, 2017 at 12:26 answered Jan 31, 2017 at 12:17 WolfieWolfie 30.7k 7 7 gold badges 30 30 silver badges 60 60 bronze badges Comments Add a comment This answer is useful 0 Save this answer. Show activity on this post. Do you need to do it numerically? This problem would have an easy analytical solution: The point (10 + rcos(th),10 + rsin(th)) is on a circle with radius R iff (10+rcos(th))^2 + (10+rsin(th))^2 == R^2 <=>200+r^2 + 2r(cos(th)+sin(th)) == R^2 <=>r^2 + 2rsqrt(2)sin(th+pi/4) + 200 - R^2 = 0 which is a quadratic equation in r. If the discriminant is positive, there are two solutions (corresponding to two intersection points), otherwise, there are none. If you work out the math, the condition for intersection is 100(sin(2th)-1)+circle_r^2 >= 0 and the roots are -10sqrt(2)sin(th+pi/4)[1,1] + sqrt(100(sin(2th)-1)+circle_r^2)[1,-1]. Here is a Matlab plot as an example for th = pi/3 and circle_r = 15. The magenta markers are calculated in closed-form using the equation shown above. Share Share a link to this answer Copy linkCC BY-SA 3.0 Improve this answer Follow Follow this answer to receive notifications answered Jan 31, 2017 at 12:18 FlorianFlorian 1,844 2 2 gold badges 12 12 silver badges 19 19 bronze badges 1 Comment Add a comment Wolfie WolfieOver a year ago Beat me to the maths! 2017-01-31T12:23:24.987Z+00:00 0 Reply Copy link Your Answer Thanks for contributing an answer to Stack Overflow! 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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2029	https://hinative.com/questions/24038612	Quality Point(s): 69 Answer: 62 Like: 35 What is the difference between To unveil and To reveal ?Feel free to just provide example sentences. Quality Point(s): 5354 Answer: 1085 Like: 756 The verbs "to unveil" and "to reveal" share the common idea of disclosing or making something known, but there are slight nuances in their usage. Here are example sentences to illustrate their differences: 1. To unveil: - The company unveiled its new product line during the press conference. - The artist will unveil her latest artwork at the gallery opening. "When we say 'unveil,' it typically refers to the act of publicly presenting or disclosing something for the first time. It often implies a sense of anticipation, ceremony, or official announcement." 2. To reveal: - The detective slowly revealed the truth behind the mysterious crime. - The author's memoir reveals intimate details about his personal life. "'Reveal' generally means to make something known or disclose information that was previously hidden, secret, or unknown. It can refer to any act of making information, facts, or truths known." While both verbs involve making something known, "unveil" tends to imply a formal or public presentation, often associated with events, announcements, or new introductions. On the other hand, "reveal" has a broader application and can be used in various contexts, including uncovering secrets, sharing information, or exposing truths. In summary, "unveil" specifically suggests the act of presenting something new or previously concealed to the public, while "reveal" is a more general term for making something known or disclosing information, regardless of the setting or circumstances. Was this answer helpful? The Language Level symbol shows a user's proficiency in the languages they're interested in. Setting your Language Level helps other users provide you with answers that aren't too complex or too simple. Has difficulty understanding even short answers in this language. Can ask simple questions and can understand simple answers. Can ask all types of general questions and can understand longer answers. Can understand long, complex answers. Show your appreciation in a way that likes and stamps can't. By sending a gift to someone, they will be more likely to answer your questions again! If you post a question after sending a gift to someone, your question will be displayed in a special section on that person’s feed. Ask native speakers questions for free Solve your problems more easily with the app!
2030	https://books.google.com/books?id=EfvZAQAAQBAJ	Inequalities - G. H. Hardy, J. E. Littlewood, G. Pólya - Google Books Sign in Hidden fields Try the new Google Books Books View sample Add to my library Try the new Google Books Check out the new look and enjoy easier access to your favorite features Try it now No thanks Try the new Google Books My library Help Advanced Book Search Good for: Web Tablet / iPad eReader Smartphone#### Features: Flowing text Scanned pages Help with devices & formats Learn more about books on Google Play Buy eBook - $99.99 Get this book in print▼ Cambridge University Press Amazon.com Barnes&Noble.com Books-A-Million IndieBound All sellers» My library My History Inequalities ============ G. H. Hardy, J. E. Littlewood, G. Pólya Cambridge University Press, Feb 25, 1988 - Mathematics - 336 pages This classic of the mathematical literature forms a comprehensive study of the inequalities used throughout mathematics. First published in 1934, it presents clearly and exhaustively both the statement and proof of all the standard inequalities of analysis. The authors were well known for their powers of exposition and were able here to make the subject accessible to a wide audience of mathematicians. More » Preview this book » Selected pages Title Page Table of Contents References Contents INTRODUCTION1 4647 43 MEAN VALUES WITH AN ARBITRARY 65 9 75 INFINITE SERIES 114 Miscellaneous theorems and examples 124 SOME APPLICATIONS OF 172 SOME THEOREMS CONCERNING 196 1 227 Thorins proof and extension 306 Copyright Common terms and phrases a₁a₂ambnargumentassertsbest possiblebilinearCalculus of Variationsconsiderconstantcontinuous and strictlyconvergentconvex functioncorrespondingcurvededuce Theoremdeduced from Theoremdefineddefinitiondenoteeffectively proportionalequalityequationexampleF₁finite numberfollows from TheoremFourierFourier seriesgeneralisationsHardy and LittlewoodHenceHilbert'sHölder's inequalityhomogeneousi₁increasing functioninfiniteintervalLebesgueLebesgue integralsmaximummeansmonotonic functionalmultilinear formsnecessary and sufficientnon-negativenull setobtainPólyapositive and finiteproof of Theoremprove Theoremproves the theoremquadratic formr+s+t=0rearrangementreplacedresultRieszsatisfiedStieltjes integralstrictly monotonicstrictly positivesufficient conditionsuffixessummationSupposeTheorem 13Theorem 9theorem concerningtheorytrueunlessupper boundvaluesW. H. Youngwritex₁zeroα₁ακΣαΣΣ Bibliographic information Title Inequalities Cambridge Mathematical Library AuthorsG. H. Hardy, J. E. Littlewood, G. Pólya Publisher Cambridge University Press, 1988 ISBN 1107647398, 9781107647398 Length 336 pages SubjectsMathematics › Algebra › Abstract Mathematics / Algebra / Abstract Mathematics / Calculus Export CitationBiBTeXEndNoteRefMan About Google Books - Privacy Policy - Terms of Service - Information for Publishers - Report an issue - Help - Google Home
2031	https://www.investopedia.com/ask/answers/what-is-the-rule-72/	Skip to content Please fill out this field. Top Stories Don't Miss the Most Important Medicare Message You’ll See This Year I’ve Saved Thousands by Refusing to Buy an iPhone. Kevin O'Leary: 'If You Don't Have Money, You Don't Have a Marriage' This Amazon Prime Member Perk Is Going Away in a Week Table of Contents Table of Contents What Is the Rule of 72? How It Works Natural Logs How to Adjust for Accuracy Using MATLAB Rule of 72 and Inflation Limitations of the Rule of 72 FAQs The Bottom Line The Rule of 72: What It Is and How to Use It in Investing By Caroline Banton Full Bio Caroline Banton has 6+ years of experience as a writer of business and finance articles. She also writes biographies for Story Terrace. Learn about our editorial policies Updated April 23, 2025 Reviewed by Andy Smith Reviewed by Andy Smith Full Bio Andy Smith is a Certified Financial Planner (CFP®), licensed realtor and educator with over 35 years of diverse financial management experience. He is an expert on personal finance, corporate finance and real estate and has assisted thousands of clients in meeting their financial goals over his career. Learn about our Financial Review Board Fact checked by Suzanne Kvilhaug Fact checked by Suzanne Kvilhaug Full Bio Suzanne is a content marketer, writer, and fact-checker. She holds a Bachelor of Science in Finance degree from Bridgewater State University and helps develop content strategies. Learn about our editorial policies What Is the Rule of 72? The Rule of 72 is an easy way to calculate how long an investment will take to double in value given a fixed annual rate of interest. Dividing 72 by the annual rate of return gives investors an estimate of how many years it will take for the initial investment to duplicate. It is a reasonably accurate estimate, especially at low interest rates. For a more accurate estimate, taking compound interest into account, you can use the rule of 69.3%. Key Takeaways The Rule of 72 is a quick way to get a useful ballpark figure. For investments without a fixed rate of return, you can instead divide 72 by the number of years you hope it will take to double your money. This will give you an estimate of the annual rate of return you’ll need to achieve that goal. The calculation is most accurate for rates of return of about 5% to 10%. For more precise outcomes, divide 69.3 by the rate of return. While not as easy to do in one’s head, it is more accurate. How the Rule of 72 Works For example, the Rule of 72 states that $1 invested at an annual fixed interest rate of 10% would take 7.2 years ((72 ÷ 10) = 7.2) to grow to $2. In reality, a 10% investment will take 7.3 years to double (1.107.3 = 2). The Rule of 72 is reasonably accurate for low rates of return. The chart below compares the numbers given by the Rule of 72 and the actual number of years it takes an investment to double. Notice that the Rule of 72 is less precise as rates of return increase. \| \| \| \| \| --- --- \| \| Rate of Return \| Rule of 72 \| Actual # of Years \| Difference (#) of Years \| \| 2% \| 36.0 \| 35 \| 1.0 \| \| 3% \| 24.0 \| 23.45 \| 0.6 \| \| 5% \| 14.4 \| 14.21 \| 0.2 \| \| 7% \| 10.3 \| 10.24 \| 0.0 \| \| 9% \| 8.0 \| 8.04 \| 0.0 \| \| 12% \| 6.0 \| 6.12 \| 0.1 \| \| 25% \| 2.9 \| 3.11 \| 0.2 \| \| 50% \| 1.4 \| 1.71 \| 0.3 \| \| 72% \| 1.0 \| 1.28 \| 0.3 \| \| 100% \| 0.7 \| 1 \| 0.3 \| The Rule of 72 and Natural Logs The Rule of 72 can estimate compounding periods using natural logarithms. In mathematics, the logarithm is the opposite concept of a power; for example, the opposite of 10³ is log base 10 of 1,000. Rule of 72=ln(e)=1where:e=2.718281828 Important e is a famous irrational number similar to pi. The most important property of the number e is related to the slope of exponential and logarithm functions, and its first few digits are 2.718281828. The natural logarithm is the amount of time needed to reach a certain level of growth with continuous compounding. The time value of money (TVM) formula is the following: Future Value=PV×(1+r)nwhere:PV=Present Valuer=Interest Raten=Number of Time Periods To see how long it will take an investment to double, state the future value as 2 and the present value as 1. 2=1×(1+r)n Simplify, and you have the following: 2=(1+r)n To remove the exponent on the right-hand side of the equation, take the natural log of each side: ln(2)=n×ln(1+r) This equation can be simplified again because the natural log of (1 + interest rate) equals the interest rate as the rate gets continuously closer to zero. In other words, you are left with: ln(2)=r×n The natural log of 2 is equal to 0.693 and, after dividing both sides by the interest rate, you have: 0.693/r=n By multiplying the numerator and denominator on the left-hand side by 100, you can express each as a percentage. This gives: 69.3/r%=n Fast Fact The number 72 has sacred significance in many religions, including Judaism, Christianity, and Islam. This has no relevance to the Rule of 72, where the number was probably chosen because it's simpler to use than the more accurate 69.3. How to Adjust the Rule of 72 for Higher Accuracy The Rule of 72 is more accurate if it is adjusted to more closely resemble the compound interest formula—which effectively transforms the Rule of 72 into the Rule of 69.3. The number 72, however, has many convenient factors, including two, three, four, six, and nine. This convenience makes it easier to use the Rule of 72 for a close approximation of compounding periods. Tip Many investors prefer to use the Rule of 69.3 rather than the Rule of 72. For maximum accuracy—particularly for continuous compounding interest rate instruments—use the Rule of 69.3. How to Calculate the Rule of 72 Using MATLAB The calculation of the Rule of 72 in the MATLAB platform requires running a simple command of “years = 72/return,” where the variable “return” is the rate of return on investment and “years” is the result for the Rule of 72. The Rule of 72 is also used to determine how long it takes for money to halve in value for a given rate of inflation. For example, if the rate of inflation is 4%, a command “years = 72/inflation” where the variable inflation is defined as “inflation = 4” gives 18 years. MATLAB, short for matrix laboratory, is a programming platform from MathWorks used for analyzing data. Rule of 72 and Inflation The Rule of 72 isn't just a useful tool for estimating how fast your investments might double. It can also be used to understand how quickly inflation erodes your purchasing power. Instead of calculating how long it takes to grow your money, you can use the same formula to see how long it takes for inflation to cut your money’s value in half. It’s a sobering perspective, but a powerful one for understanding the need to invest. Let’s say inflation is running at 3% per year. Using the Rule of 72, you divide 72 by the inflation rate (3), which gives you 24. That means that in just 24 years, your money will only buy half of what it can today, if it’s just sitting idle. Keep in mind that’s assuming a moderate inflation rate. At a higher rate of 6%, the purchasing power of your cash would be halved in just 12 years. The takeaway is that even if your money isn’t growing, the world around it keeps getting more expensive. Limitations of the Rule of 72 The biggest drawback of the Rule of 72 is that it’s only an approximation. The rule assumes compounded annual interest and works best with rates between 6% and 10%. At those mid-range percentages, the math lines up fairly closely with the actual compound interest formula. Once you go outside that range, the accuracy drops off. For example, if you’re looking at an investment with a 1% return, the Rule of 72 says it would take 72 years to double your money. However, the real number, using exact compound interest math, is about 70.5 years—not a huge difference, but still off. At the other end of the spectrum, if you’re getting a 24% return, the Rule of 72 says your money doubles in 3 years. In reality, it would double in closer to 3.2 years. These discrepancies may seem small, but they can add up over time or matter if you're trying to gauge specific portfolio balances at specific times. Also, keep in mind that the error here is roughly 6% of the months projected. Another limitation is that the Rule doesn’t take into account taxes, fees, or changing interest rates. Most real-world investments don’t have a fixed, guaranteed return. The stock market fluctuates, bond yields rise and fall, and inflation changes over time. In addition, the amount you actually take home isn't tied to your return rate; it might be eroded by those added costs, which restrict your ability to compound earnings. Lastly, the Rule of 72 assumes reinvestment of returns and no withdrawals, which isn't always realistic. In real life, you might take profits, adjust your portfolio, or deal with unexpected financial needs that interrupt the compounding process. Therefore, keep in mind there are a lot of assumptions that need to be in place for the Rule of 72 to work. Does the Rule of 72 Work for Stocks? Stocks do not have a fixed rate of return, so you cannot use the Rule of 72 to determine how long it will take to double your money. However, you still can use it to estimate what kind of average annual return you would need to double your money in a fixed amount of time. Instead of dividing 72 by the rate of return, divide by the number of years you hope it takes to double your money. For example, if you want to double your money in eight years, divide 72 by eight. This tells you that you need an average annual return of 9% to double your money in that time. What 3 Things Can the Rule of 72 Determine? There are two things the Rule of 72 can tell you with reasonable accuracy: how many years it will take to double your money and what kind of return you will need to double your money in a fixed period of time. Because you know how long it will take to double your money, it’s also easy to figure out how long it would take to quadruple your money. For example, if you can double your money in seven years, you can quadruple it in 14 years by allowing the interest to compound. Where Is the Rule of 72 Most Accurate? The Rule of 72 provides only an estimate, but that estimate is most accurate for rates of return of 5% to 10%. Looking at the chart in this article, you can see that the calculations become less precise for rates of return lower or higher than that range. The Bottom Line The Rule of 72 is a quick and easy method for determining how long it will take to double the money you're investing, assuming it has a fixed annual rate of return. While it is not precise, it does provide a ballpark figure and is easy to calculate. Investments such as stocks do not have a fixed rate of return, but the Rule of 72 still can give you an idea of the kind of return you would need to double your money in a certain amount of time. For example, to double your money in six years, you would need a rate of return of 12%. Article Sources Investopedia requires writers to use primary sources to support their work. These include white papers, government data, original reporting, and interviews with industry experts. We also reference original research from other reputable publishers where appropriate. You can learn more about the standards we follow in producing accurate, unbiased content in our editorial policy. MathWorks. “Matlab.” The offers that appear in this table are from partnerships from which Investopedia receives compensation. This compensation may impact how and where listings appear. Investopedia does not include all offers available in the marketplace. Popular Accounts from Our Partners Read more Investing Bonds Fixed Income Partner Links Take the Next Step to Invest The offers that appear in this table are from partnerships from which Investopedia receives compensation. This compensation may impact how and where listings appear. Investopedia does not include all offers available in the marketplace. 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2032	https://math.libretexts.org/Courses/Stanford_Online_High_School/Logic_for_All%3A_An_Introduction_to_Logical_Reasoning/10%3A_Predicate_Logic	Skip to main content 10: Predicate Logic Last updated : Mar 27, 2025 Save as PDF 9: More on Truth Tables 11: Modal Logic Page ID : 182581 ( \newcommand{\kernel}{\mathrm{null}\,}) \definecolorfillinmathshadegray0.9 Table of contents Part 9: Predicate Logic Why Predicate Logic Matters Propositional logic treats entire statements as indivisible units (propositions). But often we want to talk about properties of objects or relationships between objects. For example, "All cats are mammals" or "Some numbers are prime." Propositional logic can't easily express "all" or "some" — it would just see "All cats are mammals" as a single black box statement. Predicate logic (also known as first-order logic) allows us to peek inside that box. It lets us use variables and quantifiers to say things about many instances at once, greatly expanding the expressiveness of our logic. In simpler terms, predicate logic answers questions like: - How do we say every member of a category has some property? - How do we say there exists something with a certain property? These abilities are crucial in mathematics, computer science (database queries, AI), and philosophy. From Aristotle to Modern Logic Aristotle's syllogisms (e.g., "All men are mortal; Socrates is a man; therefore Socrates is mortal") were an early form of predicate logic dealing with categories and membership. Modern predicate logic (developed by Frege in 1879 and others) formalized these ideas with quantifiers and variables, laying the groundwork for much of modern mathematics and computer science. From Propositions to Predicates A predicate is like a property or relation that can be true or false of something. It’s often represented as a capital letter with parentheses, e.g. P(x) might mean "x is prime" or "x is a philosopher", depending on context. The variable x can stand for an object in the domain of discourse (the set of things we're talking about). Example: Let M(x) mean "x is mortal" and H(x) mean "x is human." The statement "All humans are mortal" can be expressed in predicate logic, whereas propositional logic could not directly capture it. Quantifiers: Universal and Existential Predicate logic introduces quantifiers to indicate how many or which elements of the domain we’re talking about: Universal Quantifier (∀): Means "for all" or "for every". ∀xP(x) says "for all x, P(x) is true." (Every x has property P.) Existential Quantifier (∃): Means "there exists" or "for at least one". ∃xP(x) says "there exists at least one x such that P(x) is true." (Some x has property P.) Let's see them in use: "All" statements (Universal) All cats are mammals. Domain: animals. Let C(x) = "x is a cat", M(x) = "x is a mammal." In logic: ∀x(C(x)→M(x)). This says: for every creature x, if x is a cat, then x is a mammal. No dogs can fly. Let D(x) = "x is a dog", F(x) = "x can fly." "No dogs can fly" can be written as: ∀x(D(x)→¬F(x)). (For every x, if x is a dog, then x cannot fly.) Alternatively, one could write ¬∃x(D(x)∧F(x)) meaning "there does not exist an x that is a dog and can fly," which is logically equivalent. "Some" statements (Existential) There exists a prime number greater than 100. Domain: numbers. Let P(x) = "x is prime", G(x) = "x > 100$. In logic: ∃x(P(x)∧G(x)). (There is at least one number x such that x is prime and x>100.) Someone in the class speaks French. Domain: people (in the class). Let F(x) = "x speaks French." ∃xF(x) says "There is at least one person x such that x speaks French." Combining Quantifiers and Logic We can make more complex statements: "Every even number greater than 2 is not prime." Domain: natural numbers. E(x) = "x is even", P(x) = "x is prime", G(x) = "x > 2$. Logic: ∀x((E(x)∧G(x))→¬P(x)). (For any number, if it's even and >2, then it is not prime.) "There is a student who is in both the chess club and the debate club." Domain: students. C(x) = "x is in the chess club", D(x) = "x is in the debate club". ∃x(C(x)∧D(x)). (At least one student is in both clubs.) Translating 'none' or 'no one' Statements like "No x satisfy condition Y" can be translated in two equivalent ways: - Universal form: ∀x(if Y(x) then false) – e.g., ∀x(Y(x)→false), which simplifies to ∀x¬Y(x) (if we mean no x at all have property Y). But more typically, "No X are Y" is translated as ∀x(X(x)→¬Y(x)). - Negated existential form: ¬∃xY(x) – "There does not exist an x with property Y." These are equivalent: saying no x has property Y is the same as saying "for all x, x does not have property Y." Domains and Scope of Quantifiers When we write ∀x or ∃x, we presume a domain of discourse for x (the set of all things x could refer to). Sometimes it's stated explicitly ("for all people x, ..."), other times it's implied by context (if we're talking about numbers, x ranges over numbers). Quantifiers have a scope similar to how parentheses or logical operators have scope. For example, in ∀x(P(x)∨Q(y)), the ∀x applies only to the P(x) part (and not to Q(y) which might involve a different variable). If we need multiple quantifiers, we can use different variables or nest them: "Every student admires some professor." Domain: people at a university. Let S(x) = "x is a student", Prof(y) = "y is a professor", A(x,y) = "x admires y". Logic: ∀x[S(x)→∃y(Prof(y)∧A(x,y))]. (For each person x, if x is a student, then there exists at least one person y who is a professor and x admires y.) Order matters for different quantifiers: ∀x∃yA(x,y) is not the same as ∃y∀xA(x,y). The first says "for each x, you can find a (potentially different) y such that A(x,y)." The second says "there is a single y that works for every x." This distinction is huge. Order of Quantifiers Pay attention to quantifier order: - ∀x∃yP(x,y): For each x, maybe different ys can witness the truth. - ∃y∀xP(x,y): There is one particular y that works for all x. These can mean very different things. For example, "For every person, there is someone whom they love" versus "There is a person whom everyone loves." Translating Between English and Predicate Logic It takes practice to capture English precisely in symbols: Each, every, all →∀ Some, there is/are, at least one →∃ None, no → use ∀ with a negation, or ¬∃ Example translations: "Every car has wheels." Domain: vehicles (or things). C(x) = "x is a car", W(x) = "x has wheels". ∀x(C(x)→W(x)). "Some cars have no wheels." C(x) = "x is a car", W(x) = "x has wheels". ∃x(C(x)∧¬W(x)). (There exists an object that is a car and does not have wheels — perhaps this is false in reality, but that's the logical form.) "Not every car has wheels." This is the negation of "every car has wheels." We can write: ¬∀x(C(x)→W(x)), which is equivalent to ∃x¬(C(x)→W(x)) which simplifies to ∃x(C(x)∧¬W(x)). So "Not all have wheels" means "Some car has no wheels" — note how English "not every" turns into an existential statement logically. "At least two people have the same birthday." (A bit trickier) Domain: people. B(x,d) = "Person x has birthday on date d". One way: ∃x∃y∃d[(x≠y)∧B(x,d)∧B(y,d)]. (There exist two different people x and y and a date d such that x and y both have birthday d.) Real-World Applications of Predicate Logic Database Queries: SQL and other query languages are built on predicate logic. A query like "SELECT FROM Employees WHERE age > 30 AND department = 'Sales'" is basically finding all x such that Age(x)>30 and Dept(x,"Sales"). Artificial Intelligence: Knowledge representation often uses first-order logic to encode facts and rules. E.g., in an expert system: ∀x(Cat(x)→Mammal(x)) encodes a general fact about the world. Mathematics: Almost all definitions and theorems in mathematics are expressed in predicate logic ("For every ε>0, there exists a δ>0 such that ...", which is a ∀∃ statement). Formal Verification: When verifying hardware or software, one often writes specifications like "for all possible inputs satisfying condition P, the output will satisfy Q" (∀x(P(x)→Q(x))) and uses automated theorem provers to check it. Common Misconceptions "∀xP(x) means the same as P(x) for a generic x." Reality: ∀xP(x) means every possible x in the domain makes P(x) true. You can't assume a specific one without stating it. It's a statement about the entire domain at once. "∃xP(x) asserts the existence but we know nothing else about that x." Reality: Yes, existential just guarantees at least one exists, but we usually can't give it a name unless we use a witness. It's a logical guarantee without identification (unless the context or proof finds a concrete example). "If ∀xP(x) is true, then ∃xP(x) must be true." Reality: True. If everything has property P, certainly at least one thing does. (In logic, ∀xP(x) implies ∃xP(x) provided the domain is nonempty. Usually we assume domains are nonempty.) "If ∃xP(x) is true, then ∀xP(x) might be true." Reality: Having one example doesn't mean all have it. E.g., "There exists a number that is even" is true, but "For all numbers, they are even" is false. Mixing up ∀ and ∃: It's common but critical to keep them straight. "Every" vs "Some" are not interchangeable. Exercises 9.1. Translate into predicate logic: "All birds can fly." (Define appropriate predicates for being a bird and being able to fly.) 9.2. Translate: "Some birds cannot fly." (You might know real examples, but just do it logically.) 9.3. Translate: "Every student in this class has taken at least one programming course." (Define predicates like S(x) for "x is a student in this class", and P(x,y) for "x has taken course y", and maybe a set or predicate for programming courses.) 9.4. Translate: "There is a prime number that is even." (Remember what the only even prime is!) 9.5. Consider the statement: "If a number is rational, then it can be written as a ratio of two integers." Express this as a logical statement with a universal quantifier. (Let R(x) = "x is rational", and define a predicate for "x can be written as m/n with m,n integers".) 9.6. Express the negation of the statement in 9.5 in a normalized way (so that negation symbol directly precedes the predicate, not the quantifier). Explain the meaning of the negation in words. 9.7. Translate: "Not every book is interesting." Then contrast it with "No book is interesting." (They are different! Provide both translations.) 9.8. Translate: "Everyone has exactly one mother." (Hint: This one is a bit advanced. It means: for every person x, there exists a person y such that y is the mother of x, and for all z, if z is a mother of x then z=y.) 9.9. Is the following statement true or false (in ordinary arithmetic)? "For every integer y, there exists an integer x such that x<y." Translate it to logic as well, then answer. 9.10. Explain in your own words the difference between ∀x∃yP(x,y) and ∃y∀xP(x,y). Give a real-life example where one is true but the other is false. Advanced: Predicate logic is a powerful tool, but with great power comes additional complexity. In further studies, you'll encounter concepts like relations, functions, multiple quantifiers, and even logical paradoxes that arise with unrestricted domains (like Russell's paradox about "the set of all sets that don't contain themselves"). Predicate logic is the gateway to these deeper explorations, including the foundations of mathematics and computer science. 9: More on Truth Tables 11: Modal Logic
2033	https://www.pearson.com/channels/college-algebra/learn/patrick/1-equations-and-inequalities/powers-of-i	Skip to main content My Courses Chemistry General Chemistry Organic Chemistry Analytical Chemistry GOB Chemistry Biochemistry Intro to Chemistry Biology General Biology Microbiology Anatomy & Physiology Genetics Cell Biology Physics Physics Math College Algebra Trigonometry Precalculus Calculus Business Calculus Statistics Business Statistics Social Sciences Psychology Health Sciences Personal Health Nutrition Business Microeconomics Macroeconomics Financial Accounting Product & Marketing Agile & Product Management Digital Marketing Project Management AI in Marketing Programming Introduction to Python Microsoft Power BI Data Analysis - Excel Introduction to Blockchain HTML, CSS & Layout Introduction to JavaScript R Programming Calculators AI Tools Study Prep Blog Study Prep Home Table of contents Skip topic navigation Skip topic navigation Review of Algebra 4h 18m Algebraic Expressions 36m + 32m + Polynomials Intro 19m + Multiplying Polynomials 36m + Factoring Polynomials 1h 2m + Radical Expressions 15m + Simplifying Radical Expressions 35m + Rationalize Denominator 15m + Rational Exponents 4m 1. Equations & Inequalities 3h 18m Linear Equations 31m + Rational Equations 21m + The Imaginary Unit 6m + Powers of i 11m + Complex Numbers 36m + Intro to Quadratic Equations 18m + The Square Root Property 11m + Completing the Square 12m + The Quadratic Formula 18m + Choosing a Method to Solve Quadratics 9m + Linear Inequalities 20m 2. Graphs of Equations 43m Graphs and Coordinates 7m + Two-Variable Equations 23m + Lines 12m 3. Functions 2h 17m Intro to Functions & Their Graphs 35m + Common Functions 5m + Transformations 45m + Function Operations 21m + Function Composition 29m 4. Polynomial Functions 1h 44m Quadratic Functions 39m + Understanding Polynomial Functions 34m + Graphing Polynomial Functions 30m + Dividing Polynomials + Zeros of Polynomial Functions 5. Rational Functions 1h 23m Introduction to Rational Functions 9m + Asymptotes 35m + Graphing Rational Functions 38m 6. Exponential & Logarithmic Functions 2h 28m Introduction to Exponential Functions 9m + Graphing Exponential Functions 25m + The Number e 8m + Introduction to Logarithms 22m + Graphing Logarithmic Functions 18m + Properties of Logarithms 27m + Solving Exponential and Logarithmic Equations 35m 7. Systems of Equations & Matrices 4h 6m Two Variable Systems of Linear Equations 1h 8m + Introduction to Matrices 1h 11m + Determinants and Cramer's Rule 1h 3m + Graphing Systems of Inequalities 42m 8. Conic Sections 2h 23m Introduction to Conic Sections 5m + Circles 15m + Ellipses: Standard Form 33m + Parabolas 36m + Hyperbolas at the Origin 40m + Hyperbolas NOT at the Origin 12m 9. Sequences, Series, & Induction 1h 22m Sequences 40m + Arithmetic Sequences 25m + Geometric Sequences 15m 10. Combinatorics & Probability 1h 45m Factorials 11m + Combinatorics 46m + Probability 47m Equations & Inequalities Powers of i Equations & Inequalities Powers of i: Videos & Practice Problems Video Lessons Practice Worksheet Topic summary Understanding the powers of the imaginary unit Math input error 1 concept Powers of i Video duration: 4m Play a video: Powers of i Video Summary The imaginary unit is defined as the square root of negative one, . When raising to various powers, a pattern emerges that simplifies calculations significantly. Understanding the first few powers of allows us to easily determine higher powers. Starting with the basic powers: Continuing this pattern, we can express higher powers of in terms of these four results. For example: This reveals a repeating cycle every four powers: . Therefore, any power of can be simplified by finding the remainder when the exponent is divided by 4: For any integer : If , then If , then If , then If , then This cyclical nature allows for quick calculations of any power of , making it a powerful tool in complex number arithmetic. 2 concept Higher Powers of i Video duration: 5m Play a video: Higher Powers of i Video Summary Understanding the powers of the imaginary unit is essential in complex number calculations. The powers of cycle through four distinct values: , , , and . This cyclical nature allows for efficient computation of higher powers without the need to manually count through each power. To simplify calculations, we can express any power of in terms of , which equals . For instance, to evaluate , we can rewrite it as . Since , it follows that . For powers that are not multiples of four, such as , we can still utilize the cyclical pattern. We start by determining how many complete sets of fit into the exponent. In this case, can be expressed as . Since , we find that . To further streamline the process, we can determine the remainder when the exponent is divided by . If the exponent is evenly divisible by , the result is . For example, can be evaluated by checking if has a remainder. Since is divisible by , . For exponents that yield a remainder, we can find the equivalent power of based on the remainder. For instance, when calculating , dividing by gives a quotient of and a remainder of . Thus, . Knowing that , we conclude that . In summary, to evaluate powers of , determine if the exponent is divisible by . If it is, the result is . If not, calculate the remainder and use it to find the corresponding power of from the cycle: , , , and . This method provides a quick and efficient way to handle complex exponentiation. 3 Problem Simplify the power of i. i 1003 A i B − 1 C − i D 1 4 Problem Simplify the power of i. i 85 A i B -1 C − i D 1 Do you want more practice? More sets Powers of i Equations & Inequalities 6 problems Topic Callie 1. Equations & Inequalities - Part 1 of 3 4 topics 11 problems Chapter Callie 1. Equations & Inequalities - Part 2 of 3 4 topics 10 problems Chapter Callie 1. Equations & Inequalities - Part 3 of 3 4 topics 10 problems Chapter Callie Go over this topic definitions with flashcards More sets Powers of i definitions Equations & Inequalities 15 Terms Take your learning anywhere! Prep for your exams on the go with video lessons and practice problems in our mobile app. Here’s what students ask on this topic: What is the imaginary unit 'i' and how do its powers cycle? The imaginary unit 'i' is defined as the square root of -1. Its powers cycle through four values: i, -1, -i, and 1. This cyclical pattern occurs every four powers, making it easier to calculate higher powers of i. Specifically, i1 = i, i2 = -1, i3 = -i, and i4 = 1. Understanding this cycle allows us to simplify calculations involving powers of i, especially when dealing with large exponents. By recognizing this repeating pattern, we can express any power of i in terms of these four values, facilitating quick and efficient computation. How can I calculate higher powers of 'i' using divisibility by 4? To calculate higher powers of 'i', check if the exponent is divisible by 4. If it is, the result is 1, as i4 = 1. If not, divide the exponent by 4 to find the remainder. The remainder determines the equivalent lower power of i: a remainder of 1 corresponds to i, 2 to -1, and 3 to -i. For example, i22 has a remainder of 2 when divided by 4, so it equals -1. Similarly, i67 has a remainder of 3, equating to -i. This method simplifies calculations by reducing the exponent to one of the four known values. What is the pattern for powers of 'i' and how does it help in calculations? The pattern for powers of 'i' is a cycle of four values: i, -1, -i, and 1. This pattern repeats every four powers, which helps simplify calculations for higher powers of 'i'. By recognizing this cycle, we can express any power of 'i' in terms of these four values, making it easier to compute large exponents. For instance, i20 can be expressed as (i4)5 = 1. This cyclical nature allows us to quickly determine the value of 'i' raised to any power by checking divisibility by 4 and using the remainder to find the corresponding value. How do you simplify i100 using the powers of 'i' cycle? To simplify i100, use the cycle of powers of 'i'. First, check if 100 is divisible by 4. Since 100 divided by 4 equals 25 with no remainder, i100 simplifies to 1, as i4 = 1. This method leverages the cyclical pattern of powers of 'i', which repeats every four powers: i, -1, -i, and 1. By recognizing this cycle, we can efficiently compute large powers of 'i' without manually cycling through each power, saving time and effort in calculations. What is the remainder method for calculating powers of 'i'? The remainder method for calculating powers of 'i' involves dividing the exponent by 4 and using the remainder to determine the equivalent lower power of 'i'. If the exponent is divisible by 4, the result is 1. If not, the remainder will be 1, 2, or 3, corresponding to i, -1, or -i, respectively. For example, dividing 22 by 4 gives a remainder of 2, so i22 equals -1. Similarly, dividing 67 by 4 gives a remainder of 3, so i67 equals -i. This method simplifies calculations by reducing the exponent to one of the four known values. Your College Algebra tutors Patrick Ford Physics and Math Lead Instructor Nick Kaneko Math Instructor Callie Rethman Math Instructor
2034	https://www.generationgenius.com/videolessons/solve-problems-with-decimals-using-models/	Solve Problems with Decimals (Using Models) Video \| 3rd, 4th & 5th Grade 1% Processing, please wait... It was processed successfully! It was processed successfully! Create a free account to unlock all videos & resources! Create a free account to unlock all resources! Create Free Account Try It Absolutely Free - Choose a role to get started TeacherParentStudent Science Lessons Math Lessons Standards About Pricing Help EN English Español bool(false) LoginCreate Free Account Choose a role: x School Home Subscribe Membership Plans Logout Subscribe Subscribe Create Free Account Choose a role: x School Home Science Lessons Math Lessons Standards About Pricing Help Login Login Login Membership Plans Manage Account Logout EN language English Español Create Free Account Preview Only Preview Only: Watch this full video Absolutely Free. 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Click for sound 13:02 1:00 min of preview left Create a free account for full access Create a free account to unlock all content! Get Full Access Solve Problems with Decimals (Using Models) LESSON MATERIALSGenerate Student Link Solve Problems with Decimals (Using Models) Lesson PDFsGenerate Student Link What you will learn from this video What you will learn We’ll add, subtract, multiply and divide decimals using models. We will also learn how to use estimation to decide if our answer is reasonable. And that this knowledge can help us feed our pets, go hiking and even grow crystals! Discussion Questions -------------------- Before Video Give the value of each digit in 5.39.ANSWER 5 wholes, 3 tenths, and 9 hundredths What is another way to name 56 hundredths?ANSWER Every 10 hundredths can be regrouped as 1 tenth. So, 56 hundredths is 5 tenths and 6 hundredths. What is another way to name 48 tenths?ANSWER Every 10 tenths can be regrouped as 1 whole. So, 48 tenths is 4 wholes and 8 tenths. Describe how to solve 63 ÷ 9.ANSWER If I divide 63 wholes equally among 9 groups, each group has 7 wholes. 63 ÷ 9 = 7. How can you use estimation to check if 58 × 32 = 1,856 makes sense?ANSWER I can round 58 to 60, and 32 to 30. 60 × 30 is 1,800, which is close to 1,856. This is how I know my answer makes sense. After Video If you are calculating 3.64 – 2.14, which number should you represent using the fraction square model? What is the second step?ANSWER;) The greater number, 3.64. Then, take away blocks that represent 2.14 in total. You want to calculate 3.4 – 0.8. You represent 3.4 using 3 wholes and 4 tenths. What else do you need to do before you can subtract 8 tenths?ANSWER I don’t have enough tenths to subtract 8 tenths, so I need to regroup 1 of the wholes as 10 tenths. Then, I have 2 wholes and 14 tenths. Now I can subtract 8 tenths from 14 tenths. The answer is what remains, 2 wholes and 6 tenths, or 2.6. How can you use skip counting to multiply 6 × 1.4?ANSWER 1.4 means 1 whole and 4 tenths. I need to multiply each number by 6, which is the same as skip counting sixes. 6 × 1 = 6. 6 × 4 tenths = 24 tenths. 24 tenths is the same as 2 wholes and 4 tenths, so altogether I have 8 wholes and 4 tenths, or 8.4. Mimi multiplies 4 by 2 wholes, 3 tenths, and 7 hundredths and gets 8 wholes, 12 tenths, and 28 hundredths. Is that the final answer?ANSWER No, Mimi needs to do some regrouping, starting with the smallest place, the hundredths place. 28 hundredths can be regrouped as 2 tenths and 8 hundredths. So, now there are 8 wholes, 12 + 2 = 14 tenths, and 8 hundredths. Next, regroup some of the tenths. 10 tenths can be regrouped as 1 whole, so the final answer is 9 wholes, 4 tenths, and 8 hundredths, or 9.48. What can you do to check if 3.8 × 5.1 = 19.38 makes sense?ANSWER I can round 3.8 to 4 and 5.1 to 5. Multiply: 4 × 5 = 20. 20 is close to 19.38, so the answer makes sense. Vocabulary ---------- Place valueDEFINE Tells us the value of a digit based on its position in a number. DecimalsDEFINE;) Numbers that allow us to show parts of a whole. Ones placeDEFINE;) Shows the number of wholes. Tenths placeDEFINE;) Shows the number of parts out of 10. Hundredths placeDEFINE;) To the right of the tenths place, and shows us the number of parts out of 100. ModelsDEFINE;) Visual tools that help us to represent numbers. RegroupDEFINE;) When we have 10 of one place value, we can regroup to create 1 of the place value on its left. Reading Material ---------------- Download as PDFDownload PDFView as Separate PageWHAT IS SOLVING PROBLEMS WITH DECIMALS? You can solve problems with decimals exactly the way you solved problems with whole numbers. Use keywords to identify if you need to add, subtract, multiply, or divide to solve. To better understand solving problems with decimals… WHAT IS SOLVING PROBLEMS WITH DECIMALS?. You can solve problems with decimals exactly the way you solved problems with whole numbers. Use keywords to identify if you need to add, subtract, multiply, or divide to solve. To better understand solving problems with decimals… LET’S BREAK IT DOWN! -------------------- ### Hiking Adesina, April, and Marcos hiked all day. They need to add up the distances they hiked to find out how far they traveled in all. The first trail was 1.7 kilometers long. The second trail was 3.2 kilometers long. To find out how much they hiked altogether, add 1.7 + 3.2. To do this, use models that represent whole and decimal numbers. We can represent 3.2 as 3 wholes and 2 tenths. We can represent 1.7 as 1 whole and 7 tenths. To add, group the wholes and group the tenths. 3 wholes plus 1 whole is 4 whole kilometers. 7 tenths plus 2 tenths is 9 tenths of a kilometer. Then we have 4 wholes and 9 tenths, or 4.9 kilometers. This means that they hiked 4.9 kilometers altogether. To check the answer, round the starting numbers to the nearest whole. 3.2 can be rounded to 3 and 1.7 can be rounded to 2. Add 3 and 2 to get 5. 5 is close to 4.9, so the answer is reasonable. Numbers with hundredths can be added the same way. If Marcos bought an energy bar for $2.34 and April bought trail mix for $1.89, how much money did they spend in all? 2.34 can be represented as 2 wholes, 3 tenths, and 4 hundredths. 1.89 can be represented as 1 whole, 8 tenths, and 9 hundredths. Adding them together, we have 3 wholes, 11 tenths, and 13 hundredths. Because we have more than 9 hundredths and tenths, we have to regroup some numbers. 10 hundredths can be regrouped to make 1 tenth, so 13 hundredths is 1 tenth and 3 hundredths. Then we have 12 tenths in total. 10 tenths can be regrouped to become 1 whole, so 12 tenths is 1 whole and 2 tenths. Altogether there are 4 wholes, 2 tenths, and 3 hundredths, or $4.23. Now you try: Add 4.58 + 2.39. Hiking Adesina, April, and Marcos hiked all day. They need to add up the distances they hiked to find out how far they traveled in all. The first trail was 1.7 kilometers long. The second trail was 3.2 kilometers long. To find out how much they hiked altogether, add 1.7 + 3.2. To do this, use models that represent whole and decimal numbers. We can represent 3.2 as 3 wholes and 2 tenths. We can represent 1.7 as 1 whole and 7 tenths. To add, group the wholes and group the tenths. 3 wholes plus 1 whole is 4 whole kilometers. 7 tenths plus 2 tenths is 9 tenths of a kilometer. Then we have 4 wholes and 9 tenths, or 4.9 kilometers. This means that they hiked 4.9 kilometers altogether. To check the answer, round the starting numbers to the nearest whole. 3.2 can be rounded to 3 and 1.7 can be rounded to 2. Add 3 and 2 to get 5. 5 is close to 4.9, so the answer is reasonable. Numbers with hundredths can be added the same way. If Marcos bought an energy bar for $2.34 and April bought trail mix for $1.89, how much money did they spend in all? 2.34 can be represented as 2 wholes, 3 tenths, and 4 hundredths. 1.89 can be represented as 1 whole, 8 tenths, and 9 hundredths. Adding them together, we have 3 wholes, 11 tenths, and 13 hundredths. Because we have more than 9 hundredths and tenths, we have to regroup some numbers. 10 hundredths can be regrouped to make 1 tenth, so 13 hundredths is 1 tenth and 3 hundredths. Then we have 12 tenths in total. 10 tenths can be regrouped to become 1 whole, so 12 tenths is 1 whole and 2 tenths. Altogether there are 4 wholes, 2 tenths, and 3 hundredths, or $4.23. Now you try: Add 4.58 + 2.39. ### Science Project Marcos, April, and Adesina grew crystals for a science project. Marcos’s crystal weighs 5.6 grams, and April’s weighs 3.5 grams. How much more does Marcos’ crystal weigh than April's? We can use decimal square models to find the answer. To find the difference, we always start with the bigger number, and we can represent 5.6 using 5 wholes and 6 tenths. To subtract, we need to take away 3.5, or 3 wholes and 5 tenths. 5 – 3 = 2 wholes. 6 tenths – 5 tenths = 1 tenth. We have 2 wholes and 1 tenth, or 2.1, remaining. Marcos’ crystal weighs 2.1 more grams than April’s. Estimate the difference to check the answer. Round 5.6 to 6 and 3.5 to 4. 6 – 4 = 2. 2 is close to 2.1, so the answer is reasonable. April’s crystal display cost $3.59. Marcos’ display cost $1.95. How much more did April’s display cost than Marcos's? Find 3.59 – 1.95 to find the difference. Use the decimal square model to show the larger number. 3.59 is 3 wholes, 5 tenths, and 9 hundredths. To subtract, cross out 1.95, or 1 whole, 9 tenths, and 5 hundredths in all. First, cross out 5 hundredths. There are 4 hundredths left. Next, we need to cross out 9 tenths, but there are only 5 tenths. Regroup 1 whole into 10 tenths. Now there are 2 wholes and 15 tenths. Subtract the 9 tenths. 15 – 9 = 6, so there are 6 tenths left. Lastly, subtract 1 whole from 2. We have 1 whole, 6 tenths, and 4 hundredths left. That means that April's display cost $1.64 more than Marcos's. Now you try: What is $6.72 – $3.45? Science Project Marcos, April, and Adesina grew crystals for a science project. Marcos’s crystal weighs 5.6 grams, and April’s weighs 3.5 grams. How much more does Marcos’ crystal weigh than April's? We can use decimal square models to find the answer. To find the difference, we always start with the bigger number, and we can represent 5.6 using 5 wholes and 6 tenths. To subtract, we need to take away 3.5, or 3 wholes and 5 tenths. 5 – 3 = 2 wholes. 6 tenths – 5 tenths = 1 tenth. We have 2 wholes and 1 tenth, or 2.1, remaining. Marcos’ crystal weighs 2.1 more grams than April’s. Estimate the difference to check the answer. Round 5.6 to 6 and 3.5 to 4. 6 – 4 = 2. 2 is close to 2.1, so the answer is reasonable. April’s crystal display cost $3.59. Marcos’ display cost $1.95. How much more did April’s display cost than Marcos's? Find 3.59 – 1.95 to find the difference. Use the decimal square model to show the larger number. 3.59 is 3 wholes, 5 tenths, and 9 hundredths. To subtract, cross out 1.95, or 1 whole, 9 tenths, and 5 hundredths in all. First, cross out 5 hundredths. There are 4 hundredths left. Next, we need to cross out 9 tenths, but there are only 5 tenths. Regroup 1 whole into 10 tenths. Now there are 2 wholes and 15 tenths. Subtract the 9 tenths. 15 – 9 = 6, so there are 6 tenths left. Lastly, subtract 1 whole from 2. We have 1 whole, 6 tenths, and 4 hundredths left. That means that April's display cost $1.64 more than Marcos's. Now you try: What is $6.72 – $3.45? ### Stacking Shelves Adesina stacks 4 drawers on top of each other. Each drawer is 0.3 meter tall. How tall is the stack? Find out by multiplying 4 by 0.3. Since multiplication is just repeated addition, 4 times 0.3 is the same as 0.3 + 0.3 + 0.3 + 0.3. Use decimal squares to model the addition. 0.3 means 3 parts out of 10. Show that 4 times. Next, count up all the tenths: there are 3, 6, 9, 12 tenths. To write this as a decimal, regroup 10 tenths to make 1 whole. That gives 1 whole and 2 tenths, or 1.2 meters. What if you stack 3 bigger shelves that are each 2.53 feet tall? To find the height of this stack, multiply 2.53 by 3. 2.53 can be modeled as 2 wholes, 5 tenths, and 3 hundredths. To multiply by 3, add each number of units 3 times. This gives 6 wholes, 15 tenths, and 9 hundredths in all. Regroup 15 tenths as 1 whole and 5 tenths. The answer is 7 wholes, 5 tenths, and 9 hundredths, or 7.59 meters tall. To use estimation to check the answer, round 2.53 to 3. 3 × 3 = 9, which is close to the answer. To multiply a decimal by a decimal, such as 0.7 times 0.3, use a model to show 0.3 as 3 parts out of 10. To show 7 tenths of 3 tenths, split each tenth into 10 equal parts, and then color only 7 of the parts in each tenth. When tenths are split into 10 parts each, there are 100 pieces in the whole. That means that each of those small squares represents one hundredth. Each model shows 7 hundredths. There are 3 models to show the original 3 tenths. So the answer is 21 hundredths, or 0.21. Now you try: Multiply 4 by 2.61 meters. Stacking Shelves Adesina stacks 4 drawers on top of each other. Each drawer is 0.3 meter tall. How tall is the stack? Find out by multiplying 4 by 0.3. Since multiplication is just repeated addition, 4 times 0.3 is the same as 0.3 + 0.3 + 0.3 + 0.3. Use decimal squares to model the addition. 0.3 means 3 parts out of 10. Show that 4 times. Next, count up all the tenths: there are 3, 6, 9, 12 tenths. To write this as a decimal, regroup 10 tenths to make 1 whole. That gives 1 whole and 2 tenths, or 1.2 meters. What if you stack 3 bigger shelves that are each 2.53 feet tall? To find the height of this stack, multiply 2.53 by 3. 2.53 can be modeled as 2 wholes, 5 tenths, and 3 hundredths. To multiply by 3, add each number of units 3 times. This gives 6 wholes, 15 tenths, and 9 hundredths in all. Regroup 15 tenths as 1 whole and 5 tenths. The answer is 7 wholes, 5 tenths, and 9 hundredths, or 7.59 meters tall. To use estimation to check the answer, round 2.53 to 3. 3 × 3 = 9, which is close to the answer. To multiply a decimal by a decimal, such as 0.7 times 0.3, use a model to show 0.3 as 3 parts out of 10. To show 7 tenths of 3 tenths, split each tenth into 10 equal parts, and then color only 7 of the parts in each tenth. When tenths are split into 10 parts each, there are 100 pieces in the whole. That means that each of those small squares represents one hundredth. Each model shows 7 hundredths. There are 3 models to show the original 3 tenths. So the answer is 21 hundredths, or 0.21. Now you try: Multiply 4 by 2.61 meters. ### Pets Adesina has a puppy and April and Marcos each have a rabbit. Adesina feeds her dog 3 times a day. She fed her dog 0.6 kg of food in one day. How much did she feed her dog for each meal? To find the answer, divide 0.6 by 3. Model 0.6 by showing 6 tenths, or 6 out of 10 parts of a whole. If 6 tenths is split into 3 equal groups, each group has 2 tenths in it. Then the answer is 0.2 kg. Marcos has 0.48 kg of rabbit food and needs to feed his rabbit 4 equal meals. To split 0.48 into 4 equal parts, show 0.48 as 4 tenths and 8 hundredths. If they are divided into 4 equal groups, each group contains 1 tenth and 2 hundredths, or 0.12 kg. Now you try: What is 0.72 kg divided by 6? Pets Adesina has a puppy and April and Marcos each have a rabbit. Adesina feeds her dog 3 times a day. She fed her dog 0.6 kg of food in one day. How much did she feed her dog for each meal? To find the answer, divide 0.6 by 3. Model 0.6 by showing 6 tenths, or 6 out of 10 parts of a whole. If 6 tenths is split into 3 equal groups, each group has 2 tenths in it. Then the answer is 0.2 kg. Marcos has 0.48 kg of rabbit food and needs to feed his rabbit 4 equal meals. To split 0.48 into 4 equal parts, show 0.48 as 4 tenths and 8 hundredths. If they are divided into 4 equal groups, each group contains 1 tenth and 2 hundredths, or 0.12 kg. Now you try: What is 0.72 kg divided by 6? Practice Word Problems ---------------------- Practice Number Problems ------------------------ Teacher Resources ----------------- Create Free Account To Unlock Teacher Resources ----------------- Lesson Plan Download PDF ### Teacher Guide Download PDF ### Answer Key Download PDF These downloadable teacher resources can help you create a full lesson around the video. These PDFs incorporate using class discussion questions, vocabulary lists, printable math worksheets, quizzes, games, and more. 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2036	https://remote2.ece.illinois.edu/~erhan/FieldsWaves/secure/notes/ECE450Sp10/350lect09.pdf	9 Poynting vector, radiated power, radiation re-sistance Consider the radiation ﬁelds of a ˆ z-polarized short-dipole antenna shown in the margin in a compact form. x y z θ φ r ˜ I(z) ˜ E = ˆ θ ˜ Eθ ˜ H = ˆ φ ˜ Hφ ˜ E × ˜ H∗ Radiation ﬁelds of short dipole: ˜ E = ˜ Eθˆ θ and ˜ H = ˜ Eθ ηo ˆ φ where ˜ Eθ = jηoIokℓsin θe−jkr 4πr and ℓ= L/2. • How much average power is radiated by the short-dipole antenna to sustain these ﬁelds, and • how can we determine this amount, Prad, by electrical measurements which can be performed at the antenna input port — the small gap at the dipole center where the dipole is connected to the source circuit (typically via some transmission line network)? To answer these questions we will calculate in this lecture the average Poynting vector of radiation ﬁelds of the dipole antenna and the “ﬂux” of the same vector computed over a sphere imagined to surround the dipole. • Recall once again that Poynting vector S ≡E × H denotes the energy transported by electromagnetic ﬁelds per unit time and per unit area normal to the vector itself. With time-harmonic ﬁelds the average value of Poynting vector can be denoted and computed as ⟨E × H⟩= 1 2Re{˜ E × ˜ H∗} in terms of ﬁeld phasors ˜ E and ˜ H. 1 • It is this quantity ⟨S⟩= ⟨E × H⟩ which is independent of the storage ﬁelds of dipole antennas and only depend on their radiation ﬁelds. • Using (see margin once again) x y z θ φ r ˜ I(z) ˜ E = ˆ θ ˜ Eθ ˜ H = ˆ φ ˜ Hφ ˜ E × ˜ H∗ Radiation ﬁelds of short dipole: ˜ E = ˜ Eθˆ θ and ˜ H = ˜ Eθ ηo ˆ φ where ˜ Eθ = jηoIokℓsin θe−jkr 4πr and ℓ= L/2. ˜ E = ˆ θ ˜ Eθ and ˜ H = ˆ φ ˜ Eθ ηo with ˜ Eθ = jηoIkℓsin θe−jkr 4πr , we ﬁnd ˜ E × ˜ H∗= ˆ θ ˜ Eθ × (ˆ φ ˜ Eθ ηo )∗= ˆ θ × ˆ φ\| ˜ Eθ\|2 ηo = ˆ r\|˜ E\|2 ηo and ⟨E × H⟩= 1 2Re{˜ E × ˜ H∗} = ˆ r\|˜ E\|2 2ηo . • Since \|˜ E\|2 = \| ˜ Eθ\|2 = η2 o\|Io\|2k2\|ℓ\|2 sin2 θ (4πr)2 = η2 o\|Io\|2\|ℓ\|2 sin2 θ 4(λr)2 , we have ⟨E × H⟩= ˆ r\|˜ E\|2 2ηo = ηo 8 \|Io\|2 \|ℓ\|2 (λr)2 sin2 θˆ r. • The expression above is the energy ﬂux density or transmitted power density of the dipole antenna as a function of distance r from the dipole and angle θ of viewing direction oﬀthe dipole axis. 2 • The average power output of the dipole — radiated power Prad — can next be obtained by computing the ﬂux of ⟨E × H⟩over any closed surface surrounding the dipole. x y z φ θ r r sin θ dr rdθ r sin θdφ Inﬁnitesimal area on a con-stant r surface is dS = (rdθ)(r sin θdφ) = r2dΩ where dΩ≡sin θdθdφ is called inﬁnitesimal solid angle. – This calculation is most easily carried out over a spherical surface of radius r having inﬁnitesimal surface elements dS = ˆ r(r sin θdφ)(rdθ) = ˆ rr2 sin θdθdφ ≡ˆ rr2dΩ, where dΩ≡sin θdθdφ (introduced to maintain a compact notation) is called a solid an-gle increment. We then note that ⟨E × H⟩· dS = ηo 8λ2\|Io\|2\|ℓ\|2 sin2 θdΩ and the ﬂux of ⟨E × H⟩is ! ⟨E × H⟩· dS " #$ % = ηo 8λ2\|Io\|2 & dΩ\|ℓ\|2 sin2 θ Prad where it is implied that & dΩ= & 2π φ=0 dφ & π θ=0 dθ sin θ. 3 – This result can be cast as Prad = 1 2Rrad\|Io\|2 where Rrad = ηo 4λ2 & dΩ\|ℓsin θ\|2 is known as radiation resistance. • If ℓis the eﬀective length of a dipole — distinct from its physical length L because of current weighting — then ℓsin θ is “how long the eﬀective length looks” when one sees it (the dipole) at an angle (see margin). Solid angle integral of the square of this “foreshortened” eﬀective length, namely x z θ φ r ℓsin 90◦ ℓsin θ ℓ Note: recall that ℓmay be a function of θ itself! & dΩ\|ℓsin θ\|2, determines the radiation resistance of the dipole antenna. Since for a short dipole ℓ= L 2 is independent of angle θ (unlike for half-wave dipole), we have & dΩ\|ℓsin θ\|2 = (L 2 )2 & dΩ\| sin θ\|2 = (L 2 )2 & 2π 0 dφ & π 0 dθ sin θ\| sin θ\|2 = (L 2 )22π & π 0 dθ sin θ\| sin θ\|2 " #$ % = 2πL2 3 . 4/3 Hence the radiation resistance of the short dipole is Rrad = ηo 4λ2 & dΩ\|ℓsin θ\|2 = ηo 4λ2 2πL2 3 = 20π2(L λ)2 Ω. 4 • Since a short dipole is constrained to have L λ ≪1, say 1 10 or smaller, Rrad will be equal to or less than about 2 Ω. • Thus, a short dipole with an input current of ˜ I(0) = Io = 1 A will have at best an average power output 1 2I2 oRrad of about 1 W. This is not quite at the level of 100s of W’s of power that typical radio stations transmit! Using antennas with higher Rrad than a short dipole1 — e.g., a half-wave dipole for which Rrad ≈73 Ω— is the best way of addressing this diﬃculty since the alternate solution of increasing Io (as needed) is not recommended because of antenna losses: – In practice, antennas appear as a circuit element with input resis-tance Ro = Rrad + Rloss where Rloss represents ohmic losses (heating of antenna wires) — an antenna consumes an average power of 1 2I2 o(Rrad + Rloss) out of which only 1 2I2 oRrad 1Short dipoles are typically employed as receiving antennas rather than transmitting antennas because of this. Receiving properties of antennas are closely related to their transmission properties, but ﬁgures of merit of antennas pertinent in transmission and reception are somewhat diﬀerent as we will learn later on in the course. 5 is the useful radiated power. Typically Rloss ∝L, whereas Rrad ∝L2 for small L, so going to longer dipoles (and learning more about them in ECE 454) really helps. • A source circuit connected to the antenna terminals “sees” the antenna (and the radiation volume with which the antenna interacts) as a two-terminal element having some impedance + -Source Circuit Antenna ZT ˜ I(0) = Io ˜ VT Zo = Ro + jXo ˜ Vo + -Zo ≡ ˜ Vo ˜ I(0) = Ro + jXo known as antenna impedance. Antenna reactance: Short dipoles have capacitive reactances, just like the line-impedance at a small dis-tance away from an open ter-mination on a transmission line. Capacitive reactance switches to an inductive one when the dipole length is about λ/2, just like the line-impedance at a distance λ/4 away from an open termination. Thus the half-wave dipole is resonant, having a zero input reactance — Zo = Rrad +j0 for an ideal half-wave dipole. In practice, resonant half-wave dipole with a length L and wire radius a has L+2a = λ/2 to a good approximation. – We have already discussed the resistive component Ro above. – Modeling the reactive component Xo requires working with storage ﬁelds of the antenna as well, matching components of total ﬁelds to proper boundary conditions imposed by the actual sur-faces of antenna wires (i.e., antenna geometry needs to be speciﬁed in detail before Xo can be determined). Antenna reactance will be examined in some detail in ECE 454 (along with methods of calculating ˜ I(z)). – We will not need to calculate antenna reactances in this course. However, it is worth mentioning that 1. antennas with Xo = 0 are known as resonant antennas, and 2. half-wave dipole is a resonant antenna (see margin note). 6
2037	https://www.teacherspayteachers.com/browse?search=percent%20worksheets	Percent Worksheets \| TPT Log InSign Up Cart is empty Total: $0.00 View Wish ListView Cart Grade Elementary Preschool Kindergarten 1st grade 2nd grade 3rd grade 4th grade 5th grade Middle school 6th grade 7th grade 8th grade High school 9th grade 10th grade 11th grade 12th grade Adult education Resource type Student practice Independent work packet Worksheets Assessment Graphic organizers Task cards Flash cards Teacher tools Classroom management Teacher manuals Outlines Rubrics Syllabi Unit plans Lessons Activities Games Centers Projects Laboratory Songs Clip art Classroom decor Bulletin board ideas Posters Word walls Printables Seasonal Holiday Black History Month Christmas-Chanukah-Kwanzaa Earth Day Easter Halloween Hispanic Heritage Month Martin Luther King Day Presidents' Day St. Patrick's Day Thanksgiving New Year Valentine's Day Women's History Month Seasonal Autumn Winter Spring Summer Back to school End of year ELA ELA by grade PreK ELA Kindergarten ELA 1st grade ELA 2nd grade ELA 3rd grade ELA 4th grade ELA 5th grade ELA 6th grade ELA 7th grade ELA 8th grade ELA High school ELA Elementary ELA Reading Writing Phonics Vocabulary Grammar Spelling Poetry ELA test prep Middle school ELA Literature Informational text Writing Creative writing Writing-essays ELA test prep High school ELA Literature Informational text Writing Creative writing Writing-essays ELA test prep Math Math by grade PreK math Kindergarten math 1st grade math 2nd grade math 3rd grade math 4th grade math 5th grade math 6th grade math 7th grade math 8th grade math High school math Elementary math Basic operations Numbers Geometry Measurement Mental math Place value Arithmetic Fractions Decimals Math test prep Middle school math Algebra Basic operations Decimals Fractions Geometry Math test prep High school math Algebra Algebra 2 Geometry Math test prep Statistics Precalculus Calculus Science Science by grade PreK science Kindergarten science 1st grade science 2nd grade science 3rd grade science 4th grade science 5th grade science 6th grade science 7th grade science 8th grade science High school science By topic Astronomy Biology Chemistry Earth sciences Physics Physical science Social studies Social studies by grade PreK social studies Kindergarten social studies 1st grade social studies 2nd grade social studies 3rd grade social studies 4th grade social studies 5th grade social studies 6th grade social studies 7th grade social studies 8th grade social studies High school social studies Social studies by topic Ancient history Economics European history Government Geography Native Americans Middle ages Psychology U.S. History World history Languages Languages American sign language Arabic Chinese French German Italian Japanese Latin Portuguese Spanish Arts Arts Art history Graphic arts Visual arts Other (arts) Performing arts Dance Drama Instrumental music Music Music composition Vocal music Special education Speech therapy Social emotional Social emotional Character education Classroom community School counseling School psychology Social emotional learning Specialty Specialty Career and technical education Child care Coaching Cooking Health Life skills Occupational therapy Physical education Physical therapy Professional development Service learning Vocational education Other (specialty) Percent Worksheets 8,700+results Sort by: Relevance Relevance Rating Rating Count Price (Ascending) Price (Descending) Most Recent Sort by: Relevance Relevance Rating Rating Count Price (Ascending) Price (Descending) Most Recent Search Grade Subject Supports Price Format All filters Filters Grade Elementary Preschool Kindergarten 1st grade 2nd grade 3rd grade 4th grade 5th grade Middle school 6th grade 7th grade 8th grade High school 9th grade 10th grade 11th grade 12th grade Higher education Adult education More Not grade specific Subject Art Coloring pages Graphic arts Visual arts Other (arts) English language arts Alphabet Balanced literacy Close reading Creative writing ELA test prep Grammar Handwriting Informational text Literature Phonics & phonological awareness Poetry Reading Reading strategies Science of reading Sight words Spelling Vocabulary Writing Writing-essays Writing-expository Other (ELA) Health Math Algebra Algebra 2 Applied math Arithmetic Basic operations Calculus Decimals Financial literacy Fractions Geometry Graphing Math test prep Measurement Mental math Money math Numbers Order of operations Place value Precalculus Statistics Telling time Other (math) Performing arts Music More Physical education Science Anatomy Astronomy Basic principles Biology Chemistry Computer science - 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Percent Coloring Worksheet Created by Beyond the Worksheet with Lindsay Gould ⭐️⭐️⭐️This Percent of a Number Coloring Worksheet is a dynamic resource designed to help students master the concept of finding percents in a fun and interactive way. This worksheet is perfect for a 30-minute class activity, providing a creative twist to math lessons during the holiday season. By solving problems to find the percent of a number and coloring based on their answers, students reinforce their understanding through visual learning. This resource focuses on finding the percent of a 6 th - 10 th Math, Other (Math) CCSS 6.RP.A.3c Also included in:6th Grade Math Curriculum Supplemental Activities Bundle $2.00 Original Price $2.00 Rated 4.86 out of 5, based on 614 reviews 4.9(614) Add to cart Wish List Part Whole Percent Word Problem Worksheets 6th Grade Created by Simone's Math Resources Part Whole Percent Word Problem Worksheets 6th Grade Included:6 worksheets with answer keys. More PercentsPercent Problems Task Cards Percent Word Problems Scavenger Hunt Bingo Common Core State Standards6.RP.3c Use ratio and rate reasoning to solve real-world and mathematical problems, Find a percent of a quantity as a rate per 100 (e.g., 30% of a 6 th Math, Other (Math) Also included in:Part, Whole, or Percent Word Problems Bundle $3.00 Original Price $3.00 Rated 4.85 out of 5, based on 112 reviews 4.9(112) Add to cart Wish List Fraction Decimal Percent Conversion Chart Practice Worksheet Created by Rise over Run Practice converting fractions, decimals, and percentages with this print & go worksheet. This activity has 15 rows of problems. Each one provides a percent, decimal, or fraction for students to fill in the equivalent values. All percentages are whole numbers. The download includes 1 page handout + answer key. Understanding how fractions, decimals, percents relate is an important concept for students to grasp to understand middle school and high school math. Extra practice in this area is o 5 th - 7 th Decimals, Fractions, Math Also included in:8th Grade Math Foundation Skills for Beginning of the Year First Week of School $1.50 Original Price $1.50 Rated 4.87 out of 5, based on 122 reviews 4.9(122) Add to cart Wish List Converting Fractions, Decimals, Percents Math Coloring by Number Worksheets Created by Cognitive Cardio Math Converting fractions, decimals, and percentages is always more fun when there's some coloring! In this set of easy-prep, paper and digital color by number activities, students solve 20 problems, self-check, and color. Students will immediately know whether or not they solved correctly - if their answer is incorrect, they won't find it on the coloring sheet! Great for centers, sub days, assessments, fun days, or fast finishers!Check out the Preview above to see how the digital version works (stu 6 th - 7 th Decimals, Fractions Also included in:6th Grade Math Warm Ups, Anchor Charts, Math Wheel Note Taking, Units, Test Prep $4.25 Original Price $4.25 Rated 4.81 out of 5, based on 220 reviews 4.8(220) Add to cart Wish List Fractions Decimals Percents (Fun Games Worksheets and Activities) Created by Miss Giraffe Fractions decimals percents math centers, worksheets, and activities to make this concept fun! Let's Roll is a great center game where students roll 2 dice to get the percent (ex: rolling a 3 and 4 would be 34%) and converting it to a fraction, a decimal, and filling in a decimal model. Recording sheet, student directions, and teacher directions are included. Match It! is a math center where students match the 4 pieces of each puzzle and then fill out their recording sheet. I have included 10 p 3 rd - 5 th Decimals, Fractions, Math $4.00 Original Price $4.00 Rated 4.83 out of 5, based on 1202 reviews 4.8(1.2K) Add to cart Wish List Converting Fractions, Decimals, and Percents Cut and Paste Worksheet Activity Created by Math With Meaning Students will practice converting between fractions, decimals, and percents with this fun activity! Students will find the equivalent fraction, decimal, and percent to complete each row of the table. They will then then find their answer from the answer choices at the bottom and glue it in the appropriate place. Please view the preview file to see make sure that this activity is appropriate for your students. An answer key is included. Please note that this activity is NOT editable. The text, p 5 th - 8 th Decimals, Fractions, Math Also included in:6th Grade Math Printable Activities Mega Bundle - 37 Activities $1.50 Original Price $1.50 Rated 4.81 out of 5, based on 98 reviews 4.8(98) Add to cart Wish List Fractions, Decimals, Percents Worksheets 6th, 7th Grade Math Task Cards Activity Created by Cognitive Cardio Math Use this Converting Fractions, Decimals, and Percents Footloose math task card activity to get students moving while they are reviewing concepts! Footloose can be used with the whole class or with small groups for center work.This engaging, easy-prep fraction, decimal, percent resource includes:30 Footloose task cards (a set with a background and a set without)Footloose gridAnswer key3 practice worksheets w/keysFootloose cards require students to:Convert fractions to decimals and percentsConvert 5 th - 7 th Decimals, Fractions, Math Test Prep CCSS 7.EE.B.3 Also included in:6th Grade Math Task Cards Footloose, Task Card Stations, Recording Sheets $3.95 Original Price $3.95 Rated 4.85 out of 5, based on 134 reviews 4.9(134) Add to cart Wish List Grade 5 Ontario Math Worksheets \| Fractions Ratios Percent \| PDF & Google Slides Created by My Call to Teach OverviewAre you looking for practical and easy to use supplement practice for the NEW Grade 5 Ontario Math Curriculum? These ready-to-print PDF and EDITABLE worksheets is exactly what you need! Note this resource compliments my digital Fractions, Ratios, Percent Unit but can also be used separately! This resource contains 6 worksheets for your students to apply what they have learned in the Number strand for Number Sense and Operations. Each worksheet specifies which Ontario Curriculum Expe 4 th - 6 th Basic Operations, Fractions, Math Also included in:Grade 5 Ontario Math Worksheets \| FULL-YEAR Bundle \| All Expectations $10.49 Original Price $10.49 Rated 4.91 out of 5, based on 32 reviews 4.9(32) Add to cart Wish List Percent Increase and Decrease Worksheet - Maze Activity Created by Amazing Mathematics Printable PDF & Digital Versions are included in this distance learning ready activity which consists of a variety of problems where students will have to calculate Percent Increase and Percent Decrease to work their way through a maze. This Google Classroom and Easel by TPT ready activity come in both digital & PDF format. Your students are sure to have a fun and engaging lesson whether you are teaching physically in the classroom or virtual! Distance learning?No problem! This activity now 7 th - 8 th Basic Operations, Math, Other (Math) CCSS 7.RP.A.3 Also included in:Algebra 1 Bundle ~ All My Algebra Products for 1 Low Price $1.50 Original Price $1.50 Rated 4.79 out of 5, based on 241 reviews 4.8(241) Add to cart Wish List Grade 6 Ontario Math Worksheets \| Fractions Ratios Percent \| PDF & Google Slides Created by My Call to Teach OverviewAre you looking for practical and easy to use supplement practice for the NEW Grade 6 Ontario Math Curriculum? These ready-to-print PDF and EDITABLE worksheets is exactly what you need! Note this resource compliments my digital Fractions, Ratios, Percent unit, but can also be used separately! This resource contains 6 worksheets for your students to apply what they have learned in the Number strand for Number Sense and Operations. Each worksheet specifies which Ontario Curriculum Exp 5 th - 7 th Basic Operations, Decimals, Fractions Also included in:Grade 6 Ontario Math Curriculum \| FULL-YEAR Worksheet Bundle \| All Expectations $10.49 Original Price $10.49 Rated 4.88 out of 5, based on 32 reviews 4.9(32) Add to cart Wish List Fraction, Decimal, and Percent Conversions Activity Coloring Worksheet Created by Jessica Barnett Math Make math more engaging by incorporating creativity and problem-solving with this fun fractions, decimals, and percent operations conversion activity! Students will practice essential conversion skills while adding an artistic touch to their work. Perfect for classwork, homework, math centers, or early finishers! WHAT'S INCLUDED:• a 10 problem version - converting between fractions, decimals and percents • a 18 problem version - includes additional practice for students who need a little more 6 th Math CCSS 6.RP.A.3c Also included in:Fraction, Decimal, and Percent Conversions Activity Worksheet Bundle $2.00 Original Price $2.00 Rated 4.9 out of 5, based on 31 reviews 4.9(31) Add to cart Wish List Percent Markup and Markdown Worksheet - Maze Activity Created by Amazing Mathematics Printable PDF, Google Slides & Easel by TPT Versions are included in this distance learning ready activity which consists of 14 problems that students must calculate the new price after a markdown or markup. This maze will help strengthen your students’ skills at calculating retail cost of an item when it has been marked up or marked down. Not all boxes are used in the maze to prevent students from just guessing the correct route. In order to complete the maze students will have to correct 6 th - 8 th Applied Math, Math, Other (Math) CCSS 7.RP.A.3 Also included in:Algebra 1 Bundle ~ All My Algebra Products for 1 Low Price $1.50 Original Price $1.50 Rated 4.74 out of 5, based on 72 reviews 4.7(72) Add to cart Wish List Percent Applications - Notes, Task Cards, Worksheets, and Assessment Created by Math on the Move UPDATED 1/30/18 - Task cards were added! 12 task cards for each concept (discount/sale price, tax/tip/commission, simple interest, percent change/percent error) This bundle contains resources used to teach the following concepts: - Discount and sale price - Tax, tips and commission - Simple interest - Percent change - Percent error Included in this set: - Six pages of guided notes (a review of finding percent of a number + one page for each of the above concepts). - Five worksheets, all of 6 th - 8 th Math CCSS 7.RP.A.3 $12.00 Original Price $12.00 $10.00 Price $10.00 Rated 4.91 out of 5, based on 144 reviews 4.9(144) Add to cart Wish List Consumer Math Skill Practice Worksheet. Budget Check Percents Cost Comparison Created by A Love for Special Learning Practice staying in budget, comparing prices, writing a check, balancing a checking account, and finding a percent of an amount of money all on 1 worksheet! These consumer math and financial literacy skills are important to learn and review at any age. With 40 pages, this resource will last you 1 full school year if used once a week! Read about why I created this resource, see a full page example, and learn how it's great for your students and saves you time!Design and graphics are appropria 3 rd - 12 th Applied Math Also included in:Consumer Math Curriculum + 4 Projects & Weekly Practice Workbook.Special Ed $13.00 Original Price $13.00 Rated 4.82 out of 5, based on 22 reviews 4.8(22) Add to cart Wish List Percent of Change and Error Activity and Worksheet Bundle Created by Jessica Barnett Math Are you looking for engaging, no prep activities to supplement a lesson on percent of change and error? This activity pack is THE RESOURCE for you! PRINT AND GOOGLE SLIDES™ VERSIONS ARE INCLUDED FOR EACH ACTIVITY!WHAT'S INCLUDED:• MATCHING ACTIVITY - 10 PROBLEMS • COLORING ACTIVITY - 10 PROBLEMS - 2 PICTURE OPTIONS • MISTORY LIB ACTIVITY - 10 PROBLEMS • PICTURE PUZZLE ACTIVITY - 20 PROBLEMS & 9 PROBLEM VERSIONS ANSWER KEYSGRAB A FULL YEAR OF 7TH GRADE MATH RESOURCES HERE!The purchase of this r 6 th - 8 th Math CCSS 7.RP.A.3 Bundle (4 products) $7.25 Original Price $7.25 $5.75 Price $5.75 Rated 4.87 out of 5, based on 74 reviews 4.9(74) Add to cart Wish List Converting Fractions Decimals &Percents Activities Coloring, Worksheets, Notes Created by Cognitive Cardio Math Converting fractions, decimals, and percentages practice is easy to implement with this variety of engaging activities. Use the activities for stations, homework, assessments, fast finishers, review days, or even sub days.This bundle includes a variety of activities and resources (some print, some digital, some both) for teaching and practicing fraction, decimal, percent conversions: Notes: Fold It Ups, Math WheelNumber linesColor by numbersTruth or Dare review gamesFootloose activitiesDecima 5 th - 6 th Decimals, Fractions, Math CCSS 7.NS.A.2d Bundle (16 products) $36.50 Original Price $36.50 $19.75 Price $19.75 Rated 4.74 out of 5, based on 94 reviews 4.7(94) Add to cart Wish List Expressions and Percents - Worksheet (7.EE.2) Created by Math on the Move With this worksheet, students practice writing equivalent expressions involving percent problems. Students can see how taking 25% off an original price is the same as paying 75% of the price. Or how adding a 15% tip is the same as multiplying the bill by 1.15. An example problem: Jack bought a bike that was 35% off the original price. Let p represent the original price. What expressions can be used to calculate the sale price? Answer: p - 0.35p and 0.65p Answer key included! This work 7 th - 9 th Algebra, Math $1.50 Original Price $1.50 Rated 4.89 out of 5, based on 112 reviews 4.9(112) Add to cart Wish List Calculating Tax & Tip Percent Proportions Activity for 7th Grade Math Worksheet Created by Math All Day In this 7th grade resource, students will calculate tax and tip in real world scenarios. Using percent proportions, students will answer 2 questions per situation to determine the tax or tip and then the final cost or a related question. Students will calculate the part, whole, or percent of each problem. The same problems are also provided on the printable version, so that students can write out their work on paper. This product includes:Self Checking Google Sheet with 8 problemsPrintable v 6 th - 8 th Math, Other (Math) CCSS 7.RP.A.3 Also included in:Calculating Tax Tip Discount Percent Proportions Activities for 7th Grade Math $3.00 Original Price $3.00 Rated 4.95 out of 5, based on 21 reviews 5.0(21) Add to cart Wish List PERCENT NOTES, REFERENCE & PRACTICE WORKSHEETS- PROPORTION Created by Teach Easy Find the Part, Whole, and Percent using the proportion method! This is a percent reference/notes sheet that will help students to learn how to do percents using the proportion method. The proportion method is described fully along with examples. Cross multiplication is covered as well. Included: 2 practice worksheets. One with 20 questions and answer key and the other is as word problem worksheet with 12 questions and answer key AND a solutions page to show each problem and the steps used to 6 th - 12 th, Adult Education, Higher Education Basic Operations, Math $6.00 Original Price $6.00 Rated 4.75 out of 5, based on 132 reviews 4.8(132) Add to cart Wish List Converting Fractions Decimals and Percents Worksheets Review Quiz Test, Practice Created by Printables and Worksheets This is a collection of converting fraction, decimal, percent worksheets. The sheets include conversion, interpreting images, and fill-in-the-blanks type of questions. In all, you can pick from over a hundred worksheets. Just print the sheets that match your preference, and you're ready to go! The problems include denominators of 100, and between 2 and 50. It can be used as a review, quiz, test, assessment, homework, or practice. These decimal worksheets suit 4th, 5th, 6th, and 7th grades and 4 th - 7 th Decimals, Fractions $3.99 Original Price $3.99 Rated 4.86 out of 5, based on 34 reviews 4.9(34) Add to cart Wish List Percent Activity Tax, Tip, and Markup Coloring Worksheet Created by Jessica Barnett Math Are you looking for something more engaging than an average worksheet? Students of all ages enjoy coloring! Pair it with math, and the work becomes fun! This 10 problem activity sheet is a great way to incorporate art into a math lesson on taxes, tips, and markups. Both a PDF for printing and a digital version created in Google Slides™ are included.★WHAT'S INCLUDED★• a 10 problem worksheet - (Fall picture) • a 10 problem worksheet - (evergreen picture) • blank coloring sheets for students to col 6 th - 8 th Math CCSS 7.RP.A.3 Also included in:7th Grade Math Activity Bundle Full Year $2.00 Original Price $2.00 Rated 4.87 out of 5, based on 45 reviews 4.9(45) Add to cart Wish List Converting Fractions Decimals Percents Activity Math Problem Solving Worksheets Created by Cognitive Cardio Math Converting fractions, decimals, and percents is more meaningful in word problem activities! Engage students in problem solving fun with this set of word problems that focuses on fraction, decimal, and percent conversions and comparisons. The included word problems help students practice:Comparing fractions, decimals, and percentsConverting fractions, decimals, and percents This easy-prep resource includes:1) Four problem solving pagesEach page presents a different situationEach situation 6 th - 7 th Decimals, Fractions, Other (Math) CCSS 7.EE.B.3 Also included in:6th Grade Math Warm Ups, Anchor Charts, Math Wheel Note Taking, Units, Test Prep $2.65 Original Price $2.65 Rated 4.79 out of 5, based on 61 reviews 4.8(61) Add to cart Wish List Valentine's Day Math Activity Worksheet Percent Change 7th Grade Math Created by Make Sense of Math Fun and engaging Valentine's Math activity! Students practice percent change with percent increase and percent decrease in this coloring activity. 20 practice problems for students to calculate the percent change of a given price. Students then color the Valentine's coloring sheet according to their answer. Included in this product:20 Percent Change QuestionsFun Valentine’s Coloring SheetAnswer KeyThis is great to keep your students' attention. CHECK OUT WHAT EDUCATORS ARE SAYINGThis was a 6 th - 8 th Decimals Also included in:Holiday Math Coloring Activities Middle School Math Bundle \| Fun Worksheets $3.00 Original Price $3.00 Rated 4.93 out of 5, based on 30 reviews 4.9(30) Add to cart Wish List Converting Fractions to Percents Color Worksheet Created by Aric Thomas 25 unique and thought out questions on converting fractions to percents. Each question corresponds to a matching answer that gets colored in to form a symmetrical design. Not only does this make it fun and rewarding for students but it also makes it easy for students and teachers to know if the worksheet has been completed correctly. Great for classwork, homework, additional practice, extra credit, and subs. The designs can also be cut out and quilted together to make a great art piece. The pdf 6 th - 9 th Arithmetic, Fractions, Math $2.50 Original Price $2.50 Rated 4.96 out of 5, based on 54 reviews 5.0(54) Add to cart Wish List 1 2 3 4 5 Showing 1-24 of 8,700+results TPT is the largest marketplace for PreK-12 resources, powered by a community of educators. Facebook Instagram Pinterest Twitter About Who we are We're hiring Press Blog Gift Cards Support Help & FAQ Security Privacy policy Student privacy Terms of service Tell us what you think Updates Get our weekly newsletter with free resources, updates, and special offers. 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2038	https://prithvijc.pythonanywhere.com/static/clique_article.pdf	A note on the search for maximal pairwise coprime subsets Prithvijit Chakrabarty Abstract: of integers. Introduction: No polynomial time algorithm to factorize integers is known. There is however, an algorithm to factorize an integer into a set of pairwise coprime numbers. This algorithm, called factor refinement, was introduced by Bach, Driscoll and Shallit in , and allowed factoring an integer into coprime numbers in quadratic time. The approach was further investigated by Bernstein in , where he introduced an algorithm to do same in approximately linear time. Indeed, in many applications, factor refinement can be used instead of integer factorization. Factor refinement is thus closely linked to integer factorization. This could be useful in investigating the complexity class of integer factorization. Instead of directly looking for a polynomial time algorithm, or proof of NP completeness of factoring, an auxiliary problem, the PC problem is introduced. This is linked to both factor refinement and integer factorization. This paper proves the intermediate result that the PC problem is NP complete. In following work, I will describe the relationship between the PC problem and integer factorization. Notations: G is a graph with vertex set V and edge set E 𝐺̅ is the complement of graph G E(G) is the edge set of graph G e(v1, v2) represents an edge from vertex v1 to v2 L :(𝑉∪𝐸) Z is the label function that maps vertices and edges to integral labels The PC problem: Pairwise coprime set: A set S is called pairwise coprime if and only if gcd(𝑝1, 𝑝2) = 1∀𝑝1, 𝑝2 ∈𝑆 The PC problem: Given a set of integers, the PC problem is to find the largest subset that is pairwise coprime. The factor refinement algorithm generates a pairwise coprime set of numbers, all of which are factors of a given integer (called the coprime base set of the integer in ). The PC problem may be related to the integer factorization problem through the fact that every integer can be decomposed into a corresponding coprime base set. NP Completeness: The PC problem is proven to be NP complete by reduction from the max clique problem. Algorithm 2: Reduction from clique to PC Input: Graph G = (V, E). Output: Set of vertex labels for the vertices of G The algorithm maintains a list of labels L for each node and edge. L = {(v1, p1), (v2, p2), (v3, p3)... (eN, pN)}, where p1, p2, p3, ..., pN are distinct primes. for e(v1,v2) in E(𝐺̅) begin: This proves the NP completeness of the search for the largest pairwise coprime subset in a set L(v1) L(v1) L(e) L(v2) L(v2) L(e) end return L The set of vertex labels L is the input to the corresponding PC problem. Theorem: The graph G contains a clique of size 'k' if and only if its corresponding label set contains a pairwise coprime subset with 'k' elements. Proof: Consider a subgraph SG = {v1, v2, v3, ..., vk}. The corresponding set of labels will be Sk = {L(v1), L(v2), L(v3), ..., L(vk)}. Algorithm 2 ensures that labels L(v1) and L(v2) have a common factor if and only if e(v1 , v2) is in E. 1. Sk is pairwise coprime: This implies that e(vi, vj) is in E for all (vi, vj) in SG. Thus, SG must be a clique. 2. Sk is not pairwise coprime: This implies that there exists at least one pair L(vi), L(vj) that are not pairwise coprime and the corresponding edge e(v1,v2) is not in E. Thus, SG will not be a clique if Sk is not pairwise coprime. Thus, G contains a clique of size 'k' if and only if S contains a pairwise coprime subset with 'k' elements. Theorem: The length of the labels generated by algorithm 1 is O(\|V\|+\|E\|). Proof: The labels of the vertices in G are initialized to \|V\| distinct primes. To restrict the size of the numbers generated as labels, we use the first \|V\| prime numbers to label the vertices and the next \|E\| primes to label the edges before running algorithm 1. Due to the labelling algorithm, the final label of a vertex vj will have the maximum value if it is connected to all the other vertices in G. In this case, L(vj) = ∏ 𝑝𝑖 ∣𝑉∣+∣𝐸∣ 𝑖=∣𝑉∣+1 < ∏ 𝑝𝑖 ∣𝑉∣+\|𝐸\| 𝑖=0 . Thus, L(v ) < prim(\|V\|+\|E\|) ∀ \|V\|,\|E\| > 0 (1) ). Consider N(x) such that N(x) = number of digits in x. If the vertex vj has the highest label value, the number of digits in L(vj) will be: N(L(vj)) = (⌊log (𝐿(𝑣𝑗))⌋+ 1) < log(𝑝𝑟𝑖𝑚(∣𝑉∣+ \|𝐸\|)). The logarithm of the primorial function is the first Chebyshev function 𝜃(∣𝑉∣+ \|𝐸\|), which grows linearly, as proved in . Thus, the number of digits in the labels of the vertices of G is generated by the labelling algorithm is O(\|V\|+\|E\|). Theorem: The reduction of the clique to the PC problem runs in polynomial time. Proof: Algorithm 2 traverses the list of edges in E and performs 2 multiplications for every edge. The number of digits of the labels is O(\|V\|+\|E\|). Using a simple multiplication algorithm running in quadratic time yields the product in O((\|V\|+\|E\|)2). Thus, the complexity of generating the label set L is O(\|E\|.(\|V\|+\|E\|)2). Conclusion and further work: classifying the complexity of integer factorization. In future papers, I will investigate the exact where prim() is the primorial function. The value of the labels is thus bounded by prim(\|V\|+\|E\| This proved the NP completeness of the PC problem, which may serve as an auxiliary to relationship of this problem with integer factorization. References: Eric Bach, James Driscoll, Jeffrey Shallit, Factor Refinement, in SODA '90 Proceedings of the first annual ACM-SIAM symposium on Discrete algorithms Pages 201-211. Daniel J. Bernstein, Factoring into coprimes in essentially linear time, ACM Journal of Algorithms, Volume 54 Issue 1, January 2005. Hardy, G. H. and Wright, E. M. The Functions θ(x) and ψ(x) and Proof that θ(x) and ψ(x) are of Order x, in An Introduction to the Theory of Numbers, 5th edition, Pages 340-342, 1979.
2039	https://sites.cc.gatech.edu/classes/AY2010/cs6505_spring/lectures/FFT.pdf	CS 6505, Computability and Algorithms Fast Fourier Transform Spring 2010; Lecturer: Santosh Vempala 1 Evaluating polynomials at many points Suppose that we want to evaluate a polynomial A(x) = a0 + a1x+· · ·+anxn at many points – say n points in all. If we were just evaluating A at one point, say x0, then we can naively perform the multiplication in O(n2) multiplications, but divide-and-conquer algorithms should make us think that we can do better. First, we can rewrite our computation of A as A (x) = a0 + x(a1 + x (a2 + x (· · · + xan))) This is known as Horner’s Rule: each coeﬃcient ai is associated with a single multiplication with x. So calculating A can be done in n multiplications for a single value of x. If we want to use this method to evaluate, say, n points, then we need O n2 multiplications. Can we do better than this? With divide-and-conquer methods, we would want to ﬁnd some way to eliminate redundant multiplications. If the xj are chosen randomly, it seems like we will have no hope for overlapping or weeding out our multiplications. What if we chose our xj cleverly so that a multiplication for one gives us the same answer for some other xi? This is at the heart of the Fast Fourier Transform: we will choose our xj so that we can cut out multiplications. Suppose we choose xj = −xi. Then all of the even monomials are the same, and the odd monomials have opposite sign. In other words, a2kx2k = a2k(−x)2k and a2k+1x2k+1 = −a2k+1(−x)2k+1 for 0 ≤k ≤n 2 . There is a set of numbers that have this property. In complex numbers, we have n solutions to the equation xn = 1, and these solutions are known as the nth roots of unity. These are denoted as 1, ω, ω2, ω3, . . . , ωn−1, where ω = e 2Πi n . There are some special properties of the nth roots of unity that we can see right away. For example, if n is even and xn = 1, then (−x)n = 1. How do we ﬁnd −x? If n is even, then we have ω n 2 = −1. So if x = ωk, then we know that ωk+ n 2 = x ∗−1 = −x. Now let’s take a look at our polynomial A(x) again. Let’s examine A (ω) and A(-ω): A (ω) = a0 + a1ω + a2ω2 + a3ω3 + . . . A ω n 2 +1 = A (−ω) = a0 −a1ω + a2ω2 −a3ω3 + . . . This looks like something that can be tackled with divide-and-conquer techniques. All of the even-exponent monomials, such as a0, a2ω2, a4 ω4, and so on, are the same in A (ω) and A (−ω), while all of the odd-exponent monomials, such as a1, a3ω3, a5ω5, and so on, are of the opposite sign in A(ω) and A(−ω).So now we can split up A into two polynomials of even and odd degree, say A0 and A1, as follows: 1 A (ω) = A0 ω2 + ωA1 ω2 A0 (ω) = a0 + a2ω + a4ω2 + · · · + adω d 2 A1 (ω) = a1 + a3ω + a5ω2 + · · · + ad−1ω d−1 2 So what has happened? We have broken the original polynomial A into two smaller polynomials A0 and A1 of degree d 2. How many distinct nth roots of unity (ωk) do we need? We don’t need to calculate all n - since we’re squaring them, we only need n 2 of them. So our original problem has now been halved: we originally were evaluating A, which has degree d, at n points, and now we’re evaluating two polynomials A0 and A1 that are of degree d 2 at n 2 points. So our recurrence for evaluating A will involve two variables: n and d. If n = d, then we have: T (n) = 2T n 2 + O (n) Then we can evaluate A over n points in time O(n log n). Now what if n ̸= d? If d = 1, then all polynomials are of degree 1 and it just takes O(n) time to evaluate them. Otherwise, we have T (n, d) = 2T n 2 , d 2 + O (n) From Horner’s Rule, we also have T (1, d) = O(d) (i.e., we’re evaluating one point of degree k). Then we can derive T (n, d) = O(d log n). Likewise, if the degree d is larger, then we could derive a running time of O(n log d). 2 Equivalence between function values and coeﬃcients So we can go from values of the polynomial, say A (1) , A (ω) , A ω2 , . . . , A ωn−1 , to the coeﬃcients a0, a1, . . . , an−1 of A (so we are assuming that d = n)? If we want to calculate all of A’s values, we could represent them in matrix form as follows:      A(1) A(ω) . . . A(ωn−1)     =        1 1 1 . . . 1 1 ω ω2 . . . ωn−1 1 ω2 ω4 . . . ω2(n−1) . . . . . . 1 ωn−1 ω2(n−1) . . . ω(n−1)(n−1)               a0 a1 a2 . . . an−1        We can write the nxn matrix of ω values as M (ω). It is also known as the Vandermonde matrix. How do we solve for the coeﬃcients of A? If we have to calculate an inverse naively, then this calculation could take a long time to compute – O n3 . However, our nxn matrix with the values of ωk has some special properties. First, note that ωn −1 = (ω −1) ωn−1 + ωn−2 + · · · + ω + 1 2 If we take ω ̸= 1, then we have ωn−1 + ωn−2 + · · · + ω + 1 = ωn−1 ω−1 = 0 because we know that ω is an nth root of unity (and so ωn = 1). This helps when we generate M(ω)−1, since the identity matrix I = M (ω) ∗M(ω)−1 will have 1s on the diagonal and 0s everywhere else. We now claim that M(ω)−1 = 1 n ∗M(−ω). We can validate this for both the diagonal and oﬀ-diagonal entries. For the kth diagonal entries, we will take the dot product of the kth row of M(ω) with the kth column of M(ω)−1. This gives us ˙ 1 n 1 ωk ω2k . . . ωk(n−1) 1 (−ω)k (−ω)2k . . . (−ω)(n−1)k = ˙ 1 n 1 ωkω2k . . . ωk(n−1) (1 ω−k ω−2k . . . ω−(n−1)k) = 1 n (1 + 1 + · · · + 1) = 1 For the oﬀ-diagonal entry at position (j,k), we multiply the j row of M (ω) with the k column of M (−ω). That gives 1 n ˙ 1 ωjω2j . . . ω(n−1)j (1 ω−k ω−2k . . . ω−(n−1)k) = 1 n 1 + ωj−k + ω2j−2k + . . . ω(n−1)(j−k) = 1 n ωn(j−k) −1 ωj−k −1 = 0 So the inverse matrix M(ω)−1 is easy to calculate. That means that, using the method we previously described, we can recover the coeﬃcients of the polynomial A in time O (n log n). This has important applications to signal analysis, where we can uncover the frequencies of a signal. 3 Multiplying Polynomials Suppose that we now have two polynomials, A(x) and B(x), of degree d, and we want to calculate C (x) = A (x) B(x). Then C (x) has degree 2d, so it is determined by any 2d + 1 points. So we can determine enough information to reconstruct C(x) from any 2d + 1 values of A(x) and B(x). So our procedure for determining C is as follows: 1. Calculate A(x) and B (x) at n = 2d + 1 points, which will require time O(n log n); 2. Calculate C (x) = A (x) B(x) at the n selected points, which will require time O(n); 3. Determine C’s coeﬃcients, which will take time O(n log n). All three of these steps have already been described in detail. We just need to select the nth roots of unity for the n selected points, and we’re done. 3 4 String Matching Suppose that we have a pattern p = pm−1pm−2 . . . p0 that we want to match against an n-character string s = sn−1sn−2 . . . s0. We could compare the m-character pattern against all possible n−(m−1) starting positions of the pattern, but that gives a running time of O(m(n −m + 1)). For example, take p = abba and s = aababbabba. Then we might have to compare abba against aaba, abab, babb, and so on. Can we do better using the FFT? Let’s assume that the pattern and string have the same binary characters, say a and b. Now let’s map a to -1 and b to 1. Then the product of the mapped pattern, abba →(-1)11(-1), with the ﬁrst four characters of the string, aaba →(-1)(-1)1(-1) gives us a dot product of (-1)(-1)+1(-1) +11 +(-1)(-1) = 2. When we do have a match, the dot product is (-1)(-1)+11+11+(-1)(-1) = 4. So when we have a match, each term in the dot product is 1 and the dot product is m. If we do not have a match, then at least one term in the dot product is -1 and so the dot product is less than m. This looks just like the polynomial multiplication that we just saw – the pattern can be translated to one polynomial, and the text string can be translated to another polynomial. We want to know if the text starting at position k agrees with the input pattern. Let the pattern’s polynomial A (x) = pm−1xm−1+pm−2xm−2+· · ·+p0 and take the m text polynomial to be B (x) = sn−1xn−1+ sn−2xn−2+· · ·+s0. Then C (x) = A (x) B(x) will not give us information about whether the pattern matches up at position k in the text – we cannot just look at a coeﬃcient of C and determine if we have m matches at the corresponding text position. Instead, we want coeﬃcient ck to tell us if position k in the text matches with position m in the pattern, if position k + 1 in the text matches with position m −1 in the pattern, and so on. So we want to know if sk = pm−1, sk+1 = pm−2, and so on until sk+m−1 = p0. Notice that the sum of the two indices is always k + m −1. Our insight is to reverse the order of one of these strings with respect to how its polynomial is created. So now let’s try A (x) = pm +pm−1x+· · ·+p0xm (i.e., A(x) with its coeﬃcients reversed) and use the same B (x) . We can see that the jth coeﬃcient of C, which we will denote cj, is as follows: cj = Σipisj−1 So we get exactly what we want: if we want to know if the text and pattern match up in the kth position, then we just need to examine ck. How does FFT help with pattern matching? We can see that we have just reduced the problem of pattern matching to polynomial multiplication: we multiplied the polynomial A(x) with B(x). The polynomial multiplication algorithm above tells us that we can do this in n log m, where n is the length of the text and m is the length of the pattern. This is not the last word on string matching, but it is an interesting application of FFT and divide-and-conquer methods in general. 4
2040	https://www.chegg.com/homework-help/questions-and-answers/21-drawing-free-body-diagram-simple-pendulum-breaking-forces-components-show-restoring-for-q73086384	Solved 2.1: By drawing a free body diagram for a simple \| Chegg.com Skip to main content Books Rent/Buy Read Return Sell Study Tasks Homework help Understand a topic Writing & citations Tools Expert Q&A Math Solver Citations Plagiarism checker Grammar checker Expert proofreading Career For educators Help Sign in Paste Copy Cut Options Upload Image Math Mode ÷ ≤ ≥ o π ∞ ∩ ∪           √  ∫              Math Math Geometry Physics Greek Alphabet Science Physics Physics questions and answers 2.1: By drawing a free body diagram for a simple pendulum and breaking the forces into components, show that the restoring force on a pendulum obeys equation 3 if we ignore friction and air resistance. (Hint: the net restoring force for a pendulum is the component of the sum of the forces that is perpendicular to the pendulum's string.) 2.2: Reconsider the Your solution’s ready to go! Our expert help has broken down your problem into an easy-to-learn solution you can count on. See Answer See Answer See Answer done loading Question: 2.1: By drawing a free body diagram for a simple pendulum and breaking the forces into components, show that the restoring force on a pendulum obeys equation 3 if we ignore friction and air resistance. (Hint: the net restoring force for a pendulum is the component of the sum of the forces that is perpendicular to the pendulum's string.) 2.2: Reconsider the Show transcribed image text There are 4 steps to solve this one.Solution Share Share Share done loading Copy link Step 1 A pendulum is a combination of a string and a mass attached to the string. The mass oscillates to at... View the full answer Step 2 UnlockStep 3 UnlockStep 4 UnlockAnswer Unlock Previous questionNext question Transcribed image text: 2.1: By drawing a free body diagram for a simple pendulum and breaking the forces into components, show that the restoring force on a pendulum obeys equation 3 if we ignore friction and air resistance. (Hint: the net restoring force for a pendulum is the component of the sum of the forces that is perpendicular to the pendulum's string.) 2.2: Reconsider the section in the introduction concerning the small angle approximation. Based on equation 4, what is the effective "spring constant," k, of a pendulum for small angles of displacement? (Hint: compare equations 4 and 1.) 2.3: For small angles, find an expression for the period of a pendulum. (Hint: compare the value of the "spring constant" you found in 2.2 and equation 2.) F = -kx (1) T = 271 k 加心 (2) where T is called the period of oscillation. If the spring is stiff (large k), the period would be small. Similarly, a larger mass leads to a longer period of oscillation. In this lab, you will measure oscillation period as a function of hanging mass. So a plot of T versus m will be a square root curve going through the origin with a fit parameter of A = 2 You will test this prediction to see if it accurately describes the "mass on a spring" system. The Physics of a Simple Pendulum and the Small Angle Approximation: A simple pendulum consists of two main components: (1) A "massless" string suspended from a pivot point, and (2) a mass or "bob" hanging from the end of the string. A pendulum, like a mass on a spring, can exhibit harmonic oscillation. It can be shown (see 3.1) that the restoring force for a simple pendulum is F = -mg sin (3) F- -mg i Not the question you’re looking for? Post any question and get expert help quickly. 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2041	https://11physicsahs.weebly.com/graphing.html	Graphing AHS SACE Stage 1 Physics Home The Basics Manipulating Equations Recording Data Analysing Data Glossary Home The Basics Manipulating Equations Recording Data Analysing Data Glossary Back SI Units Scientific Notation Prefixes Significant Figures Variables Back Proportionalities Rearranging Equations Right Angled Trigonometry Back Tables Manipulating Data in Excel Graphing Graphing in Excel Back Calculating slope Trends and Relationships Precision vs Accuracy Random Error Systematic Error Graphing Tips for drawing a graph by hand Graphing checklist Work through the following checklist when drawing a graph. 1. Put the dependent variable on the y-axis and the independent variable on the x-axis. If you are asked to plot A against B, put A on the y-axis and B on the x-axis. 2. Label the axes with the name of each variable. 3. Next to each variable name, label the units in brackets if appropriate. 4. Choose appropriate scales for each axis (see below for further details). 5. Plot each point correctly according to the major and minor grid lines provided. 6. Draw a suitable trend line. This may be straight or curved but plotted points should be evenly scattered on either side of it. In Physics, always extend any straight trend lines back to the axis. 7. A title is good but not necessary. In a practical report, always provide a captionunderneatha graph e.g.Figure 1: the period squared of a pendulum versus the mass. Choosing axes scales Find the highest data value.Choose a scale that allows your data to be graphed as large as possible in the space provided. The scale should be large enough to reach the highest value without lots of extra space. In our Physics classes, always start the scale at zero and never break it. This is important since breaking or truncating the axis will make it difficult to read the y-intercept which we often want to read for calculations. The scale of each axis may be different but each one must be consistent. If one box represents one metre at the beginning of the graph, one box always represents one metre. Choose a scale that makes it easy to plot data correctly. Setting one box equal to 7 metres would make it hard to identify what one-fifth of that box represents and will lead to misplotted points. Common mistakes Each square on a scale should be worth the same value. Both scales here are incorrect. The scales are not large enough for the space given. The trend line has not been extended back to the axis. Do not fudge a trend line to pass through each point. Here, the trend is clearly linear. Here the y axis scale does not start at 0. This makes it difficult to determine the y intercept. Never draw dot to dot trend lines. Do not curve a trend line to try and force it through the origin. Also, always mark the axis values on major grid lines, not between them (on the x axis, the 0.1, 0.3, 0.5 etc. marks are unnecessary). Do not break the scales at the start of the axis. Always start at 0 and go up in even intervals. The scale choices here also make it difficult to plot points correctly (here, each square on the x axis is worth 0.017) Interpolating and Extrapolating Besides being able to show trends between variables, plotting data on a graph allows us to predict values for which we have taken no data. When we predict values that fall within the range of data points taken, it is called interpolation. When we predict values for points outside the range of data taken, it is called extrapolation. Extrapolation over too far a range can be unreliable unless it is certain that the observed relationship between the variables continues. Watch the video to learn how to interpolate and extrapolate. When doing this for tests, always rule in the lines that you used to deduce the answer. Choosing a graph type Note that a scatter plot is not always the most appropriate graph to use. The nature of each variable affects how the data collected is best displayed. Continuous numerical variable: A numerical variable that can have any number. If dependent and independent variables are both continuous, then the data can be represented by a scatter plot with a trend line. Discrete numerical variable: A variable that can only take a finite number of values such as the number of springs connected together. These can also be represented by a scatter plot but it often does not make sense to connect the points. Otherwise, a column graph can be used. Categorical variable: Something that can be described by a label and not a number, e.g. colour, or material of construction. These can be displayed as a column graph. 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2042	https://www.chegg.com/homework-help/questions-and-answers/exercise-find-indefinite-integral-t-1-3-frac-1-t-2-2-t-respect-t--int-left-t-1-3-frac-1-t--q122230606	Solved Exercise. Find the indefinite integral of t1/3+t21+2t \| Chegg.com Skip to main content Books Rent/Buy Read Return Sell Study Tasks Homework help Understand a topic Writing & citations Tools Expert Q&A Math Solver Citations Plagiarism checker Grammar checker Expert proofreading Career For educators Help Sign in Paste Copy Cut Options Upload Image Math Mode ÷ ≤ ≥ o π ∞ ∩ ∪           √  ∫              Math Math Geometry Physics Greek Alphabet Math Calculus Calculus questions and answers Exercise. Find the indefinite integral of t1/3+t21+2t with respect to t. ∫(t1/3+t21+2t)dt=+C Your solution’s ready to go! Our expert help has broken down your problem into an easy-to-learn solution you can count on. See Answer See Answer See Answer done loading Question: Exercise. Find the indefinite integral of t1/3+t21+2t with respect to t. ∫(t1/3+t21+2t)dt=+C Show transcribed image text There are 3 steps to solve this one.Solution 100%(1 rating) Share Share Share done loading Copy link Step 1 View the full answer Step 2 UnlockStep 3 UnlockAnswer Unlock Next question Transcribed image text: Exercise. Find the indefinite integral of t 1/3+t 2 1+2 t with respect to t. ∫(t 1/3+t 2 1+2 t)d t=+C Not the question you’re looking for? Post any question and get expert help quickly. Start learning Chegg Products & Services Chegg Study Help Citation Generator Grammar Checker Math Solver Mobile Apps Plagiarism Checker Chegg Perks Company Company About Chegg Chegg For Good Advertise with us Investor Relations Jobs Join Our Affiliate Program Media Center Chegg Network Chegg Network Busuu Citation Machine EasyBib Mathway Customer Service Customer Service Give Us Feedback Customer Service Manage Subscription Educators Educators Academic Integrity Honor Shield Institute of Digital Learning © 2003-2025 Chegg Inc. All rights reserved. Cookie NoticeYour Privacy ChoicesDo Not Sell My InfoGeneral PoliciesPrivacy PolicyHonor CodeIP Rights
2043	https://skinimages.ca/skin-image/eczema-herpeticum/	Eczema Herpeticum - skin images Skip to content Home About Chair Image Contributors Working Group Founding Contributors Content Contributors Funding Sources Patient Resources Contribute an Image skin images Image Atlas Contact Us X Login Eczema Herpeticum Home Eczema Herpeticum Eczema Herpeticum Author: Kawaroreet Karwal MS2, Y, Miller-Monthrope, Dermatologist/Dermatopathologist, Toronto, Canada, 2024 Definition: Eczema herpeticum (also known as Kaposi varicelliform eruption) is a severe, potentially life-threatening skin infection caused by the herpes simplex virus (HSV) in individuals with atopic dermatitis or other underlying skin conditions. Epidemiology:It typically occurs in children and young adults but can affect individuals of any age. Symptoms: Eczema herpeticum usually presents with clusters of painful, monomorphic, punched-out erosions that rapidly spread over eczematous skin areas. The affected areas may become swollen, and the lesions can develop into pustules, vesicles, and crusts. Systemic symptoms such as fever, malaise, and lymphadenopathy may also be present. Diagnosis:Early diagnosis and treatment are crucial to prevent complications. Antiviral medications, such as acyclovir or valacyclovir, are the primary treatment options and should be administered promptly. In severe cases, hospitalization may be required for intravenous antiviral therapy and supportive care. Corticosteroids and antibiotics may be used to manage concurrent inflammation and bacterial infections. References: Eichenfield, L. F., Tom, W. L., Chamlin, S. L., Feldman, S. R., Hanifin, J. M., Simpson, E. L., & Paller, A. S. (2014). Guidelines of care for the management of atopic dermatitis. Journal of the American Academy of Dermatology, 71(1), 116-132. Wollenberg, A., Oranje, A., Deleuran, M., Simon, D., Szalai, Z., Kunz, B., … & Taïeb, A. (2016). ETFAD/EADV eczema task force position paper on diagnosis and treatment of atopic dermatitis. Journal of the European Academy of Dermatology and Venereology, 30(5), 729-747. Hebert, A. A., & Rodriguez, L. A. (2016). Management of eczema herpeticum in children. Pediatric Dermatology, 33(4), 357-361. Related Diseases Lentigines, Solar Lentigines Read More Leiomyosarcoma Read More Leiomyosarcoma Read More Keloid Read More Keloid Read More Keloid Read More SkinImages.ca is designed to establish a comprehensive and easily accessible digital database of dermatology images categorized according to skin types. Quick Links Home About Us Contact Us Quick Links Term Of Use Contribute An Image Founding Contributors Our Newsletter Subscribe FacebookTwitterYoutube Copyright© 2025 skinimages.ca All Rights Reserved.
2044	https://microbenotes.com/biochemical-test-of-klebsiella-granulomatis/	Microbe Notes Home » Biochemical Test Biochemical Test of Klebsiella granulomatis Image: Granuloma inguinale (donovanosis) in a conventional Pap test. A: Granulomatous inflammation is shown with epithelioid histiocytes and lymphocytes. B: The organisms (Donovan bodies) can be seen in thin-walled intracytoplasmic vacuoles (circle ). (Pap stain, Mag ×200 in a and ×400 in b). C: Macrophages show numerous safety pin-shaped structures with polar thickening of chromatin (Donovan bodies) and a halo around them (circle) (Pap, ×100). Image Source: PathologyOutlines.com Table of Contents Some of the characteristics are as follows: \| \| \| --- \| \| Basic Characteristics \| Properties (Klebsiella granulomatis) \| \| Capsule \| Positive (+ve) \| \| Catalase \| Positive (+ve) \| \| Citrate \| Positive (+ve) \| \| Flagella \| Negative (-ve) \| \| Gram Staining \| Gram-negative \| \| H2S \| Negative (-ve) \| \| Indole \| Negative (-ve) \| \| Motility \| Non-motile \| \| MR (Methyl Red) \| Negative (-ve) \| \| Nitrate Reduction \| Negative (-ve) \| \| Oxidase \| Positive (+ve) \| \| Shape \| Pleomorphic rods (coccobacillus) with rounded ends. Occur singly or in clusters \| \| Urease \| Positive (+ve) \| \| VP (Voges Proskauer) \| Positive (+ve) \| \| Fermentation of \| \| \| Arabinose \| Negative (-ve) \| \| Dulcitol \| Negative (-ve) \| \| Glucose \| Positive (+ve) \| \| Lactose \| Positive (+ve) \| \| Malonate \| Positive (+ve) \| \| Raffinose \| Negative (-ve) \| \| Rhamnose \| Positive (+ve) \| \| Xylose \| Positive (+ve) \| \| Enzymatic Reactions \| \| \| Lysine decarboxylase \| Positive (+ve) \| \| ONPG (β-galactosidase) \| Positive (+ve) \| Basic Characteristics Properties (Klebsiella granulomatis) Occur singly or in clusters Fermentation of Enzymatic Reactions References About Author Sagar Aryal, PhD Leave a Comment Cancel reply Comment Save my name, email, and website in this browser for the next time I comment. Topics / Categories
2045	https://en.wikipedia.org/wiki/Antibody	Published Time: Sat, 13 Sep 2025 18:53:03 GMT Antibody - 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 Donate Create account Log in [x] Personal tools Donate Create account Log in Pages for logged out editors learn more Contributions Talk Contents move to sidebar hide (Top) 1 StructureToggle Structure subsection 1.1 Antigen-binding site 1.2 Fc region 1.3 Protein structure 1.4 Antibody complexes 1.5 B cell receptors 2 ClassesToggle Classes subsection 2.1 Light chain types 2.2 In non-mammalian animals 3 Antibody–antigen interactions 4 FunctionToggle Function subsection 4.1 Activation of complement 4.2 Activation of effector cells 4.3 Natural antibodies 5 Immunoglobulin diversityToggle Immunoglobulin diversity subsection 5.1 Domain variability 5.2 V(D)J recombination 5.3 Somatic hypermutation and affinity maturation 5.4 Class switching 5.5 Specificity designations 5.6 Asymmetrical antibodies 5.7 Interchromosomal DNA Transposition 6 History 7 Medical applicationsToggle Medical applications subsection 7.1 Disease diagnosis 7.2 Disease therapy 7.3 Prenatal therapy 8 Research applications 9 RegulationsToggle Regulations subsection 9.1 Production and testing 9.2 Before clinical trials 9.3 Preclinical studies 10 Structure prediction and computational antibody design 11 Antibody mimetic 12 Binding antibody unit 13 See also 14 References 15 External links [x] Toggle the table of contents Antibody [x] 83 languages Alemannisch አማርኛ العربية অসমীয়া Azərbaycanca বাংলা Беларуская भोजपुरी Български Bosanski Català Чӑвашла Čeština Dansk Deutsch ދިވެހިބަސް Eesti Ελληνικά Español Esperanto Euskara فارسی Français Gaeilge Galego 한국어 Հայերեն हिन्दी Hrvatski Bahasa Indonesia Interlingua Íslenska Italiano עברית Jawa ಕನ್ನಡ ქართული Қазақша Kreyòl ayisyen Kurdî Кыргызча Latina Latviešu Lietuvių Magyar Македонски მარგალური Bahasa Melayu Nederlands 日本語 Norsk bokmål Norsk nynorsk Occitan ଓଡ଼ିଆ Oʻzbekcha / ўзбекча ਪੰਜਾਬੀ پښتو Plattdüütsch Polski Português Qaraqalpaqsha Română Русиньскый Русский Simple English Slovenčina Slovenščina کوردی Српски / srpski Srpskohrvatski / српскохрватски Suomi Svenska Tagalog தமிழ் ไทย Türkçe Українська اردو Tiếng Việt Winaray 吴语粵語中文 Edit 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 Print/export Download as PDF Printable version In other projects Wikimedia Commons Wikidata item Appearance move to sidebar hide From Wikipedia, the free encyclopedia Protein(s) forming a major part of an organism's immune system This article is about the class of proteins. For other uses, see Antibody (disambiguation). Each antibody binds to a specific antigen in a highly specific interaction analogous to a lock and key. An antibody (Ab), or immunoglobulin (Ig), is a large, Y-shaped protein belonging to the immunoglobulin superfamily which is used by the immune system to identify and neutralize antigens such as bacteria and viruses, including those that cause disease. Each individual antibody recognizes one or more specific antigens, and antigens of virtually any size and chemical composition can be recognized. Antigen literally means "antibody generator", as it is the presence of an antigen that drives the formation of an antigen-specific antibody. Each of the branching chains comprising the "Y" of an antibody contains a paratope that specifically binds to one particular epitope on an antigen, allowing the two molecules to bind together with precision. Using this mechanism, antibodies can effectively "tag" the antigen (or a microbe or an infected cell bearing such an antigen) for attack by cells of the immune system, or can neutralize it directly (for example, by blocking a part of a virus that is essential for its ability to invade a host cell). Antibodies may be borne on the surface of an immune cell, as in a B cell receptor, or they may exist freely by being secreted into the extracellular space. The term antibody often refers to the free (secreted) form, while the term immunoglobulin can refer to both forms. Since they are, broadly speaking, the same protein, the terms are often treated as synonymous. To allow the immune system to recognize millions of different antigens, the antigen-binding paratopes at each tip of the antibody come in an equally wide variety. The rest of an antibody's structure is much less variable; in humans, antibodies occur in five classes or isotypes: IgA, IgD, IgE, IgG, and IgM. Human IgG and IgA antibodies are also divided into discrete subclasses (IgG1, IgG2, IgG3, and IgG4; IgA1 and IgA2). The class refers to the functions triggered by the antibody (also known as effector functions), in addition to some other structural features. Antibodies from different classes also differ in where they are released in the body and at what stage of an immune response. Between species, while classes and subclasses of antibodies may be shared (at least in name), their function and distribution throughout the body may be different. For example, mouse IgG1 is closer to human IgG2 than to human IgG1 in terms of its function. The term humoral immunity is often treated as synonymous with the antibody response, describing the function of the immune system that exists in the body's humors (fluids) in the form of soluble proteins, as distinct from cell-mediated immunity, which generally describes the responses of T cells (especially cytotoxic T cells). In general, antibodies are considered part of the adaptive immune system, though this classification can become complicated. For example, natural IgM, which are made by B-1 lineage cells that have properties more similar to innate immune cells than adaptive, refers to IgM antibodies made independently of an immune response that demonstrate polyreactivity – i.e. they recognize multiple distinct (unrelated) antigens. These can work with the complement system in the earliest phases of an immune response to help facilitate clearance of the offending antigen and delivery of the resulting immune complexes to the lymph nodes or spleen for initiation of an immune response. Hence in this capacity, the functions of antibodies are more akin to that of innate immunity than adaptive. Nonetheless, in general, antibodies are regarded as part of the adaptive immune system because they demonstrate exceptional specificity (with some exceptions), are produced through genetic rearrangements (rather than being encoded directly in the germline), and are a manifestation of immunological memory. In the course of an immune response, B cells can progressively differentiate into antibody-secreting cells or into memory B cells. Antibody-secreting cells comprise plasmablasts and plasma cells, which differ mainly in the degree to which they secrete antibodies, their lifespan, metabolic adaptations, and surface markers. Plasmablasts are rapidly proliferating, short-lived cells produced in the early phases of the immune response (classically described as arising extrafollicularly rather than from a germinal center) which have the potential to differentiate further into plasma cells. Occasionally plasmablasts are mis-described as short-lived plasma cells; formally this is incorrect. Plasma cells, in contrast, do not divide (they are terminally differentiated), and rely on survival niches comprising specific cell types and cytokines to persist. Plasma cells will secrete huge quantities of antibody regardless of whether or not their cognate antigen is present, ensuring that antibody levels to the antigen in question do not fall to zero, provided the plasma cell stays alive. The rate of antibody secretion, however, can be regulated, for example, by the presence of adjuvant molecules that stimulate the immune response such as toll-like receptor ligands. Long-lived plasma cells can live for potentially the entire lifetime of the organism. Classically, the survival niches that house long-lived plasma cells reside in the bone marrow, though it cannot be assumed that any given plasma cell in the bone marrow will be long-lived. However, other work indicates that survival niches can readily be established within the mucosal tissues- though the classes of antibodies involved show a different hierarchy from those in the bone marrow. B cells can also differentiate into memory B cells which can persist for decades, similarly to long-lived plasma cells. These cells can be rapidly recalled in a secondary immune response, undergoing class switching, affinity maturation, and differentiating into antibody-secreting cells. Antibodies are central to the immune protection elicited by most vaccines and infections (although other components of the immune system certainly participate and for some diseases are considerably more important than antibodies in generating an immune response, e.g. in the case of herpes zoster). Durable protection from infections caused by a given microbe – that is, the ability of the microbe to enter the body and begin to replicate (not necessarily to cause disease) – depends on sustained production of large quantities of antibodies, meaning that effective vaccines ideally elicit persistent high levels of antibody, which relies on long-lived plasma cells. At the same time, many microbes of medical importance have the ability to mutate to escape antibodies elicited by prior infections, and long-lived plasma cells cannot undergo affinity maturation or class switching. This is compensated for through memory B cells: novel variants of a microbe that still retain structural features of previously encountered antigens can elicit memory B cell responses that adapt to those changes. It has been suggested that long-lived plasma cells secrete B cell receptors with higher affinity than those on the surfaces of memory B cells, but findings are not entirely consistent on this point. Structure [edit] Schematic structure of an antibody: two heavy chains (blue, yellow) and the two light chains (green, pink). The antigen binding site is circled. A more accurate depiction of an antibody (3D structure at RCSB PDB). Glycans in the Fc region are shown in black. Antibodies are heavy (~150 kDa) proteins of about 10 nm in size, arranged in three globular regions that roughly form a Y shape. In humans and most other mammals, an antibody unit consists of four polypeptide chains; two identical heavy chains and two identical light chains connected by disulfide bonds. Each chain is a series of domains: somewhat similar sequences of about 110 amino acids each. These domains are usually represented in simplified schematics as rectangles. Light chains consist of one variable domain V L and one constant domain C L, while heavy chains contain one variable domain V H and three to four constant domains C H 1, C H 2, ... Structurally an antibody is also partitioned into two antigen-binding fragments (Fab), containing one V L, V H, C L, and C H 1 domain each, as well as the crystallisable fragment (Fc), forming the trunk of the Y shape. In between them is a hinge region of the heavy chains, whose flexibility allows antibodies to bind to pairs of epitopes at various distances, to form complexes (dimers, trimers, etc.), and to bind effector molecules more easily. In an electrophoresis test of blood proteins, antibodies mostly migrate to the last, gamma globulin fraction. Conversely, most gamma-globulins are antibodies, which is why the two terms were historically used as synonyms, as were the symbols Ig and γ. This variant terminology fell out of use due to the correspondence being inexact and due to confusion with γ (gamma) heavy chains which characterize the IgG class of antibodies. Antigen-binding site [edit] The variable domains can also be referred to as the F V region. It is the subregion of Fab that binds to an antigen. More specifically, each variable domain contains three hypervariable regions – the amino acids seen there vary the most from antibody to antibody. When the protein folds, these regions give rise to three loops of β-strands, localized near one another on the surface of the antibody. These loops are referred to as the complementarity-determining regions (CDRs), since their shape complements that of an antigen. Three CDRs from each of the heavy and light chains together form an antibody-binding site whose shape can be anything from a pocket to which a smaller antigen binds, to a larger surface, to a protrusion that sticks out into a groove in an antigen. Typically though, only a few residues contribute to most of the binding energy. The existence of two identical antibody-binding sites allows antibody molecules to bind strongly to multivalent antigen (repeating sites such as polysaccharides in bacterial cell walls, or other sites at some distance apart), as well as to form antibody complexes and larger antigen-antibody complexes. The structures of CDRs have been clustered and classified by Chothia et al. and more recently by North et al. and Nikoloudis et al. However, describing an antibody's binding site using only one single static structure limits the understanding and characterization of the antibody's function and properties. To improve antibody structure prediction and to take the strongly correlated CDR loop and interface movements into account, antibody paratopes should be described as interconverting states in solution with varying probabilities. In the framework of the immune network theory, CDRs are also called idiotypes. According to immune network theory, the adaptive immune system is regulated by interactions between idiotypes. Fc region [edit] Main article: Fragment crystallizable region The Fc region (the trunk of the Y shape) is composed of constant domains from the heavy chains. Its role is in modulating immune cell activity: it is where effector molecules bind to, triggering various effects after the antibody Fab region binds to an antigen.Effector cells (such as macrophages or natural killer cells) bind via their Fc receptors (FcR) to the Fc region of an antibody, while the complement system is activated by binding the C1q protein complex. IgG or IgM can bind to C1q, but IgA cannot, therefore IgA does not activate the classical complement pathway. Another role of the Fc region is to selectively distribute different antibody classes across the body. In particular, the neonatal Fc receptor (FcRn) binds to the Fc region of IgG antibodies to transport it across the placenta, from the mother to the fetus. In addition to this, binding to FcRn endows IgG with an exceptionally long half-life relative to other plasma proteins of 3-4 weeks. IgG3 in most cases (depending on allotype) has mutations at the FcRn binding site which lower affinity for FcRn, which are thought to have evolved to limit the highly inflammatory effects of this subclass. Antibodies are glycoproteins, that is, they have carbohydrates (glycans) added to conserved amino acid residues. These conserved glycosylation sites occur in the Fc region and influence interactions with effector molecules. Protein structure [edit] The N-terminus of each chain is situated at the tip. Each immunoglobulin domain has a similar structure, characteristic of all the members of the immunoglobulin superfamily: it is composed of between 7 (for constant domains) and 9 (for variable domains) β-strands, forming two beta sheets in a Greek key motif. The sheets create a "sandwich" shape, the immunoglobulin fold, held together by a disulfide bond. Antibody complexes [edit] Some antibodies form complexes that bind to multiple antigen molecules. Secreted antibodies can occur as a single Y-shaped unit, a monomer. However, some antibody classes also form dimers with two Ig units (as with IgA), tetramers with four Ig units (like teleost fish IgM), or pentamers with five Ig units (like shark IgW or mammalian IgM, which occasionally forms hexamers as well, with six units). IgG can also form hexamers, though no J chain is required. IgA tetramers and pentamers have also been reported. Antibodies also form complexes by binding to antigen: this is called an antigen-antibody complex or immune complex. Small antigens can cross-link two antibodies, also leading to the formation of antibody dimers, trimers, tetramers, etc. Multivalent antigens (e.g., cells with multiple epitopes) can form larger complexes with antibodies. An extreme example is the clumping, or agglutination, of red blood cells with antibodies in blood typing to determine blood groups: the large clumps become insoluble, leading to visually apparent precipitation. B cell receptors [edit] Main article: B-cell receptor The membrane-bound form of an antibody may be called a surface immunoglobulin (sIg) or a membrane immunoglobulin (mIg). It is part of the B cell receptor (BCR), which allows a B cell to detect when a specific antigen is present in the body and triggers B cell activation. The BCR is composed of surface-bound IgD or IgM antibodies and associated Ig-α and Ig-β heterodimers, which are capable of signal transduction. A typical human B cell will have 50,000 to 100,000 antibodies bound to its surface. Upon antigen binding, they cluster in large patches, which can exceed 1 micrometer in diameter, on lipid rafts that isolate the BCRs from most other cell signaling receptors. These patches may improve the efficiency of the cellular immune response. In humans, the cell surface is bare around the B cell receptors for several hundred nanometers, which further isolates the BCRs from competing influences. Classes [edit] Antibodies can come in different varieties known as isotypes or classes. In humans there are five antibody classes known as IgA, IgD, IgE, IgG, and IgM, which are further subdivided into subclasses such as IgA1, IgA2. The prefix "Ig" stands for immunoglobulin, while the suffix denotes the type of heavy chain the antibody contains: the heavy chain types α (alpha), γ (gamma), δ (delta), ε (epsilon), μ (mu) give rise to IgA, IgG, IgD, IgE, IgM, respectively. The distinctive features of each class are determined by the part of the heavy chain within the hinge and Fc region. The classes differ in their biological properties, functional locations and ability to deal with different antigens, as depicted in the table. For example, IgE antibodies are responsible for an allergic response consisting of histamine release from mast cells, often a sole contributor to asthma (though other pathways exist as do symptoms very similar to yet not technically asthma). The variable region of these antibodies bind to allergic antigen, for example house dust mite particles, while its Fc region (in the ε heavy chains) binds to Fc receptor ε on a mast cell, triggering its degranulation: the release of molecules stored in its granules. Antibody isotypes of humans \| Class \| Subclasses \| Description \| --- \| IgA \| 2 \| Found in mucosal areas, such as the gut, respiratory tract and urogenital tract, and prevents colonization by pathogens. Also found in saliva, tears, and breast milk. Early clinical studies suggest that IgA isotype antibodies have potential as anti-cancer therapeutics, demonstrating the ability to reduce tumor growth. \| \| IgD \| 1 \| Functions mainly as an antigen receptor on B cells that have not been exposed to antigens. It has been shown to activate basophils and mast cells to produce antimicrobial factors. In addition, IgD has also been reported to induce the release of immunoactivity and pro-inflammatory mediators. \| \| IgE \| 1 \| IgE antibodies are the least abundant class of immunoglobulin. They can engage Fc receptors on monocytes and macrophages to activate various effector cell populations. Binds to allergens and triggers histamine release from mast cells and basophils, and is involved in allergy. Humans and other animals evolved IgE to protect against parasitic worms, though in the present, IgE is primarily related to allergies and asthma. \| \| IgG \| 4 \| In its four forms, provides the majority of antibody-based immunity against invading pathogens. The only antibody capable of crossing the placenta to give passive immunity to the fetus. IgG is the most commonly used molecular format in current antibody drugs because it neutralizes infectious agents and activates the complement system to engage immune cells. \| \| IgM \| 1 \| Expressed on the surface of B cells (monomer) and in a secreted form (pentamer) with very high avidity. Eliminates pathogens in the early stages of B cell-mediated (humoral) immunity before there is sufficient IgG. IgM also is a pro-inflammatory antibody that serves as the primary defense and effectively stimulates the complement system with specialized immune functions, including higher avidity and steric hindrance, allowing it to neutralize viruses. \| The antibody isotype of a B cell changes during cell development and activation. Immature B cells, which have never been exposed to an antigen, express only the IgM isotype in a cell surface bound form. The B lymphocyte, in this ready-to-respond form, is known as a "naive B lymphocyte." The naive B lymphocyte expresses both surface IgM and IgD. The co-expression of both of these immunoglobulin isotypes renders the B cell ready to respond to antigen. B cell activation follows engagement of the cell-bound antibody molecule with an antigen, causing the cell to divide and differentiate into an antibody-producing cell called a plasma cell. This requires cytokines from T helper cells, unless antigen cross-links B cell receptors. In this activated form, the B cell starts to produce antibody in a secreted form rather than a membrane-bound form. Activated B cells that encounter certain signaling molecules undergo immunoglobulin class switching, also known as isotope switching, which causes the production of antibodies to change from IgM or IgD to the other antibody isotypes, IgE, IgA, or IgG. Light chain types [edit] Further information: Immunoglobulin light chain In mammals there are two types of immunoglobulin light chain, which are called lambda (λ) and kappa (κ). However, there is no known functional difference between them, and both can occur with any of the five major types of heavy chains. Each antibody contains two identical light chains: both κ or both λ. Proportions of κ and λ types vary by species and can be used to detect abnormal proliferation of B cell clones. Other types of light chains, such as the iota (ι) chain, are found in other vertebrates like sharks (Chondrichthyes) and bony fishes (Teleostei). In non-mammalian animals [edit] In most placental mammals, the structure of antibodies is generally similar. Jawed fish appear to be the most primitive animals that are able to make antibodies similar to those of mammals, although many features of their adaptive immunity appeared somewhat earlier. Cartilaginous fish (such as sharks) produce heavy-chain-only antibodies (i.e., lacking light chains) which moreover feature longer chain pentamers (with five constant units per molecule). Camelids (such as camels, llamas, alpacas) are also notable for producing heavy-chain-only antibodies. Antibody classes not found in mammals \| Class \| Types \| Description \| --- \| IgY \| \| Found in birds and reptiles; related to mammalian IgG. \| \| IgW \| \| Found in sharks and skates; related to mammalian IgD. \| \| IgT/Z \| \| Found in teleost fish \| Antibody–antigen interactions [edit] The antibody's paratope interacts with the antigen's epitope. An antigen usually contains different epitopes along its surface arranged discontinuously, and dominant epitopes on a given antigen are called determinants.[citation needed] Antibody and antigen interact by spatial complementarity (lock and key). The molecular forces involved in the Fab-epitope interaction are weak and non-specific – for example electrostatic forces, hydrogen bonds, hydrophobic interactions, and van der Waals forces. This means binding between antibody and antigen is reversible, and the antibody's affinity towards an antigen is relative rather than absolute. Relatively weak binding also means it is possible for an antibody to cross-react with different antigens of different relative affinities.[citation needed] Function [edit] Further information: Immune system Antibodies (A) and pathogens (B) free roam in the blood. The antibodies bind to pathogens, and can do so in different formations such as: opsonization, neutralisation, and agglutination. A phagocyte (C) approaches the pathogen, and the Fc region (D) of the antibody binds to one of the Fc receptors (E) of the phagocyte. Phagocytosis occurs as the pathogen is ingested. The main categories of antibody action include the following:[citation needed] Neutralisation, in which neutralizing antibodies block parts of the surface of a bacterial cell or virion to render its attack ineffective Agglutination, in which antibodies "glue together" foreign cells into clumps that are attractive targets for phagocytosis Precipitation, in which antibodies "glue together" serum-soluble antigens, forcing them to precipitate out of solution in clumps that are attractive targets for phagocytosis Complement activation (fixation), in which antibodies that are latched onto a foreign cell encourage complement to attack it with a membrane attack complex, which leads to the following: Lysis of the foreign cell Encouragement of inflammation by chemotactically attracting inflammatory cells More indirectly, an antibody can signal immune cells to present antibody fragments to T cells, or downregulate other immune cells to avoid autoimmunity.[citation needed] Activated B cells differentiate into either antibody-producing cells called plasma cells that secrete soluble antibody or memory cells that survive in the body for years afterward in order to allow the immune system to remember an antigen and respond faster upon future exposures. At the prenatal and neonatal stages of life, the presence of antibodies is provided by passive immunization from the mother. Early endogenous antibody production varies for different kinds of antibodies, and usually appear within the first years of life. Since antibodies exist freely in the bloodstream, they are said to be part of the humoral immune system. Circulating antibodies are produced by clonal B cells that specifically respond to only one antigen (an example is a viruscapsid protein fragment). Antibodies contribute to immunity in three ways: They prevent pathogens from entering or damaging cells by binding to them; they stimulate removal of pathogens by macrophages and other cells by coating the pathogen; and they trigger destruction of pathogens by stimulating other immune responses such as the complement pathway. Antibodies will also trigger vasoactive amine degranulation to contribute to immunity against certain types of antigens (helminths, allergens). The secreted mammalian IgM has five Ig units. Each Ig unit (labeled 1) has two epitope binding Fab regions, so IgM is capable of binding up to 10 epitopes. Activation of complement [edit] Antibodies that bind to surface antigens (for example, on bacteria) will attract the first component of the complement cascade with their Fc region and initiate activation of the "classical" complement system. This results in the killing of bacteria in two ways. First, the binding of the antibody and complement molecules marks the microbe for ingestion by phagocytes in a process called opsonization; these phagocytes are attracted by certain complement molecules generated in the complement cascade. Second, some complement system components form a membrane attack complex to assist antibodies to kill the bacterium directly (bacteriolysis). Activation of effector cells [edit] To combat pathogens that replicate outside cells, antibodies bind to pathogens to link them together, causing them to agglutinate. Since an antibody has at least two paratopes, it can bind more than one antigen by binding identical epitopes carried on the surfaces of these antigens. By coating the pathogen, antibodies stimulate effector functions against the pathogen in cells that recognize their Fc region. Those cells that recognize coated pathogens have Fc receptors, which, as the name suggests, interact with the Fc region of IgA, IgG, and IgE antibodies. The engagement of a particular antibody with the Fc receptor on a particular cell triggers an effector function of that cell; phagocytes will phagocytose, mast cells and neutrophils will degranulate, natural killer cells will release cytokines and cytotoxic molecules; that will ultimately result in destruction of the invading microbe. The activation of natural killer cells by antibodies initiates a cytotoxic mechanism known as antibody-dependent cell-mediated cytotoxicity (ADCC) – this process may explain the efficacy of monoclonal antibodies used in biological therapies against cancer. The Fc receptors are isotype-specific, which gives greater flexibility to the immune system, invoking only the appropriate immune mechanisms for distinct pathogens. Natural antibodies [edit] Humans and higher primates also produce "natural antibodies" that are present in serum before viral infection. Natural antibodies have been defined as antibodies that are produced without any previous infection, vaccination, other foreign antigen exposure or passive immunization. These antibodies can activate the classical complement pathway leading to lysis of enveloped virus particles long before the adaptive immune response is activated. Antibodies are produced exclusively by B cells in response to antigens where initially, antibodies are formed as membrane-bound receptors, but upon activation by antigens and helper T cells, B cells differentiate to produce soluble antibodies. Many natural antibodies are directed against the disaccharide galactose α(1,3)-galactose (α-Gal), which is found as a terminal sugar on glycosylated cell surface proteins, and generated in response to production of this sugar by bacteria contained in the human gut. These antibodies undergo quality checks in the endoplasmic reticulum (ER), which contains proteins that assist in proper folding and assembly. Rejection of xenotransplantated organs is thought to be, in part, the result of natural antibodies circulating in the serum of the recipient binding to α-Gal antigens expressed on the donor tissue. Immunoglobulin diversity [edit] Virtually all microbes can trigger an antibody response. Successful recognition and eradication of many different types of microbes requires diversity among antibodies; their amino acid composition varies allowing them to interact with many different antigens. It has been estimated that humans generate about 10 billion different antibodies, each capable of binding a distinct epitope of an antigen. Although a huge repertoire of different antibodies is generated in a single individual, the number of genes available to make these proteins is limited by the size of the human genome. Several complex genetic mechanisms have evolved that allow vertebrate B cells to generate a diverse pool of antibodies from a relatively small number of antibody genes. Domain variability [edit] The complementarity determining regions of the heavy chain are shown in red (PDB: 1IGT) The chromosomal region that encodes an antibody is large and contains several distinct gene loci for each domain of the antibody—the chromosome region containing heavy chain genes (IGH@) is found on chromosome 14, and the loci containing lambda and kappa light chain genes (IGL@ and IGK@) are found on chromosomes 22 and 2 in humans. One of these domains is called the variable domain, which is present in each heavy and light chain of every antibody, but can differ in different antibodies generated from distinct B cells. Differences between the variable domains are located on three loops known as hypervariable regions (HV-1, HV-2 and HV-3) or complementarity-determining regions (CDR1, CDR2 and CDR3). CDRs are supported within the variable domains by conserved framework regions. The heavy chain locus contains about 65 different variable domain genes that all differ in their CDRs. Combining these genes with an array of genes for other domains of the antibody generates a large cavalry of antibodies with a high degree of variability. This combination is called V(D)J recombination and discussed below. V(D)J recombination [edit] Further information: V(D)J recombination Simplified overview of V(D)J recombination of immunoglobulin heavy chains Somatic recombination of immunoglobulins, also known as V(D)J recombination, involves the generation of a unique immunoglobulin variable region. The variable region of each immunoglobulin heavy or light chain is encoded in several pieces—known as gene segments (subgenes). These segments are called variable (V), diversity (D) and joining (J) segments. V, D and J segments are found in Ig heavy chains, but only V and J segments are found in Ig light chains. Multiple copies of the V, D and J gene segments exist, and are tandemly arranged in the genomes of mammals. In the bone marrow, each developing B cell will assemble an immunoglobulin variable region by randomly selecting and combining one V, one D and one J gene segment (or one V and one J segment in the light chain). As there are multiple copies of each type of gene segment, and different combinations of gene segments can be used to generate each immunoglobulin variable region, this process generates a huge number of antibodies, each with different paratopes, and thus different antigen specificities. The rearrangement of several subgenes (i.e. V2 family) for lambda light chain immunoglobulin is coupled with the activation of microRNA miR-650, which further influences biology of B-cells.[citation needed] RAG proteins play an important role with V(D)J recombination in cutting DNA at a particular region. Without the presence of these proteins, V(D)J recombination would not occur. After a B cell produces a functional immunoglobulin gene during V(D)J recombination, it cannot express any other variable region (a process known as allelic exclusion) thus each B cell can produce antibodies containing only one kind of variable chain. Somatic hypermutation and affinity maturation [edit] Further information: Somatic hypermutation and Affinity maturation Following activation with antigen, B cells begin to proliferate rapidly. In these rapidly dividing cells, the genes encoding the variable domains of the heavy and light chains undergo a high rate of point mutation, by a process called somatic hypermutation (SHM). SHM results in approximately one nucleotide change per variable gene, per cell division. As a consequence, any daughter B cells will acquire slight amino acid differences in the variable domains of their antibody chains.[citation needed] This serves to increase the diversity of the antibody pool and impacts the antibody's antigen-binding affinity. Some point mutations will result in the production of antibodies that have a weaker interaction (low affinity) with their antigen than the original antibody, and some mutations will generate antibodies with a stronger interaction (high affinity). B cells that express high affinity antibodies on their surface will receive a strong survival signal during interactions with other cells, whereas those with low affinity antibodies will not, and will die by apoptosis. Thus, B cells expressing antibodies with a higher affinity for the antigen will outcompete those with weaker affinities for function and survival allowing the average affinity of antibodies to increase over time. The process of generating antibodies with increased binding affinities is called affinity maturation. Affinity maturation occurs in mature B cells after V(D)J recombination, and is dependent on help from helper T cells. Class switching [edit] Mechanism of class switch recombination that allows isotype switching in activated B cells Isotype or class switching is a biological process occurring after activation of the B cell, which allows the cell to produce different classes of antibody (IgA, IgE, or IgG). The different classes of antibody, and thus effector functions, are defined by the constant (C) regions of the immunoglobulin heavy chain. Initially, naive B cells express only cell-surface IgM and IgD with identical antigen binding regions. Each isotype is adapted for a distinct function; therefore, after activation, an antibody with an IgG, IgA, or IgE effector function might be required to effectively eliminate an antigen. Class switching allows different daughter cells from the same activated B cell to produce antibodies of different isotypes. Only the constant region of the antibody heavy chain changes during class switching; the variable regions, and therefore antigen specificity, remain unchanged. Thus the progeny of a single B cell can produce antibodies, all specific for the same antigen, but with the ability to produce the effector function appropriate for each antigenic challenge. Class switching is triggered by cytokines; the isotype generated depends on which cytokines are present in the B cell environment. Class switching occurs in the heavy chain gene locus by a mechanism called class switch recombination (CSR). This mechanism relies on conserved nucleotide motifs, called switch (S) regions, found in DNA upstream of each constant region gene (except in the δ-chain). The DNA strand is broken by the activity of a series of enzymes at two selected S-regions. The variable domain exon is rejoined through a process called non-homologous end joining (NHEJ) to the desired constant region (γ, α or ε). This process results in an immunoglobulin gene that encodes an antibody of a different isotype. Specificity designations [edit] An antibody can be called monospecific if it has specificity for a single antigen or epitope, or bispecific if it has affinity for two different antigens or two different epitopes on the same antigen. A group of antibodies can be called polyvalent (or unspecific) if they have affinity for various antigens or microorganisms.Intravenous immunoglobulin, if not otherwise noted, consists of a variety of different IgG (polyclonal IgG). In contrast, monoclonal antibodies are identical antibodies produced by a single B cell. Asymmetrical antibodies [edit] Heterodimeric antibodies, which are also asymmetrical antibodies, allow for greater flexibility and new formats for attaching a variety of drugs to the antibody arms. One of the general formats for a heterodimeric antibody is the "knobs-into-holes" format. This format is specific to the heavy chain part of the constant region in antibodies. The "knobs" part is engineered by replacing a small amino acid with a larger one. It fits into the "hole", which is engineered by replacing a large amino acid with a smaller one. What connects the "knobs" to the "holes" are the disulfide bonds between each chain. The "knobs-into-holes" shape facilitates antibody dependent cell mediated cytotoxicity. Single-chain variable fragments (scFv) are connected to the variable domain of the heavy and light chain via a short linker peptide. The linker is rich in glycine, which gives it more flexibility, and serine/threonine, which gives it specificity. Two different scFv fragments can be connected together, via a hinge region, to the constant domain of the heavy chain or the constant domain of the light chain. This gives the antibody bispecificity, allowing for the binding specificities of two different antigens. The "knobs-into-holes" format enhances heterodimer formation but does not suppress homodimer formation.[citation needed] To further improve the function of heterodimeric antibodies, many scientists are looking towards artificial constructs. Artificial antibodies are largely diverse protein motifs that use the functional strategy of the antibody molecule, but are not limited by the loop and framework structural constraints of the natural antibody. Being able to control the combinational design of the sequence and three-dimensional space could transcend the natural design and allow for the attachment of different combinations of drugs to the arms.[citation needed] Heterodimeric antibodies have a greater range in shapes they can take and the drugs that are attached to the arms do not have to be the same on each arm, allowing for different combinations of drugs to be used in cancer treatment. Pharmaceuticals are able to produce highly functional bispecific, and even multispecific, antibodies. The degree to which they can function is impressive given that such a change of shape from the natural form should lead to decreased functionality.[citation needed] Interchromosomal DNA Transposition [edit] Antibody diversification typically occurs through somatic hypermutation, class switching, and affinity maturation targeting the BCR gene loci, but on occasion more unconventional forms of diversification have been documented. For example, in the case of malaria caused by Plasmodium falciparum, some antibodies from those who had been infected demonstrated an insertion from chromosome 19 containing a 98-amino acid stretch from leukocyte-associated immunoglobulin-like receptor 1, LAIR1, in the elbow joint. This represents a form of interchromosomal transposition. LAIR1 normally binds collagen, but can recognize repetitive interspersed families of polypeptides (RIFIN) family members that are highly expressed on the surface of P. falciparum-infected red blood cells. In fact, these antibodies underwent affinity maturation that enhanced affinity for RIFIN but abolished affinity for collagen. These "LAIR1-containing" antibodies have been found in 5-10% of donors from Tanzania and Mali, though not in European donors. European donors did show 100-1000 nucleotide stretches inside the elbow joints as well, however. This particular phenomenon may be specific to malaria, as infection is known to induce genomic instability. History [edit] See also: History of immunology The first use of the term "antibody" occurred in a text by Paul Ehrlich. The term Antikörper (the German word for antibody) appears in the conclusion of his article "Experimental Studies on Immunity", published in October 1891, which states that, "if two substances give rise to two different Antikörper, then they themselves must be different". However, the term was not accepted immediately and several other terms for antibody were proposed; these included Immunkörper, Amboceptor, Zwischenkörper, substance sensibilisatrice, copula, Desmon, philocytase, fixateur, and Immunisin. The word antibody has formal analogy to the word antitoxin and a similar concept to Immunkörper (immune body in English). As such, the original construction of the word contains a logical flaw; the antitoxin is something directed against a toxin, while the antibody is a body directed against something. Angel of the West (2008) by Julian Voss-Andreae is a sculpture based on the antibody structure published by E. Padlan. Created for the Florida campus of the Scripps Research Institute, the antibody is placed into a ring referencing Leonardo da Vinci'sVitruvian Man thus highlighting the similarity of the antibody and the human body. The study of antibodies began in 1890 when Emil von Behring and Kitasato Shibasaburō described antibody activity against diphtheria and tetanus toxins. Von Behring and Kitasato put forward the theory of humoral immunity, proposing that a mediator in serum could react with a foreign antigen. His idea prompted Paul Ehrlich to propose the side-chain theory for antibody and antigen interaction in 1897, when he hypothesized that receptors (described as "side-chains") on the surface of cells could bind specifically to toxins– in a "lock-and-key" interaction– and that this binding reaction is the trigger for the production of antibodies. Other researchers believed that antibodies existed freely in the blood and, in 1904, Almroth Wright suggested that soluble antibodies coated bacteria to label them for phagocytosis and killing; a process that he named opsoninization. Michael Heidelberger In the 1920s, Michael Heidelberger and Oswald Avery observed that antigens could be precipitated by antibodies and went on to show that antibodies are made of protein. The biochemical properties of antigen-antibody-binding interactions were examined in more detail in the late 1930s by John Marrack. The next major advance was in the 1940s, when Linus Pauling confirmed the lock-and-key theory proposed by Ehrlich by showing that the interactions between antibodies and antigens depend more on their shape than their chemical composition. In 1948, Astrid Fagraeus discovered that B cells, in the form of plasma cells, were responsible for generating antibodies. Further work concentrated on characterizing the structures of the antibody proteins. A major advance in these structural studies was the discovery in the early 1960s by Gerald Edelman and Joseph Gally of the antibody light chain, and their realization that this protein is the same as the Bence-Jones protein described in 1845 by Henry Bence Jones. Edelman went on to discover that antibodies are composed of disulfide bond-linked heavy and light chains. Around the same time, antibody-binding (Fab) and antibody tail (Fc) regions of IgG were characterized by Rodney Porter. Together, these scientists deduced the structure and complete amino acid sequence of IgG, a feat for which they were jointly awarded the 1972 Nobel Prize in Physiology or Medicine. The Fv fragment was prepared and characterized by David Givol. While most of these early studies focused on IgM and IgG, other immunoglobulin isotypes were identified in the 1960s: Thomas Tomasi discovered secretory antibody (IgA); David S. Rowe and John L. Fahey discovered IgD; and Kimishige Ishizaka and Teruko Ishizaka discovered IgE and showed it was a class of antibodies involved in allergic reactions. In a landmark series of experiments beginning in 1976, Susumu Tonegawa showed that genetic material can rearrange itself to form the vast array of available antibodies. Medical applications [edit] Disease diagnosis [edit] Detection of particular antibodies is a very common form of medical diagnostics, and applications such as serology depend on these methods. For example, in biochemical assays for disease diagnosis, a titer of antibodies directed against Epstein–Barr virus or Lyme disease is estimated from the blood. If those antibodies are not present, either the person is not infected or the infection occurred a very long time ago, and the B cells generating these specific antibodies have naturally decayed.[citation needed] In clinical immunology, levels of individual classes of immunoglobulins are measured by nephelometry (or turbidimetry) to characterize the antibody profile of patient. Elevations in different classes of immunoglobulins are sometimes useful in determining the cause of liver damage in patients for whom the diagnosis is unclear. For example, IgM levels are often elevated in patients with primary biliary cirrhosis, whereas IgA deposition along hepatic sinusoids can suggest alcoholic liver disease. Autoimmune disorders can often be traced to antibodies that bind the body's own epitopes; many can be detected through blood tests. Antibodies directed against red blood cell surface antigens in immune mediated hemolytic anemia are detected with the Coombs test. The Coombs test is also used for antibody screening in blood transfusion preparation and also for antibody screening in antenatal women. Practically, several immunodiagnostic methods based on detection of complex antigen-antibody are used to diagnose infectious diseases, for example ELISA, immunofluorescence, Western blot, immunodiffusion, immunoelectrophoresis, and magnetic immunoassay. Over-the-counter home pregnancy tests rely on human chorionic gonadotropin (hCG)-directed antibodies. New dioxaborolane chemistry enables radioactive fluoride (18 F) labeling of antibodies, which allows for positron emission tomography (PET) imaging of cancer. Disease therapy [edit] Targeted monoclonal antibody therapy is employed to treat diseases such as rheumatoid arthritis,multiple sclerosis,psoriasis, and many forms of cancer including non-Hodgkin's lymphoma,colorectal cancer, head and neck cancer and breast cancer. Some immune deficiencies, such as X-linked agammaglobulinemia and hypogammaglobulinemia, result in partial or complete lack of antibodies. These diseases are often treated by inducing a short-term form of immunity called passive immunity. Passive immunity is achieved through the transfer of ready-made antibodies in the form of human or animal serum, pooled immunoglobulin or monoclonal antibodies, into the affected individual. Prenatal therapy [edit] Rh factor, also known as Rh D antigen, is an antigen found on red blood cells; individuals that are Rh-positive (Rh+) have this antigen on their red blood cells and individuals that are Rh-negative (Rh–) do not. During normal childbirth, delivery trauma or complications during pregnancy, blood from a fetus can enter the mother's system. In the case of an Rh-incompatible mother and child, consequential blood mixing may sensitize an Rh- mother to the Rh antigen on the blood cells of the Rh+ child, putting the remainder of the pregnancy, and any subsequent pregnancies, at risk for hemolytic disease of the newborn. Rho(D) immune globulin antibodies are specific for human RhD antigen. Anti-RhD antibodies are administered as part of a prenatal treatment regimen to prevent sensitization that may occur when a Rh-negative mother has a Rh-positive fetus. Treatment of a mother with Anti-RhD antibodies prior to and immediately after trauma and delivery destroys Rh antigen in the mother's system from the fetus. This occurs before the antigen can stimulate maternal B cells to "remember" Rh antigen by generating memory B cells. Therefore, her humoral immune system will not make anti-Rh antibodies, and will not attack the Rh antigens of the current or subsequent babies. Rho(D) Immune Globulin treatment prevents sensitization that can lead to Rh disease, but does not prevent or treat the underlying disease itself. Research applications [edit] Immunofluorescence image of the eukaryoticcytoskeleton. Microtubules as shown in green, are marked by an antibody conjugated to a green fluorescing molecule, FITC. Specific antibodies are produced by injecting an antigen into a mammal, such as a mouse, rat, rabbit, goat, sheep, or horse for large quantities of antibody. Blood isolated from these animals contains polyclonal antibodies—multiple antibodies that bind to the same antigen—in the serum, which can now be called antiserum. Antigens are also injected into chickens for generation of polyclonal antibodies in egg yolk. To obtain antibody that is specific for a single epitope of an antigen, antibody-secreting lymphocytes are isolated from the animal and immortalized by fusing them with a cancer cell line. The fused cells are called hybridomas, and will continually grow and secrete antibody in culture. Single hybridoma cells are isolated by dilution cloning to generate cell clones that all produce the same antibody; these antibodies are called monoclonal antibodies. Polyclonal and monoclonal antibodies are often purified using Protein A/G or antigen-affinity chromatography. In research, purified antibodies are used in many applications. Antibodies for research applications can be found directly from antibody suppliers, or through use of a specialist search engine. Research antibodies are most commonly used to identify and locate intracellular and extracellular proteins. Antibodies are used in flow cytometry to differentiate cell types by the proteins they express; different types of cells express different combinations of cluster of differentiation molecules on their surface, and produce different intracellular and secretable proteins. They are also used in immunoprecipitation to separate proteins and anything bound to them (co-immunoprecipitation) from other molecules in a cell lysate, in Western blot analyses to identify proteins separated by electrophoresis, and in immunohistochemistry or immunofluorescence to examine protein expression in tissue sections or to locate proteins within cells with the assistance of a microscope. Proteins can also be detected and quantified with antibodies, using ELISA and ELISpot techniques. Antibodies used in research are some of the most powerful, yet most problematic reagents with a tremendous number of factors that must be controlled in any experiment including cross reactivity, or the antibody recognizing multiple epitopes and affinity, which can vary widely depending on experimental conditions such as pH, solvent, state of tissue etc. Multiple attempts have been made to improve both the way that researchers validate antibodies and ways in which they report on antibodies. Researchers using antibodies in their work need to record them correctly in order to allow their research to be reproducible (and therefore tested, and qualified by other researchers). Less than half of research antibodies referenced in academic papers can be easily identified. Papers published in F1000 in 2014 and 2015 provide researchers with a guide for reporting research antibody use. The RRID paper, is co-published in 4 journals that implemented the RRIDs Standard for research resource citation, which draws data from the antibodyregistry.org as the source of antibody identifiers (see also group at Force11). Antibody regions can be used to further biomedical research by acting as a guide for drugs to reach their target. Several application involve using bacterial plasmids to tag plasmids with the Fc region of the antibody such as pFUSE-Fc plasmid.[citation needed] Regulations [edit] Production and testing [edit] There are several ways to obtain antibodies, including in vivo techniques like animal immunization and various in vitro approaches, such as the phage display method. Traditionally, most antibodies are produced by hybridoma cell lines through immortalization of antibody-producing cells by chemically induced fusion with myeloma cells. In some cases, additional fusions with other lines have created "triomas" and "quadromas". The manufacturing process should be appropriately described and validated. Validation studies should at least include: The demonstration that the process is able to produce in good quality (the process should be validated) The efficiency of the antibody purification (all impurities and virus must be eliminated) The characterization of purified antibody (physicochemical characterization, immunological properties, biological activities, contaminants, ...) Determination of the virus clearance studies Before clinical trials [edit] Product safety testing: Sterility (bacteria and fungi), in vitro and in vivo testing for adventitious viruses, murineretrovirus testing..., product safety data needed before the initiation of feasibility trials in serious or immediately life-threatening conditions, it serves to evaluate dangerous potential of the product. Feasibility testing: These are pilot studies whose objectives include, among others, early characterization of safety and initial proof of concept in a small specific patient population (in vitro or in vivo testing).[citation needed] Preclinical studies [edit] Testing cross-reactivity of antibody: to highlight unwanted interactions (toxicity) of antibodies with previously characterized tissues. This study can be performed in vitro (reactivity of the antibody or immunoconjugate should be determined with a quick-frozen adult tissues) or in vivo (with appropriates animal models).[citation needed] Preclinicalpharmacology and toxicity testing: preclinical safety testing of antibody is designed to identify possible toxicity in humans, to estimate the likelihood and severity of potential adverse events in humans, and to identify a safe starting dose and dose escalation, when possible. Animal toxicity studies: Acute toxicity testing, repeat-dose toxicity testing, long-term toxicity testing Pharmacokinetics and pharmacodynamics testing: Use for determinate clinical dosages, antibody activities, evaluation of the potential clinical effects Structure prediction and computational antibody design [edit] The importance of antibodies in health care and the biotechnology industry demands knowledge of their structures at high resolution. This information is used for protein engineering, modifying the antigen binding affinity, and identifying an epitope, of a given antibody. X-ray crystallography is one commonly used method for determining antibody structures. However, crystallizing an antibody is often laborious and time-consuming. Computational approaches provide a cheaper and faster alternative to crystallography, but their results are more equivocal, since they do not produce empirical structures. Online web servers such as Web Antibody Modeling (WAM) and Prediction of Immunoglobulin Structure (PIGS) enable computational modeling of antibody variable regions. Rosetta Antibody is a novel antibody F V region structure prediction server, which incorporates sophisticated techniques to minimize CDR loops and optimize the relative orientation of the light and heavy chains, as well as homology models that predict successful docking of antibodies with their unique antigen. However, describing an antibody's binding site using only one single static structure limits the understanding and characterization of the antibody's function and properties. To improve antibody structure prediction and to take the strongly correlated CDR loop and interface movements into account, antibody paratopes should be described as interconverting states in solution with varying probabilities. The ability to describe the antibody through binding affinity to the antigen is supplemented by information on antibody structure and amino acid sequences for the purpose of patent claims. Several methods have been presented for computational design of antibodies based on the structural bioinformatics studies of antibody CDRs. There are a variety of methods used to sequence an antibody including Edman degradation, cDNA, etc.; albeit one of the most common modern uses for peptide/protein identification is liquid chromatography coupled with tandem mass spectrometry (LC-MS/MS). High volume antibody sequencing methods require computational approaches for the data analysis, including de novo sequencing directly from tandem mass spectra and database search methods that use existing protein sequence databases. Many versions of shotgun protein sequencing are able to increase the coverage by utilizing CID/HCD/ETD fragmentation methods and other techniques, and they have achieved substantial progress in attempt to fully sequence proteins, especially antibodies. Other methods have assumed the existence of similar proteins, a known genome sequence, or combined top-down and bottom up approaches. Current technologies have the ability to assemble protein sequences with high accuracy by integrating de novo sequencingpeptides, intensity, and positional confidence scores from database and homology searches. Antibody mimetic [edit] Antibody mimetics are organic compounds, like antibodies, that can specifically bind antigens. They consist of artificial peptides or proteins, or aptamer-based nucleic acid molecules with a molar mass of about 3 to 20 kDa. Antibody fragments, such as Fab and nanobodies are not considered as antibody mimetics. Common advantages over antibodies are better solubility, tissue penetration, stability towards heat and enzymes, and comparatively low production costs. Antibody mimetics have been developed and commercialized as research, diagnostic and therapeutic agents. Binding antibody unit [edit] BAU (binding antibody unit, often as BAU/mL) is a measurement unit defined by the WHO for the comparison of assays detecting the same class of immunoglobulins with the same specificity. See also [edit] Affimer Anti-mitochondrial antibodies Anti-nuclear antibodies Antibody mimetic Aptamer Colostrum ELISA Humoral immunity Immunology Immunosuppressive drug Intravenous immunoglobulin (IVIg) Magnetic immunoassay Microantibody Monoclonal antibody Neutralizing antibody Optimer Ligand Secondary antibodies Single-domain antibody Slope spectroscopy Surrobody Synthetic antibody Western blot normalization References [edit] ^ abcdefghiJaneway C (2001). Immunobiology (5th ed.). Garland Publishing. 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Mike's Immunoglobulin Structure/Function Page at University of Cambridge Antibodies as the PDB molecule of the month Discussion of the structure of antibodies at RCSB Protein Data Bank A hundred years of antibody therapy History and applications of antibodies in the treatment of disease at University of Oxford How Lymphocytes Produce Antibody from Cells Alive! \| v t e Lymphocyticadaptive immune system and complement \| \| Lymphoid \| \| Antigens \| Antigen Superantigen Allergen Antigenic variation Hapten Epitope Linear Conformational Mimotope Antigen presentation/professional APCs: Dendritic cell Macrophage B cell Immunogen \| \| Antibodies \| Antibody Monoclonal antibodies Polyclonal antibodies Autoantibody Microantibody Polyclonal B cell response Allotype Isotype Idiotype Immune complex Paratope \| \| Immunity vs. tolerance \| Action: Immunity Autoimmunity Alloimmunity Allergy Hypersensitivity Inflammation Cross-reactivity Co-stimulation Inaction: Tolerance Central Peripheral Clonal anergy Clonal deletion T-cell depletion Tolerance in pregnancy Immunodeficiency Immune privilege \| \| Immunogenetics \| Affinity maturation Somatic hypermutation Clonal selection V(D)J recombination Junctional diversity Immunoglobulin class switching MHC/HLA \| \| \| Lymphocytes \| Cellular T cell Humoral B cell NK cell \| \| Substances \| Cytokines Opsonin Cytolysin \| \| v t e Globular proteins \| \| Serumglobulins \| \| Alpha globulins \| \| serpins: \| alpha-1 (Alpha 1-antichymotrypsin, Alpha 1-antitrypsin) alpha-2 (Alpha 2-antiplasmin) Antithrombin (Heparin cofactor II) \| \| carrier proteins: \| alpha-1 (Transcortin) alpha-2 (Ceruloplasmin) Retinol binding protein \| \| other: \| alpha-1 (Orosomucoid) alpha-2 (alpha-2-Macroglobulin, Haptoglobin) \| \| \| Beta globulins \| \| carrier proteins: \| Sex hormone-binding globulin Transferrin \| \| other: \| Angiostatin Hemopexin Beta-2 microglobulin Factor H Plasminogen Properdin \| \| \| Gamma globulin \| Immunoglobulins \| \| Other \| Fibronectin (fFN: Fetal fibronectin) Macroglobulin/Microglobulin Transcobalamin Edestin \| \| \| Other globulins \| Beta-lactoglobulin Lactoferrin Thyroglobulin Alpha-lactalbumin 11S globulin family 7S seed storage protein \| \| Albumins \| \| Egg white \| Conalbumin Ovalbumin Avidin \| \| Serum albumin \| Human serum albumin Bovine serum albumin Prealbumin \| \| Other \| C-reactive protein Lactalbumin (Alpha-lactalbumin) Parvalbumin Ricin Vitamin D-binding protein Extracellular matrix protein 1 \| \| \| see alsodisorders of globin and globulin proteins \| \| v t e Antibodies \| \| Main types \| IgA IgD IgE IgG IgM IgY \| \| Chains \| Light chain IGK@ IGL@ surrogate IGLL1 Heavy chain IGH@ IGHM Heavy-chain antibody/IgNAR \| \| v t e Autoantibodies \| \| Anti-nuclear antibody \| PBC: Anti-gp210 Anti-p62 Anti-sp100 ENA: Anti-topoisomerase/Scl-70 Anti-Jo1 ENA4 Anti-Sm Anti-nRNP Anti-Ro Anti-La Anti-centromere Anti-dsDNA Anti-histone \| \| Anti-mitochondrial antibody \| Anti-cardiolipin \| \| Anti-cytoplasm antibody \| Anti-neutrophil cytoplasmic C-ANCA P-ANCA Anti-smooth muscle Anti-actin Anti-TPO/Antimicrosomal \| \| Cell membrane \| Anti-ganglioside Anti-GBM Anti-glutamate \| \| Extracellular \| Anti-thrombin Lupus anticoagulant Coeliac disease: Anti-transglutaminase Anti-gliadin not autoantibody RA Rheumatoid factor Anti-citrullinated peptide \| \| Multiple locations \| Anti-phospholipid Anti-apolipoprotein \| \| Authority control databases \| \| National \| Germany France BnF data Czech Republic \| \| Other \| Encyclopedia of Modern Ukraine Yale LUX \| Retrieved from " Categories: Antibodies Glycoproteins Immunology Reagents for biochemistry Hidden categories: CS1 maint: article number as page number CS1 maint: location missing publisher CS1 German-language sources (de) Webarchive template wayback links Articles with short description Short description is different from Wikidata Use dmy dates from March 2020 All articles with unsourced statements Articles with unsourced statements from July 2024 Commons category link is on Wikidata This page was last edited on 4 September 2025, at 20:14(UTC). 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2046	https://emedicine.medscape.com/article/331864-treatment	Scleroderma Treatment & Management Updated: Mar 01, 2023 Author: Abhishek Nandan, MD; Chief Editor: Herbert S Diamond, MD more...;) Share Print Feedback Approach Considerations Current treatment of systemic sclerosis is directed toward managing complications and providing symptomatic relief. In addition, a range of disease-modifying treatments have been investigated. Disease-modifying treatment aims at inhibiting tissue fibrosis and vascular and immune system alterations, which are the three crucial components of disease pathogenesis. [131, 132, 133, 134, 135] Therapies targeting cytokine signalling, including interleukin-6 (IL-6), and other immune-inflammatory therapies, have produced promising results. To date, the US Food and Drug Administration (FDA) has approved nintedanib and as well as tocilizumab for refractory, progressive intersitial lung disease due to systemic sclerosis. A phase 3 trial of tocilizumab in systemic sclerosis failed to meet its primary endpoint of improvement in modified Rodnan skin score; however, tocilizumab therapy did result in preservation of lung function. After 48 weeks, decline in lung function from baseline, as defined by changes in predicted forced vital capacity (FVC), had occurred 50.5% of patients in the tocilizumab arm, versus 70.3% of those in the placebo arm (P = 0.015). In patients with systemic sclerosis–related interstitial lung disease, a clinically meaningful decline of 10% or more in lung function was seen in 24.5% of placebo recipients, compared with 8.6% of tocilizumab recipients. Supporting the FVC results, high-resolution computed tomography showed less progression of lung fibrosis with tocilizumab than with placebo. [137, 138, 139] No placebo-controlled studies have demonstrated clear superiority for any drug except for a modest benefit from use of methotrexate [140, 141] and improvement in scleroderma-related interstitial lung disease with cyclophosphamide. [142, 143] Numerous uncontrolled prospective and retrospective trials along with post-hoc analysis have suggested a beneficial effect from mycophenolate mofetil. [144, 145, 146] Retrospective uncontrolled studies also supported a beneficial role for D-penicillamine, but a large high-dose versus low-dose controlled trial failed to demonstrate benefits of the higher dose versus the lower dose. Other agents are currently being studied for skin and lung involvement. For example, trials of rituximab have yielded promising results, with improvement of skin fibrosis and prevention of worsening lung fibrosis. [149, 150, 151, 152] The tyrosine kinase inhibitor nintedanib is approved for slowing the rate of decline in pulmonary function in adults with systemic sclerosis–related interstitial lung disease. Phototherapy using longer-wavelength ultraviolet A (UVA) light (ie, UVA1, 340-400 nm) has proved beneficial for cutaneous lesions in scleroderma. UVA1 inhibits the inflammatory process and can soften former sclerotic skin lesions. However, data on this technique remain limited, and the most effective dose has yet to be determined. No serious systemic effects of UVA1 phototherapy have been reported. Skin pigmentation or tanning, which can persist for months, is the most common acute adverse effect; uncommon acute adverse effects include reactivation of herpes simplex, cholinergic urticaria, and transient and reversible changes in the appearance of moles. The risk of long-term adverse effects, particularly skin cancer, has not been determined. Transplantation Hematopoietic stem cell transplantation (HSCT) has been shown to be effective. However, it is associated with a high rate of procedure-related complications. [154, 155, 156, 157] In the prospective Autologous Stem Cell Transplantation International Scleroderma (ASTIS) trial, a phase 3 comparison of autologous HSCT with 12 successive monthly intravenous pulses of cyclophosphamide in 156 patients with early diffuse cutaneous systemic sclerosis, HCST was associated with higher treatment-related mortality than in the first year after treatment. However, HCST conferred a significant long-term survival benefit. In ASTIS during the first year, 13 events, including 8 treatment-related deaths, occurred in the HSCT group and 8 events with no treatment-related deaths occurred in the control group (16.5% vs. 10.4%, respectively). At 4 years, however, 15 events had occurred cumulatively in the HSCT group compared with 20 events in the control group (19% vs 26%, respectively). In the Scleroderma: Cyclophosphamide Or Transplantation (SCOT) trial, a randomized, open-label, phase 2 trial in 75 patients with severe scleroderma, autologous HSCT improved event-free and overall survival, at a cost of increased expected toxicity, compared with 12 monthly infusions of cyclophosphamide. Rates of treatment-related death and post-transplantation use of disease-modifying antirheumatic drugs (DMARDs) were lower than those in previous reports of nonmyeloablative transplantation. In SCOT participants who received a transplant or completed 9 or more doses of cyclophosphamide, event-free survival was 79% in the transplantation group and 50% in the cyclophosphamide group (P = 0.02) at 54 months and 74% vs 47%, respectively, at 72 months (P = 0.03), By 54 months, 9% of the participants in the transplantation group had initiated DMARDs, as compared with 44% of those in the cyclophosphamide group (P = 0.001). Treatment-related mortality in the transplantation group was 3% at 54 months and 6% at 72 months; no treatment-related deaths occurred in the cyclophosphamide group. A review by Sullivan et al of HSCT for patients with scleroderma and pulmonary involvement suggested that indications for referral for HSCT in this population may include the following : Diffuse cutaneous systemic sclerosis with internal organ involvement Age < 65 years Disease duration < 5 years Modified Rodnan skin thickness score (MRSS) >15 Early pulmonary involvement, documented on high-resolution computed tomography (HRCT) and pulmonary function tests (PFTs) These authors recommended the following as PFT criteria for pulmonary involvement : Forced vital capacity (FVC) or hemoglobin‐adjusted diffusing capacity of the lungs for carbon monoxide (DLCO) between 80% and 45% predicted A decline in FVC of > 10% or DLCO of > 15% on serial testing Failure to respond to initial therapy on serial PFT monitoring Recommended exclusion criteria for HSCT include the following : FVC or hemoglobin‐adjusted DLCO of < 45% predicted Pulmonary arterial hypertension Cardiac insufficiency or involvement with systemic sclerosis Kidney insufficiency Prior cyclophosphamide treatment of > 6 months’ duration Skin Fibrosis Numerous experimental drugs or interventions have been investigated for treatment of skin induration and fibrosis in systemic sclerosis. Interventions that have demonstrated benefit include the following: D-penicillamine [147, 148] Bovine collagen - Possible benefit in late-phase disease Methotrexate [140, 141] Mycophenolate mofetil [144, 145, 146] Allogeneic bone marrow transplantation [149, 150, 151, 152, 154, 155, 156] Interventions that have failed to demonstrate significant benefit for treatment of skin induration and fibrosis in systemic sclerosis include the following: Human relaxin Interferon-alpha Anti–transforming growth factor beta antibodies Next: Pruritus Pruritus Treatment measures for pruritus include the following : Moisturizers Histamine 1 (H1) and histamine 2 (H2) blockers Tricyclic antidepressants Trazodone Previous Next: Pruritus Raynaud Phenomenon Raynaud phenomenon can be treated with the following agents [50, 51, 166] : Calcium channel blockers (increasing the dose to tolerance) Prazosin Prostaglandin derivatives (eg, prostaglandin E1) Dipyridamole Aspirin Topical nitrates Sildenafil, an inhibitor of phosphodiesterase 5 (PDE-5), has been approved for treatment of pulmonary hypertension. In addition, it has been shown to be effective and well tolerated in patients with Raynaud phenomenon. [167, 168, 169] In the event of thrombosis and vascular flow compromise, a tissue plasminogen activator, heparin, and urokinase may be necessary. In very severe cases, patients may benefit from intravenous iloprost or related prostanoids. Some patients may require pharmacologic cervical sympathectomy or surgical digital sympathectomy. [171, 172, 173] Digital ulcers Repeated episodes of Raynaud phenomenon in individuals with systemic sclerosis may result in digital ulcers. Bosentan, a dual endothelin receptor antagonist approved for treatment of systemic sclerosis–associated pulmonary hypertension, may curtail the formation of new digital ulcers. [174, 175] Combination therapy with iloprost and bosentan has also shown benefit in reducing new digital ulcers. Ambrisentan and other endothelin receptor antagonists have also shown beneficial effects in preliminary or open-label studies. [177, 178] Previous Next: Pruritus Gastrointestinal Involvement Treatments for gastrointestinal symptoms of systemic sclerosis are listed below. [56, 179, 180] Gastroesophageal reflux disease: Antacids Histamine 2 (H2) blockers Reflux and aspiration precautions Proton pump inhibitors Gastroparesis: Prokinetic agents Octreotide Colonic / anorectal dysmotility: Stool softeners Laxatives Some patients with scleroderma may also develop small intestinal bacterial overgrowth (SIBO). This manifestation generally improves with the addiition of rifaximin. Previous Next: Pruritus Pulmonary Fibrosis/Alveolitis Although there is some controversy regarding the beneficial effects of immunosuppressive therapy in idiopathic pulmonary fibrosis, numerous studies support the use of these agents in systemic sclerosis–associated interstitial lung disease. [182, 183] Pulmonary fibrosis in systemic sclerosis has been successfully treated with cyclophosphamide, either orally or in intravenous pulses. [184, 142, 143] Several nonrandomized studies showed benefit from mycophenolate mofetil. [144, 145, 146, 185, 186] The Scleroderma Lung Study II (SLS-II), a double-blind, parallel group, randomized controlled trial, found that a 2-year course of mycophenolate mofetil was better tolerated and associated with less toxicity than 12 months of oral cyclophosphamide, and yielded comparable significant improvements in lung function, dyspnea, lung imaging, and skin disease. Nintedanib was approved by the FDA in 2019 to slow the rate of decline in pulmonary function in patients who have interstitial lung disease associated with scleroderma. Nintedanib is a tyrosine kinase inhibitor that targets growth factors (eg, vascular endothelial growth factor receptor [VEGFR], fibroblast growth factor receptor [FGFR], platelet-derived growth factor receptor [PDGFR] 1-3, colony-stimulating factor 1 receptor [CSFIR]) that are implicated in the pathogenesis of interstitial lung diseases. Approval of nintedanib was based on results from the phase 3 SENSCIS trial (n=576). Results showed that nintedanib slowed the loss of pulmonary function by 41 mL/year in patients with systemic sclerosis–related interstitial lung disease relative to placebo, as measured by forced vital capacity (FVC) over a 52-week period. Tocilizumab, an anti-IL-6 inhibitor has also shown promise for the treatment of interstitial lung disease due to scleroderma. In the focuSSed trial with open-label extension, the decline in FVC over 48 months was 14 mL in the tocilizumab group compared with 255 mL in the placebo group. Previous Next: Pruritus Pulmonary Hypertension Numerous agents have been approved by the FDA for the treatment of pulmonary arterial hypertension (PAH). These include the following: Prostaglandin derivatives such as epoprostenol, treprostinil, beraprost, and iloprost Phosphodiesterase type 5 (PDE-5) inhibitors such as sildenafil and tadalafil Endothelin receptor antagonists such as bosentan, ambrisentan and macitentan The combination of ambrisentan and tadalafil was approved by the FDA in 2015 as up-front therapy for PAH to reduce the risk of worsening disease and improve exercise ability. In a prospective multicenter open-label trial by Hassoun et al, this combination significantly improved hemodynamics, right ventricular structure and function, and functional status in treatment-naïve systemic sclerosis patients with PAH. One study reported that warfarin provided no significant benefit in either systemic sclerosis–associated or idiopathic pulmonary arterial hypertension. Previous Next: Pruritus Scleroderma Renal Crisis Patients with diffuse, rapidly progressive skin involvement have the highest risk of developing scleroderma renal crisis. Renal crisis occurs in about 10% of all patients with systemic sclerosis. Renal crisis is observed within 4 years of diagnosis in about 75% of patients but may develop as late as 20 years after diagnosis. Renal crises are slightly more common in Blacks than in Whites, and men have a greater risk than women. The presence of RNA polymerase III antibodies increases the risk for renal crisis. Scleroderma renal crisis that is not treated promptly and aggressively invariably leads to kidney failure requiring dialysis or kidney transplantation, or even death. Consequently, it is critical to check blood pressure, monitor serum creatinine, and start angiotensin-converting enzyme (ACE) inhibitors at the earliest signs of hypertension in at-risk patients. The preferred ACE inhibitor of choice is captopril, owing to its short half-life and ease to titrate quickly, which aid in managing the hypertensive urgency that is common in patients presenting with scleroderma renal crisis. High doses of corticosteroids (> 20 mg) should be avoided in patients with systemic sclerosis owing to an increased risk of developing renal crisis. Previous Next: Pruritus Musculoskeletal Symptoms Carpal tunnel entrapment symptoms may require local corticosteroid injections although frequently these symptoms resolve spontaneously. Myositis may be treated cautiously with corticosteroids (first choice), or with methotrexate or azathioprine in corticosteroid-resistant cases or when there are contraindications to corticosteroid use. Doses of prednisone greater than 40 mg/d are associated with a higher incidence of scleroderma renal crisis. Use of methotrexate should be guarded if there is severe or active baseline pulmonary interstitial disease. Arthralgias can be treated with acetaminophen. In cases of more severe arthritis or inflammatory arthritis, low-dose prednisone (10 mg) may be considered for the short term. Hydroxychloroquine is generally the preferred first-line therapy for long-term treatment. Methotrexate, biologics (such as tocilizumab or abatacept), and Janus kinase inhibitors (such as tofacitinib) may also be considered in refractory cases. Physical therapy has been shown to be beneficial in improving hand function in patients with scleroderma. [193, 194, 195] In a randomized controlled trial in 24 patients with systemic sclerosis, Maitland's joint mobilization and therapeutic exercises, twice weekly for 12 weeks, improved hand functionality, reduced pain in the hands and wrists, increased range of motion, and improved quality of life. In another study, a 4-week complex, supervised rehabilitation protocol in 27 patients improved hand and overall function for up to 6 months, compared with a control group of 24 patients prescribed a home exercise program alone. However, the improvement was not sustained after 6 months, suggesting that the rehabilitation program should be repeated every 3-6 months. Previous Next: Pruritus Pregnancy Pregnancy in women with systemic sclerosis is considered a high risk because of a higher risk of pregnancy loss and higher complication rates, but a diagnosis of systemic sclerosis without pulmonary hypertension is not an absolute contraindication for pregnancy. However, maternal mortality during pregnancy is increased in patients with pulmonary hypertension and some consider pulmonary hypertension to be a contraindication for pregnancy. A study of 50 patients (67 pregnancies) showed that 18% miscarried, 26% delivered preterm, and 55% delivered at full term. Pregnancy risk is greatest in those who have had the disease for less than 4 years and who also have diffuse cutaneous involvement. Some systemic sclerosis symptoms may increase during pregnancy (eg, edema, arthralgias, gastroesophageal reflux disease [GERD]). Skin manifestations are not reported to worsen. Raynaud symptoms may improve during pregnancy, only to worsen after delivery. Certain medications, such as D-penicillamine, mycophenolate mofetil, cytotoxic agents, and angiotensin-converting enzyme (ACE) inhibitors, should be discontinued prior to pregnancy Previous Next: Pruritus Surgical Care Digital sympathectomy and botulinum toxin injections may be used in patients with severe Raynaud phenomenon who have an unrelenting acute attack and who are threatened by digital loss. Many ulcers require management by a wound care specialist. Debridement or amputation may be required in severe ischemic or infected digital lesions. Digital sympathectomy may be indicated in severe cases. Hand surgery may be performed to correct severe flexion contractures. Removal of severely painful, draining or infected calcinotic deposits is occasionally required. Surgical fundoplication may be required to treat severe esophageal reflux with complicating aspiration pneumonitis. Laser ablation or even gastric antrectomy may be required to control persistent bleeding caused by gastric antral vascular ectasia (GAVE). Episodes of acute abdominal pain need to be evaluated with extreme care to avoid the misdiagnosis of an acute abdomen, since occasionally patients with systemic sclerosis present with symptoms of acute abdomen that resolve with proper medical treatment and do not require surgical intervention (pseudo acute abdomen). Previous Next: Pruritus Diet There are no specific dietary recommendations in patients with systemic sclerosis. Appropriate caloric intake should be encouraged. The following points may be considered: Patients with esophageal involvement should avoid hard solid foods. Patients with intestinal hypomotility may benefit from high-fiber diets. Vitamin deficiencies and malabsorption should be addressed in patients with frequent or severe bacterial overgrowth. Large doses of vitamin C (> 1000 mg/d) should be avoided because this stimulates collagen formation and may enhance fibrotic tissue deposition. A study of 51 Romanian patients with systemic sclerosis found that 25hydroxy-hydroxyvitamin D serum levels were insufficient in approximately two thirds of cases, and inadequate in almost one quarter. Vitamin D levels did not correlate with extent of skin involvement, but low levels did seem to be related to more aggressive disease with multivisceral and severe organ involvement, especially pulmonary and cardiac. Previous Next: Pruritus Activity Recommendations regarding activity include the following: Ensure that the patient maintains a core body temperature to try to minimize the occurrence of Raynaud phenomenon episodes. Assist the patient in avoiding contamination of any skin wound caused by ischemic lesions or calcinosis. Digital ulcers must be kept clean and dry. Instruct the patient to perform continuous physical and occupational therapy to maintain joint range of motion and to minimize or delay joint contractures. Encourage patients with pulmonary fibrosis to receive pulmonary rehabilitation treatment. Previous Next: Pruritus Consultations All patients with systemic sclerosis should be treated in conjunction with an experienced rheumatologist who has a full understanding of the disease, the complications of the therapies, and the frequently serious adverse effects. Consultation with an experienced practitioner can help avoid potentially fatal pulmonary, vascular, or renal complications. Patients with systemic sclerosis may need to be treated by other subspecialists, depending on the involvement of specific organ systems (eg, cardiologist, pulmonologist, gastroenterologist, nephrologist, endocrinologist, vascular surgeon, wound care specialist, hand surgeon). Kidney and lung transplantation are performed in specialized centers for patients with end-stage kidney or lung involvement. Previous Next: Pruritus Long-term Monitoring Evaluate the patient every 3-6 months, depending on the disease activity and progression. Serial skin scoring (also known as the modified Rodnan skin score) is useful for monitoring skin changes over time. Patients with scleroderma need annual ECGs (to screen for cardiac arrhythmias) and serial pulmonary function tests (PFTs) to screen for interstitital lung disease/pulmonary hypertension. 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Validity, reliability, and feasibility of durometer measurements of scleroderma skin disease in a multicenter treatment trial. Arthritis Rheum. 2008 May 15. 59(5):699-705. [QxMD MEDLINE Link].[Full Text]. Media Gallery Scleroderma: Tightening of the skin in the face, with a characteristic beaklike facies and paucity of wrinkles. Scleroderma: Sclerodactyly with digital ulceration, loss of skin creases, joint contractures, and sparse hair. Scleroderma: Anterior chest demonstrating salt-and-pepper hypopigmentation and diffuse hyperpigmentation in a white woman. A radiograph of the distal digits demonstrating calcinosis and distal phalanx reabsorption (acral osteolysis). Scleroderma: Fingernail capillary bed demonstrating capillary dropout with large dilated vessels. Lung biopsy demonstrating severe interstitial fibrosis and medial fibrosis and smooth muscle hyperplasia of a pulmonary arteriole compatible with pulmonary hypertension. Lung biopsy demonstrating expansion of the interstitium of the lung by fibrous tissue along with chronic inflammatory cells. Scleroderma: Barium swallow demonstrating reflux into the distal esophagus, as well as an accordion appearance in the duodenum. Skin biopsy showing extensive fibrosis. The biopsy has a square morphology, which reflects the rigidity of the tissue biopsy specimen due to striking pan-dermal sclerosis. In addition, the fibrosing reaction extends into the panniculus. The number of adnexal structures is reduced, another characteristic feature of scleroderma. A significant inflammatory cell infiltrate is not observed. This is in contradistinction to morphea, in which a prominent inflammatory cell infiltrate is present. Skin biopsy showing severe fibrosis. The fibrosis reflects a widening of collagen bundles in concert with an increase in the number of collagen fibers. Note the superimposed deposition of the newly synthesized delicate collagen bundles interposed between the preexisting collagen bundles, the latter appearing wide and manifesting a hyalinized morphology. Overall scheme illustrating a proposed sequence of events involved in tissue fibrosis and fibroproliferative vasculopathy in systemic sclerosis. An unknown causative agent induces activation of immune and inflammatory cells in genetically predisposed hosts, resulting in chronic inflammation. Activated inflammatory and immune cells secrete cytokines, chemokines, and growth factors, which cause fibroblast activation, differentiation of endothelial and epithelial cells into myofibroblasts, and recruitment of fibrocytes from the bone marrow and the peripheral blood. The activated myofibroblasts produce exaggerated amounts of extracellular matrix, resulting in tissue fibrosis. of 11 Tables Table 1. ACR/EULAR Revised Systemic Sclerosis Classification Criteria;) Table 1. ACR/EULAR Revised Systemic Sclerosis Classification Criteria \| \| \| \| --- \| Item \| Sub-item(s) \| Score \| \| Skin thickening of the fingers of both hands extending proximally to the metacarpophalangeal joints (presence of this criterion is sufficient criterion for SSc classification) \| None \| 9 \| \| Skin thickening of the fingers (count the higher score only) \| Puffy fingers \| 2 \| \| Sclerodactyly (distal to the metacarpophalangeal joints but proximal to the proximal interphalangeal joints) \| 4 \| \| Fingertip lesions (count the higher score only) \| Digital tip ulcers \| 2 \| \| Fingertip pitting scars \| \| \| \| None \| 2 \| \| Abnormal nailfold capillaries \| None \| 2 \| \| Pulmonary arterial hypertension and/or interstitial lung disease (maximum score is 2) \| Pulmonary arterial hypertension \| 2 \| \| Interstitial lung disease \| 2 \| \| Raynaud phenomenon \| None \| 3 \| \| Systemic sclerosis–related autoantibodies (maximum score is 3) \| \| 3 \| \| Anti–topoisomerase I \| 3 \| \| Anti–RNA polymerase III \| 3 \| \| The total score is determined by adding the maximum score in each category. Patients with a total score ≥ 9 are classified as having definite systemic sclerosis (modified from van den Hoogen F, Khanna D, Fransen J, et al. 2013 classification criteria for systemic sclerosis: an American College of Rheumatology/European League against Rheumatism collaborative initiative. Arthritis Rheum. Nov 2013;65(11):2737-47. ) \| \| \| Back to List encoded search term (Scleroderma) and Scleroderma What to Read Next on Medscape Log in or register for free to unlock more Medscape content Unlimited access to our entire network of sites and services Log in or Register
2047	https://en.wikipedia.org/wiki/Entropy_of_activation	Jump to content Entropy of activation Čeština Deutsch Français Magyar Українська Edit links From Wikipedia, the free encyclopedia In chemical kinetics, the entropy of activation of a reaction is one of the two parameters (along with the enthalpy of activation) that are typically obtained from the temperature dependence of a reaction rate constant, when these data are analyzed using the Eyring equation of the transition state theory. The standard entropy of activation is symbolized ΔS‡ and equals the change in entropy when the reactants change from their initial state to the activated complex or transition state (Δ = change, S = entropy, ‡ = activation). Importance [edit] Entropy of activation determines the preexponential factor A of the Arrhenius equation for temperature dependence of reaction rates. The relationship depends on the molecularity of the reaction: for reactions in solution and unimolecular gas reactions : A = (ekBT/h) exp(ΔS‡/R), while for bimolecular gas reactions : A = (e2kBT/h) (RT/p) exp(ΔS‡/R). In these equations e is the base of natural logarithms, h is the Planck constant, kB is the Boltzmann constant and T the absolute temperature. R′ is the ideal gas constant. The factor is needed because of the pressure dependence of the reaction rate. R′ = 8.3145×10−2 (bar·L)/(mol·K). The value of ΔS‡ provides clues about the molecularity of the rate determining step in a reaction, i.e. the number of molecules that enter this step. Positive values suggest that entropy increases upon achieving the transition state, which often indicates a dissociative mechanism in which the activated complex is loosely bound and about to dissociate. Negative values for ΔS‡ indicate that entropy decreases on forming the transition state, which often indicates an associative mechanism in which two reaction partners form a single activated complex. Derivation [edit] It is possible to obtain entropy of activation using Eyring equation. This equation is of the form where: = reaction rate constant = absolute temperature = enthalpy of activation = gas constant = transmission coefficient = Boltzmann constant = R/NA, NA = Avogadro constant = Planck constant = entropy of activation This equation can be turned into the form The plot of versus gives a straight line with slope from which the enthalpy of activation can be derived and with intercept from which the entropy of activation is derived. References [edit] ^ Laidler, K.J. and Meiser J.H. Physical Chemistry (Benjamin/Cummings 1982) p. 381–382 ISBN 0-8053-5682-7 ^ Laidler and Meiser p. 365 ^ James H. Espenson Chemical Kinetics and Reaction Mechanisms (2nd ed., McGraw-Hill 2002), p. 156–160 ISBN 0-07-288362-6 Retrieved from " Category: Chemical kinetics
2048	https://www.youtube.com/watch?v=7dRlpi8SZQ4	Altitude Geometric Mean Theorem Mario's Math Tutoring 458000 subscribers 1052 likes Description 98221 views Posted: 22 Feb 2016 Learn how to use the Altitude Geometric Mean Theorem in this free math video tutorial by Mario's Math Tutoring. Timestamps: 00:00 Intro 0:09 What is the Geometric Mean 1:08 Using Similar Triangles to Show Why the Altitude Geometric Mean Theorem Works 1:49 Example 1 Solving for the Altitude 2:17 Example 2 Solving for a Portion of the Hypotenuse 2:54 Example 3 3:26 Further Exploration of How to Compare the 3 Similar Triangles Formed When an Altitude is Drawn from the Right Angle to the Hypotenuse Related Videos: Leg Geometric Mean Theorem Organized List of My Video Lessons to Help You Raise Your Scores & Pass Your Class. Videos Arranged by Math Subject as well as by Chapter/Topic. (Bookmark the Link Below) ➡️JOIN the channel as a CHANNEL MEMBER at the "ADDITIONAL VIDEOS" level to get access to my math video courses(Algebra 1, Algebra 2/College Algebra, Geometry, and PreCalculus), midterm & final exam reviews, ACT and SAT prep videos and more! (Over 390+ videos) 72 comments Transcript: Intro geometric mean okay the geometric mean are these two values and the extremes are these two values and there's something called the means extremes What is the Geometric Mean property you probably know it as the cross multiplication where this product equals this product okay but what we're going to do here is we're going to solve for the altitude or one of these pieces of the hypotenuse when you drop an altitude to the hypotenuse of a right triangle it splits this hypotenuse into two parts a and B and what happens is you hand up with three similar triangles a small one this medium one and this over all right triangle and these triangles are all similar to one another the corresponding angles are congruent the corresponding sides are going to be proportional and what happens is you end up having a proportion that you can set up this triangle and this triangle or the triangle is really compared and I'll just show you real quick if you take this triangle okay and you rotate it and you take this triangle and you rotate it so you have them in the same orientation same direction here you've got a and the Using Similar Triangles to Show Why the Altitude Geometric Mean Theorem Works altitude here you've got V and the altitude and you can see you can set up a proportion so a is 2 the altitude as the altitude is to B so you're comparing the base to the base as the height is to the height so these two triangles are similar so the corresponding sides are proportional but practicality just practically speaking how do we use this well let's go over here and look at some examples if we want to find this altitude here X all we do is we set up a proportion we say 4 is 2 X as X is 2 9 so the altitude Example 1 Solving for the Altitude okay that's the geometric mean that's these guys these are the means these are the extremes 4 & 9 cross multiply so you get x squared equals 36 and take the square root of both sides and you get x equals 6 so that's going to be the altitude okay this one same thing we have a right triangle we drop an altitude to the hypotenuse to the base here this altitude splits the hypotenuse up into Example 2 Solving for a Portion of the Hypotenuse two pieces axon 16 so we know that all three of these triangles formed are similar so we can set up a proportion so same thing access to 8 as 8 is 216 okay so again you can see the altitude is the mean okay these are the means and these are the extremes so if we cross multiply we get 16 x equals 64 divided by 16 and we get 4 okay last example sometimes what people do is they'll give you this whole length so by taking 29 minus 25 we see that this is 4 okay and let's just verify that this Example 3 works so you can see that 4 is 210 as 10 is 225 100 equals 100 so you can see the altitude here is the geometric mean of the two parts of the hypotenuse that it divides it up into
2049	http://www.pressure-drop.com/Online-Calculator/rauh.html	Roughness of pipes SF Pressure Drop Online-Calculator: Roughness of pipes Aluminium, drawn/pressed new 0.0013 - 0.0015 mm Aluminium, drawn/pressed used to 0.03 mm Asbestos-cement new, smooth 0.03 - 0.1 mm Brass, drawn/pressed new 0.0013 - 0.0014 mm Brass, drawn/pressed used to 0.03 mm Cast iron average city severage 1.2 mm Cast iron incrusted to 3.0 mm Cast iron new, bituminized 0.10 - 0.13 mm Cast iron new, with skin 0.2 - 0.6 mm Cast iron operating several years, cleaned 1.5 mm Cast iron slightly rusty 1.0 - 1.5 mm Clay new, clay tile 9.0 mm Clay, Drainage-pipe new, calcined 0.7 mm Concrete new, medium rough 1.0 - 2.0 mm Concrete new, rough 2.0 - 3.0 mm Concrete new, smooth 0.3 - 0.8 mm Concrete operating several years 0.2 - 0.3 mm Concrete, Centrifugal-new, smooth plastered 0.1 - 0.15 mm Concrete, Centrifugal-new, without plaster 0.2 - 0.8 mm Concrete, Steel-new, smooth 01. - 0.15 mm Copper, drawn/pressed new 0.0013 - 0.0015 mm Copper, drawn/pressed used to 0.03 mm Glass, drawn/pressed new 0.0013 - 0.0015 mm Glass, drawn/pressed used to 0.03 mm Plastic, drawn/pressed new 0.0013 - 0.0015 mm Plastic, drawn/pressed used to 0.03 mm Rubber new, smoot 0.0016 mm Steel after long operation cleaned 0.15 - 0.20 mm Steel homogeneous corrosion pits 0.15 mm Steel intensely incrusted 2.0 - 4.0 mm Steel slightly rusty and incrusted 0.15 - 0.40 mm Steel, longitudinal welded new, bituminized 0.01 - 0.05 mm Steel, longitudinal welded new, galvanized 0.008 mm Steel, longitudinal welded new, rolling skin 0.04 - 0.1 mm Steel, weldless new, comm.size galvanized 0.10 - 0.16 mm Steel, weldless new, neatly galvanized 0.07 - 0.10 mm Steel, weldless new, pickled 0.03 - 0.04 mm Steel, weldless new, rolling skin 0.02 - 0.06 mm Steel, weldless new, unpickled 0.03 - 0.06 mm Stoneware 0.25 mm Wood after long operating 0.1 mm Wood new 0.2 - 1.0 mm
2050	https://uroonkolojibulteni.com/articles/a-rare-side-effect-of-intravesical-bacillus-calmette-guerin-therapy-reactive-arthritis/uob.915	A Rare Side Effect of Intravesical Bacillus Calmette-Guérin Therapy: Reactive Arthritis - Bulletin of Urooncology Abstracting and Indexing Web Site of Urooncology Association E-ISSN: 2667-4610 About Us Journal History Aims and Scope Abstracting and Indexing Editorial Board Recruitment of Editors Roles and Responsibilities Editorial Board Journal Policies Advertising Allegations of Misconduct Appeals and Complaints Archiving and Preservation Article Processing Charges (APC) Artificial Intelligence (AI) Authorship and Contributorship Conflicts of Interest Copyright and Licensing Data Availability and Reproducibility Diversity, Equity, Inclusion, and Accessibility (DEIA) Editorial Oversight Ethical Misconduct Open Access Peer Review Policy Post Publication Discussions Research Ethics Instructions to Authors Instructions to Authors Instructions for Reviewers Ethical Policy Peer Review Process Contact Us A Rare Side Effect of Intravesical Bacillus Calmette-Guérin Therapy: Reactive Arthritis PDF Cite Share Request Case Report VOLUME: 17 ISSUE: 1 P: 33 - 35 March 2018 A Rare Side Effect of Intravesical Bacillus Calmette-Guérin Therapy: Reactive Arthritis Bull Urooncol 2018;17(1):33-35 DOI: 10.4274/uob.915 Bahattin Kızılgök 1 Volkan İzol 1 Eren Erken 2 Mustafa Zühtü Tansuğ 1 Author Information Copyright & Licence Information Galenos & Bull Urooncol Disclaimer Çukurova University Faculty of Medicine, Department of Urology, Adana, Turkey Çukurova University Faculty of Medicine, Department of Internal Diseases, Division of Rheumatology, Adana, Turkey No information available. No information available Received Date: 15.09.2017 Accepted Date: 04.10.2017 Publish Date: 30.03.2018 PDF Cite Share Request ABSTRACTIntroductionDiscussion ABSTRACT Approximately 70-80% of bladder cancers are superficial tumors and not muscle invasive. Complete transurethral resection of the bladder tumour (TUR-BT) is the standard approach to these patients. Intravesical treatments such as adriamycin, doxorubicin, epirubicin, mitomycin-c and Bacillus Calmette-Guérin (BCG) may be performed after TUR-BT in order to prevent further recurrence or progression. BCG is generally used in high-risk patients and causes local or systemic side effects in less than 5% of patients. Osteoarticular side effects are very rare and usually manifest as joint pain and arthritis (%0.5-1). In this case report, we present the management of reactive arthritis in a patient treated with intravesical BCG for bladder cancer. Keywords: Intravesical Bacillus Calmette-Guérin, bladder cancer, reactive arthritis Introduction Bladder cancer is the seventh most common cancer in males worldwide, and eleventh most common when both genders are considered. It is 3-4 times more common in men than in women, especially after 60 years of age. Histopathologically, more than 90% of bladder cancers are transitional cell carcinomas and 70-80% are muscle non-invasive. Approximately 70% of non-invasive bladder cancers are Ta tumors, 20% are T1 tumors, and 10% are carcinoma in situ (1). The treatment approach in these cases is complete transurethral resection of bladder tumor (TUR-BT). A well-performed TUR-BT allows correct staging and administration of intravesical treatment, which can prevent recurrence or progression (2,3,4). These intravesical treatments include adriamycin, doxorubicin, epirubicin, mitomycin-c, and Bacillus Calmette-Guérin (BCG). BCG is obtained from an attenuated strain of Mycobacterium bovis(5). Intravesical BCG application, performed since 1976, is believed to have an antitumor effect (5,6,7). Severe local or systemic side effects related to BCG occur in less than 5% of patients. Local side effects include cystitis findings, hematuria, granulomatous prostatitis, and epididymuritis. Systemic side effects may manifest as fever, allergic reaction, sepsis, arthralgia, or arthritis (8,9). Of these adverse systemic effects, osteoarticular involvement is rare, occurring in 1-5% of cases. Osteoarticular side effects usually manifest clinically with joint pain and arthritis (0.5-1%) (10,11). In this case report, we present a patient who developed reactive arthritis after undergoing TUR-BT for non-muscle-invasive bladder cancer and intravesical BCG treatment. Case Report Ultrasound examination in a 43-year-old male patient with hematuria revealed a pelvic mass approximately 3 cm in diameter. In cystoscopy, a papillary formation about 3 cm in size was observed on the left sidewall and was resected. Following TUR-BT, T1G3 was identified in histopathological examination and intravesicular BCG therapy was planned. The first instillation was done on postoperative day 18. After the fifth instillation, the patient developed complaints of fever (38.2 °C), fatigue, and pain, redness, and tenderness in his right big toe. The following day he developed pain in the right knee and shoulder joints and was readmitted to the hospital. After consultation with the rheumatology department, laboratory analysis of complete blood count, acute phase reactants, complete urinalysis, urine culture, and serological markers was requested. Complete blood count revealed leukocytosis (white blood cell count: 14.2x10 3/µL), acute phase reactants were elevated (C-reactive protein: 15.7 mg/dL), and erythrocyte sedimentation rate was 18 mm/L. Urinalysis and urine culture were normal. Serologic examination was negative for rheumatoid factor negative and positive for human leukocyte antigen (HLA)-B27. Anteroposterior X-rays of the right foot and knee were taken. X-ray to investigate the cause of swelling in the knee revealed no evidence other than edema in the soft tissue (Figure 1). The orthopedics and traumatology department was consulted and magnetic resonance imaging (MRI) of the right knee joint was requested. MRI of the right knee joint revealed fluid accumulation around the medial and collateral ligaments. Samples of the fluid were obtained under sterile conditions. On microscopic examination, 10-12 leukocytes were observed in each field. Cultures of the fluid were negative. Therefore, a diagnosis of septic arthritis was excluded. Reactive arthritis secondary to intravesical BCG treatment was suspected due to fever, oligoarthritis, and HLA-B27 positivity. The patient was treated with prednisolone (32 mg/day for 7 days, followed by 24 mg/day), betamethasone (once a week), and diclofenac sodium. Steroid therapy was tapered and discontinued after 3 months, and the patient’s complaints resolved completely (Figure 2). The patient provided informed consent for his case to be presented in this report. Discussion Osteoarticular side effects rarely develop due to intravesical BCG therapy. In a 2006 review, Tinazzi et al. (6) screened 48 articles and classified 61 autoimmune complications developed after intravesical BCG treatment. They reported that 64% of the patients developed joint pain and arthritis, 24% had Reiter’s syndrome, 4% had arthritis and fever, 2% had psoriatic arthritis, and 2% had Sjogren’s syndrome. Reactive arthritis secondary to intravesical BCG treatment is usually seen in the fifth and sixth decades. It typically occurs after the fourth or fifth instillation. Onset occurred after the fifth instillation in our case. It is characterized by asymmetric oligoarticular involvement. The most commonly involved joints are the knee, wrist, and ankle. In addition to arthritis, patients may exhibit dactylitis, urethritis, and uveitis (7,12,13,14). Elevated acute phase reactants, inflammatory response in the synovial fluid, and negative mycobacterial cultures of synovial fluid are expected findings of laboratory studies. Other findings should raise suspicion of septic arthritis. Blood and urine analyses negative for infection support a diagnosis of reactive arthritis (7,9,11). We suspected reactive arthritis in our case due to the asymmetric articular involvement as well as the patient’s high acute phase reactants levels and our findings of 10-12 leukocytes/field and negative cultures of fluid obtained from the knee. The mechanism by which reactive arthritis develops secondary to intravesical BCG therapy is not yet clear. In a case report published in 2002, Pardalidis et al. (15) claimed that mycobacterial heat shock protein 65 shared similar homology to human cartilage tissue, and suggested that this may cause cross-reaction and increase cytokine production. The increase in cytokine release activates CD8+ T cells, and cellular immunity may lead to tissue damage (15). This autoimmune response is more common after intravesical BCG treatment in individuals who are positive for HLA-B27 and B7 (6). About 60% of patients are HLA-B27 positive, as was our patient. Reactive arthritis secondary to intravesical BCG therapy resolves with treatment in most cases. The condition may become chronic in small proportion of patients, but they can be treated with non-steroidal anti-inflammatory drugs (NSAIDs) and corticosteroids either alone or in combination, and if necessary, immunosuppressive treatments such as methotrexate can be used (6,7). Antituberculous therapy may also be needed in some patients who do not respond to treatment (7). The condition resolved in our patient after a successful three months of treatment with combined corticosteroid and NSAID. Reactive arthritis is a rare side effect of intravesical BCG therapy. Despite the low incidence of osteoarticular side effects, it should be remembered that such reactions are more likely to occur in HLA-B27 and HLA-B7 positive patients. In the event of osteoarticular side effects, the patient should be evaluated in collaboration with relevant departments, particularly rheumatology, to determine whether the clinical presentation is inflammatory in nature, to rule out septic arthritis, and to initiate treatment as early as possible. Ethics Informed Consent: It was taken. Peer-review: Internally peer-reviewed. Authorship Contributions Surgical and Medical Practices: V.İ., Concept: B.K., V.İ., Design: B.K., Data Collection or Processing: B.K., Analysis or Interpretation: M.Z.T., E.E., Literature Search: B.K., Writing: B.K. Conflict of Interest:No conflict of interest was declared by the authors. Financial Disclosure:The authors declared that this study received no financial support. About US Aims and Scope Editorial Board Instructions to Authors Contact Us 2024 ©️ Galenos Publishing House
2051	https://www.khanacademy.org/science/ka-chemistry-grade-12/xde7d06e720e32944:electrochemistry	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
2052	https://www.cliffsnotes.com/study-notes/21062878	Epidemiology-exam-practice-questions (pdf) - CliffsNotes Lit NotesStudy GuidesDocumentsQ&AAsk AI Chat PDF Log InSign Up Literature NotesStudy GuidesDocumentsHomework QuestionsChat PDFLog InSign Up Epidemiology-exam-practice-questions .pdf School University of GuelphWe aren't endorsed by this school Course POPM 4040 Subject Health Science Date Oct 15, 2024 Pages 19 Uploaded by NickPetro16 Download Helpful Unhelpful Download Helpful Unhelpful Home/ Health Science This is a preview Want to read all 19 pages? Go Premium today. View Full Document Already Premium? Sign in here Epidemiology Exam Practice Questions Epidemiology F (University of Guelph) Scan to open on Studocu Studocu is not sponsored or endorsed by any college or university Epidemiology Exam Practice Questions Epidemiology F (University of Guelph) Scan to open on Studocu Studocu is not sponsored or endorsed by any college or university Downloaded by Nicholas Petruzella (nicholas.petruzella@gmail.com)lOMoARcPSD\|12497200 Practice Exam 1)On August 12 th, Paolo is bitten by a mosquito and infected with the Zika virus. Five days later (August 17 th ), he would test positive for the virus using a serum IgM antibody test. Ten days after that (August 27 th), Paolo begins to feel ill (fever, chills, muscle soreness). Fifteen days later (September 11 th ), Paolo feels healthy again, but will still test positive for the antibody for another 100 days. What is the incubation period of Zika? a.5 days b.15 days c.30 days d.130 days 2)You want to randomly sample 20 cows from a herd of 800. You decide to line the animals up and send them through a chute into another paddock, and you select every 40 th cow for your study. What type of sampling is this? a.Systematic random sampling b.Simple random sampling c.Multi-stage sampling d.Cluster sampling 3)Wealth is negatively associated with obesity in Mexico City. Which of the following about the association between wealth and obesity is true? a.OR>0 b.RD>1 c.RR<1 d.PAR>0 4)The Black Death was an outbreak of bubonic plague from 1346-53 that originated in Central Asia, then travelled along the Silk Road to Crimea and the rest of Europe. It killed an estimated 75 to 200 million people. Which of the following disease patterns best describes the Black Death? a.Endemic b.Epidemic c.Sporadic d.Pandemic 5)Which of the following accurately describes the difference between the terms "infected" and "diseased"? a.Malaria is an infection but not a disease b.A person whose blood contains the malaria protozoa is infected, whereas someone displaying clinical symptoms of malaria is diseased c."Infected" and "diseased" are synonyms d.People with the zombie virus in The Walking Dead are diseased, and when they are reanimated as zombies they are infected 6)Which of the following is the WHO definition of health? a.A state of complete physical, mental, and social well-being, and not merely the absence of disease or infirmity b.The lack of derangement in the function of the whole body of the host or any of its part c.The lack of physiological/psychological disfunction Downloaded by Nicholas Petruzella (nicholas.petruzella@gmail.com)lOMoARcPSD\|12497200 Why is this page out of focus? Because this is a premium document. Subscribe to unlock this document and more. d.A state of physical well-being, including the absence of infection or illness 7)Who discovered that scurvy is caused by vitamin C deficiency based on randomized trials with British seamen? a.James Lind b.Percival Pott c.John Snow d.Cathy Bauman 8)HIV is of particular concern for pregnant women, who can pass the virus to their unborn children. What is this an example of? a.Vector-borne disease b.Vertical transmission of disease c.Indirect transmission of disease d.Horizontal transmission of disease 9)West Nile virus originates in birds, is picked up by mosquitoes, and is transported to humans, who are infected when the mosquito bites them. What are the bird, mosquito, and human (respectively) in this scenario? a.Disease pool, fomite, host b.Environment, agent, host c.Reservoir, vector, host d.Disease, indirect contact, infected 10) Donald Trump was just elected president of the US, but most polls predicted otherwise. Some experts believe that this is due to pollsters disproportionately sampling people from cities (and ignoring rural areas), where populations were more likely to support Hillary Clinton. What type of bias is this? a.Confounding bias b.Selection bias c.Nonresponse bias d.Misclassification bias 11) Joe Biden was just elected president of the US, but the margin was much narrower than the polls predicted. Some experts believe this is due to Trump supporters being more likely to refuse to disclose their voting intentions (refused to disclose who they would for or if they were going to vote)to pollsters. What type of bias is this? a.Confounding bias b.Selection bias c.Information bias d.Reporting bias Downloaded by Nicholas Petruzella (nicholas.petruzella@gmail.com)lOMoARcPSD\|12497200 Why is this page out of focus? Because this is a premium document. Subscribe to unlock this document and more. 12) Donald Trump was just elected president of the US, but most polls predicted otherwise. You are a political analyst and you're trying to determine how rural vs. urban households responded to polls. You conclude that rural voters are more likely to lie to pollsters and tell them they were going to vote for Clinton, when in fact they voted for Trump (compared to urban voters). What type of bias would this lead to? a.Confounding bias b.Differential misclassification bias c.Non-differential misclassification bias d.Selection bias 13) You are an epidemiologist who is tired of questions about the US election, so you decide to conduct a study the association between physical activity and cardiovascular disease. Your supervisor tells you to stratify your analysis based on age. Why would he tell you this? a.Older people are less likely to respond to surveys (nonresponse bias) b.Young people are more likely to lie about physical activity (misclassification bias) c.Age is likely associated with both cardiovascular disease and physical activity and may affect the association of interest (confounding bias) d.Older people have a higher mortality rate than younger people (healthy worker effect) 14) Herpes simplex virus is present in about 75% of the adult population. It is generally asymptomatic, but when combined with stress, lowered immune function (e.g. during infection with flu or cold), or exposure to sunlight, it can cause cold sores to form on the lips and mouth. Herpes cold sores cannot occur in the absence of the virus. Herpes simplex virus is therefore: a.Sufficient cause b.Necessary cause and sufficient cause c.Necessary cause but not sufficient cause d.Neither necessary cause nor sufficient cause 15) There are three sufficient causes of a disease: Sufficient cause Component cause A (P=present, N=not present)Component cause B (P=present, N=not present)Componen t cause C (P=present,N=not present)Component cause D (P=present, N=not present) PAF Sufficient cause #1 P N P P 30% Sufficient cause #2 P P P N 35% Sufficient cause #3 P P N P 35% What proportion of this disease (in the general population) could you remove if you eliminated component cause A? a.30% b.60% c.65% Downloaded by Nicholas Petruzella (nicholas.petruzella@gmail.com)lOMoARcPSD\|12497200 Why is this page out of focus? Because this is a premium document. Subscribe to unlock this document and more. d.100% 16) You are testing the effectiveness of a new, relatively inexpensive diagnostic test for Lyme disease. You test 1000 people. The actual prevalence of Lyme disease among the study group is 100. The apparent prevalence using the new test is 132, of which 83 have been previously diagnosed with Lyme disease (and thus actually have it). Seventeen people who have Lyme disease test negative on the new diagnostic test. What are the sensitivity and specificity of the new test? a.1.69 and 0.02 b.0.83 and 0.95 c.0.63 and 0.98 d.0.13 and 0.1 Disease +-Totals Test+ - Totals 17) You are testing the effectiveness of a new diagnostic test for prostate cancer. Here are the results of your analysis: Gold Standard +-Totals New Test+92 15 - Totals 120 1300 What is the PPV of the above test? a.86% b.76.7% c.9.2% d.8.2% 18) Is the above test better at ruling in or ruling out a disease? a.Ruling out b.Ruling in c.It's the same at both d.It's impossible to know 19) What is the apparent prevalence of prostate cancer among the study population (using the new test)? a.8.2% b.76.7% c.7.1% d.9.2% Downloaded by Nicholas Petruzella (nicholas.petruzella@gmail.com)lOMoARcPSD\|12497200 Why is this page out of focus? Because this is a premium document. Subscribe to unlock this document and more. 20) You are in charge of a screening program for breast cancer, which is easy and inexpensive to treat, but may be fatal if not diagnosed early. You have access to existing hospital databases, and you can use up to two diagnostic tests on anyone selected for screening. Which of the following screening protocols would you suggest? a.Passive surveillance b.A protocol involving parallel testing c.A protocol involving series testing d.Use whichever test has highest sensitivity 21) You are screening an at-risk Aboriginal population for mercury poisoning following a mining disaster on the Athabasca River. You have two tests: One is a symptoms check- list, is not invasive, and has a sensitivity of 0.8 and a specificity of 0.7. The other is a blood test, which is invasive, but has a sensitivity of 0.95 and a specificity of 0.9. Since the treatment for mercury poisoning is invasive and expensive, you decide to use the tests in series on 5000 people. What is the net sensitivity of the protocol? Note that 1000 people actually have mercury poisoning. a.0.97 b.0.83 c.0.76 d.0.8 Mercury Poison +-Totals Symptoms+ - Totals Mercury Poison +-Totals Blood Test+ - Totals 22) Which of the following is true about parallel testing? a.Only those who test positive on one test are subjected to the next test b.Only those subjects that test positive on both tests are considered outcome positive c.Parallel testing implements the same test multiple times d.Parallel testing increases net sensitivity e.Parallel testing increases net specificity Downloaded by Nicholas Petruzella (nicholas.petruzella@gmail.com)lOMoARcPSD\|12497200 Why is this page out of focus? Because this is a premium document. Subscribe to unlock this document and more. Page 1 of 19 Students also studied W01 Syllabus Quiz_ Steps for Success_ Epidemiology.pdf 1/8/24, 9:33 AM W01 Syllabus Quiz: Steps for Success: Epidemiology W01 Syllabus Quiz: Steps for Success Due Jan 13 at 11:59pm Points 15 Questions 15 Available Jan 8 at 12am - Jan 13 at 11:59pm Time Limit None Instructions To succeed in this class, it is e Brigham Young University, Idaho ECON 370 RCIS Exam Practice Questions and Answers 2023 Graded A.docx RCIS Exam Practice Questions and Answers 2023 Graded A What is the formula for calculating cardiac output? a. CO= PA- 1SVC b. CO= AO x PA c. CO= HR x SV Ans- c. CO= HR x SV Stoke Volume is _? a. Related to preload b. Related to afterload c. The same as ej Romanian-American University ACC MISC 9-2 Textbook Questions.docx Aja Burns 9-2 Textbook Questions: Chapter 9 IHP-515-Q4466 Population-Based Epidemiology 24TW4 Southern New Hampshire University 1. Calculate the etiologic fraction when the RR for disease associated with a given exposure is Etiologic Fraction: (RR-1) RR ( Southern New Hampshire University PHE 690 Fundamentals of Epi 2025 Spring Homework 3.pdf Fundamentals of Epidemiology, 280.350, Spring 2025 Homework Assignment 3 Johns Hopkins University Fundamentals of Epidemiology 280.350 Spring 2025 Homework Assignment 3 1. 2. A case-control study was designed to determine if coffee drinking (exposed: >3 c Johns Hopkins University AS.280 280.350 Principles Of Epidemiology.pdf Principles of epidemiology Principles of epidemiology Topic 1: Overview of Epidemiology Objectives: 1. 2. 3. 4. Discuss historically significant epidemiological advances. Evaluate the three types of prevention used in public health interventions. Explore Galen College of Nursing NSG 410 Study Guide #5 (1).doc Course: PH 2000 Assignment: Study Guide #5, Chapter 5 & Quiz #2 Due: See syllabus for due date Instructions: For this assignment, read Chapter #5 of your textbook, refer to the glossary when necessary, and answer the following questions as a way of taking Appalachian State University PH 2000 Study your doc or PDF Upload your materials to get instant AI help Try for free Students also studied W01 Syllabus Quiz_ Steps for Success_ Epidemiology.pdf 1/8/24, 9:33 AM W01 Syllabus Quiz: Steps for Success: Epidemiology W01 Syllabus Quiz: Steps for Success Due Jan 13 at 11:59pm Points 15 Questions 15 Available Jan 8 at 12am - Jan 13 at 11:59pm Time Limit None Instructions To succeed in this class, it is e Brigham Young University, Idaho ECON 370 RCIS Exam Practice Questions and Answers 2023 Graded A.docx RCIS Exam Practice Questions and Answers 2023 Graded A What is the formula for calculating cardiac output? a. CO= PA- 1SVC b. CO= AO x PA c. CO= HR x SV Ans- c. CO= HR x SV Stoke Volume is _? a. Related to preload b. Related to afterload c. The same as ej Romanian-American University ACC MISC 9-2 Textbook Questions.docx Aja Burns 9-2 Textbook Questions: Chapter 9 IHP-515-Q4466 Population-Based Epidemiology 24TW4 Southern New Hampshire University 1. Calculate the etiologic fraction when the RR for disease associated with a given exposure is Etiologic Fraction: (RR-1) RR ( Southern New Hampshire University PHE 690 Fundamentals of Epi 2025 Spring Homework 3.pdf Fundamentals of Epidemiology, 280.350, Spring 2025 Homework Assignment 3 Johns Hopkins University Fundamentals of Epidemiology 280.350 Spring 2025 Homework Assignment 3 1. 2. A case-control study was designed to determine if coffee drinking (exposed: >3 c Johns Hopkins University AS.280 280.350 Principles Of Epidemiology.pdf Principles of epidemiology Principles of epidemiology Topic 1: Overview of Epidemiology Objectives: 1. 2. 3. 4. Discuss historically significant epidemiological advances. Evaluate the three types of prevention used in public health interventions. Explore Galen College of Nursing NSG 410 Study Guide #5 (1).doc Course: PH 2000 Assignment: Study Guide #5, Chapter 5 & Quiz #2 Due: See syllabus for due date Instructions: For this assignment, read Chapter #5 of your textbook, refer to the glossary when necessary, and answer the following questions as a way of taking Appalachian State University PH 2000 Other related materials Data_Analysis_Cholesterol_Data_Submission.pdf Data Analysis Lab Instructions You will be using Microsoft Excel to complete this assignment. We would recommend watching the following Excel tutorial videos before you begin: ● #1 Getting Started; #1 Getting Started transcript ● #2 Finding Stuff and Gett Arizona State University BIO 181 c2c80dbdfa0df9de29ddffaee969daf23d2cbef8.pdf PRACTICE QUESTIONS GENETICS AND EVOLUTION MULTIPLE CHOICE QUESTIONS 1. 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What mechanisms come into play in Asthma and COPD?(THE MECHANISMS FOR ASMA ARE CUNY College of Staten Island BIO 382 CliffsNotes study guides are written by real teachers and professors, so no matter what you're studying, CliffsNotes can ease your homework headaches and help you score high on exams. Quick Links Literature NotesStudy GuidesDocumentsHomework Questions Company About CliffsNotesContact us Do Not Sell My Personal Information Legal Service TermsPrivacy policyCopyright, Community Guidelines & other legal resourcesHonor CodeDisclaimer CliffsNotes, a Learneo, Inc. business © Learneo, Inc. 2025 AI homework help Explanations instantly Do Not Sell or Share 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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2053	https://hellothinkster.com/math-tutor/fractions/what-is-0.4-as-a-fraction	These coaching plans come with a learning guarantee and two tutors - a dedicated math coach for 1:1 live tutoring & an expert AI Learning Lab coach). Get access to world-class curriculum, homework help, and continuous personalization. Designed to make your child math confident for life! Our high school live tutoring packs match students with a dedicated math tutor for help with school topics, test prep, and homework help. Book a call to find out all the ways we can help your child! Book Call Math Tutor Explains: What is 0.4 as a fraction? Numbers can be represented in various forms, including decimals, percentages, and fractions. With guidance from a math tutor, mastering these conversions becomes even easier. In this article, we’ll show you how to convert a decimal to a fraction, helping you build confidence and accuracy in your math skills! Math Tutor's Solution: 0.4 as a fraction is 2/5 Converting 0.4 to a fraction, Step-by-Step from a Math Tutor Step 1: The first step to converting 0.4 to a fraction is to re-write 0.4 in the form p/q where p and q are both positive integers. To start with, 0.4 can be written as simply 0.4/1 to technically be written as a fraction. Step 2: Next, we will count the number of fractional digits after the decimal point in 0.4, which in this case is 1. For however many digits after the decimal point there are, we will multiply the numerator and denominator of 0.4/1 each by 10 to the power of that many digits. For instance, for 0.45, there are 2 fractional digits so we would multiply by 100; or for 0.324, since there are 3 fractional digits, we would multiply by 1000. So, in this case, we will multiply the numerator and denominator of 0.4/1 each by 10: 0.4×101×10=410\frac{0.4 × 10}{1 × 10} = \frac{4}{10}1×100.4×10=104 Step 3: Now the last step is to simplify the fraction (if possible) by finding similar factors and cancelling them out, which leads to the following answer: 410=25\frac{4}{10} = \frac{2}{5}104=52 Math Tutor Suggests: Try to Convert Other Values to Fractions Become a pro at converting decimals or percentages to fractions by exploring some examples from our math tutors, like the ones below: What is 88.51 as a fraction? What is 701.7 as a fraction? What is 668.5 as a fraction? What is 32.82 as a fraction? What is 6.258 as a fraction? Download FREE Math Resources Take advantage of our free downloadable resources and study materials for at-home learning. 8 Math Hacks and Tricks to Turn Your ‘Okay’ Math Student Into a Math Champion! One thing we teach our students at Thinkster is that there are multiple ways to solve a math problem. This helps our students learn to think flexibly and non-linearly. How to Make Sure Your Child is Highly Successful and Becomes a Millionaire As a parent, you hope your child is extremely successful and likely become the next Gates, Zuckerberg, or Meg Whitman. To set your child on the right path, there are many skills and traits that you can start building and nurturing now. Doing so plants the seeds for future success. Want to Prepare for Success and Boost Math Skills? Access even more math problems! Grab your FREE copy of "10 Critical Thinking Math Questions for Grades K - Geometry." Equip your child with the knowledge of how to solve must-know questions from every grade. Math Tutoring to Boost Your Child’s Math Skills & Scores by 90% in Just 3 Months – Guaranteed! Does your child struggle with math homework or understanding tricky math concepts? Do they do okay in math, but express excitement to learn new material or advanced math? A Thinkster Math tutor provides one-to-one support to help elementary, middle school, and high school students build confidence and master math subjects like K-8 math, pre-algebra, algebra, geometry, calculus, and more. Our expert math tutors customize math lessons to your child’s unique needs, making learning math fun and effective. We help students improve grades, develop strong critical thinking skills through solving word problems, excel in standardized tests, and develop strong problem-solving skills. Our expert math tutors are ready to help make your child a champion and develop strong math mastery! Sign up for our 7-day free trial and get the best math tutor for your child today!
2054	https://en.wikipedia.org/wiki/Parabola_(disambiguation)	Parabola (disambiguation) - 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 Arches 2 Music 3 Other uses 4 See also Parabola (disambiguation) [x] 10 languages Čeština Euskara Français Italiano Latina Magyar Oʻzbekcha / ўзбекча Polski Slovenčina Slovenščina Edit 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 Edit interlanguage links Print/export Download as PDF Printable version In other projects Wikimedia Commons Wikidata item From Wikipedia, the free encyclopedia Look up parabola, parábola, paràbola, or parabolă in Wiktionary, the free dictionary. Look up parabole, parabolé, or parabolë in Wiktionary, the free dictionary. A parabola is a mathematical curve. Parabola or Parabole may also refer to: Arches [edit] Parabolic arch Music [edit] Parabola (album), an album by Gil Evans "Parabola" (song), a song by Tool "Parabola", a song by Jawbreaker from Bivouac Other uses [edit] Parabola (magazine), a magazine published by The Society for the Study of Myth and Tradition Parabola (moth), a genus of moth Parabola (operating system), a GNU/Linux-libre distribution Parabola Allegory, a Rosicrucian allegory attributed to Hinricus Madathanus See also [edit] Parable Parabola of safety Topics referred to by the same term This disambiguation page lists articles associated with the title Parabola. If an internal link led you here, you may wish to change the link to point directly to the intended article. Retrieved from " Category: Disambiguation pages Hidden categories: Short description is different from Wikidata All article disambiguation pages All disambiguation pages This page was last edited on 23 May 2022, at 05:08(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 Parabola (disambiguation) 10 languagesAdd topic
2055	https://artofproblemsolving.com/wiki/index.php/Coordinate_system?srsltid=AfmBOorbDJcIYxr_0MKgZ5Lk-bYPoUHehdU7zxwb2N2cxRl2Ugsa2j8c	Art of Problem Solving Coordinate system - 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 Coordinate system Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Coordinate system The coordinate system is often used in geometry. The plane determined by a horizontal number line, called the x-axis, and a vertical number line, called the y-axis, intersecting at a point called the origin. Each point in the coordinate plane can be specified by an ordered pair of numbers, (x,y). The cordinate system is organized to 4 quadrants. In the first quadrant, both (x,y) are positive. In the second quadrant x is negative, while y is positive.In the third quadrant, both (x,y) are negative. Finally, in the fourth quadrant x is positive while y is negative. To find the slope of a line on the coordinate system with points we need to compute . The slope is usually expressed as . Also, there is the point-slope form which states for some real numbers b,x,y. Also, if two lines are perpendicular, then product of the slopes is -1. (For example, the slope could be -3/4 and 4/3.) The coordinate system is also widely useful were right triangle, since there is 90 degrees. Related: Cartesian coordinate system Wikipedia: A Cartesian coordinate system is a coordinate system that specifies each point uniquely in a plane by a pair of numerical coordinates, which are the signed distances to the point from two fixed perpendicular directed lines, measured in the same unit of length. Each reference line is called a coordinate axis or just axis of the system, and the point where they meet is its origin, usually at ordered pair (0, 0). The coordinates can also be defined as the positions of the perpendicular projections of the point onto the two axes, expressed as signed distances from the origin. One can use the same principle to specify the position of any point in three-dimensional space by three Cartesian coordinates, its signed distances to three mutually perpendicular planes (or, equivalently, by its perpendicular projection onto three mutually perpendicular lines). In general, n Cartesian coordinates (an element of real n-space) specify the point in an n-dimensional Euclidean space for any dimension n. These coordinates are equal, up to sign, to distances from the point to n mutually perpendicular hyperplanes. Cartesian coordinate system with a circle of radius 2 centered at the origin marked in red. The equation of a circle is (x − a)2 + (y − b)2 = r2 where a and b are the coordinates of the center (a, b) and r is the radius. The invention of Cartesian coordinates in the 17th century by René Descartes (Latinized name: Cartesius) revolutionized mathematics by providing the first systematic link between Euclidean geometry and algebra. Using the Cartesian coordinate system, geometric shapes (such as curves) can be described by Cartesian equations: algebraic equations involving the coordinates of the points lying on the shape. For example, a circle of radius 2, centered at the origin of the plane, may be described as the set of all points whose coordinates x and y satisfy the equation x2 + y2 = 4. Cartesian coordinates are the foundation of analytic geometry, and provide enlightening geometric interpretations for many other branches of mathematics, such as linear algebra, complex analysis, differential geometry, multivariate calculus, group theory and more. A familiar example is the concept of the graph of a function. Cartesian coordinates are also essential tools for most applied disciplines that deal with geometry, including astronomy, physics, engineering and many more. They are the most common coordinate system used in computer graphics, computer-aided geometric design and other geometry-related data processing Retrieved from " Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
2056	https://link.springer.com/article/10.1007/s11269-008-9351-8	Advertisement Impact of Dam Construction on Water Quality and Water Self-Purification Capacity of the Lancang River, China 3097 Accesses 159 Citations Explore all metrics Abstract Along with the sequent completion of Manwan and Dachaoshan Dam, the river continuum of the middle and lower reaches of the Lancang River was separated into three types of segments: reservoir, below-dam segment and downstream flowing segment. The long-term series of water quality and river flow data over 20 years were analyzed in order to study the impact of dam construction and operation on water quality and water self-purification capacity of these different river segments. From pre-dam period to the first 7 years after Manwan Dam had been accomplished, the water quality of Manwan Reservoir became worse due to the accumulation of pollutants, and then to the next 5 years the water quality became better in virtue of the water self-purification of the reservoir. The cooperative operation of Manwan and Dachaoshan Dam had cumulatively positive impacts on water quality of their below-dam segment but no impacts on that of downstream flowing segment. From pre-dam period to the first 7 years after the closure of Manwan Dam, the water self-purification capacity of Xiaowan–Manwan segment for BOD5, CODMn and NH3–N decreased. Also, the water self-purification capacity of Manwan–Dachaoshan segment for BOD5 and CODMn decreased but for NH3–N increased. However, the water self-purification capacity of Jinghong–Ganlanba segment changed contrary to Manwan–Dachaoshan segment. In general, the construction of Manwan Dam negatively affected the water self-purification capacity of reservoir and below-dam segment but impose little impact on that of downstream flowing segment. This study suggested that it is necessary to pay attention to the effect of complicated temporal and spatial characteristics of dam on aquatic ecosystem. This is a preview of subscription content, log in via an institution to check access. Access this article Subscribe and save Buy Now Price excludes VAT (USA) Tax calculation will be finalised during checkout. Instant access to the full article PDF. Institutional subscriptions Similar content being viewed by others Influences of anthropogenic activities and topography on water quality in the highly regulated Huai River basin, China Assessing changes in flow and water quality emerging from hydropower development and operation in the Sesan River Basin of the Lower Mekong Region Water Quality in Main Dam Reservoirs in Poland Explore related subjects References Agthe DE, Billings RB, Ince S (2000) Integrating market solutions into government flood control policies. Water Resour Manage 14:247–256. doi:10.1023/A:1008133825980 Article Google Scholar Ben Alaya A, Souissi A, Tarhouni J, Ncib K (2003) Optimization of Nebhana Reservoir water allocation by stochastic dynamic programming. Water Resour Manage 17:259–272. doi:10.1023/A:1024721507339 Article Google Scholar Chen YC, Liu ZW (2005) Analysis and evaluation of water environmental carrying capacity in Three Gorges Reservoir. 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China GuoLiang Wei & Bing Li Search author on:PubMed Google Scholar Search author on:PubMed Google Scholar Search author on:PubMed Google Scholar Search author on:PubMed Google Scholar Search author on:PubMed Google Scholar Search author on:PubMed Google Scholar Search author on:PubMed Google Scholar Corresponding author Correspondence to BaoShan Cui. Rights and permissions Reprints and permissions About this article Cite this article Wei, G., Yang, Z., Cui, B. et al. Impact of Dam Construction on Water Quality and Water Self-Purification Capacity of the Lancang River, China. Water Resour Manage 23, 1763–1780 (2009). Download citation Received: 25 August 2007 Accepted: 09 September 2008 Published: 07 November 2008 Issue Date: July 2009 DOI: Share this article Anyone you share the following link with will be able to read this content: Sorry, a shareable link is not currently available for this article. 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2057	https://math.libretexts.org/Bookshelves/Precalculus/Book%3A_Precalculus__An_Investigation_of_Functions_(Lippman_and_Rasmussen)/03%3A_Polynomial_and_Rational_Functions./3.01%3A_Power_Functions	Published Time: 2018-12-04T02:39:07Z 3.1: Power Functions - 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 3: Polynomial and Rational Functions. 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Home 2. Bookshelves 3. Precalculus & Trigonometry 4. Precalculus - An Investigation of Functions (Lippman and Rasmussen) 5. 3: Polynomial and Rational Functions. 6. 3.1: Power Functions Expand/collapse global location Precalculus - An Investigation of Functions (Lippman and Rasmussen) Front Matter 1: Functions 2: Linear Functions 3: Polynomial and Rational Functions. 4: Exponential and Logarithmic Functions 5: Trigonometric Functions of Angles 6: Periodic Functions 7: Trigonometric Equations and Identities 8: Further Applications of Trigonometry 9: Conics Back Matter 3.1: Power Functions Last updated Jul 13, 2022 Save as PDF 3: Polynomial and Rational Functions. 3.1E: Power Functions (Exercises) Page ID 13840 David Lippman & Melonie Rasmussen The OpenTextBookStore ( \newcommand{\kernel}{\mathrm{null}\,}) Table of contents 1. Characteristics of Power Functions 2. Polynomials 3. Short Run Behavior 4. Important Topics of this Section A square is cut out of cardboard, with each side having length L L. If we wanted to write a function for the area of the square, with L L as the input and the area as output, you may recall that the area of a rectangle can be found by multiplying the length times the width. Since our shape is a square, the length & the width are the same, giving the formula: A(L)=L⋅L=L 2(3.1.1)(3.1.1)A(L)=L⋅L=L 2 Likewise, if we wanted a function for the volume of a cube with each side having some length L L, you may recall volume of a rectangular box can be found by multiplying length by width by height, which are all equal for a cube, giving the formula: V(L)=L⋅L⋅L=L 3(3.1.2)(3.1.2)V(L)=L⋅L⋅L=L 3 These two functions are examples of power functions, functions that are some power of the variable. Definition: Power Function A power function is a function that can be represented in the form f(x)=x p(3.1.3)(3.1.3)f(x)=x p Where the base is a variable and the exponent, p p, is a number. Example 3.1.1 3.1.1 Which of our toolkit functions are power functions? (This chapter is part of Precalculus: An Investigation of Functions© Lippman & Rasmussen 2017, and contains content remixed with permission from College Algebra© Stitz & Zeager 2013. This material is licensed under a Creative Commons CC-BY-SA license.) Solutions The constant and identity functions are power functions, since they can be written as f(x)=x 0 f(x)=x 0 and f(x)=x 1 f(x)=x 1 respectively. The quadratic and cubic functions are both power functions with whole number powers: f(x)=x 2 f(x)=x 2 and f(x)=x 3 f(x)=x 3. The reciprocal and reciprocal squared functions are both power functions with negative whole number powers since they can be written as f(x)=x−1 f(x)=x−1 and f(x)=x−2 f(x)=x−2. The square and cube root functions are both power functions with fractional powers since they can be written as f(x)=x 1 2 f(x)=x 1 2 or f(x)=x 1 3 f(x)=x 1 3 Exercise 3.1.1 3.1.1 What point(s) do the toolkit power functions have in common? Answer (0, 0) and (1, 1) are common to all power functions. Characteristics of Power Functions Shown to the right are the graphs of f(x)=x 2 f(x)=x 2, f(x)=x 4 f(x)=x 4, and f(x)=x 6 f(x)=x 6, all even whole number powers. Notice that all these graphs have a fairly similar shape, very similar to the quadratic toolkit, but as the power increases the graphs flatten somewhat near the origin, and become steeper away from the origin. To describe the behavior as numbers become larger and larger, we use the idea of infinity. The symbol for positive infinity is ∞∞, and −∞−∞ for negative infinity. When we say that “x x approaches infinity”, which can be symbolically written as x→∞x→∞, we are describing a behavior – we are saying that x x is getting large in the positive direction. With the even power functions, as the x x becomes large in either the positive or negative direction, the output values become very large positive numbers. Equivalently, we could describe this by saying that as x x approaches positive or negative infinity, the f(x)f(x) values approach positive infinity. In symbolic form, we could write: as x→±∞x→±∞, f(x)→∞f(x)→∞. Shown here are the graphs of f(x)=x 3 f(x)=x 3, f(x)=x 5 f(x)=x 5, and f(x)=x 7 f(x)=x 7, all odd whole number powers. Notice all these graphs look similar to the cubic toolkit, but again as the power increases the graphs flatten near the origin and become steeper away from the origin. For these odd power functions, as x x approaches negative infinity, f(x)f(x) approaches negative infinity. As x x approaches positive infinity, f(x)f(x) approaches positive infinity. In symbolic form we write: as x→−∞x→−∞, f(x)→−∞f(x)→−∞ and as x→∞x→∞, f(x)→∞f(x)→∞. Definition: long run behavior The behavior of the graph of a function as the input takes on large negative values,x→−∞x→−∞, and large positive values, x→∞x→∞, is referred to as the long run behavior of the function. Example 3.1.2 3.1.2 Describe the long run behavior of the graph of f(x)=x 8 f(x)=x 8. Solution Since f(x)=x 8 f(x)=x 8 has a whole, even power, we would expect this function to behave somewhat like the quadratic function. As the input gets large positive or negative, we would expect the output to grow without bound in the positive direction. In symbolic form, as x→±∞x→±∞, f(x)→∞f(x)→∞. Example 3.1.3 3.1.3 Describe the long run behavior of the graph of f(x)=−x 9 f(x)=−x 9 Solution Since this function has a whole odd power, we would expect it to behave somewhat like the cubic function. The negative in front of the x 9 x 9 will cause a vertical reflection, so as the inputs grow large positive, the outputs will grow large in the negative direction, and as the inputs grow large negative, the outputs will grow large in the positive direction. In symbolic form, for the long run behavior we would write: as x→∞x→∞, f(x)→−∞f(x)→−∞and as x→−∞x→−∞, f(x)→∞f(x)→∞. You may use words or symbols to describe the long run behavior of these functions. Exercise 3.1.2 3.1.2 Describe in words and symbols the long run behavior of f(x)=−x 4 f(x)=−x 4 Answer As x x approaches positive and negative infinity, f(x)f(x) approaches negative infinity: as x→±∞x→±∞, f(x)→−∞f(x)→−∞ because of the vertical flip. Treatment of the rational and radical forms of power functions will be saved for later. Polynomials An oil pipeline bursts in the Gulf of Mexico, causing an oil slick in a roughly circular shape. The slick is currently 24 miles in radius, but that radius is increasing by 8 miles each week. If we wanted to write a formula for the area covered by the oil slick, we could do so by composing two functions together. The first is a formula for the radius, r r, of the spill, which depends on the number of weeks, w, that have passed. Hopefully you recognized that this relationship is linear: r(w)=24+8 w r(w)=24+8 w We can combine this with the formula for the area, A A, of a circle: A(r)=π r 2 A(r)=π r 2 Composing these functions gives a formula for the area in terms of weeks: A(w)=A(r(w))=A(24+8 w)=π(24+8 w)2 A(w)=A(r(w))=A(24+8 w)=π(24+8 w)2 Multiplying this out gives the formula A(w)=576 π+384 π w+64 π w 2 A(w)=576 π+384 π w+64 π w 2 This formula is an example of a polynomial. A polynomial is simply the sum of terms each consisting of a vertically stretched or compressed power function with non-negative whole number power. terminology of polynomial functions A polynomial is function that can be written as f(x)=a 0+a 1 x+a 2 x 2+⋯+a n x n(3.1.4)(3.1.4)f(x)=a 0+a 1 x+a 2 x 2+⋯+a n x n Each of the a i a i constants are called coefficients and can be positive, negative, or zero, and be whole numbers, decimals, or fractions. A term of the polynomial is any one piece of the sum, that is any a i x i a i x i. Each individual term is a transformed power function. The degree of the polynomial is the highest power of the variable that occurs in the polynomial. The leading term is the term containing the highest power of the variable: the term with the highest degree. The leading coefficient is the coefficient of the leading term. Because of the definition of the “leading” term we often rearrange Equation 3.1.43.1.4 so that the powers are descending. f(x)=a n x n+.....+a 2 x 2+a 1 x+a 0(3.1.5)(3.1.5)f(x)=a n x n+.....+a 2 x 2+a 1 x+a 0 Example 3.1.4 3.1.4 Identify the degree, leading term, and leading coefficient of these polynomials: f(x)=3+2 x 2−4 x 3 f(x)=3+2 x 2−4 x 3 g(t)=5 t 5−2 t 3+7 t g(t)=5 t 5−2 t 3+7 t h(p)=6 p−p 3−2 h(p)=6 p−p 3−2 Solution For the function f(x)f(x), the degree is 3, the highest power on x x. The leading term is the term containing that power, −4 x 3−4 x 3. The leading coefficient is the coefficient of that term, -4. For g(t)g(t), the degree is 5, the leading term is 5 t 5 5 t 5, and the leading coefficient is 5. For h(p)h(p), the degree is 3, the leading term is −p 3−p 3, so the leading coefficient is -1. Definition: long run behavior of polynomials For any polynomial, the long run behavior of the polynomial will match the long run behavior of the leading term. Example 3.1.5 3.1.5 What can we determine about the long run behavior and degree of the equation for the polynomial graphed here? Solution Since the output grows large and positive as the inputs grow large and positive, we describe the long run behavior symbolically by writing: as x→∞x→∞, f(x)→∞f(x)→∞. Similarly, as x→−∞x→−∞, f(x)→−∞f(x)→−∞. In words, we could say that as x x values approach infinity, the function values approach infinity, and as x x values approach negative infinity the function values approach negative infinity. We can tell this graph has the shape of an odd degree power function which has not been reflected, so the degree of the polynomial creating this graph must be odd, and the leading coefficient would be positive. Exercise 3.1.3 3.1.3 Given the function f(x)=0.2(x−2)(x+1)(x−5)f(x)=0.2(x−2)(x+1)(x−5) use your algebra skills to write the function in standard polynomial form (as a sum of terms) and determine the leading term, degree, and long run behavior of the function. Answer The leading term is 0.2 x 3 0.2 x 3, so it is a degree 3 polynomial.As x x approaches infinity (or gets very large in the positive direction) f(x)f(x) approaches infinity; as x x approaches negative infinity (or gets very large in the negative direction) f(x)f(x) approaches negative infinity. (Basically the long run behavior is the same as the cubic function). Short Run Behavior Characteristics of the graph such as vertical and horizontal intercepts and the places the graph changes direction are part of the short run behavior of the polynomial. Like with all functions, the vertical intercept is where the graph crosses the vertical axis, and occurs when the input value is zero. Since a polynomial is a function, there can only be one vertical intercept, which occurs at the point (0,a 0)(0,a 0). The horizontal intercepts occur at the input values that correspond with an output value of zero. It is possible to have more than one horizontal intercept. Horizontal intercepts are also called zeros, or roots of the function. Example 3.1.6 3.1.6 Given the polynomial function f(x)=(x−2)(x+1)(x−4)f(x)=(x−2)(x+1)(x−4), written in factored form for your convenience, determine the vertical and horizontal intercepts. Solution The vertical intercept occurs when the input is zero. f(0)=(0−2)(0+1)(0−4)=8 f(0)=(0−2)(0+1)(0−4)=8 The graph crosses the vertical axis at the point (0, 8). The horizontal intercepts occur when the output is zero. 0=(x−2)(x+1)(x−4)0=(x−2)(x+1)(x−4) when x x = 2, -1, or 4. f(x)f(x) has zeros, or roots, at x x = 2, -1, and 4. The graph crosses the horizontal axis at the points (2, 0), (-1, 0), and (4, 0) Notice that the polynomial in the previous example, which would be degree three if multiplied out, had three horizontal intercepts and two turning points – places where the graph changes direction. We will now make a general statement without justifying it – the reasons will become clear later in this chapter. intercepts and turning points of polynomials A polynomial of degree n n will have: At most n n horizontal intercepts. An odd degree polynomial will always have at least one. At most n−1 n−1 turning points Example 3.1.7 3.1.7 What can we conclude about the graph of the polynomial shown here? Solution Based on the long run behavior, with the graph becoming large positive on both ends of the graph, we can determine that this is the graph of an even degree polynomial. The graph has 2 horizontal intercepts, suggesting a degree of 2 or greater, and 3 turning points, suggesting a degree of 4 or greater. Based on this, it would be reasonable to conclude that the degree is even and at least 4, so it is probably a fourth degree polynomial. Exercise 3.1.4 3.1.4 Given the function f(x)=0.2(x−2)(x+1)(x−5)f(x)=0.2(x−2)(x+1)(x−5), determine the short run behavior. Answer Horizontal intercepts are (2, 0) (-1, 0) and (5, 0), the vertical intercept is (0, 2) and there are 2 turns in the graph. Important Topics of this Section Power Functions Polynomials Coefficients Leading coefficient Term Leading Term Degree of a polynomial Long run behavior Short run behavior This page titled 3.1: Power Functions is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by David Lippman & Melonie Rasmussen (The OpenTextBookStore) via source content that was edited to the style and standards of the LibreTexts platform. Back to top 3: Polynomial and Rational Functions. 3.1E: Power Functions (Exercises) Was this article helpful? Yes No Recommended articles 5.3: Power Functions and Polynomial FunctionsSuppose a certain species of bird thrives on a small island. The population can be estimated using a polynomial function. We can use this model to est... 5.2: Power Functions and Polynomial FunctionsSuppose a certain species of bird thrives on a small island. The population can be estimated using a polynomial function. We can use this model to est... 1.2: Basic Classes of FunctionsWe begin by reviewing the basic properties of linear and quadratic functions, and then generalize to include higher-degree polynomials. By combining r... Article typeSection or PageAuthorDavid Lippman & Melonie RasmussenLicenseCC BY-SALicense Version4.0Show Page TOCno Tags power function source@ © Copyright 2025 Mathematics LibreTexts Powered by CXone Expert ® ? The LibreTexts libraries arePowered by NICE CXone Expertand are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. Privacy Policy. Terms & Conditions. Accessibility Statement.For more information contact us atinfo@libretexts.org. Support Center How can we help? Contact Support Search the Insight Knowledge Base Check System Status× contents readability resources tools ☰ 3: Polynomial and Rational Functions. 3.1E: Power Functions (Exercises)
2058	https://math.stackexchange.com/questions/626251/is-there-a-general-algebraic-way-of-solving-this-seating-arrangement-problem	linear algebra - Is there a general, algebraic way of solving this seating arrangement problem? - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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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 Is there a general, algebraic way of solving this seating arrangement problem? Ask Question Asked 11 years, 9 months ago Modified11 years, 9 months ago Viewed 1k times This question shows research effort; it is useful and clear 0 Save this question. Show activity on this post. The following question is from the "OECD Pisa questions" [I found it via a link from the guardian newspaper]. ``` It is Alan's birthday and he is having a party. Seven other people will attend: Amy, Brad, Beth, Charles, Debbie, Emily and Frances. Everyone will sit around the circular dining table. The seating arrangement must meet the following conditions: • Amy and Alan sit together • Brad and Beth sit together • Charles sits next to either Debbie or Emily • Frances sits next to Debbie • Amy and Alan do not sit next to either Brad or Beth • Brad does not sit next to Charles or Frances • Debbie and Emily do not sit next to each other • Alan does not sit next to either Debbie or Emily • Amy does not sit next to Charles Arrange the guests around the table to meet all of the conditions listed above. ``` If I were to use each name as a variable name, and their seat number as the value of that variable (wrapping 8's to 1's - yes I realize there's still an issue around that); and arbitrarily assigning Alan = 1, we have: Alan = 1 Amy = (2 or 7) Brad = (Beth + 1) or (Beth - 1) Charles = (Debbie + 1) or (Debbie - 1) or (Emily + 1) or (Emily-1) ...and now we get into a bunch more or's and equations stating inequalities. I'm not really sure if I could solve it this way...or whether this is even be a good way of looking at the problem. When I solved it by trial-and-error, I treated the pairs that had to sit together as one, and just swapped their positions whenever some inconsistency came up. I think I was basically just solving via brute force. Is there a better way? I'm not really interested in the actual solution per se (I found a solution through trial and error in a few minutes), but an algebraic way of finding a solution to these types of problems in general. And by these types of problems, I guess I mean this class of word problems. Is this still just a bunch of linear equations when it involves inequalities and logical operators? linear-algebra Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications asked Jan 3, 2014 at 19:44 GerratGerrat 97 7 7 bronze badges Add a comment\| 3 Answers 3 Sorted by: Reset to default This answer is useful 2 Save this answer. Show activity on this post. One way to simplify the circular part is to seat (say) Amy and Alan, chop the table between them, and then unroll the table. So Alan sits in space 1 (say) and Amy in space 8. You can do this because it doesn't matter where Alan sits, so you can pick space 1. It also doesn't matter whether Amy sits to his right or his left, so pick one of those. Now you have hard boundary conditions for the other six, and it's clear who's sitting next to whom without needing to "wrap around the end of the table." From here on, though, the "polarity" of each pair matters, since we've broken the symmetry of the problem. Whether Brad is to Beth's right or to her left matters now. Now, Alan seems pretty picky. There are only two people left that he can sit next to: Charles or Frances. The breadth of the search is reduced considerably if you tackle this one next; you've eliminated a whole bunch of possibilities. If Charles or Frances is in space 2, then Brad cannot be in space 3. Which means that Beth cannot be in space 4. To sum up, look for the "long poles" in the tent to reduce your search space, and simplify out things by symmetry. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Jan 3, 2014 at 20:46 answered Jan 3, 2014 at 20:16 JohnJohn 26.6k 3 3 gold badges 41 41 silver badges 63 63 bronze badges 4 Yup. I'm really just looking for the "mathy" way of solving it.Gerrat –Gerrat 2014-01-03 20:32:00 +00:00 Commented Jan 3, 2014 at 20:32 I deleted that last sentence. I should stop selling my answers short like that.John –John 2014-01-03 20:47:17 +00:00 Commented Jan 3, 2014 at 20:47 This isn't really what I'm looking for, but you get a +1 for the neat trick in your first sentence. Nice.Gerrat –Gerrat 2014-01-03 21:01:46 +00:00 Commented Jan 3, 2014 at 21:01 I am curious to know how I may adapt similar concept to fit into an algorithm for dynamic seating in places like cafes where we merge and split tables...bonCodigo –bonCodigo 2014-06-17 01:03:25 +00:00 Commented Jun 17, 2014 at 1:03 Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. These problems are best solved with methods of discrete mathematics (combinations, permutations, etc). You see many of these type of questions in coding problems (projecteuler.net is a good example) where one is to write an efficient algorithm to solve the problem at hand. Dynamic programming is usually used to solve such problems. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Jan 3, 2014 at 20:15 abden003abden003 101 1 1 bronze badge 4 This is a PISA problem, but a programming problem.copper.hat –copper.hat 2014-01-03 20:18:35 +00:00 Commented Jan 3, 2014 at 20:18 Did you mean "not a programming problem"?abden003 –abden003 2014-01-03 20:23:06 +00:00 Commented Jan 3, 2014 at 20:23 Yes, thanks, I meant 'not, not 'but'.copper.hat –copper.hat 2014-01-03 20:26:11 +00:00 Commented Jan 3, 2014 at 20:26 When these problems get complicated they can only be solved (in reasonable time) with computers. To solve these problems you need a well thought out algorithm involving a lot of math and cleverness (a naive algorithm would be exponential or n!, which would take too long even on a modern computer). The link between the computer and algorithm is programming. That is why I felt it was appropriate to mention it.abden003 –abden003 2014-01-03 20:35:39 +00:00 Commented Jan 3, 2014 at 20:35 Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. Premises: Amy and Alan sit together Brad and Beth sit together Charles sits next to either Debbie or Emily Frances sits next to Debbie Amy and Alan do not sit next to either Brad or Beth Brad does not sit next to Charles or Frances Debbie and Emily do not sit next to each other Alan does not sit next to either Debbie or Emily Amy does not sit next to Charles Additionally (implicitly), we have: Alan != Amy != Brad != Charles != Frances != Emily != Beth != Debbie Let's designate the seats 1-8, and seat Alan at 1. We'll just start filling in seats that are consistent with the premises and backtrack when we hit an inconsistency: Alan = 1 (arbitrary, so I assigned) Amy = 8 (neat trick John mentioned) Try: Brad = 2...fail (#5) Try: Charles = 2: (next #3 says pick between Debbie and Emily): Try: Debbie = 3, Frances = 4 (from #4), Try: Brad = 5...fail (#6) Try: Beth = 5, Brad = 6 (from #2), Emily = 7 (last seat) ...success!!! Ok, much, much faster than I expected (I think this run just got lucky and didn't have to backtrack). So we have Alan-Charles-Debbie-Frances-Beth-Brad-Emily-Amy (and back to Alan, in a circle). Double checking the original premises, this satisfies. I like this method, as it is algorithmic (but I'd still prefer a (simple) method that solved a system of equations involving inequalities and booleans if such a thing existed). Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Jan 3, 2014 at 21:55 GerratGerrat 97 7 7 bronze badges 1 Ok, I guess one way of categorizing this type of problem is a Constraint satisfaction problem, and backtracking is actually a half decent way to solve these!Gerrat –Gerrat 2014-01-03 22:04:42 +00:00 Commented Jan 3, 2014 at 22:04 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 linear-algebra See similar questions with these tags. 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2059	https://satprepget800.com/2016/11/23/exponential-growth-decay/	Exponential Growth and Decay Home Resources About Contact Books Blog Join the Newsletter Learn strategies to help you get a perfect score and into your dream school with the Get 800 Newsletter. Exponential Growth and Decay The current version of the SAT gives problems on exponential growth and decay. Here are the basics that you should know if you want to get a perfect SAT score: A general exponential function has the form f(t) = a(1 +r)ct, where a=f(0) is the initial amount and r is the growth rate. If r> 0, then we have exponential growth and if r< 0 we have exponential decay. Examples:(1) The exponential function f(t) = 300(2)t can be used to model a population with a growth rate of 1 = 100% each year that begins with 300 specimens. The growth rate of 100% tells us that the population doubles each year. (2) The exponential function f(t) = 50(3)2t can be used to model a population with a growth rate of 2 = 200% every 6 months that begins with 50 specimens. The growth rate of 200% tells us that the population triples. Since c= 2, the tripling occurs every 1/2 year or 6 months. (3) The exponential function f(t) = 120(0.75)t/3 can be used to model a substance which is decaying at a rate of 1 – 0.75 = 0.25 = 25% every 3 years. The initial amount of the substance might be 120 grams. Since c= 1/3, the 25% decay occurs every 3 years. (4) A quantity that continually doubles over a fixed time period can be modeled by the exponential function f(t) = a(2)t/d where a is the quantity at time t= 0, and d is the doubling time in years. Now try the following SAT math problem. I will post a solution after Thanksgiving, but feel free to post your own solutions in the comments meanwhile. Level 4 SAT Exponential Growth On January 1, 2015, a family living on an island releases their two pet rabbits into the wild. Due to the short gestation period of rabbits, and the fact that the rabbits have no natural predators on this island, the rabbit population doubles each month. If P represents the rabbit population years after January 1, 2015, then which of the following equations best models the rabbit population on this island over time? A)P= 2 t/12 + 1 B)P= 2 t+ 1 C)P= 2 12 t D)P= 2 12 t + 1 More SAT and ACT Math Problems with Explanations If you are preparing for the SAT or ACT, you may want to take a look at the Get 800 collection of test prep books. And if you liked this article, please share it with your Facebook friends: Comments comments
2060	https://mathspace.co/textbooks/syllabuses/Syllabus-949/topics/Topic-18867/subtopics/Subtopic-255081/	Book a Demo Topics Conic Sections & Parametric Equations 9.01 Review: Completing the square 9.02 Parabolas 9.03 Circles 9.04 Ellipses 9.05 Hyperbolas 9.06 Eccentricity and identifying conic sections LessonPractice 9.07 Parametric equations 9.08 Applications of parametric equations Log inSign upBook a Demo United States of America Algebra 1- Precalculus 9.06 Eccentricity and identifying conic sections Lesson What is Mathspace About Mathspace
2061	https://www.youtube.com/watch?v=-xGANPeP3iw	Art of Problem Solving: Radical Equations Art of Problem Solving 103000 subscribers 14 likes Description 700 views Posted: 9 Jan 2025 Art of Problem Solving's Richard Rusczyk gets us started with solving equations including radical expressions. This video is part of our AoPS Algebra curriculum. Take your math skills to the next level with our Algebra materials: 📚 AoPS Introduction to Algebra Textbook: 🖥️ AoPS Introduction to Algebra A Course (Textbook Chapters 1-13): 🖥️ AoPS Introduction to Algebra B Course (Textbook Chapters 10-22): 🔔 Subscribe to our channel for more engaging math videos and updates Transcript:
2062	https://theoremoftheweek.wordpress.com/2010/08/18/theorem-34-halls-marriage-theorem/	Theorem of the week Theorem 34: Hall’s marriage theorem I was recently given a box of chocolates. This was a very nice surprise, but closer inspection revealed that I don’t like all of the flavours in the box. There were twelve chocolates, and I like six of them. I thought that perhaps I could invite some other people to share the chocolates with me, hoping that they’d take the ones I don’t like. Naturally, if I do that I have to make sure that everyone gets one chocolate each, to be fair. But I expect that some of them will like only certain chocolates too. So I thought that I’d ask eleven people which of these chocolates they like. That would give me twelve lists of acceptable chocolates (including my own). But will I be able to assign chocolates to people so that everybody gets a chocolate they like? If not, I’ll have to revise my guest list. Well, I could try assigning theoretical chocolates to people in advance, to check. But that sounds terribly time-consuming. Is there some easy way in which I could tell whether it will be possible? Are there any circumstances in which I could immediately tell that it would be impossible? Here’s a scenario in which it’s obviously impossible. Twelve people, twelve chocolates. So no spares (on either side). So if everybody tells me that, like me, they don’t like Scotch whisky chocolates, we’re going to have a problem: there’ll be twelve people fighting over the other eleven chocolates. I should uninvite one person, and find someone who likes whisky in their chocolate. Can I say more? Well, we could extend the above idea. I only like six of the chocolates. If there are six other people who also only like those same six chocolates, we’re going to have a disagreement: seven people, six chocolates. Disaster! Hopefully you’re starting to get the idea. Let’s be a bit more general. For any collection of k people, I need to know that they (between them) like at least k different chocolates, don’t I? So we have a necessary condition: any collection of k people must like at least k different chocolates. If that condition isn’t satisfied, then I’m not going to be able to share the chocolates to everyone’s satisfaction. Is that also a sufficient condition? That is, if we know that condition is satisfied, can we be sure that I can share the chocolates, or might there be some other problem? This week’s theorem tells us that this is both a necessary and a sufficient condition. Theorem (Hall) I can match chocolates to people so that everyone gets a chocolate they like if, and only if, for each k (between 1 and 12), each possible group of k people likes at least k different chocolates between them. OK, Hall did not write about chocolates. I’ve usually come across this theorem in graph theory, where it talks about finding matchings in bipartite graphs, and where it’s usually phrased in terms of marrying men and women (which is why it’s called Hall’s marriage theorem). But it can also be phrased in the language of group theory (in a way that Wikipedia seems to make particularly hard to understand). It’s one of those theorems where one direction is pretty easy (it didn’t take us long to prove that the condition was necessary), but the other direction takes a bit more work (it’s certainly not obvious that the condition is sufficient). I wrote about this phenomenon in Theorem 1 on this blog (for example). I think it’s pretty interesting that the condition is indeed sufficient here, because it really isn’t obvious. I think I’m going to skip the proof here. There are various ways of proving the result, including induction (which is the proof that the Wikipedia page describes). You should be able to find a proof in a graph theory textbook, or just try searching online (or try to prove it for yourself!). There are some very nice proofs, and they’re not terribly difficult. Where next? It turns out that it’s possible to generalise this result. For example, what would happen if I decided to invite just five people, so that we each get two chocolates? What would the necessary and sufficient condition be then? (You might think of this as the polygamous version of Hall’s theorem!) I’ll leave this as a nice exercise for you. Of course, I have no intention of allocating chocolates in this way. I intend to eat the ones that I like, and then pass the rest on to some lucky person (who may or may not like Scotch whisky and Earl Grey tea chocolates). But that doesn’t make for a good blog post! Share this: Related This entry was posted on August 18, 2010 at 7:57 am and is filed under Cambridge Maths Tripos, II Graph Theory, Theorem. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site. 4 Responses to “Theorem 34: Hall’s marriage theorem” hey, nice blog…really like it and added to bookmarks. keep up with good work […] been a bit vague here. The only key idea that I haven’t mentioned, though, is the use of Hall’s marriage theorem to show that it’s possible to do the `joining’ I’ve […] Thanks a lot for avoiding the term marriage or matching men an women! At least I could finally understand the idea of this theorem without getting my head on anoying analogies. I love this post. You have a good sense of humour. 🙂 Leave a comment Δ Create a free website or blog at WordPress.com. Entries (RSS) and Comments (RSS).
2063	https://courses.lumenlearning.com/slcc-elementaryalgebra/chapter/3-1-intro-functions/	3.2: Intro to Functions \| Elementary Algebra Skip to main content Elementary Algebra Module 3: Linear Functions and Their Graphs Search for: 3.2: Intro to Functions section 3.2 Learning Objectives 3.2: Introduction to Functions Determine if a relation is a function Identify the domain and range of a set of ordered pairs Evaluate a function written in function notation Find the domain of a rational function There are many kinds of relations. A relation is simply a correspondence between sets of values or information. Think about members of your family and their ages. The pairing of each member of your family and their age is a relation. Each family member can be paired with an age in the set of ages of your family members. Another example of a relation is the pairing of a state with its U.S. senators. Each state can be matched with two individuals who have each been elected to serve as a senator. In turn, each senator can be matched with one specific state that he or she represents. Both of these are real-life examples of relations. Determine if a relation is a function The first value of a relation is an input value and the second value is the output value. A function is a specific type of relation in which each input value has one and only one output value. An input is the independent value, and the output is the dependent value, as it depends on the value of the input. Which set of values make up the input values and output values can affect whether or not the relation is a function. Notice in the first table below, where the input is “name” and the output is “age,” each input matches with exactly one output. This is an example of a function. \| Family Member’s Name (Input) \| Family Member’s Age (Output) \| --- \| \| Nellie \| 13 13 \| \| Marcos \| 11 11 \| \| Esther \| 46 46 \| \| Samuel \| 47 47 \| \| Nina \| 47 47 \| \| Paul \| 47 47 \| \| Katrina \| 21 21 \| \| Andrew \| 16 16 \| \| Maria \| 13 13 \| \| Ana \| 81 81 \| Compare this with the next table where the input is “age” and the output is “name.” Some of the inputs result in more than one output. This is an example of a correspondence that is not a function. \| Family Member’s Age (Input) \| Family Member’s Name (Output) \| --- \| \| 11 11 \| Marcos \| \| 13 13 \| Nellie, Maria \| \| 16 16 \| Andrew \| \| 21 21 \| Katrina \| \| 46 46 \| Esther \| \| 47 47 \| Samuel, Nina, Paul \| \| 81 81 \| Ana \| Now let us look at some other examples to determine whether the relations are functions or not and under what circumstances. Remember that a relation is a function if there is only one output for each input. Example 1 Fill in the table. \| Input \| Output \| Function? \| Why or why not? \| --- --- \| \| Name of senator \| Name of state \| \| \| \| Name of state \| Name of senator \| \| \| \| Time elapsed \| Height of a tossed ball \| \| \| \| Height of a tossed ball \| Time elapsed \| \| \| \| Number of cars \| Number of tires \| \| \| \| Number of tires \| Number of cars \| \| \| Show Solution \| Input \| Output \| Function? \| Why or why not? \| --- --- \| \| Name of senator \| Name of state \| Yes \| For each input, there will only be one output because a senator only represents one state. \| \| Name of state \| Name of senator \| No \| For each state that is an input, 2 names of senators would result because each state has two senators. \| \| Time elapsed \| Height of a tossed ball \| Yes \| At a specific time, the ball has one specific height. \| \| Height of a tossed ball \| Time elapsed \| No \| Remember that the ball was tossed up and fell down. So for a given height, there could be two different times when the ball was at that height. The input height can result in more than one output. \| \| Number of cars \| Number of tires \| Yes \| For any input of a specific number of cars, there is one specific output representing the number of tires. \| \| Number of tires \| Number of cars \| Yes \| For any input of a specific number of tires, there is one specific output representing the number of cars (assuming each car has all four of its tires). \| Relations can be written as ordered pairs of numbers or as numbers in a table of values. Unless stated otherwise, the first number of an ordered pair is the x-value and is the input, while the second number is the y-value and is the output.By examining the inputs (x-coordinates) and outputs (y-coordinates), you can determine whether or not the relation is a function. Remember, in a function, each input has only one output. Identify the domain and range of a set of ordered pairs There is a name for the set of input values and another name for the set of output values for a function. The set of input values is called the domain of the function. The set of output values is called the range of the function. If you have a set of ordered pairs, you can find the domain by listing all of the input values, which are the x-coordinates. To find the range, list all of the output values, which are the y-coordinates. Consider the following set of ordered pairs: {(−2,0),(0,6),(2,12),(4,18)}{(−2,0),(0,6),(2,12),(4,18)} You have the following: Domain:{−2,0,2,4}Range:{0,6,12,18}Domain:{−2,0,2,4}Range:{0,6,12,18} Now try it yourself. Example 2 List the domain and range for the following table of values where x is the input and y is the output. Then determine if the relation is a function. \| x \| y \| --- \| \| −3−3 \| 4 4 \| \| −2−2 \| 4 4 \| \| −1−1 \| 4 4 \| \| 2 2 \| 4 4 \| \| 3 3 \| 4 4 \| Show Solution The domain describes all the inputs, and we can use set notation with brackets { } to make the list. Domain:{−3,−2,−1,2,3}Domain:{−3,−2,−1,2,3} The range describes all the outputs. Range:{4}Range:{4} We only listed 4 4 once because it is not necessary to list a number every time it appears in a set. Each input has only one output, and the fact that it is the same output (4) does not matter. Therefore, this relation is a function. A helpful way to visualize this is through an “arrow diagram.” We list our domain and range (and as mentioned above with our range, do not list an element more than once in either set). Then, for each input in the range, we draw an arrow to the output it maps to in the range. This table of values corresponds to the arrow diagram shown below. Since each input maps to only one output (equivalently, there is only one arrow coming from each input), the relation is a function. Answer Domain:{−3,−2,−1,2,3}Domain:{−3,−2,−1,2,3} Range:{4}Range:{4} The relation is a function. In the following video we provide another example of identifying whether a table of values represents a function as well as determining the domain and range of each. Example 3 Define the domain and range for the following set of ordered pairs, and determine whether the relation given is a function. {(−3,−6),(−2,−1),(1,0),(1,5),(2,0)}{(−3,−6),(−2,−1),(1,0),(1,5),(2,0)} Show Solution We list all of the input values as the domain. The input values are represented first in the ordered pair as a matter of convention. Domain: {−3,−2,1,2−3,−2,1,2} Note how we did not enter repeated values more than once; it is not necessary. The range is the list of outputs for the relation; they are entered second in the ordered pair. Range: {−6,−1,0,5−6,−1,0,5} Organizing the ordered pairs in a table can help you tell whether this relation is a function. By definition, the inputs in a function have only one output. \| x \| y \| --- \| \| −3−3 \| −6−6 \| \| −2−2 \| −1−1 \| \| 1 1 \| 0 0 \| \| 1 1 \| 5 5 \| \| 2 2 \| 0 0 \| The relation is not a function because the input 1 1 has two outputs:0 0 and 5 5. Again, we could use an arrow diagram to see this visually. Making sure to only list the input 1 once in our domain, we get the following diagram: Because the input of 1 maps to two different outputs (two arrows coming from one input), this confirms that the relation is not a function. Answer Domain: {−3,−2,1,2−3,−2,1,2} Range: {−6,−1,0,5−6,−1,0,5} The relation is not a function. In the following video, we show how to determine whether a relation is a function and how to find the domain and range. Summary: Determining Whether a Relation is a Function Identify the input values – this is your domain. Identify the output values – this is your range. If each value in the domain leads to only one value in the range, classify the relationship as a function. If any value in the domain leads to two or more values in the range, do not classify the relationship as a function. Function notation Some people think of functions as “mathematical machines.” Imagine you have a machine that changes a number according to a specific rule such as “multiply by 3 3 and add 2 2” or “divide by 5 5, add 25 25, and multiply by −1−1.” If you put a number into the machine, a new number will pop out the other end having been changed according to the rule. The number that goes in is called the input, and the number that is produced is called the output. You can also call the machine “f” for function. If you put x into the machine, f(x), comes out. Mathematically speaking, x is the input, or the “independent variable,” and f(x) is the output, or the “dependent variable,” since it depends on the value of x. f(x)=4 x+1 f(x)=4 x+1 is written in function notation and is read “f of x equals 4 x 4 x plus 1” It represents the following situation: A function named f acts upon an input, x, and produces f(x) which is equal to 4 x+1 4 x+1. This is the same as the equation y=4 x+1 y=4 x+1. Function notation gives you more flexibility because you do not have to use y y for every equation. Instead, you could use f(x)f(x) or g(x)g(x)or even c(x)c(x). This can be a helpful way to distinguish equations of functions when you are dealing with more than one at a time. Using Function Notation Once we determine that a relationship is a function, we need to display and define the functional relationship so that we can understand it, use it, and possibly even program it into a computer. There are various ways of representing functions. A standard function notation is one representation that facilitates working with functions. A General Note: Function Notation The notation y=f(x)y=f(x) defines a function named f f. This is read as “y y is a function of x x.” The letter x x represents the input value, or independent variable. The letter y y or f(x)f(x), represents the output value, or dependent variable. Example 4 Use function notation to represent a function whose input is the name of a month and output is the number of days in that month. Show Solution The number of days in a month is a function of the name of the month, so if we name the function f f, we write f(month)=days f(month)=days or f(m)=d f(m)=d. The name of the month is the input to a “rule” that associates a specific number (the output) with each input. For example, f(March)=31 f(March)=31, because March has 31 31 days. The notation d=f(m)d=f(m) reminds us that the number of days, d d (the output), is dependent on the name of the month, m m (the input). Note that the inputs to a function do not have to be numbers; function inputs can be names of people, labels of geometric objects, or any other element that determines some kind of output. However, most of the functions we will work with here will have numbers as inputs and outputs. Example 5 A function f(y)=N f(y)=N gives the number of police officers, N N, in a town in year y y. What does f(2005)=300 f(2005)=300 represent? Show Solution When we read f(2005)=300 f(2005)=300, we see that the input year is 2005. The value for the output, the number of police officers, N N, is 300. The statement f(2005)=300 f(2005)=300 tells us that in the year 2005 there were 300 300 police officers in the town.The notation N=f(y)N=f(y) reminds us that the number of police officers, N N (the output), is dependent on the year, y y (the input). In the following videos we show two more examples of how to express a relationship using function notation. Evaluating a function written in function notation Throughout this course, you have been and will continue working with algebraic equations. Many of these equations are functions. For example, y=4 x+1 y=4 x+1 is an equation that represents a function. When you input values for x, you can determine a single output for y. In this case, if you substitute x=10 x=10 into the equation you will find that y must be 41 41; there is no other value of y that would make the equation true. Rather than using the variable y, the equations of functions can be written using function notation. Function notation is very useful when you are working with more than one function at a time and substituting more than one value in for x. Equations written using function notation can also be evaluated. With function notation, you might see the following: Given f(x)=4 x+1 f(x)=4 x+1 , find f(2)f(2). You read this problem like this: “given f f of x x equals 4 x 4 x plus one, find f f of 2 2.” While the notation and wording is different, the process of evaluating a function is the same as evaluating an expression at a specific value. In both cases, you substitute 2 2 for x x, multiply it by 4 4 and add 1 1, simplifying to get 9 9. In this function, an input of 2 2 results in an output of 9 9. f(x)=4 x+1 f(2)=4(2)+1=8+1=9 f(x)=4 x+1 f(2)=4(2)+1=8+1=9 You can simply apply what you already know about evaluating expressions to evaluate a function. It is important to note that the parentheses that are part of function notation do not mean multiply. The notation f(x)f(x)does not mean f f multiplied by x x. Instead, the notation means “f f of x x” or “the function of x x .” To evaluate the function, take the value given for x x , and substitute that value in for x x in the expression. Let us look at a couple of examples. Example 6 Given f(x)=3 x–4 f(x)=3 x–4,find f(5)f(5). Show Solution Substitute 5 5 in for x in the function. f(5)=3(5)−4 f(5)=3(5)−4 Simplify the expression on the right side of the equation. f(5)=15−4 f(5)=15−4 f(5)=11 f(5)=11 Functions can be evaluated for negative values of x, too. Keep in mind the rules for integer operations. Example 7 Given p(x)=2 x 2+5 p(x)=2 x 2+5, find p(−3)p(−3). Show Solution Substitute −3−3 in for x in the function. p(−3)=2(−3)2+5 p(−3)=2(−3)2+5 Simplify the expression on the right side of the equation. p(−3)=2(9)+5 p(−3)=2(9)+5 p(−3)=18+5 p(−3)=18+5 p(−3)=23 p(−3)=23 You may also be asked to evaluate a function for more than one value as shown in the example that follows. Example 8 Given f(x)=\|4 x−3\|f(x)=\|4 x−3\|, find f(0)f(0), f(2)f(2), and f(−1)f(−1). Show Solution Treat each of these like three separate problems. In each case, you substitute the value in for x and simplify. Start with x=0 x=0. f(0)=\|4(0)−3\|=\|−3\|=3 f(0)=\|4(0)−3\|=\|−3\|=3 f(0)=3 f(0)=3 Evaluate for x=2 x=2. f(2)=\|4(2)−3\|=\|5\|=5 f(2)=\|4(2)−3\|=\|5\|=5 f(2)=5 f(2)=5 Evaluate for x=−1 x=−1. f(−1)=\|4(−1)−3\|=\|−7\|=7 f(−1)=\|4(−1)−3\|=\|−7\|=7 f(−1)=7 f(−1)=7 Next we look at an example where one of the inputs leads to a problem. Example 9 Given f(x)=2 x x+4 f(x)=2 x x+4, find f(0)f(0), f(3)f(3), and f(−4)f(−4). Show Solution We treat this like the previous problems, replacing x x by the given inputs. For the first two, we get f(0)=2(0)0+4 f(0)=2(0)0+4 f(0)=0 4 f(0)=0 4 f(0)=0 f(0)=0 and f(3)=2(3)3+4 f(3)=2(3)3+4 f(3)=6 7 f(3)=6 7 However, if we try plugging in x=−4 x=−4, we encounter an issue since division by 0 is undefined. f(−4)=2(−4)−4+4 f(−4)=2(−4)−4+4 f(−4)=−8 0 f(−4)=−8 0 f(−4)f(−4) is undefined Finding the domain of a rational function The previous example has an important implication about the domain of this fractional function, called arational function. Expanding upon our earlier definition of domain, thedomain of a function is the set of input values that lead to valid output values. It follows that for the function f(x)=2 x x+4 f(x)=2 x x+4, x=−4 x=−4 is not in the domain. Moreover, we can see that −4−4 is the only value of x x that results in a zero denominator, so the domain is all x x-values except x=−4 x=−4. In set-builder notation, we would write the domain as {x\|x≠−4}{x\|x≠−4}. We conclude this section with an example that explicitly asks for the domain of a rational function. Example 10 Find the domain of f(x)=1 3 x+4 f(x)=1 3 x+4. Show Solution As seen in the previous example, the problem we must avoid is a zero denominator. However, for this function, it may be less obvious what value of x x results in a zero denominator. To find this, we set the denominator equal to zero and solve. 3 x+4=0 3 x+4=0 −4−4––––––––––––−4−4 _ 3 x 3=−4 3 3 x 3=−4 3 x=−4 3 x=−4 3 Recall that we don’t want the denominator to be zero, so this is the only value for x x that is not in the domain. In set-builder notation, we would write this as D={x\|x≠−4 3}D={x\|x≠−4 3} Candela Citations CC licensed content, Shared previously Modified from Beginning and Intermediate Algebra. Authored by: Tyler Wallace. License: CC BY: Attribution Provided by: Openstax. Located at: License: CC BY: Attribution Ex: Function Notation Application Problem. Authored by: James Sousa (Mathispower4u.com) . Located at: License: CC BY: Attribution Function Notation Application. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. Located at: License: CC BY: Attribution Provided by: OpenStax. Located at: License: CC BY: Attribution. License Terms: Download for free at : Authored by: James Sousa (Mathispower4u.com) . Located at: License: CC BY: Attribution Licenses and Attributions CC licensed content, Shared previously Modified from Beginning and Intermediate Algebra. Authored by: Tyler Wallace. License: CC BY: Attribution Provided by: Openstax. Located at: License: CC BY: Attribution Ex: Function Notation Application Problem. Authored by: James Sousa (Mathispower4u.com) . Located at: License: CC BY: Attribution Function Notation Application. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. Located at: License: CC BY: Attribution Provided by: OpenStax. Located at: License: CC BY: Attribution. License Terms: Download for free at : Authored by: James Sousa (Mathispower4u.com) . Located at: License: CC BY: Attribution PreviousNext
2064	https://www.nagwa.com/en/videos/390194892059/	Question Video: Investigating the Relation between the Orbital Speed and the Orbital Radius for Objects in Circular Orbits Physics • First Year of Secondary School Which line on the graph shows the relation between orbital speed and orbital radius for objects moving along circular orbits due to gravity? Video Transcript Which line on the graph shows the relation between orbital speed and orbital radius for objects moving along circular orbits due to gravity? The question asks us to identify which of the curves on this graph, which has orbital radius on the horizontal axis and orbital speed on the vertical axis, corresponds to the correct functional relationship between those two quantities. Recall that by equating the centripetal force and the gravitational force for an object in a circular orbit. We find that the orbital speed is equal to the square root of the universal gravitation constant times the mass of the body being orbited divided by the radius of the orbit. Since our question is only about the relationship between orbital speed and orbital radius, we can treat the mass of the body being orbited, that is capital 𝑀, as constant. Then, since capital 𝐺 is also constant, we can rewrite our formula as orbital speed is equal to the square root of a constant divided by the orbital radius, where we’ve used the fact that a constant times a constant is just another constant. Now, using the fact that the square root of a constant is just another constant, we can rewrite this form as orbital speed is equal to a constant divided by the square root of the orbital radius. We don’t actually care about the identity of this constant. So, we can rewrite this equation as a qualitative proportionality relationship. And that relationship is that 𝑣 is proportional to one over the square root of 𝑟. We’ve written the relationship this way to focus on the connection between orbital speed and orbital radius and avoid getting confused by constants that are not relevant to this particular question. Let’s now use this functional form to make predictions for the orbital speed corresponding to large orbital radii, that is the right edge of the graph, and to small orbital radii, that is the left edge of the graph. We can then match those predictions to the appropriate line. Let’s see what happens as 𝑟 increases. As the orbital radius gets larger, the square root of the orbital radius also gets larger. So one over the square root of the orbital radius gets smaller because as the denominator of a fraction grows, the size of the fraction shrinks. So when orbital radius increases, we expect orbital speed to decrease. Although because orbital radius can never be infinite, the orbital speed can never be zero. Looking at the graph, clearly, the green line doesn’t work because it never changes, so it isn’t decreasing with increasing radius. The blue, orange, and red lines all decrease with increasing radius. But it also can’t be the red line because the red line reaches zero. And we know that the orbital speed can never be zero. Let’s now see what happens as the orbital radius shrinks. As 𝑟 decreases, so does the square root of 𝑟. So, the denominator of our fraction is shrinking. And so, the fraction itself is growing. And since the fraction is proportional to orbital speed, the orbital speed is also growing. We also know that as the denominator of a fraction gets closer and closer to zero, the value of that fraction increases without limit. So here, too, we expect the orbital speed to increase without limit as the orbital radius gets closer to zero. The orange curve has a maximum value as the radius approaches zero. So, the orange curve is not the right curve because the orbital speed doesn’t increase without limit. This leaves the blue line as the correct answer. Indeed, the blue line shows an orbital speed that gets smaller and smaller as the radius increases and gets larger and larger without limit as the radius decreases. So, the answer is that the blue line shows the correct relationship between orbital speed and orbital radius for objects moving in circular orbits due to gravity. Lesson Menu Join Nagwa Classes Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher! Nagwa is an educational technology startup aiming to help teachers teach and students learn. Company Content Copyright © 2025 Nagwa All Rights Reserved Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy
2065	https://pressbooks.cuny.edu/oerversiondonotedit/chapter/memory/	Memory – Psychology 2e OpenStax Updated 2025 Skip to content Menu Primary Navigation Home Read Sign in Search in book: Search Book Contents Navigation Contents Preface Acknowledgements and Ancillary Materials 1.Introduction to Psychology 2.Psychological Research 3.Biopsychology 4.States of Consciousness 5.Sensation and Perception 6.Learning 7.Thinking and Intelligence 8.Memory 9.Lifespan Development 10.Emotion and Motivation 11.Personality 12.Social Psychology 13.Industrial-Organizational Psychology 14.Stress, Lifestyle, and Health 15.Psychological Disorders 16.Therapy and Treatment References Psychology 2e OpenStax Updated 2025 8 Figure 8.1 Photographs can trigger our memories and bring past experiences back to life. (credit: modification of work by Cory Zanker) CHAPTER OUTLINE 8.1 How Memory Functions 8.2 Parts of the Brain Involved with Memory 8.3 Problems with Memory 8.4 Ways to Enhance Memory INTRODUCTION:We may be top-notch learners, but if we don’t have a way to store what we’ve learned, what good is the knowledge we’ve gained? Take a few minutes to imagine what your day might be like if you could not remember anything you had learned. You would have to figure out how to get dressed. What clothing should you wear, and how do buttons and zippers work? You would need someone to teach you how to brush your teeth and tie your shoes. Who would you ask for help with these tasks, since you wouldn’t recognize the faces of these people in your house? Wait . . . is this even your house? Uh oh, your stomach begins to rumble and you feel hungry. You’d like something to eat, but you don’t know where the food is kept or even how to prepare it. Oh dear, this is getting confusing. Maybe it would be best just go back to bed. A bed . . . what is a bed? We have an amazing capacity for memory, but how, exactly, do we process and store information? Are there different kinds of memory, and if so, what characterizes the different types? How, exactly, do we retrieve our memories? And why do we forget? This chapter will explore these questions as we learn about memory. 8.1 How Memory Functions LEARNING OBJECTIVES By the end of this section, you will be able to: Discuss the three basic functions of memory Describe the three stages of memory storage Describe and distinguish between procedural and declarative memory and semantic and episodic memory Memory is an information processing system; therefore, we often compare it to a computer.Memory is the set of processes used to encode, store, and retrieve information over different periods of time (Figure 8.2). Figure 8.2 Encoding involves the input of information into the memory system. Storage is the retention of the encoded information. Retrieval, or getting the information out of memory and back into awareness, is the third function. LINK TO LEARNING: Watch this video of unexpected facts about memoryto learn more. Encoding We get information into our brains through a process calledencoding, which is the input of information into the memory system. Once we receive sensory information from the environment, our brains label or code it. We organize the information with other similar information and connect new concepts to existing concepts. Encoding information occurs through automatic processing and effortful processing. If someone asks you what you ate for lunch today, more than likely you could recall this information quite easily. This is known as automatic processing, or the encoding of details like time, space, frequency, and the meaning of words. Automatic processing is usually done without any conscious awareness. Recalling the last time you studied for a test is another example of automatic processing. But what about the actual test material you studied? It probably required a lot of work and attention on your part in order to encode that information. This is known as effortful processing(Figure 8.3). Figure 8.3 When you first learn new skills such as driving a car, you have to put forth effort and attention to encode information about how to start a car, how to brake, how to handle a turn, and so on. Once you know how to drive, you can encode additional information about this skill automatically. (credit: Robert Couse-Baker) What are the most effective ways to ensure that important memories are well encoded? Even a simple sentence is easier to recall when it is meaningful (Anderson, 1984). Read the following sentences (Bransford & McCarrell, 1974), then look away and count backwards from 30 by threes to zero, and then try to write down the sentences (no peeking back at this page!). The notes were sour because the seams split. The voyage wasn’t delayed because the bottle shattered. The haystack was important because the cloth ripped. How well did you do? By themselves, the statements that you wrote down were most likely confusing and difficult for you to recall. Now, try writing them again, using the following prompts: bagpipe, ship christening, and parachutist. Next count backwards from 40 by fours, then check yourself to see how well you recalled the sentences this time. You can see that the sentences are now much more memorable because each of the sentences was placed in context. Material is far better encoded when you make it meaningful. There are three types of encoding. The encoding of words and their meaning is known assemantic encoding. It was first demonstrated by William Bousfield (1935) in an experiment in which he asked people to memorize words. The 60 words were actually divided into 4 categories of meaning, although the participants did not know this because the words were randomly presented. When they were asked to remember the words, they tended to recall them in categories, showing that they paid attention to the meanings of the words as they learned them. Visual encodingis the encoding of images, and acoustic encoding is the encoding of sounds, words in particular. To see how visual encoding works, read over this list of words:car, level, dog, truth, book, value. If you were asked later to recall the words from this list, which ones do you think you’d most likely remember? You would probably have an easier time recalling the words car, dog, and book, and a more difficult time recalling the words level, truth, and value. Why is this? Because you can recall images (mental pictures) more easily than words alone. When you read the words car, dog, and book you created images of these things in your mind. These are concrete, high-imagery words. On the other hand, abstract words like level, truth, and value are low-imagery words. High-imagery words are encoded both visually and semantically (Paivio, 1986), thus building a stronger memory. Now let’s turn our attention to acoustic encoding. You are driving in your car and a song comes on the radio that you haven’t heard in at least 10 years, but you sing along, recalling every word. In the United States, children often learn the alphabet through song, and they learn the number of days in each month through rhyme:“ Thirty days hath September, / April, June, and November; / All the rest have thirty-one, / Save February, with twenty-eight days clear, / And twenty-nine each leap year.” These lessons are easy to remember because of acoustic encoding. We encode the sounds the words make. This is one of the reasons why much of what we teach young children is done through song, rhyme, and rhythm. Which of the three types of encoding do you think would give you the best memory of verbal information? Some years ago, psychologists Fergus Craik and Endel Tulving (1975) conducted a series of experiments to find out. Participants were given words along with questions about them. The questions required the participants to process the words at one of the three levels. The visual processing questions included such things as asking the participants about the font of the letters. The acoustic processing questions asked the participants about the sound or rhyming of the words, and the semantic processing questions asked the participants about the meaning of the words. After participants were presented with the words and questions, they were given an unexpected recall or recognition task. Words that had been encoded semantically were better remembered than those encoded visually or acoustically. Semantic encoding involves a deeper level of processing than the shallower visual or acoustic encoding. Craik and Tulving concluded that we process verbal information best through semantic encoding, especially if we apply what is called the self-reference effect. The self-reference effect is the tendency for an individual to have better memory for information that relates to oneself in comparison to material that has less personal relevance (Rogers, Kuiper, & Kirker, 1977). Could semantic encoding be beneficial to you as you attempt to memorize the concepts in this chapter? Storage Once the information has been encoded, we have to somehow retain it. Our brains take the encoded information and place it in storage.Storageis the creation of a permanent record of information. In order for a memory to go into storage (i.e., long-term memory), it has to pass through three distinct stages:Sensory Memory,Short-Term Memory, and finally Long-Term Memory. These stages were first proposed by Richard Atkinson and Richard Shiffrin(1968). Their model of human memory (Figure 8.4), called Atkinson and Shiffrin's model, is based on the belief that we process memories in the same way that a computer processes information. Figure 8.4 According to the Atkinson-Shiffrin model of memory, information passes through three distinct stages in order for it to be stored in long-term memory. Atkinson and Shiffrin’s model is not the only model of memory. Baddeley and Hitch (1974) proposed a working memory model in which short-term memory has different forms. In their model, storing memories in short-term memory is like opening different files on a computer and adding information. The working memory files hold a limited amount of information. The type of short-term memory (or computer file) depends on the type of information received. There are memories in visual-spatial form, as well as memories of spoken or written material, and they are stored in three short-term systems: a visuospatial sketchpad, an episodic buffer (Baddeley, 2000), and a phonological loop. According to Baddeley and Hitch, a central executive part of memory supervises or controls the flow of information to and from the three short-term systems, and the central executive is responsible for moving information into long-term memory. Sensory Memory In the Atkinson-Shiffrin model, stimuli from the environment are processed first insensory memory: storage of brief sensory events, such as sights, sounds, and tastes. It is very brief storage—up to a couple of seconds. We are constantly bombarded with sensory information. We cannot absorb all of it, or even most of it. And most of it has no impact on our lives. For example, what was your professor wearing the last class period? As long as the professor was dressed appropriately, it does not really matter what she was wearing. Sensory information about sights, sounds, smells, and even textures, which we do not view as valuable information, we discard. If we view something as valuable, the information will move into our short-term memory system. Short-Term Memory Short-term memory (STM)is a temporary storage system that processes incoming sensory memory. The terms short-term and working memory are sometimes used interchangeably, but they are not exactly the same. Short-term memory is more accurately described as a component of working memory. Short-term memory takes information from sensory memory and sometimes connects that memory to something already in long-term memory. Short-term memory storage lasts 15 to 30 seconds. Think of it as the information you have displayed on your computer screen, such as a document, spreadsheet, or website. Then, information in STM goes to long-term memory (you save it to your hard drive), or it is discarded (you delete a document or close a web browser). Rehearsalmoves information from short-term memory to long-term memory. Active rehearsal is a way of attending to information to move it from short-term to long-term memory. During active rehearsal, you repeat (practice) the information to be remembered. If you repeat it enough, it may be moved into long-term memory. For example, this type of active rehearsal is the way many children learn their ABCs by singing the alphabet song. Alternatively, elaborative rehearsal is the act of linking new information you are trying to learn to existing information that you already know. For example, if you meet someone at a party and your phone is dead but you want to remember his phone number, which starts with area code 203, you might remember that your uncle Abdul lives in Connecticut and has a 203 area code. This way, when you try to remember the phone number of your new prospective friend, you will easily remember the area code. Craik and Lockhart (1972) proposed the levels of processing hypothesis that states the deeper you think about something, the better you remember it. You may find yourself asking, “How much information can our memory handle at once?” To explore the capacity and duration of your short-term memory, have a partner read the strings of random numbers (Figure 8.5) out loud to you, beginning each string by saying, “Ready?” and ending each by saying, “Recall,” at which point you should try to write down the string of numbers from memory. Figure 8.5 Work through this series of numbers using the recall exercise explained above to determine the longest string of digits that you can store. Note the longest string at which you got the series correct. For most people, the capacity will probably be close to 7 plus or minus 2. In 1956, George Miller reviewed most of the research on the capacity of short-term memory and found that people can retain between 5 and 9 items, so he reported the capacity of short-term memory was the “magic number” 7 plus or minus 2. However, more contemporary research has found working memory capacity is 4 plus or minus 1 (Cowan, 2010). Generally, recall is somewhat better for random numbers than for random letters (Jacobs, 1887) and also often slightly better for information we hear (acoustic encoding) rather than information we see (visual encoding) (Anderson, 1969). Memory trace decay and interference are two factors that affect short-term memory retention. Peterson and Peterson (1959) investigated short-term memory using the three-letter sequences called trigrams (e.g., CLS) that had to be recalled after various time intervals between 3 and 18 seconds. Participants remembered about 80% of the trigrams after a 3-second delay, but only 10% after a delay of 18 seconds, which caused them to conclude that short-term memory decayed in 18 seconds. During decay, the memory trace becomes less activated over time, and the information is forgotten. However, Keppel and Underwood (1962) examined only the first trials of the trigram task and found that proactive interference also affected short-term memory retention. During proactive interference, previously learned information interferes with the ability to learn new information. Both memory trace decay and proactive interference affect short-term memory. Once the information reaches long-term memory, it has to be consolidated at both the synaptic level, which takes a few hours, and into the memory system, which can take weeks or longer. Long-term Memory Long-term memory (LTM)is the continuous storage of information. Unlike short-term memory, long-term memory storage capacity is believed to be unlimited. It encompasses all the things you can remember that happened more than just a few minutes ago. One cannot really consider long-term memory without thinking about the way it is organized. Really quickly, what is the first word that comes to mind when you hear “peanut butter”? Did you think of jelly? If you did, you probably have associated peanut butter and jelly in your mind. It is generally accepted that memories are organized in semantic (or associative) networks (Collins & Loftus, 1975). A semantic network consists of concepts, and as you may recall from what you’ve learned about memory, concepts are categories or groupings of linguistic information, images, ideas, or memories, such as life experiences. Although individual experiences and expertise can affect concept arrangement, concepts are believed to be arranged hierarchically in the mind (Anderson & Reder, 1999; Johnson & Mervis, 1997, 1998; Palmer, Jones, Hennessy, Unze, & Pick, 1989; Rosch, Mervis, Gray, Johnson, & Boyes-Braem, 1976; Tanaka & Taylor, 1991). Related concepts are linked, and the strength of the link depends on how often two concepts have been associated. Semantic networks differ depending on personal experiences. Importantly for memory, activating any part of a semantic network also activates the concepts linked to that part to a lesser degree. The process is known as spreading activation (Collins & Loftus, 1975). If one part of a network is activated, it is easier to access the associated concepts because they are already partially activated. When you remember or recall something, you activate a concept, and the related concepts are more easily remembered because they are partially activated. However, the activations do not spread in just one direction. When you remember something, you usually have several routes to get the information you are trying to access, and the more links you have to a concept, the better your chances of remembering. There are two types of long-term memory:explicit and implicit(Figure 8.6). Understanding the difference between explicit memory and implicit memory is important because aging, particular types of brain trauma, and certain disorders can impact explicit and implicit memory in different ways.Explicit memories are those we consciously try to remember, recall, and report. For example, if you are studying for your chemistry exam, the material you are learning will be part of your explicit memory. In keeping with the computer analogy, some information in your long-term memory would be like the information you have saved on the hard drive. It is not there on your desktop (your short-term memory), but most of the time you can pull up this information when you want it. Not all long-term memories are strong memories, and some memories can only be recalled using prompts. For example, you might easily recall a fact, such as the capital of the United States, but you might struggle to recall the name of the restaurant at which you had dinner when you visited a nearby city last summer. A prompt, such as that the restaurant was named after its owner, might help you recall the name of the restaurant. Explicit memory is sometimes referred to as declarative memory, because it can be put into words. Explicit memory is divided into episodic memory and semantic memory. LINK TO LEARNING: View this video that explains short-term and long-term memoryto learn more about how memories are stored and retrieved. Episodic memoryis information about events we have personally experienced (i.e., an episode). For instance, the memory of your last birthday is an episodic memory. Usually, episodic memory is reported as a story. The concept of episodic memory was first proposed about in the 1970s (Tulving, 1972). Since then, Tulving and others have reformulated the theory, and currently scientists believe that episodic memory is memory about happenings in particular places at particular times—the what, where, and when of an event (Tulving, 2002). It involves recollection of visual imagery as well as the feeling of familiarity (Hassabis & Maguire, 2007).Semantic memoryis knowledge about words, concepts, and language-based knowledge and facts. Semantic memory is typically reported as facts. Semantic means having to do with language and knowledge about language. For example, answers to the following questions like “what is the definition of psychology” and “who was the first African American president of the United States” are stored in your semantic memory. Implicit memories are long-term memories that are not part of our consciousness. Although implicit memories are learned outside of our awareness and cannot be consciously recalled, implicit memory is demonstrated in the performance of some task (Roediger, 1990; Schacter, 1987). Implicit memory has been studied with cognitive demand tasks, such as performance on artificial grammars (Reber, 1976), word memory (Jacoby, 1983; Jacoby & Witherspoon, 1982), and learning unspoken and unwritten contingencies and rules (Greenspoon, 1955; Giddan & Eriksen, 1959; Krieckhaus & Eriksen, 1960). Returning to the computer metaphor, implicit memories are like a program running in the background, and you are not aware of their influence. Implicit memories can influence observable behaviors as well as cognitive tasks. In either case, you usually cannot put the memory into words that adequately describe the task. There are several types of implicit memories, including procedural, priming, and emotional conditioning. Figure 8.6 There are two components of long-term memory: explicit and implicit. Explicit memory includes episodic and semantic memory. Implicit memory includes procedural memory and things learned through conditioning. Implicitprocedural memoryis often studied using observable behaviors (Adams, 1957; Lacey & Smith, 1954; Lazarus & McCleary, 1951). Implicit procedural memory stores information about the way to do something, and it is the memory for skilled actions, such as brushing your teeth, riding a bicycle, or driving a car. You were probably not that good at riding a bicycle or driving a car the first time you tried, but you were much better after doing those things for a year. Your improved bicycle riding was due to learning balancing abilities. You likely thought about staying upright in the beginning, but now you just do it. Moreover, you probably are good at staying balanced, but cannot tell someone the exact way you do it. Similarly, when you first learned to drive, you probably thought about a lot of things that you just do now without much thought. When you first learned to do these tasks, someone may have told you how to do them, but everything you learned since those instructions that you cannot readily explain to someone else as the way to do it is implicit memory. Implicit priming is another type of implicit memory (Schacter, 1992). During priming exposure to a stimulus affects the response to a later stimulus. Stimuli can vary and may include words, pictures, and other stimuli to elicit a response or increase recognition. For instance, some people really enjoy picnics. They love going into nature, spreading a blanket on the ground, and eating a delicious meal. Now, unscramble the following letters to make a word. AETPL What word did you come up with? Chances are good that it was “plate.” Had you read, “Some people really enjoy growing flowers. They love going outside to their garden, fertilizing their plants, and watering their flowers,” you probably would have come up with the word “petal” instead of plate. Do you recall the earlier discussion of semantic networks? The reason people are more likely to come up with “plate” after reading about a picnic is that plate is associated (linked) with picnic. Plate was primed by activating the semantic network. Similarly, “petal” is linked to flower and is primed by flower. Priming is also the reason you probably said jelly in response to peanut butter. Implicit emotional conditioning is the type of memory involved in classically conditioned emotion responses (Olson & Fazio, 2001). These emotional relationships cannot be reported or recalled but can be associated with different stimuli. For example, specific smells can cause specific emotional responses for some people. If there is a smell that makes you feel positive and nostalgic, and you don’t know where that response comes from, it is an implicit emotional response. Similarly, most people have a song that causes a specific emotional response. That song’s effect could be an implicit emotional memory (Yang, Xu, Du, Shi, & Fang, 2011). EVERYDAY CONNECTION Can You Remember Everything You Ever Did or Said? Episodic memories are also called autobiographical memories. Let’s quickly test your autobiographical memory. What were you wearing exactly five years ago today? What did you eat for lunch on April 10, 2009? You probably find it difficult, if not impossible, to answer these questions. Can you remember every event you have experienced over the course of your life—meals, conversations, clothing choices, weather conditions, and so on? Most likely none of us could even come close to answering these questions; however, American actress Marilu Henner, best known for the television show Taxi, can remember. She has an amazing and highly superior autobiographical memory (Figure 8.7). Figure 8.7 Marilu Henner’s super autobiographical memory is known as hyperthymesia. (credit: Mark Richardson) Very few people can recall events in this way; right now, fewer than 20 have been identified as having this ability, and only a few have been studied (Parker, Cahill & McGaugh 2006). And although hyperthymesia normally appears in adolescence, two children in the United States appear to have memories from well before their tenth birthdays. LINK TO LEARNING: Watch this video about superior autobiographical memoryfrom the television news show 60 Minutes to learn more. Retrieval So you have worked hard to encode (via effortful processing) and store some important information for your upcoming final exam. How do you get that information back out of storage when you need it? The act of getting information out of memory storage and back into conscious awareness is known asretrieval. This would be similar to finding and opening a paper you had previously saved on your computer’s hard drive. Now it’s back on your desktop, and you can work with it again. Our ability to retrieve information from long-term memory is vital to our everyday functioning. You must be able to retrieve information from memory in order to do everything from knowing how to brush your hair and teeth, to driving to work, to knowing how to perform your job once you get there. There are three ways you can retrieve information out of your long-term memory storage system: recall, recognition, and relearning.Recallis what we most often think about when we talk about memory retrieval: it means you can access information without cues. For example, you would use recall for an essay test.Recognitionhappens when you identify information that you have previously learned after encountering it again. It involves a process of comparison. When you take a multiple-choice test, you are relying on recognition to help you choose the correct answer. Here is another example. Let’s say you graduated from high school 10 years ago, and you have returned to your hometown for your 10-year reunion. You may not be able to recall all of your classmates, but you recognize many of them based on their yearbook photos. The third form of retrieval isrelearning, and it’s just what it sounds like. It involves learning information that you previously learned. Whitney took Spanish in high school, but after high school she did not have the opportunity to speak Spanish. Whitney is now 31, and her company has offered her an opportunity to work in their Mexico City office. In order to prepare herself, she enrolls in a Spanish course at the local community center. She’s surprised at how quickly she’s able to pick up the language after not speaking it for 13 years; this is an example of relearning. 8.1 TEST YOURSELF Summary: Summary: 8.1 How Memory Functions Memory is a system or process that stores what we learn for future use. Our memory has three basic functions: encoding, storing, and retrieving information. Encoding is the act of getting information into our memory system through automatic or effortful processing. Storage is retention of the information, and retrieval is the act of getting information out of storage and into conscious awareness through recall, recognition, and relearning. The idea that information is processed through three memory systems is called the Atkinson-Shiffrin model of memory. First, environmental stimuli enter our sensory memory for a period of less than a second to a few seconds. Those stimuli that we notice and pay attention to then move into short-term memory. According to the Atkinson-Shiffrin model, if we rehearse this information, then it moves into long-term memory for permanent storage. Other models like that of Baddeley and Hitch suggest there is more of a feedback loop between short-term memory and long-term memory. Long-term memory has a practically limitless storage capacity and is divided into implicit and explicit memory. 8.2 Parts of the Brain Involved with Memory LEARNING OBJECCTIVES By the end of this section, you will be able to: Explain the brain functions involved in memory Recognize the roles of the hippocampus, amygdala, and cerebellum Are memories stored in just one part of the brain, or are they stored in many different parts of the brain? Karl Lashley began exploring this problem, about 100 years ago, by making lesions in the brains of animals such as rats and monkeys. He was searching for evidence of the engram: the group of neurons that serve as the “physical representation of memory” (Josselyn, 2010). First, Lashley (1950) trained rats to find their way through a maze. Then, he used the tools available at the time—in this case a soldering iron—to create lesions in the rats’ brains, specifically in the cerebral cortex. He did this because he was trying to erase the engram, or the original memory trace that the rats had of the maze. Lashley did not find evidence of the engram, and the rats were still able to find their way through the maze, regardless of the size or location of the lesion. Based on his creation of lesions and the animals’ reaction, he formulated theequipotentiality hypothesis: if part of one area of the brain involved in memory is damaged, another part of the same area can take over that memory function (Lashley, 1950). Although Lashley’s early work did not confirm the existence of the engram, modern psychologists are making progress locating it. For example, Eric Kandel has spent decades studying the synapse and its role in controlling the flow of information through neural circuits needed to store memories (Mayford, Siegelbaum, & Kandel, 2012). Many scientists believe that the entire brain is involved with memory. However, since Lashley’s research, other scientists have been able to look more closely at the brain and memory. They have argued that memory is located in specific parts of the brain, and specific neurons can be recognized for their involvement in forming memories. The main parts of the brain involved with memory are the amygdala, the hippocampus, the cerebellum, and the prefrontal cortex (Figure 8.8). Figure 8.8 The amygdala is involved in fear and fear memories. The hippocampus is associated with declarative and episodic memory as well as recognition memory. The cerebellum plays a role in processing procedural memories, such as how to play the piano. The prefrontal cortex appears to be involved in remembering semantic tasks. The Amygdala First, let’s look at the role of the amygdala in memory formation. The main job of the amygdala is to regulate emotions, such as fear and aggression (Figure 8.8). The amygdala plays a part in how memories are stored because storage is influenced by stress hormones. For example, one researcher experimented with rats and the fear response (Josselyn, 2010). Using Pavlovian conditioning, a neutral tone was paired with a foot shock to the rats. This produced a fear memory in the rats. After being conditioned, each time they heard the tone, they would freeze (a defense response in rats), indicating a memory for the impending shock. Then the researchers induced cell death in neurons in the lateral amygdala, which is the specific area of the brain responsible for fear memories. They found the fear memory faded (became extinct). Because of its role in processing emotional information, the amygdala is also involved in memory consolidation: the process of transferring new learning into long-term memory. The amygdala seems to facilitate encoding memories at a deeper level when the event is emotionally arousing. LINK TO LEARNING:In this TED Talk called “A Mouse. A Laser Beam. A Manipulated Memory,” Steve Ramirez and Xu Liu from MIT talk about using laser beams to manipulate fear memory in rats. Find out why their work caused a media frenzy once it was published in Science. The Hippocampus Another group of researchers also experimented with rats to learn how the hippocampus functions in memory processing (Figure 8.8). They created lesions in the hippocampi of the rats, and found that the rats demonstrated memory impairment on various tasks, such as object recognition and maze running. They concluded that the hippocampus is involved in memory, specifically normal recognition memory as well as spatial memory (when the memory tasks are like recall tests) (Clark, Zola, & Squire, 2000). Another job of the hippocampus is to project information to cortical regions that give memories meaning and connect them with other memories. It also plays a part in memory consolidation: the process of transferring new learning into long-term memory. Injury to this area leaves us unable to process new declarative memories. One famous patient, known for years only as H. M., had both his left and right temporal lobes (hippocampi) removed in an attempt to help control the seizures he had been suffering from for years (Corkin, Amaral, González, Johnson, & Hyman, 1997). As a result, his declarative memory was significantly affected, and he could not form new semantic knowledge. He lost the ability to form new memories, yet he could still remember information and events that had occurred prior to the surgery. The Cerebellum and Prefrontal Cortex Although the hippocampus seems to be more of a processing area for explicit memories, you could still lose it and be able to create implicit memories (procedural memory, motor learning, and classical conditioning), thanks to your cerebellum(Figure 8.8). For example, one classical conditioning experiment is to accustom subjects to blink when they are given a puff of air to the eyes. When researchers damaged the cerebellums of rabbits, they discovered that the rabbits were not able to learn the conditioned eye-blink response (Steinmetz, 1999; Green & Woodruff-Pak, 2000). Other researchers have used brain scans, including positron emission tomography (PET) scans, to learn how people process and retain information. From these studies, it seems the prefrontal cortex is involved. In one study, participants had to complete two different tasks: either looking for the letter a in words (considered a perceptual task) or categorizing a noun as either living or non-living (considered a semantic task) (Kapur et al., 1994). Participants were then asked which words they had previously seen. Recall was much better for the semantic task than for the perceptual task. According to PET scans, there was much more activation in the left inferior prefrontal cortex in the semantic task. In another study, encoding was associated with left frontal activity, while retrieval of information was associated with the right frontal region (Craik et al., 1999). Neurotransmitters There also appear to be specific neurotransmitters involved with the process of memory, such as epinephrine, dopamine, serotonin, glutamate, and acetylcholine (Myhrer, 2003). There continues to be discussion and debate among researchers as to which neurotransmitter plays which specific role (Blockland, 1996). Although we don’t yet know which role each neurotransmitter plays in memory, we do know that communication among neurons via neurotransmitters is critical for developing new memories. Repeated activity by neurons leads to increased neurotransmitters in the synapses and more efficient and more synaptic connections. This is how memory consolidation occurs. It is also believed that strong emotions trigger the formation of strong memories, and weaker emotional experiences form weaker memories; this is called arousal theory(Christianson, 1992). For example, strong emotional experiences can trigger the release of neurotransmitters, as well as hormones, which strengthen memory; therefore, our memory for an emotional event is usually better than our memory for a non-emotional event. When humans and animals are stressed, the brain secretes more of the neurotransmitter glutamate, which helps them remember the stressful event (McGaugh, 2003). This is clearly evidenced by what is known as the flashbulb memory phenomenon. Aflashbulb memoryis an exceptionally clear recollection of an important event (Figure 8.9). Where were you when you first heard about the 9/11 terrorist attacks? Most likely you can remember where you were and what you were doing. In fact, a Pew Research Center (2011) survey found that for those Americans who were age 8 or older at the time of the event, 97% can recall the moment they learned of this event, even a decade after it happened. Figure 8.9 Most people can remember where they were when they first heard about the 9/11 terrorist attacks. This is an example of a flashbulb memory: a record of an atypical and unusual event that has very strong emotional associations. (credit: Michael Foran) DIG DEEPER Inaccurate and False Memories Even flashbulb memories for important events can have decreased accuracy with the passage of time. For example, on at least three occasions, when asked how he heard about the terrorist attacks of 9/11, President George W. Bush responded inaccurately. In January 2002, less than 4 months after the attacks, the then sitting President Bush was asked how he heard about the attacks. He responded: I was sitting there, and my Chief of Staff—well, first of all, when we walked into the classroom, I had seen this plane fly into the first building. There was a TV set on. And you know, I thought it was pilot error and I was amazed that anybody could make such a terrible mistake. (Greenberg, 2004, p. 2) Contrary to what President Bush stated, no one saw the first plane hit, except people on the ground near the twin towers. Video footage of the first plane was not recorded because it was a normal Tuesday morning, until the first plane hit. Memory is not like a video recording. Human memory, even flashbulb memories, can be frail. Different parts of them, such as the time, visual elements, and smells, are stored in different places. When something is remembered, these components have to be put back together for the complete memory, which is known as memory reconstruction. Each component creates a chance for an error to occur.False memory is remembering something that did not happen. Research participants have recalled hearing a word, even though they never heard the word (Roediger & McDermott, 2000). Do you remember where you were when you heard about a historic or perhaps a tragic event? Who were you with and what were you doing? What did you talk about? Can you contact those people you were with? Do they have the same memories as you or do they have different memories? 8.2 TEST YOURSELF Summary: 8.2 Part of the Brain Involved with Memory Beginning with Karl Lashley, researchers and psychologists have been searching for the engram, which is the physical trace of memory. Lashley did not find the engram, but he did suggest that memories are distributed throughout the entire brain rather than stored in one specific area. Now we know that three brain areas do play significant roles in the processing and storage of different types of memories: cerebellum, hippocampus, and amygdala. The cerebellum’s job is to process procedural memories; the hippocampus is where new memories are encoded; the amygdala helps determine what memories to store, and it plays a part in determining where the memories are stored based on whether we have a strong or weak emotional response to the event. Strong emotional experiences can trigger the release of neurotransmitters, as well as hormones, which strengthen memory, so that memory for an emotional event is usually stronger than memory for a non-emotional event. This is shown by what is known as the flashbulb memory phenomenon: our ability to remember significant life events. However, our memory for life events (autobiographical memory) is not always accurate. 8.3 Problems with Memory LEARNING OBJECTIVES By the end of this section, you will be able to: Compare and contrast the two types of amnesia Discuss the unreliability of eyewitness testimony Discuss encoding failure Discuss the various memory errors Compare and contrast the two types of interference You may pride yourself on your amazing ability to remember the birthdates and ages of all of your friends and family members, or you may be able recall vivid details of your 5th birthday party at Chuck E. Cheese’s. However, all of us have at times felt frustrated, and even embarrassed, when our memories have failed us. There are several reasons why this happens. Amnesia Amnesiais the loss of long-term memory that occurs as the result of disease, physical trauma, or psychological trauma. Endel Tulving (2002) and his colleagues at the University of Toronto studied K. C. for years. K. C. suffered a traumatic head injury in a motorcycle accident and then had severe amnesia. Tulving writes, the outstanding fact about K.C.’s mental make-up is his utter inability to remember any events, circumstances, or situations from his own life. His episodic amnesia covers his whole life, from birth to the present. The only exception is the experiences that, at any time, he has had in the last minute or two. (Tulving, 2002, p. 14) Anterograde Amnesia There are two common types of amnesia: anterograde amnesia and retrograde amnesia (Figure 8.10). Anterograde amnesia is commonly caused by brain trauma, such as a blow to the head. With anterograde amnesia, you cannot remember new information, although you can remember information and events that happened prior to your injury. The hippocampus is usually affected (McLeod, 2011). This suggests that damage to the brain has resulted in the inability to transfer information from short-term to long-term memory; that is, the inability to consolidate memories. Many people with this form of amnesia are unable to form new episodic or semantic memories, but are still able to form new procedural memories (Bayley & Squire, 2002). This was true of H. M., which was discussed earlier. The brain damage caused by his surgery resulted in anterograde amnesia. H. M. would read the same magazine over and over, having no memory of ever reading it—it was always new to him. He also could not remember people he had met after his surgery. If you were introduced to H. M. and then you left the room for a few minutes, he would not know you upon your return and would introduce himself to you again. However, when presented the same puzzle several days in a row, although he did not remember having seen the puzzle before, his speed at solving it became faster each day (because of relearning) (Corkin, 1965, 1968). Figure 8.10 This diagram illustrates the timeline of retrograde and anterograde amnesia. Memory problems that extend back in time before the injury and prevent retrieval of information previously stored in long-term memory are known as retrograde amnesia. Conversely, memory problems that extend forward in time from the point of injury and prevent the formation of new memories are called anterograde amnesia. Retrograde Amnesia Retrograde amnesiais loss of memory for events that occurred prior to the trauma. People with retrograde amnesia cannot remember some or even all of their past. They have difficulty remembering episodic memories. What if you woke up in the hospital one day and there were people surrounding your bed claiming to be your spouse, your children, and your parents? The trouble is you don’t recognize any of them. You were in a car accident, suffered a head injury, and now have retrograde amnesia. You don’t remember anything about your life prior to waking up in the hospital. This may sound like the stuff of Hollywood movies, and Hollywood has been fascinated with the amnesia plot for nearly a century, going all the way back to the film Garden of Lies from 1915 to more recent movies such as the Jason Bourne spy thrillers. However, for real-life sufferers of retrograde amnesia, like former NFL football player Scott Bolzan, the story is not a Hollywood movie. Bolzan fell, hit his head, and deleted 46 years of his life in an instant. He is now living with one of the most extreme cases of retrograde amnesia on record. LINK TO LEARNING: View the video story about Scott Bolzan’s amnesia and his attempts to get his life backto learn more. Memory Construction and Reconstruction The formulation of new memories is sometimes calledconstruction, and the process of bringing up old memories is calledreconstruction. Yet as we retrieve our memories, we also tend to alter and modify them. A memory pulled from long-term storage into short-term memory is flexible. New events can be added and we can change what we think we remember about past events, resulting in inaccuracies and distortions. People may not intend to distort facts, but it can happen in the process of retrieving old memories and combining them with new memories (Roediger & DeSoto, 2015). Suggestibility When someone witnesses a crime, that person’s memory of the details of the crime is very important in catching the suspect. Because memory is so fragile, witnesses can be easily (and often accidentally) misled due to the problem of suggestibility.Suggestibilitydescribes the effects of misinformation from external sources that leads to the creation of false memories. In the fall of 2002, a sniper in the DC area shot people at a gas station, leaving Home Depot, and walking down the street. These attacks went on in a variety of places for over three weeks and resulted in the deaths of ten people. During this time, as you can imagine, people were terrified to leave their homes, go shopping, or even walk through their neighborhoods. Police officers and the FBI worked frantically to solve the crimes, and a tip hotline was set up. Law enforcement received over 140,000 tips, which resulted in approximately 35,000 possible suspects (Newseum, n.d.). Most of the tips were dead ends, until a white van was spotted at the site of one of the shootings. The police chief went on national television with a picture of the white van. After the news conference, several other eyewitnesses called to say that they too had seen a white van fleeing from the scene of the shooting. At the time, there were more than 70,000 white vans in the area. Police officers, as well as the general public, focused almost exclusively on white vans because they believed the eyewitnesses. Other tips were ignored. When the suspects were finally caught, they were driving a blue sedan. As illustrated by this example, we are vulnerable to the power of suggestion, simply based on something we see on the news. Or we can claim to remember something that in fact is only a suggestion someone made. It is the suggestion that is the cause of the false memory. Eyewitness Misidentification Even though memory and the process of reconstruction can be fragile, police officers, prosecutors, and the courts often rely on eyewitness identification and testimony in the prosecution of criminals. However, faulty eyewitness identification and testimony can lead to wrongful convictions (Figure 8.11). Figure 8.11 In studying cases where DNA evidence has exonerated people from crimes, the Innocence Project discovered that eyewitness misidentification is the leading cause of wrongful convictions (Benjamin N. Cardozo School of Law, Yeshiva University, 2009). How does this happen? In 1984, Jennifer Thompson, then a 22-year-old college student in North Carolina, was brutally raped at knifepoint. As she was being raped, she tried to memorize every detail of her rapist’s face and physical characteristics, vowing that if she survived, she would help get him convicted. After the police were contacted, a composite sketch was made of the suspect, and Jennifer was shown six photos. She chose two, one of which was of Ronald Cotton. After looking at the photos for 4–5 minutes, she said, “Yeah. This is the one,” and then she added, “I think this is the guy.” When questioned about this by the detective who asked, “You’re sure? Positive?” She said that it was him. Then she asked the detective if she did OK, and he reinforced her choice by telling her she did great. These kinds of unintended cues and suggestions by police officers can lead witnesses to identify the wrong suspect. The district attorney was concerned about her lack of certainty the first time, so she viewed a lineup of seven men. She said she was trying to decide between numbers 4 and 5, finally deciding that Cotton, number 5, “Looks most like him.” He was 22 years old. By the time the trial began, Jennifer Thompson had absolutely no doubt that she was raped by Ronald Cotton. She testified at the court hearing, and her testimony was compelling enough that it helped convict him. How did she go from, “I think it’s the guy” and it “Looks most like him,” to such certainty? Gary Wells and Deah Quinlivan (2009) assert it’s suggestive police identification procedures, such as stacking lineups to make the defendant stand out, telling the witness which person to identify, and confirming witnesses choices by telling them “Good choice,” or “You picked the guy.” After Cotton was convicted of the rape, he was sent to prison for life plus 50 years. After 4 years in prison, he was able to get a new trial. Jennifer Thompson once again testified against him. This time Ronald Cotton was given two life sentences. After serving 11 years in prison, DNA evidence finally demonstrated that Ronald Cotton did not commit the rape, was innocent, and had served over a decade in prison for a crime he did not commit. LINK TO LEARNING: Watch this first video about Ronald Cotton who was falsely convictedand then watch this second video about the task of his accuser to learn more about the fallibility of memory. DIG DEEPER Preserving Eyewitness Memory: The Elizabeth Smart Case Contrast the Cotton case with what happened in the Elizabeth Smart case. When Elizabeth was 14 years old and fast asleep in her bed at home, she was abducted at knifepoint. Her nine-year-old sister, Mary Katherine, was sleeping in the same bed and watched, terrified, as her beloved older sister was abducted. Mary Katherine was the sole eyewitness to this crime and was very fearful. In the following weeks, the Salt Lake City police and the FBI proceeded with caution with Mary Katherine. They did not want to implant any false memories or mislead her in any way. They did not show her police line-ups or push her to do a composite sketch of the abductor. They knew if they corrupted her memory, Elizabeth might never be found. For several months, there was little or no progress on the case. Then, about 4 months after the kidnapping, Mary Katherine first recalled that she had heard the abductor’s voice prior to that night (he had worked exactly one day as a handyman at the family’s home) and then she was able to name the person whose voice it was. The family contacted the press and others recognized him—after a total of nine months, the suspect was caught and Elizabeth Smart was returned to her family. The Misinformation Effect Cognitive psychologist Elizabeth Loftus has conducted extensive research on memory. She has studied false memories as well as recovered memories of childhood sexual abuse. Loftus also developed themisinformation effect paradigm, which holds that after exposure to additional and possibly inaccurate information, a person may misremember the original event. According to Loftus, an eyewitness’s memory of an event is very flexible due to the misinformation effect. To test this theory, Loftus and John Palmer (1974) asked 45 U.S. college students to estimate the speed of cars using different forms of questions (Figure 8.12). The participants were shown films of car accidents and were asked to play the role of the eyewitness and describe what happened. They were asked, “About how fast were the cars going when they (smashed, collided, bumped, hit, contacted) each other?” The participants estimated the speed of the cars based on the verb used. Participants who heard the word “smashed” estimated that the cars were traveling at a much higher speed than participants who heard the word “contacted.” The implied information about speed, based on the verb they heard, had an effect on the participants’ memory of the accident. In a follow-up one week later, participants were asked if they saw any broken glass (none was shown in the accident pictures). Participants who had been in the “smashed” group were more than twice as likely to indicate that they did remember seeing glass. Loftus and Palmer demonstrated that a leading question encouraged them to not only remember the cars were going faster, but to also falsely remember that they saw broken glass. Figure 8.12 When people are asked leading questions about an event, their memory of the event may be altered. (credit a: modification of work by Rob Young) Controversies over Repressed and Recovered Memories Other researchers have described how whole events, not just words, can be falsely recalled, even when they did not happen. The idea that memories of traumatic events could be repressed has been a theme in the field of psychology, beginning with Sigmund Freud, and the controversy surrounding the idea continues today. Recall of false autobiographical memories is calledfalse memory syndrome. This syndrome has received a lot of publicity, particularly as it relates to memories of events that do not have independent witnesses—often the only witnesses to the abuse are the perpetrator and the victim (e.g., sexual abuse). On one side of the debate are those who have recovered memories of childhood abuse years after it occurred. These researchers argue that some children’s experiences have been so traumatizing and distressing that they must lock those memories away in order to lead some semblance of a normal life. They believe that repressed memories can be locked away for decades and later recalled intact through hypnosis and guided imagery techniques (Devilly, 2007). Research suggests that having no memory of childhood sexual abuse is quite common in adults. For instance, one large-scale study conducted by John Briere and Jon Conte (1993) revealed that 59% of 450 men and women who were receiving treatment for sexual abuse that had occurred before age 18 had forgotten their experiences. Ross Cheit (2007) suggested that repressing these memories created psychological distress in adulthood. The Recovered Memory Project was created so that victims of childhood sexual abuse can recall these memories and allow the healing process to begin (Cheit, 2007; Devilly, 2007). On the other side, Loftus has challenged the idea that individuals can repress memories of traumatic events from childhood, including sexual abuse, and then recover those memories years later through therapeutic techniques such as hypnosis, guided visualization, and age regression. Loftus is not saying that childhood sexual abuse doesn’t happen, but she does question whether or not those memories are accurate, and she is skeptical of the questioning process used to access these memories, given that even the slightest suggestion from the therapist can lead to misinformation effects. For example, researchers Stephen Ceci and Maggie Brucks (1993, 1995) asked three-year-old children to use an anatomically correct doll to show where their pediatricians had touched them during an exam. Fifty-five percent of the children pointed to the genital/anal area on the dolls, even when they had not received any form of genital exam. Ever since Loftus published her first studies on the suggestibility of eyewitness testimony in the 1970s, social scientists, police officers, therapists, and legal practitioners have been aware of the flaws in interview practices. Consequently, steps have been taken to decrease suggestibility of witnesses. One way is to modify how witnesses are questioned. When interviewers use neutral and less leading language, children more accurately recall what happened and who was involved (Goodman, 2006; Pipe, 1996; Pipe, Lamb, Orbach, & Esplin, 2004). Another change is in how police lineups are conducted. It’s recommended that a blind photo lineup be used. This way the person administering the lineup doesn’t know which photo belongs to the suspect, minimizing the possibility of giving leading cues. Additionally, judges in some states now inform jurors about the possibility of misidentification. Judges can also suppress eyewitness testimony if they deem it unreliable. Forgetting “I’ve a grand memory for forgetting,” quipped Robert Louis Stevenson.Forgettingrefers to loss of information from long-term memory. We all forget things, like a loved one’s birthday, someone’s name, or where we put our car keys. As you’ve come to see, memory is fragile, and forgetting can be frustrating and even embarrassing. But why do we forget? To answer this question, we will look at several perspectives on forgetting. Encoding Failure Sometimes memory loss happens before the actual memory process begins, which is encoding failure. We can’t remember something if we never stored it in our memory in the first place. This would be like trying to find a book on your e-reader that you never actually purchased and downloaded. Often, in order to remember something, we must pay attention to the details and actively work to process the information (effortful encoding). Lots of times we don’t do this. For instance, think of how many times in your life you’ve seen a penny. Can you accurately recall what the front of a U.S. penny looks like? When researchers Raymond Nickerson and Marilyn Adams (1979) asked this question, they found that most Americans don’t know which one it is. The reason is most likely encoding failure. Most of us never encode the details of the penny. We only encode enough information to be able to distinguish it from other coins. If we don’t encode the information, then it’s not in our long-term memory, so we will not be able to remember it. Figure 8.13 Can you tell which coin, (a), (b), (c), or (d) is the accurate depiction of a US nickel? The correct answer is (c). Memory Errors Psychologist Daniel Schacter (2001), a well-known memory researcher, offers seven ways our memories fail us. He calls them the seven sins of memory and categorizes them into three groups: forgetting, distortion, and intrusion (Table 8.1). Table 8.1 Let’s look at the first sin of the forgetting errors:transience, which means that memories can fade over time. Here’s an example of how this happens. Nathan’s English teacher has assigned his students to read the novel To Kill a Mockingbird. Nathan comes home from school and tells his mom he has to read this book for class. “Oh, I loved that book!” she says. Nathan asks her what the book is about, and after some hesitation, she says, “Well . . . I know I read the book in high school, and I remember that one of the main characters is named Scout, and her father is an attorney, but I honestly don’t remember anything else.” Nathan wonders if his mother actually read the book, and his mother is surprised she can’t recall the plot. What is going on here is storage decay: unused information tends to fade with the passage of time. In 1885, German psychologist Hermann Ebbinghaus analyzed the process of memorization. First, he memorized lists of nonsense syllables. Then he measured how much he learned (retained) when he attempted to relearn each list. He tested himself over different periods of time from 20 minutes later to 30 days later. The result is his famous forgetting curve (Figure 8.14). Due to storage decay, an average person will lose 50% of the memorized information after 20 minutes and 70% of the information after 24 hours (Ebbinghaus, 1885/1964). Your memory for new information decays quickly and then eventually levels out. Figure 8.14 The Ebbinghaus forgetting curve shows how quickly memory for new information decays. Are you constantly losing your cell phone? Have you ever driven back home to make sure you turned off the stove? Have you ever walked into a room for something, but forgotten what it was? You probably answered yes to at least one, if not all, of these examples—but don’t worry, you are not alone. We are all prone to committing the memory error known as absentmindedness, which describes lapses in memory caused by breaks in attention or our focus being somewhere else. Cynthia, a psychologist, recalls a time when she recently committed the memory error of absentmindedness. When I was completing court-ordered psychological evaluations, each time I went to the court, I was issued a temporary identification card with a magnetic strip which would open otherwise locked doors. As you can imagine, in a courtroom, this identification is valuable and important and no one wanted it to be lost or be picked up by a criminal. At the end of the day, I would hand in my temporary identification. One day, when I was almost done with an evaluation, my daughter’s day care called and said she was sick and needed to be picked up. It was flu season, I didn’t know how sick she was, and I was concerned. I finished up the evaluation in the next ten minutes, packed up my briefcase, and rushed to drive to my daughter’s day care. After I picked up my daughter, I could not remember if I had handed back my identification or if I had left it sitting out on a table. I immediately called the court to check. It turned out that I had handed back my identification. Why could I not remember that? (personal communication, September 5, 2013) When have you experienced absentmindedness? “I just streamed this movie called Oblivion, and it had that famous actor in it. Oh, what’s his name? He’s been in all of those movies, like The Shawshank Redemption and The Dark Knight trilogy. I think he’s even won an Oscar. Oh gosh, I can picture his face in my mind, and hear his distinctive voice, but I just can’t think of his name! This is going to bug me until I can remember it!” This particular error can be so frustrating because you have the information right on the tip of your tongue. Have you ever experienced this? If so, you’ve committed the error known as[pb_glossary id=”862″]blocking[/pb_glossary]: you can’t access stored information (Figure 8.15). Figure 8.15 Blocking is also known as tip-of-the-tongue (TOT) phenomenon. The memory is right there, but you can’t seem to recall it, just like not being able to remember the name of that very famous actor, Morgan Freeman. (credit: modification of work by D. Miller) Now let’s take a look at the three errors of distortion: misattribution, suggestibility, and bias. Misattribution happens when you confuse the source of your information. Let’s say Alejandra was dating Lucia and they saw the first Hobbit movie together. Then they broke up and Alejandra saw the second Hobbit movie with someone else. Later that year, Alejandra and Lucia get back together. One day, they are discussing how the Hobbit books and movies are different and Alejandra says to Lucia, “I loved watching the second movie with you and seeing you jump out of your seat during that super scary part.” When Lucia responded with a puzzled and then angry look, Alejandra realized she’d committed the error of misattribution. What if someone is a victim of rape shortly after watching a television program? Is it possible that the victim could actually blame the rape on the person she saw on television because of misattribution? This is exactly what happened to Donald Thomson. Australian eyewitness expert Donald Thomson appeared on a live TV discussion about the unreliability of eyewitness memory. He was later arrested, placed in a lineup and identified by a victim as the man who had raped her. The police charged Thomson although the rape had occurred at the time he was on TV. They dismissed his alibi that he was in plain view of a TV audience and in the company of the other discussants, including an assistant commissioner of police. . . . Eventually, the investigators discovered that the rapist had attacked the woman as she was watching TV—the very program on which Thomson had appeared. Authorities eventually cleared Thomson. The woman had confused the rapist’s face with the face that she had seen on TV. (Baddeley, 2004, p. 133) The second distortion error is suggestibility. Suggestibility is similar to misattribution, since it also involves false memories, but it’s different. With misattribution you create the false memory entirely on your own, which is what the victim did in the Donald Thomson case above. With suggestibility, it comes from someone else, such as a therapist or police interviewer asking leading questions of a witness during an interview. Memories can also be affected bybias, which is the final distortion error. Schacter (2001) says that your feelings and view of the world can actually distort your memory of past events. There are several types of bias: Stereotypical bias involves racial and gender biases. For example, when Asian American and European American research participants were presented with a list of names, they more frequently incorrectly remembered typical African American names such as Jamal and Tyrone to be associated with the occupation basketball player, and they more frequently incorrectly remembered typical White names such as Greg and Howard to be associated with the occupation of politician (Payne, Jacoby, & Lambert, 2004). Egocentric bias involves enhancing our memories of the past (Payne et al., 2004). Did you really score the winning goal in that big soccer match, or did you just assist? Hindsight bias happens when we think an outcome was inevitable after the fact. This is the “I knew it all along” phenomenon. The reconstructive nature of memory contributes to hindsight bias (Carli, 1999). We remember untrue events that seem to confirm that we knew the outcome all along. Have you ever had a song play over and over in your head? How about a memory of a traumatic event, something you really do not want to think about? When you keep remembering something, to the point where you can’t “get it out of your head” and it interferes with your ability to concentrate on other things, it is calledpersistence. It’s Schacter’s seventh and last memory error. It’s actually a failure of our memory system because we involuntarily recall unwanted memories, particularly unpleasant ones (Figure 8.16). For instance, you witness a horrific car accident on the way to work one morning, and you can’t concentrate on work because you keep remembering the scene. Figure 8.16 Many veterans of military conflicts involuntarily recall unwanted, unpleasant memories. (credit: Department of Defense photo by U.S. Air Force Tech. Sgt. Michael R. Holzworth) Interference Sometimes information is stored in our memory, but for some reason it is inaccessible. This is known as interference, and there are two types: proactive interference and retroactive interference (Figure 8.17). Have you ever gotten a new phone number or moved to a new address, but right after you tell people the old (and wrong) phone number or address? When the new year starts, do you find you accidentally write the previous year? These are examples ofproactive interference: when old information hinders the recall of newly learned information.Retroactive interferencehappens when information learned more recently hinders the recall of older information. For example, this week you are studying about memory and learn about the Ebbinghaus forgetting curve. Next week you study lifespan development and learn about Erikson’s theory of psychosocial development, but thereafter have trouble remembering Ebbinghaus’s work because you can only remember Erickson’s theory. Figure 8.17 Sometimes forgetting is caused by a failure to retrieve information. This can be due to interference, either retroactive or proactive. 8.3 TEST YOURSELF Summary: 8.3 Problems with Memory All of us at times have felt dismayed, frustrated, and even embarrassed when our memories have failed us. Our memory is flexible and prone to many errors, which is why eyewitness testimony has been found to be largely unreliable. There are several reasons why forgetting occurs. In cases of brain trauma or disease, forgetting may be due to amnesia. Another reason we forget is due to encoding failure. We can’t remember something if we never stored it in our memory in the first place. Schacter presents seven memory errors that also contribute to forgetting. Sometimes, information is actually stored in our memory, but we cannot access it due to interference. Proactive interference happens when old information hinders the recall of newly learned information. Retroactive interference happens when information learned more recently hinders the recall of older information. 8.4 Ways to Enhance Memory LEARNING OBJECTIVES By the end of this section, you will be able to: Recognize and apply memory-enhancing strategies Recognize and apply effective study techniques Most of us suffer from memory failures of one kind or another, and most of us would like to improve our memories so that we don’t forget where we put the car keys or, more importantly, the material we need to know for an exam. In this section, we’ll look at some ways to help you remember better, and at some strategies for more effective studying. Memory-Enhancing Strategies What are some everyday ways we can improve our memory, including recall? To help make sure information goes from short-term memory to long-term memory, you can usememory-enhancing strategies. One strategy isrehearsal, or the conscious repetition of information to be remembered (Craik & Watkins, 1973). Think about how you learned your multiplication tables as a child. You may recall that 6 x 6 = 36, 6 x 7 = 42, and 6 x 8 = 48. Memorizing these facts is rehearsal. Another strategy ischunking: you organize information into manageable bits or chunks (Bodie, Powers, & Fitch-Hauser, 2006). Chunking is useful when trying to remember information like dates and phone numbers. Instead of trying to remember 5205550467, you remember the number as 520-555-0467. So, if you met an interesting person at a party and you wanted to remember his phone number, you would naturally chunk it, and you could repeat the number over and over, which is the rehearsal strategy. LINK TO LEARNING: Try this fun activity that employs a memory-enchancing strategyto learn more. You could also enhance memory by usingelaborative rehearsal: a technique in which you think about the meaning of new information and its relation to knowledge already stored in your memory (Tigner, 1999). Elaborative rehearsal involves both linking the information to knowledge already stored and repeating the information. For example, in this case, you could remember that 520 is an area code for Arizona and the person you met is from Arizona. This would help you better remember the 520 prefix. If the information is retained, it goes into long-term memory. Mnemonic devicesare memory aids that help us organize information for encoding (Figure 8.18). They are especially useful when we want to recall larger bits of information such as steps, stages, phases, and parts of a system (Bellezza, 1981). Brian needs to learn the order of the planets in the solar system, but he’s having a hard time remembering the correct order. His friend Kelly suggests a mnemonic device that can help him remember. Kelly tells Brian to simply remember the name Mr. VEM J. SUN, and he can easily recall the correct order of the planets:Mercury,Venus,Earth,Mars,Jupiter,Saturn,Uranus, andNeptune. You might use a mnemonic device to help you remember someone’s name, a mathematical formula, or the order of mathematical operations. Figure 8.18 This is a knuckle mnemonic to help you remember the number of days in each month. Months with 31 days are represented by the protruding knuckles and shorter months fall in the spots between knuckles. (credit: modification of work by Cory Zanker) If you have ever watched the television show Modern Family, you might have seen Phil Dunphy explain how he remembers names: The other day I met this guy named Carl. Now, I might forget that name, but he was wearing a Grateful Dead t-shirt. What’s a band like the Grateful Dead? Phish. Where do fish live? The ocean. What else lives in the ocean? Coral. Hello, Co-arl. (Wrubel & Spiller, 2010) It seems the more vivid or unusual the mnemonic, the easier it is to remember. The key to using any mnemonic successfully is to find a strategy that works for you. LINK TO LEARNING: Joshua Foer is a science writer who “accidentally” won the U.S. Memory Championships. Watch his TEDTalk, titled “Feats of Memory Anyone Can Do,” in which he explains a mnemonic device called the memory palaceto learn more. Some other strategies that are used to improve memory include expressive writing and saying words aloud. Expressive writing helps boost your short-term memory, particularly if you write about a traumatic experience in your life. Masao Yogo and Shuji Fujihara (2008) had participants write for 20-minute intervals several times per month. The participants were instructed to write about a traumatic experience, their best possible future selves, or a trivial topic. The researchers found that this simple writing task increased short-term memory capacity after five weeks, but only for the participants who wrote about traumatic experiences. Psychologists can’t explain why this writing task works, but it does. What if you want to remember items you need to pick up at the store? Simply say them out loud to yourself. A series of studies (MacLeod, Gopie, Hourihan, Neary, & Ozubko, 2010) found that saying a word out loud improves your memory for the word because it increases the word’s distinctiveness. Feel silly, saying random grocery items aloud? This technique works equally well if you just mouth the words. Using these techniques increased participants’ memory for the words by more than 10%. These techniques can also be used to help you study. How to Study Effectively Based on the information presented in this chapter, here are some strategies and suggestions to help you hone your study techniques (Figure 8.19). The key with any of these strategies is to figure out what works best for you. Figure 8.19 Memory techniques can be useful when studying for class. (credit: Barry Pousman) Use elaborative rehearsal: In a famous article, Fergus Craik and Robert Lockhart (1972) discussed their belief that information we process more deeply goes into long-term memory. Their theory is called levels of processing. If we want to remember a piece of information, we should think about it more deeply and link it to other information and memories to make it more meaningful. For example, if we are trying to remember that the hippocampus is involved with memory processing, we might envision a hippopotamus with excellent memory and then we could better remember the hippocampus. Apply the self-reference effect: As you go through the process of elaborative rehearsal, it would be even more beneficial to make the material you are trying to memorize personally meaningful to you. In other words, make use of the self-reference effect. Write notes in your own words. Write definitions from the text, and then rewrite them in your own words. Relate the material to something you have already learned for another class, or think how you can apply the concepts to your own life. When you do this, you are building a web of retrieval cues that will help you access the material when you want to remember it. Use distributed practice: Study across time in short durations rather than trying to cram it all in at once. Memory consolidation takes time, and studying across time allows time for memories to consolidate. In addition, cramming can cause the links between concepts to become so active that you get stuck in a link, and it prevents you from accessing the rest of the information that you learned. Rehearse, rehearse, rehearse: Review the material over time, in spaced and organized study sessions. Organize and study your notes, and take practice quizzes/exams. Link the new information to other information you already know well. Study efficiently: Students are great highlighters, but highlighting is not very efficient because students spend too much time studying the things they already learned. Instead of highlighting, use index cards. Write the question on one side and the answer on the other side. When you study, separate your cards into those you got right and those you got wrong. Study the ones you got wrong and keep sorting. Eventually, all your cards will be in the pile you answered correctly. Be aware of interference: To reduce the likelihood of interference, study during a quiet time without interruptions or distractions (like television or music). Keep moving: Of course you already know that exercise is good for your body, but did you also know it’s also good for your mind? Research suggests that regular aerobic exercise (anything that gets your heart rate elevated) is beneficial for memory (van Praag, 2008). Aerobic exercise promotes neurogenesis: the growth of new brain cells in the hippocampus, an area of the brain known to play a role in memory and learning. Get enough sleep: While you are sleeping, your brain is still at work. During sleep the brain organizes and consolidates information to be stored in long-term memory (Abel & Bäuml, 2013). Make use of mnemonic devices: As you learned earlier in this chapter, mnemonic devices often help us to remember and recall information. There are different types of mnemonic devices, such as the acronym. An acronym is a word formed by the first letter of each of the words you want to remember. For example, even if you live near one, you might have difficulty recalling the names of all five Great Lakes. What if I told you to think of the word Homes? HOMES is an acronym that represents Huron, Ontario, Michigan, Erie, and Superior: the five Great Lakes. Another type of mnemonic device is an acrostic: you make a phrase of all the first letters of the words. For example, if you are taking a math test and you are having difficulty remembering the order of operations, recalling the following sentence will help you: “Please Excuse My Dear Aunt Sally,” because the order of mathematical operations is Parentheses, Exponents, Multiplication, Division, Addition, Subtraction. There also are jingles, which are rhyming tunes that contain keywords related to the concept, such as i before e, except after c. 8.4 TEST YOURSELF Summary: 8.4 Ways to Enhance Memory There are many ways to combat the inevitable failures of our memory system. Some common strategies that can be used in everyday situations include mnemonic devices, rehearsal, self-referencing, and adequate sleep. These same strategies also can help you to study more effectively. Critical Thinking Questions Compare and contrast implicit and explicit memory. According to the Atkinson-Shiffrin model, name and describe the three stages of memory. Compare and contrast the two ways in which we encode information. What might happen to your memory system if you sustained damage to your hippocampus? Compare and contrast the two types of interference. Compare and contrast the two types of amnesia. What is the self-reference effect, and how can it help you study more effectively? You and your roommate spent all of last night studying for your psychology test. You think you know the material; however, you suggest that you study again the next morning an hour prior to the test. Your roommate asks you to explain why you think this is a good idea. What do you tell them? Personal Application Questions Describe something you have learned that is now in your procedural memory. Discuss how you learned this information. Describe something you learned in high school that is now in your semantic memory. Describe a flashbulb memory of a significant event in your life. Which of the seven memory errors presented by Schacter have you committed? Provide an example of each one. Jurors place a lot of weight on eyewitness testimony. Imagine you are an attorney representing a defendant who is accused of robbing a convenience store. Several eyewitnesses have been called to testify against your client. What would you tell the jurors about the reliability of eyewitness testimony? Create a mnemonic device to help you remember a term or concept from this chapter. What is an effective study technique that you have used? How is it similar to/different from the strategies suggested in this chapter? Study Guide Outline: Memory Chapter 8. (Note: Any of the information below can be placed on flash cards or used to create a flowchart or a concept map.) 1. How Memory Functions A. Overview Memory is the system we use to encode, store, and retrieve information—similar to how a computer processes data. B. Encoding: Getting Information In Definition: Input of information into the memory system. Encoding information occurs through: Automatic Processing: Unconscious (e.g., time, space, what you ate for lunch). Effortful Processing: Requires focus and attention (e.g., studying for a test). Types of Encoding: Semantic: Meaning of words and concepts Example: Grouping related words together Visual: Mental Images Example: Remembering “dog” better than “truth” Acoustic: Sound (especially words) Example: Remembering a rhyme or song Semantic encoding is the most effective for learning verbal information—especially if it’s personally meaningful (self-reference effect). C. Storage: Keeping Information Definition: Creating a permanent record of information. Based on Atkinson &Shiffrin’s 3-stage model: Sensory Memory: Brief input from senses (lasts 1–2 seconds). Short-Term Memory (STM): Temporary (15–30 seconds), limited (5–9 items). Long-Term Memory (LTM): Lasts potentially forever, unlimited capacity. Working Memory (Baddeley &Hitch) STM isn’t just one system—it includes: Visuospatial Sketchpad (visual info) Phonological Loop (sound info) Episodic Buffer (links to long-term memory) Central Executive (controls and organizes info) D. Short-Term Memory (STM) Holds current thoughts/info. Capacity: ~7 items (±2), or possibly ~4 (newer research). Rehearsal: Maintenance (Active): Repetition (e.g., repeating phone number). Elaborative: Linking new info to existing knowledge (e.g., area code = uncle’s). Interference and Decay: Decay: Info fades quickly if not rehearsed (~18 seconds). Proactive Interference: Old info blocks new info. E. Long-Term Memory (LTM) Unlimited and durable storage. Organized in semantic networks (related concepts linked). Spreading activation: Triggering one idea activates related ones. Types of Long-Term Memory: \| Type \| Description \| Example \| --- \| Explicit (Declarative) \| Conscious recall \| Facts, experiences \| \| – Semantic \| Facts, meanings \| “Capital of France is Paris” \| \| – Episodic \| Personal events \| “My 18th birthday party” \| \| Implicit \| Unconscious recall \| Riding a bike, classical conditioning \| \| – Procedural \| Motor skills \| Typing, brushing teeth \| \| – Priming \| Earlier exposure affects response \| Recognizing a word faster if seen before \| \| – Emotional Conditioning \| Learned emotional responses \| Feeling anxious at dentist’s office \| F. Retrieval Retrieval is the process of getting information out of long-term memory and bringing it back into conscious awareness. Think of it like opening a saved file on your computer—you stored it earlier, and now you’re accessing it to use again. Types: Recall: Definition: Accessing information without any cues. Example: Answering an essay question on a test. Key Idea: You must “pull” the info from memory on your own. Recognition: Definition: Identifying previously learned information when you see it again. Example: Choosing the correct answer on a multiple-choice test. Key Idea: Involves comparison—you recognize the right answer from options. Relearning: Definition: Learning something again that you previously learned. Example: Taking a Spanish class years after high school and finding it easier than expected. Key Idea: Shows how memory can be retained even if not actively used for years. 2. Parts of the Brain Involved with Memory A. Are Memories Stored in One Part of the Brain? No, memories are stored in many parts of the brain. Early research by Karl Lashley: Tried to find the engram (physical trace of memory) using lesions on rats’ cerebral cortex. Equipotentiality Hypothesis: If one area is damaged, other areas can take over memory functions. Modern research shows specific brain areas are involved in different types of memory. B. Key Brain Structures Involved in Memory Amygdala Function: Regulates emotion (fear, aggression). Role in Memory: Helps store emotional memories. Involved in memory consolidation, especially for emotionally charged events. Damage can erase fear memories (e.g., tone-shock conditioning in rats). Hippocampus Function: Involved in forming new memories. Role in Memory: Supports recognition memory and spatial memory. Connects new memories with existing knowledge. Crucial for declarative memory (facts/events). Case of H.M.: Removal of hippocampi led to loss of ability to form new memories, but old memories remained. Cerebellum Function: Coordinates movement and motor skills. Role in Memory: Involved in implicit memories (like skills and conditioned responses). Example: Rabbits with cerebellum damage could not learn a conditioned eye-blink. Prefrontal Cortex Function: Involved in thinking, decision-making, and memory tasks. Role in Memory: Active during semantic tasks (understanding meaning). PET scans show: Left frontal lobe: encoding information. Right frontal lobe: retrieving information. C. Neurotransmitters and Memory Key neurotransmitters: Epinephrine, dopamine, serotonin, glutamate, acetylcholine. Function: Help neurons communicate, which is essential for memory formation. Repeated neural activity strengthens synapses – basis of memory consolidation. D. Arousal and Memory Arousal Theory Strong emotional experiences = strong memories. Weak emotions = weaker memories. Emotional events trigger neurotransmitter and hormone release, enhancing memory. Flashbulb Memory: Vivid, detailed memory of an emotionally significant event. Example: Remembering where you were during 9/11. Supported by studies showing long-lasting emotional memory recall. 3. Problems with Memory A. What is Amnesia? Definition: Loss of long-term memory due to disease, injury, or trauma. Example: Patient K.C. could not recall any personal life events—only retained experiences from the last minute or two. B. Types of Amnesia Anterograde Amnesia: Inability to form new long-term memories after trauma. Hippocampus usually affected. Example: H.M. couldn’t form new memories but could improve on puzzles through procedural memory. Retrograde Amnesia: Inability to recall past memories from before trauma. Affects episodic memory. Example: Scott Bolzan lost 46 years of memory after a head injury. C. Memory Construction and Reconstruction Construction: Forming new memories. Reconstruction: Recalling and potentially altering existing memories. Memory is flexible—can unintentionally change when combined with new information. D. Suggestibility and False Memories Suggestibility: Memory can be influenced by external information, leading to false memories. Example: D.C. sniper case—focus on a white van led to thousands of false tips due to suggestion. False Memory: A distorted or fabricated recollection of an event. Often occurs during retrieval when suggestion alters memory content. E. Eyewitness Misidentification Eyewitness memory can be inaccurate, leading to wrongful convictions. Causes: Leading questions, police reinforcement, non-blind lineups. Example: Ronald Cotton was wrongly convicted based on misidentification. F. The Misinformation Effect Introduced by Elizabeth Loftus Misinformation Effect Paradigm: After being given misleading info, people may misremember the original event. Study: Car crash video—wording like “smashed” vs. “contacted” affected speed estimates and memory of broken glass (even when there was none). G. Repressed and Recovered Memories False Memory Syndrome: Remembering an event (like abuse) that didn’t actually happen. Controversy: Can traumatic memories be repressed and later recovered? Supporters: Say trauma can be blocked and later recalled (e.g., through hypnosis). Critics (Loftus): Warn that suggestion during therapy can create inaccurate memories. Example: Children misidentified genital exams with dolls due to suggestive questioning. Modern Safeguards Use neutral questioning. Blind lineups in police work. Judges may inform jurors about risks of misidentification or exclude unreliable testimony. H. Forgetting Forgetting is the loss of information from long-term memory. Memory is not perfect — it’s fragile and prone to errors and failure. I. Encoding Failure Occurs before memory storage — information never enters long-term memory. Example: Most people can’t recall what a penny looks like in detail — we never paid enough attention to encode it. Effortful encoding is needed for strong memories. J. Memory Errors: Memory researcher Daniel Schacter identified seven ways memory fails (See Table 8.1 for a summary of the information found below), grouped into: Forgetting Distortion Intrusion Forgetting Errors Transience Memory fades over time (storage decay). Example: Forgetting details of a book you read in high school. Ebbinghaus’s Forgetting Curve: 50% of learned info is lost after 20 minutes. 70% is lost after 24 hours. Absentmindedness Lapses in memory due to inattention. Example: Forgetting if you returned your ID badge because your mind was on your sick child. Blocking “Tip of the tongue” phenomenon — you know the info but can’t retrieve it. Example: Trying to remember an actor’s name you know well. Distortion Errors Misattribution Confusing the source of a memory. Example: Thinking you saw a movie with one person, when it was someone else. Famous case: Donald Thomson was wrongly identified as a rapist because the victim confused him with someone on TV. Suggestibility False memories form due to external influence, such as leading questions. Differs from misattribution, which is internally generated. Bias Current beliefs or feelings distort memory Stereotypical bias: Racial/gender assumptions. Egocentric bias: Inflating your role in past events. Hindsight bias: Thinking “I knew it all along” after an event. Intrusion Error Persistence Unwanted, intrusive memories keep recurring. Often involves trauma or emotionally charged memories. Example: Witnessing a car crash and being unable to focus afterward. K. Interference: Memory Competition Proactive Interference Old info blocks new learning Example: Giving out your old address instead of your new one. Retroactive Interference New info blocks access to old memories. Example: Learning new psychology terms makes it hard to remember previously studied concepts. 4. Ways to Enhance Memory A. Memory Enhancing Strategies Rehearsal Definition:Repeating information over and over to remember it. Example:Repeating multiplication facts like 6 × 6 = 36. Chunking Definition:Breaking down information into smaller, manageable units. Example:Phone number 5205550467 becomes 520-555-0467. Elaborative Rehearsal Definition:Linking new info to existing knowledge and giving it meaning. Example:Connecting area code 520 to someone being from Arizona. Mnemonic Devices Definition:Memory aids that help encode and recall large amounts of information. Examples: Acronym:HOMES for the Great Lakes (Huron, Ontario, Michigan, Erie, Superior). Acrostic:Please Excuse My Dear Aunt Sally for math operations. Jingles:Rhymes like “i before e, except after c.” Expressive Writing Definition:Writing about personal experiences to improve short-term memory. Note:Works best when writing about emotional or traumatic events. Saying Words Aloud Definition:Speaking or mouthing words boosts distinctiveness and recall. Example:Saying grocery items aloud improves memory of them. B. How to Study Effectively Use Elaborative Rehearsal Deeply process info and link it to what you already know. Example: Picture a “hippopotamus with great memory” to remember the hippocampus. Apply the Self-Reference Effect Make the info personally meaningful. Example: Relate psychology terms to your daily life or other classes. Use Distributed Practice Study a little over time (not all at once). Promotes memory consolidation and reduces overload. Rehearse Regularly Review info in spaced sessions. Use practice quizzes and connect ideas. Study Efficiently Avoid over-highlighting. Instead: Use flashcards. Focus on what you get wrong and keep practicing. Minimize Interference Study in a quiet, distraction-free environment. Avoid multitasking (like watching TV or listening to music). Exercise Regular aerobic exercise (like jogging or biking) improves brain health and memory. Promotes growth of new brain cells in the hippocampus. Sleep Well Sleep helps your brain consolidate and store memories. Use Mnemonics (Again !) Great for memorizing lists, orders, or systems. Types: Acronyms, acrostics, rhymes, jingles. Access for free at definition set of processes used to encode, store, and retrieve information over different periods of time ×Close definition input of information into the memory system ×Close definition encoding of informational details like time, space, frequency, and the meaning of words ×Close definition encoding of information that takes effort and attention ×Close definition input of words and their meaning ×Close definition input of images ×Close definition input of sounds, words, and music ×Close definition creation of a permanent record of information ×Close definition memory model that states we process information through three systems: sensory memory, short-term memory, and long-term memory ×Close definition storage of brief sensory events, such as sights, sounds, and tastes ×Close definition holds about seven bits of information before it is forgotten or stored, as well as information that has been retrieved and is being used ×Close definition repetition of information to be remembered ×Close definition thinking about the meaning of new information and its relation to knowledge already stored in your memory ×Close definition old information hinders the recall of newly learned information ×Close definition continuous storage of information ×Close definition memories we consciously try to remember and recall ×Close definition memories that are not part of our consciousness ×Close definition type of declarative memory that contains information about events we have personally experienced, also known as autobiographical memory ×Close definition type of declarative memory about words, concepts, and language-based knowledge and facts ×Close definition type of long-term memory for making skilled actions, such as how to brush your teeth, how to drive a car, and how to swim ×Close definition act of getting information out of long-term memory storage and back into conscious awareness ×Close definition accessing information without cues ×Close definition identifying previously learned information after encountering it again, usually in response to a cue ×Close definition learning information that was previously learned ×Close definition physical trace of memory ×Close definition some parts of the brain can take over for damaged parts in forming and storing memories ×Close definition exceptionally clear recollection of an important event ×Close definition recall of false autobiographical memories ×Close definition loss of long-term memory that occurs as the result of disease, physical trauma, or psychological trauma ×Close definition loss of memory for events that occur after the brain trauma ×Close definition loss of memory for events that occurred prior to brain trauma ×Close definition formulation of new memories ×Close definition process of bringing up old memories that might be distorted by new information ×Close definition effects of misinformation from external sources that leads to the creation of false memories ×Close definition after exposure to additional and possibly inaccurate information, a person may misremember the original event ×Close definition recall of false autobiographical memories ×Close definition loss of information from long-term memory ×Close definition memory error in which unused memories fade with the passage of time ×Close definition lapses in memory that are caused by breaks in attention or our focus being somewhere else ×Close definition memory error in which you confuse the source of your information ×Close definition how feelings and view of the world distort memory of past events ×Close definition failure of the memory system that involves the involuntary recall of unwanted memories, particularly unpleasant ones ×Close definition information learned more recently hinders the recall of older information ×Close definition techniques to help make sure information goes from short-term memory to long-term memory ×Close definition organizing information into manageable bits or chunks ×Close definition memory aids that help organize information for encoding ×Close definition tendency for an individual to have better memory for information that relates to oneself in comparison to material that has less personal relevance ×Close definition Previous/next navigation Previous: Thinking and Intelligence Next: Lifespan Development Back to top License Psychology 2e OpenStax Updated 2025 Copyright © 2020 by Openstax is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted. 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2066	https://stackify.com/solid-design-liskov-substitution-principle/	Stackify is now BMC. Read theBlog SOLID Design Principles Explained: The Liskov Substitution Principle with Code Examples By: Thorben \| April 11, 2018 The Open/Closed Principle, which I explained in a previous article, is one of the key concepts in OOP that enables you to write robust, maintainable and reusable software components. But following the rules of that principle alone is not enough to ensure that you can change one part of your system without breaking other parts. Your classes and interfaces also need to follow the Liskov Substitution Principle to avoid any side-effects. The Liskov Substitution Principle is the 3rd of Robert C. Martin‘s famous SOLID design principles: Single Responsibility Principle Open/Closed Principle Liskov Substitution Principle Interface Segregation Principle Dependency Inversion It extends the Open/Closed Principle by focusing on the behavior of a superclass and its subtypes. As I will show you in this article, this is at least as important but harder to validate that the structural requirements of the Open/Closed Principle. Definition of the Liskov Substitution Principle The Liskov Substitution principle was introduced by Barbara Liskov in her conference keynote “Data abstraction” in 1987. A few years later, she published a paper with Jeanette Wing in which they defined the principle as: Let Φ(x) be a property provable about objects x of type T. Then Φ(y) should be true for objects y of type S where S is a subtype of T. OK, let’s be honest. Such a scientific definition might be necessary, but it doesn’t help a lot in our daily work as software developers. So, what does it mean for our code? The Liskov Substitution Principle in practical software development The principle defines that objects of a superclass shall be replaceable with objects of its subclasses without breaking the application. That requires the objects of your subclasses to behave in the same way as the objects of your superclass. You can achieve that by following a few rules, which are pretty similar to the design by contract concept defined by Bertrand Meyer. An overridden method of a subclass needs to accept the same input parameter values as the method of the superclass. That means you can implement less restrictive validation rules, but you are not allowed to enforce stricter ones in your subclass. Otherwise, any code that calls this method on an object of the superclass might cause an exception, if it gets called with an object of the subclass. Similar rules apply to the return value of the method. The return value of a method of the subclass needs to comply with the same rules as the return value of the method of the superclass. You can only decide to apply even stricter rules by returning a specific subclass of the defined return value, or by returning a subset of the valid return values of the superclass. Enforcing the Liskov Substitution Principle If you decide to apply this principle to your code, the behavior of your classes becomes more important than its structure. Unfortunately, there is no easy way to enforce this principle. The compiler only checks the structural rules defined by the Java language, but it can’t enforce a specific behavior. You need to implement your own checks to ensure that your code follows the Liskov Substitution Principle. In the best case, you do this via code reviews and test cases. In your test cases, you can execute a specific part of your application with objects of all subclasses to make sure that none of them causes an error or significantly changes its performance. You can try to do similar checks during a code review. But what’s even more important is that you check that you created and executed all the required test cases. Okay, enough theory. Let’s take a look at an example Making coffee with the Liskov Substitution Principle Most articles about the Liskov Substitution Principle use an example in which they implement a Rectangle and a Square class to show that you break the design principle if your Square class extends the Rectangle class. But that example is a little bit boring. There are already lots of articles about it, and I have never implemented an application that just requires a set of simple geometric shapes. So, let’s create an example that’s a little bit more fun. I enjoy drinking a good cup of coffee in the morning, and I want to show you a simple application that uses different kinds of coffee machines to brew a cup of coffee. You might already know very similar examples from my previous articles about the Single Responsibility Principle or the Open/Closed Principle. You can get all source files of this example at If you enjoy coffee as much as I do, you most likely used several different coffee machines in the past. There are relatively basic ones that you can use to transform one or two scoops of ground coffee and a cup of water into a nice cup of filter coffee. And there are others that include a grinder to grind your coffee beans and you can use to brew different kinds of coffee, like filter coffee and espresso. If you decide to implement an application that automatically brews a cup of coffee every morning so that you don’t have to get out of bed before it’s ready, you might decide to model these coffee machines as two classes with the methods addCoffee and brewCoffee. A basic coffee machine The BasicCoffeeMachine can only brew filter coffee. So, the brewCoffee method checks if the provided CoffeeSelection value is equal to FILTER_COFFEE before it calls the private brewFilterCoffee method to create and return a CoffeeDrink object. ``` public class BasicCoffeeMachine { private Map configMap; private Map groundCoffee; private BrewingUnit brewingUnit; public BasicCoffeeMachine(Map coffee) { this.groundCoffee = coffee; this.brewingUnit = new BrewingUnit(); this.configMap = new HashMap(); this.configMap.put(CoffeeSelection.FILTER_COFFEE, new Configuration(30, 480)); } public CoffeeDrink brewCoffee(CoffeeSelection selection) throws CoffeeException { switch (selection) { case FILTER_COFFEE: return brewFilterCoffee(); default: throw new CoffeeException( "CoffeeSelection [" + selection + "] not supported!"); } } private CoffeeDrink brewFilterCoffee() { Configuration config = configMap.get(CoffeeSelection.FILTER_COFFEE); // get the coffee GroundCoffee groundCoffee = this.groundCoffee.get( CoffeeSelection.FILTER_COFFEE); // brew a filter coffee return this.brewingUnit.brew(CoffeeSelection.FILTER_COFFEE, groundCoffee, config.getQuantityWater()); } public void addCoffee(CoffeeSelection sel, GroundCoffee newCoffee) throws CoffeeException { GroundCoffee existingCoffee = this.groundCoffee.get(sel); if (existingCoffee != null) { if (existingCoffee.getName().equals(newCoffee.getName())) { existingCoffee.setQuantity( existingCoffee.getQuantity() + newCoffee.getQuantity()); } else { throw new CoffeeException( "Only one kind of coffee supported for each CoffeeSelection."); } } else { this.groundCoffee.put(sel, newCoffee); } } } ``` The addCoffee method expects a CoffeeSelection enum value and a GroundCoffee object. It uses the CoffeeSelection as the key of the internal groundCoffee Map. These are the most important parts of the BasicCoffeeMachine class. Let’s take a look at the PremiumCoffeeMachine. A premium coffee machine The premium coffee machine has an integrated grinder, and the internal implementation of the brewCoffee method is a little more complex. But you don’t see that from the outside. The method signature is identical to the one of the BasicCoffeeMachine class. ``` public class PremiumCoffeeMachine { private Map<CoffeeSelection, Configuration> configMap; private Map<CoffeeSelection, CoffeeBean> beans; private Grinder grinder; private BrewingUnit brewingUnit; public PremiumCoffeeMachine(Map<CoffeeSelection, CoffeeBean> beans) { this.beans = beans; this.grinder = new Grinder(); this.brewingUnit = new BrewingUnit(); this.configMap = new HashMap<>(); this.configMap.put(CoffeeSelection.FILTER_COFFEE, new Configuration(30, 480)); this.configMap.put(CoffeeSelection.ESPRESSO, new Configuration(8, 28)); } @Override public CoffeeDrink brewCoffee(CoffeeSelection selection) throws CoffeeException { switch(selection) { case ESPRESSO: return brewEspresso(); case FILTER_COFFEE: return brewFilterCoffee(); default: throw new CoffeeException( "CoffeeSelection [" + selection + "] not supported!"); } } private CoffeeDrink brewEspresso() { Configuration config = configMap.get(CoffeeSelection.ESPRESSO); // grind the coffee beans GroundCoffee groundCoffee = this.grinder.grind( this.beans.get(CoffeeSelection.ESPRESSO), config.getQuantityCoffee()); // brew an espresso return this.brewingUnit.brew(CoffeeSelection.ESPRESSO, groundCoffee, config.getQuantityWater()); } private CoffeeDrink brewFilterCoffee() { Configuration config = configMap.get(CoffeeSelection.FILTER_COFFEE); // grind the coffee beans GroundCoffee groundCoffee = this.grinder.grind( this.beans.get(CoffeeSelection.FILTER_COFFEE), config.getQuantityCoffee()); // brew a filter coffee return this.brewingUnit.brew(CoffeeSelection.FILTER_COFFEE, groundCoffee, config.getQuantityWater()); } public void addCoffee(CoffeeSelection sel, CoffeeBean newBeans) throws CoffeeException { CoffeeBean existingBeans = this.beans.get(sel); if (existingBeans != null) { if (existingBeans.getName().equals(newBeans.getName())) { existingBeans.setQuantity( existingBeans.getQuantity() + newBeans.getQuantity()); } else { throw new CoffeeException( "Only one kind of coffee supported for each CoffeeSelection."); } } else { this.beans.put(sel, newBeans); } } } ``` But that’s not the case for the addCoffee method. It expects an object of type CoffeeBean instead of an object of type GroundCoffee. If you add a shared superclass or an interface that gets implemented by the BasicCoffeeMachine and the PremiumCoffeeMachine class, you will need to decide how to handle this difference. Introducing a shared interface You can either create another abstraction, e.g., Coffee, as the superclass of CoffeeBean and GroundCoffee and use it as the type of the method parameter. That would unify the structure of both addCoffee methods, but require additional validation in both methods. The addCoffee method of the BasicCoffeeMachine class would need to check that the caller provided an instance of GroundCoffee, and the addCoffee implementation of the PremiumCoffeeMachine would require an instance of CoffeeBean. This would obviously break the Liskov Substitution Principle because the validation would fail if you provide a BasicCoffeeMachine object instead of a PremiumCoffeeMachine and vice versa. The better approach is to exclude the addCoffee method from the interface or superclass because you can’t interchangeably implement it. The brewCoffee method, on the other hand, could be part of a shared interface or a superclass, as long as the superclass or interface only guarantees that you can use it to brew filter coffee. The input parameter validation of both implementations accept the CoffeeSelection value FILTER_COFFEE. The addCoffee method of the PremiumCoffeeMachine class also accepts the enum value ESPRESSO. But as I explained at the beginning of this article, the different subclasses may implement less restrictive validation rules. Summary The Liskov Substitution Principle is the third of Robert C. Martin’s SOLID design principles. It extends the Open/Closed principle and enables you to replace objects of a parent class with objects of a subclass without breaking the application. This requires all subclasses to behave in the same way as the parent class. To achieve that, your subclasses need to follow these rules: Don’t implement any stricter validation rules on input parameters than implemented by the parent class. Apply at the least the same rules to all output parameters as applied by the parent class. With APM, server health metrics, and error log integration, improve your application performance with Stackify Retrace. Try your free two week trial today Related posts: How Spring Boot Can Level Up your Spring Application Top Java Tools: 63 Tools to Power Every Phase of Java Development How to Specify and Handle Exceptions in Java What to Do About Java Memory Leaks: Tools, Fixes, and More Finally Getting the Most out of the Java Thread Pool Improve Your Code with Retrace APM Stackify's APM tools are used by thousands of .NET, Java, PHP, Node.js, Python, & Ruby developers all over the world. Explore Retrace's product features to learn more. 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2067	https://pubmed.ncbi.nlm.nih.gov/40234111/	Familial chylomicronemia syndrome: An expert clinical review from the National Lipid Association - PubMed Clipboard, Search History, and several other advanced features are temporarily unavailable. Skip to main page content An official website of the United States government Here's how you know The .gov means it’s official. Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you’re on a federal government site. The site is secure. The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely. 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Epub 2025 Mar 22. Familial chylomicronemia syndrome: An expert clinical review from the National Lipid Association Fiza Javed1,Robert A Hegele2,Abhimanyu Garg3,Nivedita Patni4,Daniel Gaudet5,Lauren Williams6,Mohamed Khan7,Qingyang Li7,Zahid Ahmad8 Affiliations Expand Affiliations 1 Department of Medicine, Faculty of Health Sciences, McMaster University, Hamilton, ON, Canada (Dr Javed). 2 Department of Medicine and Robarts Research Institute, Schulich School of Medicine and Dentistry, Western University, London, ON, Canada (Dr Hegele). 3 Section of Nutrition and Metabolic Diseases, Division of Endocrinology, Department of Internal Medicine and the Center for Human Nutrition, UT Southwestern Medical Center, Dallas, TX, USA (Dr Garg). 4 Division of Pediatric Endocrinology, Department of Pediatrics, UT Southwestern Medical Center, Dallas, TX, USA (Dr Patni). 5 ECOGENE-21 Department of Medicine, Université de Montréal, Chicoutimi, QC, Canada (Dr Gaudet). 6 Department of Pediatric Cardiology, Baylor Scott & White McLane Children's Medical Center, Temple, TX, USA (Ms Williams). 7 FCS Foundation, San Diego, CA, USA (Mr Khan and Mr Li). 8 Section of Nutrition and Metabolic Diseases, Division of Endocrinology, Department of Internal Medicine, UT Southwestern Medical Center, Dallas, TX, USA (Dr Ahmad). Electronic address: Zahid.ahmad@utsouthwestern.edu. PMID: 40234111 DOI: 10.1016/j.jacl.2025.03.013 Item in Clipboard Review Familial chylomicronemia syndrome: An expert clinical review from the National Lipid Association Fiza Javed et al. J Clin Lipidol.2025 May-Jun. Show details Display options Display options Format J Clin Lipidol Actions Search in PubMed Search in NLM Catalog Add to Search . 2025 May-Jun;19(3):382-403. doi: 10.1016/j.jacl.2025.03.013. Epub 2025 Mar 22. Authors Fiza Javed1,Robert A Hegele2,Abhimanyu Garg3,Nivedita Patni4,Daniel Gaudet5,Lauren Williams6,Mohamed Khan7,Qingyang Li7,Zahid Ahmad8 Affiliations 1 Department of Medicine, Faculty of Health Sciences, McMaster University, Hamilton, ON, Canada (Dr Javed). 2 Department of Medicine and Robarts Research Institute, Schulich School of Medicine and Dentistry, Western University, London, ON, Canada (Dr Hegele). 3 Section of Nutrition and Metabolic Diseases, Division of Endocrinology, Department of Internal Medicine and the Center for Human Nutrition, UT Southwestern Medical Center, Dallas, TX, USA (Dr Garg). 4 Division of Pediatric Endocrinology, Department of Pediatrics, UT Southwestern Medical Center, Dallas, TX, USA (Dr Patni). 5 ECOGENE-21 Department of Medicine, Université de Montréal, Chicoutimi, QC, Canada (Dr Gaudet). 6 Department of Pediatric Cardiology, Baylor Scott & White McLane Children's Medical Center, Temple, TX, USA (Ms Williams). 7 FCS Foundation, San Diego, CA, USA (Mr Khan and Mr Li). 8 Section of Nutrition and Metabolic Diseases, Division of Endocrinology, Department of Internal Medicine, UT Southwestern Medical Center, Dallas, TX, USA (Dr Ahmad). Electronic address: Zahid.ahmad@utsouthwestern.edu. PMID: 40234111 DOI: 10.1016/j.jacl.2025.03.013 Item in Clipboard Full text links Cite Display options Display options Format Abstract Familial chylomicronemia syndrome (FCS) is a rare Mendelian autosomal recessive disorder (MIM 238600) characterized by extreme and sustained hypertriglyceridemia due to profound reduction of lipoprotein lipase (LPL) activity. This expert opinion statement synthesizes current knowledge on the definition, pathophysiology, genetics, prevalence, diagnosis, and management of FCS. FCS typically manifests at a young age with persistent severe hypertriglyceridemia-defined as ≥10 mmol/L (≥885 mg/dL), or ≥1000 mg/dL (≥11.2 mmol/L) depending on region and whether Systeme International (SI) units are utilized-in the absence of secondary factors, resistance to conventional lipid-lowering therapies, and a high lifetime risk of acute pancreatitis. It is caused by biallelic pathogenic variants in the LPL gene encoding LPL, or 1 of 4 other related genes that encode proteins that interact with LPL. Affected individuals require a strict, lifelong very low-fat diet with <15% of energy from fat. Emerging therapies inhibiting apolipoprotein C-III show promise in reducing serum triglycerides and pancreatitis risk in patients with FCS. A multidisciplinary approach, encompassing dietary management, pharmacotherapy, and patient education, is pivotal in mitigating the significant morbidity associated with FCS. Keywords: Acute pancreatitis; Chylomicronemia; Familial chylomicronemia syndrome; Hyperlipoproteinemia type 1; Hypertriglyceridemia; Hypertriglyceridemic pancreatitis; Lipemia; Lipoprotein lipase; Lipoprotein lipase deficiency; Triglyceride. Copyright © 2025 National Lipid Association. Published by Elsevier Inc. All rights reserved. PubMed Disclaimer Conflict of interest statement Declaration of competing interest F.J., M.K., Q.L. reports no conflicts of interest. A.G. consults for Chiesi, Regeneron, and Zvelt Pharmaceuticals, and has received grant support from Chiesi, Regeneron, Quintiles, Akcea Pharmaceuticals, and Intercept Pharmaceuticals. AG has NIH funding related to treatment options for FCS R33-HL66373 (Garg) NHLBI. N.P. has consulted for Chiesi, and has NIH funding related to treatment options for FCS R33-HL66373 (Garg) NHLBI. D.G. has received funding and/or consulting fees from Alnylam, Amgen, Chiesi (Amryt), Applied therapeutics, Arrowhead, Astra Zeneca, Boehringer-Ingelheim, CRISPR Therapeutics, Eli Lilly, Esperion, Flagship Pioneering, Ionis, Kowa, National Research Council Canada, New Amsterdam Pharma, Novartis, Novo Nordisk, Pfizer, Regeneron, Sanofi, Ultragenyx, Uniqure, and Verve therapeutics. L.W. has no conflicts of interest to declare. R.A.H. reports grants or contracts from Ionis, Arrowhead, Novartis, Amryt; consulting fees from Arrowhead, Amgen, Acasti Pharma, Aegerion Pharmaceuticals, Akcea Therapeutics/Ionis Pharmaceuticals, HLS Therapeutics, Novartis, Pfizer, Regeneron, Sanofi, and Ultragenyx; honoraria from Amgen, Sanofi, HLS Therapeutics, and Ionis; and participation on a data safety monitoring board or advisory board for Novartis. Z.A. reports institutional grants or contracts from the US Department of Defense, the NIH, and Ionis. Chair of metreleptin adjudication committee for Amyrt Pharmaceuticals. Similar articles Differential effects of volanesorsen on apoC-III, triglycerides, and pancreatitis in familial chylomicronemia syndrome diagnosed by genetic or nongenetic criteria.Tsimikas S, Ginsberg HN, Alexander VJ, Karwatowska-Prokopczuk E, Dibble A, Li L, Witztum JL, Hegele RA.Tsimikas S, et al.J Clin Lipidol. 2025 May-Jun;19(3):422-431. doi: 10.1016/j.jacl.2024.12.018. Epub 2024 Dec 21.J Clin Lipidol. 2025.PMID: 39848842 Clinical Trial. Familial Chylomicronemia Syndrome: A Clinical Guide For Endocrinologists.Falko JM.Falko JM.Endocr Pract. 2018 Aug;24(8):756-763. doi: 10.4158/EP-2018-0157.Endocr Pract. 2018.PMID: 30183397 Review. Familial chylomicronemia syndrome caused by 2 genetic variants in the APOA5 gene: Severe hypertriglyceridemia that complicates pregnancy.Gutiérrez J, Castaño P, Fariña G, Berg G, Gálvez JM, Nogueira JP.Gutiérrez J, et al.J Clin Lipidol. 2025 May-Jun;19(3):701-706. doi: 10.1016/j.jacl.2024.12.020. Epub 2025 Jan 6.J Clin Lipidol. 2025.PMID: 40023744 Etiology and emerging treatments for familial chylomicronemia syndrome.Spagnuolo CM, Hegele RA.Spagnuolo CM, et al.Expert Rev Endocrinol Metab. 2024 Jul;19(4):299-306. doi: 10.1080/17446651.2024.2365787. Epub 2024 Jun 12.Expert Rev Endocrinol Metab. 2024.PMID: 38866702 Review. Epidemiology and longitudinal course of chylomicronemia: Insights from NHANES and a large health care system.Saadatagah S, Naderian M, Larouche M, Gaudet D, Kullo IJ, Ballantyne CM.Saadatagah S, et al.J Clin Lipidol. 2025 May-Jun;19(3):432-441. doi: 10.1016/j.jacl.2025.02.008. Epub 2025 Mar 28.J Clin Lipidol. 2025.PMID: 40155283 See all similar articles Cited by Recognition and management of persistent chylomicronemia: A joint expert clinical consensus by the National Lipid Association and the American Society for Preventive Cardiology.Saadatagah S, Larouche M, Naderian M, Nambi V, Brisson D, Kullo IJ, Duell PB, Michos ED, Shapiro MD, Watts GF, Gaudet D, Ballantyne CM.Saadatagah S, et al.Am J Prev Cardiol. 2025 Mar 28;22:100978. doi: 10.1016/j.ajpc.2025.100978. eCollection 2025 Jun.Am J Prev Cardiol. 2025.PMID: 40242365 Free PMC article. Publication types Review Actions Search in PubMed Search in MeSH Add to Search MeSH terms Humans Actions Search in PubMed Search in MeSH Add to Search Hyperlipoproteinemia Type I / diagnosis Actions Search in PubMed Search in MeSH Add to Search Hyperlipoproteinemia Type I / epidemiology Actions Search in PubMed Search in MeSH Add to Search Hyperlipoproteinemia Type I / genetics Actions Search in PubMed Search in MeSH Add to Search Hyperlipoproteinemia Type I / therapy Actions Search in PubMed Search in MeSH Add to Search Lipoprotein Lipase / genetics Actions Search in PubMed Search in MeSH Add to Search Lipoprotein Lipase / metabolism Actions Search in PubMed Search in MeSH Add to Search Substances Lipoprotein Lipase Actions Search in PubMed Search in MeSH Add to Search Supplementary concepts Familial hyperchylomicronemia syndrome Actions Search in PubMed Search in MeSH Add to Search Related information Cited in Books MedGen PubChem Compound (MeSH Keyword) LinkOut - more resources Full Text Sources ClinicalKey Elsevier Science Full text links[x] Elsevier Science [x] Cite Copy Download .nbib.nbib Format: Send To Clipboard Email Save My Bibliography Collections Citation Manager [x] NCBI Literature Resources MeSHPMCBookshelfDisclaimer The PubMed wordmark and PubMed logo are registered trademarks of the U.S. Department of Health and Human Services (HHS). 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2068	https://www.vhlab.umn.edu/atlas/cardiac-veins/index.shtml	\| \| \| University of Minnesota 612-625-5000 \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- --- --- --- --- \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- \| \| \| \| --- \| \| \| \| \| \| Location: The coronary system is composed of arteries, arterioles, capillaries, venules and veins. From the innumerable cardiac capillaries, blood flows back to the cardiac chambers through venules, which in turn coalesce into the cardiac veins. Most cardiac veins collect and return blood to the right atrium through the coronary sinus; there may or not be a Thebesian valve covering the ostium of the coronary sinus. The major venous vessels of the human heart are: coronary sinus, the anterior interventricular veins, left marginal veins, posterior veins of the left ventricle, and the posterior interventricular veins (see also the Coronary System Tutorial). Function: The cardiac veins returns deoxygenated blood (containing metabolic waste products) from the myocardium to the right atrium. This blood then flows back to the lungs for reoxygenation and removal of carbon dioxide. Importance in cardiovascular diseases: Cardiac veins contain valves preventing back flow; a Thebesian valve may or may not cover the ostium of the coronary sinus. Typically cardiac veins are free of atherosclerotic plaques. Importance in device delivery: Left heart pacing can be achieved by placing leads into the cardiac veins through the coronary sinus, which is located within the right atrium. While cannulating the coronary sinus ostium can be challenging due to the presence of the Thebesian valve, venous valves within the coronary venous system may hinder advancement of guide wires, catheters or pacing leads. Stem cell or other biologic therapies can be delivered to the heart via the coronary vasculature. Localización: El sistema coronario está compuesto por arterias, arteriolas, capilares, vénulas y venas. Desde los innumerables capilares cardíacos, la sangre regresa a las cámaras cardíacas a través de las vénulas, que a su vez se fusionan en las venas cardíacas. La mayoría de las venas cardíacas devuelven la sangre a la aurícula derecha a través del seno coronario; puede haber o no una válvula de Tebes que cubre el ostium del seno coronario. Los principales vasos venosos del corazón humano son: el seno coronario, la vena interventricular anterior, la vena marginal izquierda, las venas posteriores del ventrículo izquierdo y las venas interventriculares posteriores (ver también el Tutorial del sistema coronario). Función: Las venas cardíacas devuelven sangre desoxigenada (contiene productos de desecho metabólico) del miocardio a la aurícula derecha. Esta sangre luego regresa a los pulmones para la oxigenación y eliminación del dióxido de carbono. Importancia en las enfermedades cardiovasculares: Las venas cardíacas contienen válvulas que evitan el reflujo; la válvula de Tebesio puede cubrir o no el ostium del seno coronario. Normalmente, las venas cardíacas están libres de placas ateroscleróticas. Importancia en implante de dispositivos: La estimulación cardíaca izquierda se puede lograr implantando electrodos en las venas cardíacas a través del seno coronario, ubicado dentro de la aurícula derecha. Si bien la canulación del ostium del seno coronario puede ser un desafío debido a la presencia de la válvula de Tebesio, las válvulas venosas dentro del sistema venoso coronario pueden obstaculizar el avance de los alambres guía, catéteres o cables de estimulación. Las terapias con células madre u otras terapias biológicas se pueden administrar a través de la vasculatura coronaria. \| \| \| \| \| Download movie: Cardiac-Veins-startimage.mp4 \| \| \| \| \| \| © 2025 Regents of the University of Minnesota. All rights reserved. The University of Minnesota is an equal opportunity educator and employer. Privacy Statement \|
2069	https://cs.stackexchange.com/questions/80774/prove-that-hitting-set-is-np-complete	complexity theory - Prove that Hitting Set is NP-Complete - Computer Science Stack Exchange Join Computer Science 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 Prove that Hitting Set is NP-Complete Ask Question Asked 8 years ago Modified5 years, 7 months ago Viewed 22k times This question shows research effort; it is useful and clear 4 Save this question. Show activity on this post. The Hitting Set Problem (HS) is defined as follow. Let (C,k(C,k) C={S 1,S 2,...,S m}C={S 1,S 2,...,S m} collection of subset of S i.e. S i⊆S,∀i S i⊆S,∀i k∈N We want to know if exists S′⊂S where \|S′\|<k such that S i∩S′≠∅, i=1,2,...,m. Prove that HS is Np-Complete. Hint: When \|S i\|=2, HS becomes Vertex Cover, already know to be NP-complete. complexity-theory np-complete decision-problem information-theory Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Improve this question Follow Follow this question to receive notifications asked Sep 2, 2017 at 17:44 NickNick 169 1 1 gold badge 1 1 silver badge 12 12 bronze badges 1 4 What did you try? Where did you get stuck? We're happy to help you understand the concepts but just solving exercises for you is unlikely to achieve that. You might find this page helpful in improving your question.D.W. –D.W.♦ 2017-09-03 02:46:48 +00:00 Commented Sep 3, 2017 at 2:46 Add a comment\| 3 Answers 3 Sorted by: Reset to default This answer is useful 9 Save this answer. Show activity on this post. 3SAT is reduced to the Hitting Set problem. Given a 3SAT ϕ having m clauses and n variables, define S={x 1,…x n,¯x 1,…,¯x n} S i={y 1,y 2,y 3},if y 1,y 2,y 3∈S and(y 1∨y 2∨y 3)is a clause.S x={x,¯x},for all variable x.k=n Assume ϕ is satisfiable, then there a is true-value assignment for n variables. So, add x to S′ if x=t r u e, otherwise add ¯x to S′. Now, assume the HS problem has a solution S′. Then since for every variable x, S′∩S x≠∅, S′ has at least n literals of the form x or ¯x for each variable x. Furthermore, no x and ¯x may belong to S′ at the same time since the size of S′ is at most n. Also, for each clause C i, S′∩S i≠∅ and so for each clause we select y i∈S′∩S i, and set x=t r u e if y i=x and x=f a l s e if y i=¯x. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Improve this answer Follow Follow this answer to receive notifications edited Sep 2, 2017 at 22:35 answered Sep 2, 2017 at 18:15 fade2blackfade2black 9,895 2 2 gold badges 26 26 silver badges 36 36 bronze badges 1 can you show a positive and a negative instance?Nick –Nick 2017-11-03 02:37:18 +00:00 Commented Nov 3, 2017 at 2:37 Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. It is easy to show that Hitting Set is in NP. A solution just needs to exhibit the set H – one can easily verify in polynomial time whether H is of size k and intersects each of the sets B1, . . . , Bm. We reduce from vertex cover. Consider an instance of the vertex cover problem – graph G = (V, E) and a positive integer k. We map it to an instance of the hitting set problem as follows. The set A is the set of vertices V . For every edge e ∈ E, we have a set Se consisting of the two end-points of e. It is easy to see that a set of vertices S is a vertex cover of G iff the corresponding elements form a hitting set in the hitting set instance. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications answered Feb 22, 2020 at 23:12 NehaNeha 11 1 1 bronze badge 2 Welcome to ComputerScience@SE. Interesting use of special symbol ∈ - it is common to use MathJax, allowing, e.g., proper subscripting: B 1,…,B m.greybeard –greybeard 2020-02-22 23:31:04 +00:00 Commented Feb 22, 2020 at 23:31 what if size of Se is greater than 2? Will it work?Dhruvil Amin –Dhruvil Amin 2020-11-19 12:14:42 +00:00 Commented Nov 19, 2020 at 12:14 Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. The Hitting Set Problem is equivalent to the Set Cover Problem, that is defined as follow: INPUT: a collection C of subsets of a finite set S. SOLUTION: A set cover for S, i.e., a subset C′⊆C such that every element in S belongs to at least one member of C′. Given k does it exist a set cover C′ such that \|C′\|≤k? This looks like a homework, so I let you figure how, given an instance of Set Cover you can convert it to an equivalent instance of Hitting Set. Otherwise, if you want to follow the hint, you first show that Hitting Set in in N P, that is given a solution you can verify it in polynomial-time. Then, you show that given a graph G=(V,E), finding a vertex cover of size k is equivalent to find a hitting set of size k. That is, if \|E\|=m, you can build a hitting set instance with the same amount of sets. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Improve this answer Follow Follow this answer to receive notifications edited Sep 2, 2017 at 18:17 answered Sep 2, 2017 at 18:01 abcabc 1,675 2 2 gold badges 12 12 silver badges 22 22 bronze badges Add a comment\| Your Answer Thanks for contributing an answer to Computer Science Stack Exchange! Please be sure to answer the question. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. Making statements based on opinion; back them up with references or personal experience. Use MathJax to format equations. MathJax reference. To learn more, see our tips on writing great answers. 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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 Report this ad Linked 4Hitting Set Problem with non-minimal Greedy Algorithm Related 5Is the set partitioning problem NP-complete? 3Reduce Set problem to SAT 2Is this variant of set cover NP-complete? 1Reduce EXACT 3-SET COVER to a Crossword Puzzle 6Partitioning bag of sets such that each set in a group has a unique element 2A variant of hitting set problem? 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2070	https://air-source.com/blog/industrial-and-commercial-acetylene-gas-applications/	Skip to content Gas Delivery for Los Angeles and Orange County Get a quote Industrial and Commercial Acetylene Gas Applications Compressed Air Acetylene is a colorless gas that has a strong garlic-like odor. When combined with oxygen, it becomes very hot and is classified as the hottest fuel gas. It is also the only fuel gas that can weld steel, making it the preferred choice for welders. Air Source Industries is a trusted provider of acetylene gas for a variety of commercial and industrial applications in the local Long Beach area. Acetylene can be produced three different ways: It can be produced by the reaction of water and calcium carbide, by partial combustion of oxygen and methane, and by passing hydrocarbon through an electric arc. To get a better understanding on how acetylene gas works, we’ll take a look at some of the most common uses of acetylene and how they are applied in various industries. Welding and Cutting Acetylene gas is commonly used for welding and cutting in a process known as oxy fuel cutting. Acetylene is commonly used because it produces the highest flame temperature that can reach up to 3,160 degrees Celsius (5,720 degrees Fahrenheit). The flame temperature aids in speeding up the cutting process due to the quick piercing movements and is considered a safe fuel gas to use when handled properly. During the welding process, acetylene is known for reducing the area of the Heat Affected Zone (HAZ), which is the base area of a piece of metal that has not been melted, but altered by the heat. Because of its high temperature and precise cutting ability, acetylene is the ideal gas for oxy fuel welding and cutting. With its quality welding abilities, it’s the perfect fuel gas to get the job done in a timely and efficient manner, thus saving you money in the long run. Portable Lighting Back in the late 1800’s, people used acetylene to create light using portable gas lamps called carbides. They were used to illuminate homes, cars, mines and sometimes small cities. When acetylene is burned with just the right amount of air or oxygen, it reacts with a pure white light, which is why it was used to illuminate areas that did not have electricity. Carbide lamps work by the reaction of mixing calcium carbide and water to produce acetylene. Carbide lamps contain an upper chamber which control the amount of water that drips through and mixes with the calcium carbide. The amount of acetylene produced directly affects the magnitude of the flame, which is why the chamber is an important piece of the carbide lamp. Carbide lamps are still used today in some coal mining industries in areas without safety laws. Chemical Production Acetylene gas is most commonly used for the production of other chemicals used in perfumes, vitamins, polymers, solvents and other materials due to its versatility and ease of reactive control in comparison to other gases. It combines nicely with a variety of elements and compounds, making it the most important of starting materials for organic synthesis. Acetylene Safety & Handling Although acetylene is among the safer gases to use when handled correctly, it is still highly combustible. That’s why it is very important to be knowledgeable on the safety and handling of the gas. When transporting, make sure the cylinder is settled upright to avoid any hazards. After the gases settle within the cylinder and are ready to release, make sure you dispense the gas slowly to avoid any combustion or malfunctions that could occur. As always, please make sure you are knowledgeable on the safety regulations or have a professional technician from Air Source Industries deliver the cylinder for you. CONCLUSION If you have a business or project that requires the use of acetylene gas, we carry it in commercial and atomic absorption (AA) grades. For assistance with gas specifications, please call us directly at (562) 426-4017 or get a quote online if you already know what you need. We also offer next day delivery services for local businesses in LA and Orange County. Sources: Share Article Tags: acetylene, commercial, industrial Gases Acetylene Air Argon Carbon Dioxide Helium High Purity Gas Supply Mixes Nitrogen Nitrous Oxide Oxygen Acetylene Air Argon Carbon Dioxide Helium High Purity Gas Supply Mixes Nitrogen Nitrous Oxide Oxygen Services Cylinder Rentals & Refills Cylinder Repairs & Testing Delivery Equipment Online Ordering Cylinder Rentals & Refills Cylinder Repairs & Testing Delivery Equipment Online Ordering Industries Served Automotive Nitrous Oxide Medical Gases for Clinics Medical Gases for Dentists Medical Gases for Physicians Medical Gases for Veterinarians Medical Gases Restaurant/Food Grade Gases Automotive Nitrous Oxide Medical Gases for Clinics Medical Gases for Dentists Medical Gases for Physicians Medical Gases for Veterinarians Medical Gases Restaurant/Food Grade Gases About Us Testimonials Meet the Team Case Studies Testimonials Meet the Team Case Studies Air Source Industries 3976 Cherry AvenueLong Beach, CA 90807 Phone (562) 426-4017 Fax 562.490.9477 in@ce.com © 2015 – 2025 Air Source Industries. 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2071	https://coconote.app/notes/3d5eb1bf-0d0b-409c-94f1-b07c7b64392c	Pressure Measurement in Fluids \| Coconote Coconote AI notes AI voice & video notes Try for free 💧 Pressure Measurement in Fluids Sep 17, 2025 Overview This lecture covers pressure measurement in fluids, focusing on types of manometers, their working principles, advantages, limitations, and a numerical example using a U-tube manometer. Pressure Measurement Devices Pressure is measured using manometers (studied in fluid mechanics) and mechanical gauges (studied in metrology). Manometers are glass tubes containing a manometric fluid and work on the hydrostatic law: Pressure = ρgh. Types of Manometers Simple Manometer: Measures pressure at a single point; includes piezometer and U-tube manometer. Differential Manometer: Measures pressure difference between two points in a fluid system. Piezometer Simplest manometer; a glass tube connected to a pipe or vessel and open to the atmosphere. Measures pressure using the rise (h) in fluid column: Pressure = ρgh, where ρ is fluid density in the pipe. Limitations: Only measures positive (gauge) pressure. Cannot measure vacuum (negative) or high pressures. Not suitable for measuring gas pressure or pressure of very light liquids (requires impractically long tube). U-Tube Manometer Consists of a U-shaped tube, one end connected to the pressure point, the other open to the atmosphere. Uses a manometric fluid denser than the fluid in the vessel, allowing measurement of higher and both positive and negative pressures. Pressure measurement uses the hydrostatic law and reference to the lower level of manometric fluid. Manometric Equation and Numerical Example Apply hydrostatic law and sign convention: pressure increases downward (+), decreases upward (–). Reference the lower level of manometric fluid for measurements. Example: Calculate pressure using known specific gravities and heights of columns, applying the manometric equation. Key Terms & Definitions Manometer — Device using liquid columns to measure fluid pressure. Hydrostatic Law — Pressure at a point in a fluid = ρgh. Piezometer — Simple manometer with only the system fluid, measuring positive pressures. U-Tube Manometer — U-shaped device using a denser manometric fluid to measure pressure differences. Gauge Pressure — Pressure relative to atmospheric pressure. Absolute Pressure — Total pressure including atmospheric pressure. Action Items / Next Steps Review the numerical example calculation for clarity. Prepare for upcoming topics: Micro Manometer and Differential Manometer in the next lecture. Practice related numerical problems for exam preparation. 📄 Full transcript Overview This lecture covers pressure measurement in fluids, focusing on types of manometers, their working principles, advantages, limitations, and a numerical example using a U-tube manometer. Pressure Measurement Devices Pressure is measured using manometers (studied in fluid mechanics) and mechanical gauges (studied in metrology). Manometers are glass tubes containing a manometric fluid and work on the hydrostatic law: Pressure = ρgh. Types of Manometers Simple Manometer: Measures pressure at a single point; includes piezometer and U-tube manometer. Differential Manometer: Measures pressure difference between two points in a fluid system. Piezometer Simplest manometer; a glass tube connected to a pipe or vessel and open to the atmosphere. Measures pressure using the rise (h) in fluid column: Pressure = ρgh, where ρ is fluid density in the pipe. Limitations: Only measures positive (gauge) pressure. Cannot measure vacuum (negative) or high pressures. Not suitable for measuring gas pressure or pressure of very light liquids (requires impractically long tube). U-Tube Manometer Consists of a U-shaped tube, one end connected to the pressure point, the other open to the atmosphere. Uses a manometric fluid denser than the fluid in the vessel, allowing measurement of higher and both positive and negative pressures. Pressure measurement uses the hydrostatic law and reference to the lower level of manometric fluid. Manometric Equation and Numerical Example Apply hydrostatic law and sign convention: pressure increases downward (+), decreases upward (–). Reference the lower level of manometric fluid for measurements. Example: Calculate pressure using known specific gravities and heights of columns, applying the manometric equation. Key Terms & Definitions Manometer — Device using liquid columns to measure fluid pressure. Hydrostatic Law — Pressure at a point in a fluid = ρgh. Piezometer — Simple manometer with only the system fluid, measuring positive pressures. U-Tube Manometer — U-shaped device using a denser manometric fluid to measure pressure differences. Gauge Pressure — Pressure relative to atmospheric pressure. Absolute Pressure — Total pressure including atmospheric pressure. Action Items / Next Steps Review the numerical example calculation for clarity. Prepare for upcoming topics: Micro Manometer and Differential Manometer in the next lecture. Practice related numerical problems for exam preparation.
2072	https://math.stackexchange.com/questions/4189462/proof-of-fx-0-implies-text-function-is-increasing-proof-verification?noredirect=1&lq=1	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 Proof of $f'(x) > 0 \implies \text{ Function is increasing }$ proof verification [duplicate] Ask Question Asked Modified 4 years, 2 months ago Viewed 310 times -1 $\begingroup$ I tried to prove the theorem myself, that if for every $x$ in the domain $f'(x)$ is positive, then the function $f$ is strictly increasing. This is how it went: $$f(x_1) < f(x_2)$$ $$f(x) < f(x+h), h>0 $$ $$f(x) < f(x) + f'(x)\cdot h + O(h), \lim_{h \rightarrow 0} \frac{O(h)}{h} = 0$$ $$0 < f'(x) \cdot h + O(h) $$ $$0 < f'(x) + \frac{O(h)}{h}$$ Now here is where I am not sure about the correctness of what I am about to do next...to take the limit of both sides and then: $$ 0 < f'(x) + 0$$ Is the proof correct ? real-analysis calculus solution-verification Share edited Jul 3, 2021 at 17:04 VLCVLC asked Jul 3, 2021 at 16:57 VLCVLC 2,5431010 silver badges1818 bronze badges $\endgroup$ 2 1 $\begingroup$ This isn't a proof, it's a list of formulas with no explanation of how they are connected. Furthermore, you appear to be reasoning backwards. Don't start with what you're supposed to prove. Start with what you know, and reason from there to what you're supposed to prove. So start with "Assume that $f'(x) > 0$ for all $x$ in some interval $I$. Suppose $x_1, x_2 \in I$ and $x_1 < x_2$." Now explain (in sentences, not just a list of formulas) why it must be the case that $f(x_1) < f(x_2)$. $\endgroup$ Dan Velleman – Dan Velleman 2021-07-03 17:05:43 +00:00 Commented Jul 3, 2021 at 17:05 $\begingroup$ What about the Lagrange mean value theorem? $\endgroup$ Matcha Latte – Matcha Latte 2021-07-03 17:09:11 +00:00 Commented Jul 3, 2021 at 17:09 Add a comment \| 1 Answer 1 Reset to default 1 $\begingroup$ You actually attempted to prove the converse of the statement (counterexample given below). In order to prove the desired statement, you would start with $f'(x) > 0$ for all $x$ in the domain of $f$, and try to conclude $f(x_1) < f(x_2)$ for all $x_2 > x_1$. Share edited Jul 3, 2021 at 17:44 answered Jul 3, 2021 at 17:06 Jeff ChengJeff Cheng 15666 bronze badges $\endgroup$ 1 2 $\begingroup$ Increasing function doesn't imply $f'>0$. $x^3$ is even strictly increasing, yet $f'(0)=0$. $\endgroup$ Matcha Latte – Matcha Latte 2021-07-03 17:17:44 +00:00 Commented Jul 3, 2021 at 17:17 Add a comment \| Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions real-analysis calculus solution-verification See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Linked 9 Positive derivative implies increasing without Mean Value Theorem Related $f'$ strictly increases and $f'(c)=0$. There exist $x_1 < c < x_2$ such that $f'(c)=\frac{f(x_2)-f(x_1)}{x_2 - x_1}$ 2 Showing that a function is strictly increasing in $(a,b)$ 2 Bound for $\prod_i^\infty(1+x_i)$ if $\sum_i^\infty x_i$ converges to $L$. 1 Show $\|\mathbb R\| = \|(0, 1)\|$ 1 [Let $f$ be continuous in $a,b)$ with no mimimum at all open interval and tend to infinity. Prove it is strictly increasing. 2 Inverse of strictly increasing function is continuous (proof verification) 1 [Solution verification of $f$ convex, positive for $x>0$ and $f(0)=0$ implies $f$ increasing in $0,\infty)$. $f:\mathbb{R}\to \mathbb{R}$ is a strictly increasing function, such that, $0 2 Example of unbounded, strictly increasing, strictly concave real-valued function on the real line Hot Network Questions Exchange a file in a zip file quickly How to use \zcref to get black text Equation? What "real mistakes" exist in the Messier catalog? What meal can come next? 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2073	https://chemequations.com/en/?s=KI+%2B+H2SO4+%3D+I2+%2B+K2SO4+%2B+H2S+%2B+H2O	Chemical Equations online! Chemical Equations online! Submit Advanced search Error: KI H2SO4 = I2 K2SO4 H2S H2O ^ Unexpected character, expected the character '+' or the end of the equation Discover more Sodium bicarbonate Hydrogen hydroxide Calcium carbonate calcium carbonate Caustic soda Carbonate de calcium Carbonato de calcio oxidane Sulfuric acid Sodium hydrogen carbonate Found a bug?Do you know what to improve? © Michal Punčochář
2074	https://www.biointeractive.org/taxonomy/term/283	Rock Pocket Mice \| HHMI BioInteractive Skip to main content User MenuBioInteractive HomeMenu Community Classroom Resources Teaching Tools Professional Development Partner Content Site Search Log in Create Account Español HHMI BioInteractive Español Site Search Log in Create Account Community Classroom Resources Teaching Tools Professional Development Partner Content How Do Some Cells Affect Mouse Color? In this phenomenon-driven activity, students investigate how cells are signaled to make melanin and explain how mutations in melanin pathway proteins affect the coat color of various organisms. Cell Biology Genetics Lessons High School — General High School — AP/IB College What Causes Different Fur Colors? In this inquiry-based activity, students investigate the phenomenon of fur colors in rock pocket mice to connect genotypes to phenotypes and molecular genetics to evolution. Genetics Evolution Lessons High School — General High School — AP/IB Color Variation Over Time in Rock Pocket Mouse Populations This activity allows students to collect and analyze data on the evolution of coat color in rock pocket mouse populations living on differently colored substrates. Evolution Card Activities High School — General High School — AP/IB Molecular Genetics of Color Mutations in Rock Pocket Mice In these activities, students transcribe and translate portions of the rock pocket mouse Mc1r gene to further explore the genetic variations responsible for different coat colors as described in the short film Making of the Fittest: Natural Selection and Adaptation. Genetics Evolution Lessons High School — General High School — AP/IB College Developing an Explanation for Mouse Fur Color In this activity, students collect and analyze evidence for each of the major conditions for evolution by natural selection to develop an explanation for how populations change over time. Evolution Science Practices Lessons High School — AP/IB College Interactive Assessment for Natural Selection and Adaptation Several questions are embedded within the short film The Making of the Fittest: Natural Selection and Adaptation, which uses the rock pocket mouse as a living example of natural selection. Genetics Evolution Interactive Videos High School — General High School — AP/IB College The Making of the Fittest: Natural Selection and Adaptation This film describes natural selection and adaptation in populations of rock pocket mice living in the American Southwest. Genetics Evolution Short Films High School — General High School — AP/IB College Natural Selection and Evolution of Rock Pocket Mouse Populations This activity supports concepts covered in the film Natural Selection and Adaptation. Students analyze genetic sequence data and draw conclusions about the evolution of coat-color phenotypes in the rock pocket mouse. Genetics Evolution Lessons High School — General High School — AP/IB College Biochemistry and Cell Signaling Pathway of the Mc1r Gene This activity supports concepts presented in the short film Natural Selection and Adaptation about the evolution of different fur colors among rock pocket mouse populations. The activity challenges students to analyze partial DNA sequences of the Mc1r gene to identify the effects of mutations on the MC1R protein pathway. Biochemistry & Molecular Biology Genetics Lessons High School — AP/IB College Allele and Phenotype Frequencies in Rock Pocket Mouse Populations This activity reinforces concepts of variation and natural selection covered in the short film Natural Selection and Adaptation. Students apply the Hardy-Weinberg equation to real data collected on rock pocket mouse populations. Evolution Science Practices Lessons High School — General High School — AP/IB College Pagination Currently on page 1 1 Next page› Subscribe to Rock Pocket Mice HHMI BioInteractive About Us Our Research Our Team Our Advisors FAQ Contact Us Facebook Instagram YouTube Newsletter Signup HHMI.org Terms of Use Privacy Policy Accessibility This site uses cookies or similar technologies. To learn more, please review our Privacy Policy and Cookie Notice. By continuing to browse the site, you agree to these uses. More info Accept
2075	https://crei.cat/wp-content/uploads/users/pages/gov_%20size%20and%20macro.pdf	European Economic Review 38 (1994) 117-l 32. North-Holland Government stability size and macroeconomic Jordi Gali Columbia University. New York, USA Received July 1990, final version received March 1993 We analyze the effects of government size on output variability in the context of a RBC model in which government size is parameterized by the income tax rate and the share of government purchases in output. The model implies that: (i) income taxes are destabilizing, and (ii) for most specifications considered, government purchases are stabilizing. We compare those predictions with the results of simple cross-country regressions using data for 22 OECD countries. The estimated relationship between empirical indicators of government size and measures of GDP variability appears far stronger than the model predicts, and often has the opposite sign. 1. Introduction The present paper investigates the relationship between government size and macroeconomic stability in the context of a real business cycle (RBC) model. Focus on the previous relationship can be traced back to the traditional Keynesian literature on ‘automatic stabilizers’ [e.g., Burns (1960), Baily (1978), DeLong and Summers (1986)]. That literature points to the increase in income taxes, government purchases, unemployment insurance benetits and other governmental programs as some of the ‘structural changes’ underlying the greater stability of the U.S. economy after World War. II. Underlying the notion of automatic stabilizers found in the Keynesian literature there is a framework in which households face some sort of liquidity constraints that link consumption to current disposable income, and where short-run fluctuations are demand-driven. In the present paper we revisit the concept of automatic stabilizers by studying its potential role in the context of the basic RBC model, the central paradigm of modern Correspondence to: Jordi Gali, 607 Uris Hall, Graduate School of Business, Columbia University, New York, NY 10027, USA. Thanks are due to Glenn Hubbard, four different referees, and workshop participants at the NBER Summer Institute, I.M.F. and U. Pomneu Fabra for helnful comments. and Joon-Ho Hahm for expert research assistance. Part of the present research was funded by a CBS Faculty Grant. 0014-2921/94/$07.00 ~3 1994-Elsevier Science B.V. All rights reserved SSDI 0014-2921(93)E0045-M 118 J. Gali, Government size und macroeconomic stability macroeconomics.’ We pose the following basic question: do income taxes and government purchases behave as automatic stabilizers in the basic, technology shock-driven, RBC model? In other words, can the government alter the intensity of the business cycle by changing its own ‘size’, when technology shocks are the source of economic fluctuations? In section 2 we attempt to provide an answer to those questions, by augmenting an otherwise standard RBC model with a government sector. More precisely, we introduce two fiscal parameters in the model: (i) an income tax rate, and (ii) the share of government purchases in output. Then we analyze the effects of variations in the model’s fiscal parameters on output variability, as measured by the standard deviation of either percent devi- ations of output from its steady-state value or output growth rates, depending on the specification.’ We show that, for most specifications considered, both a low tax rate and a high share of government purchases are associated with low output variability. The magnitude of the predicted effects is, however, very small. In section 3 we confront those predictions with evidence based on postwar data corresponding to 22 OECD countries. We seek to characterize a possible correlation between (i) average share of government revenues and purchases in GDP (the empirical counterpart to our fiscal variables) and (ii) the standard deviations of GDP growth or detrended log GDP (our measures of output variability). Given the theoretical results of section 2, we view this exercise as an attempt to assess the ability of the basic RBC model to match dimensions of the data other than the time series dimension, e.g. differences in the characteristics of business cycles across countries. Though calibrated versions of simple RBC models are capable of generating time series with properties that roughly match those of actual U.S. macroecono- mic time series, the number of stylized facts which current RBC models are able to replicate is very limited and, as Danthine and Donaldson (1993) have emphasized, any progress in modeling must necessarily involve confronting existing models with an increasingly richer set of stylized facts3 The empirical results of section 3 point to the presence of a significant relationship between fiscal variables and measures of GDP variability. ‘Expositions of the basic RBC model can be found in Prescott (1986). For a survey of the literature see King et al. (1988) and Plosser (1989). ‘Danthine and Donaldson (1985) and Greenwood and Huffman (1991) contain related exercises involving the effect of changes in tax rates on output variability in a RBC model, but neither paper looks at government purchases. See below for a discussion of their models and results, as well as for other relevant references. jA similar approach underlies some of the recent developments in growth theory, where alternatives to the standard neoclassical growth model have been developed in response to the latter’s inability to account for some cross-country evidence (e.g., the lack of convergence of income levels). As Lucas (1988) points out, that critical assessment has taken place despite the fact that the neoclassical model is consistent with most features of long-term U.S. time series, the main aspect of the data it was originally meant to explain. J. tiali, tiovernmmt size and macroeconomic stability 119 Nevertheless, the sign and magnitude of the effect detected empirically cannot always be reconciled with the predictions of the standard RBC model. This is true even when we control for policy variability factors which, though ignored by the model, may potentially affect output variability while being correlated with our fiscal measures. Section 4 discusses the implications of our results and concludes. 2. Government size and macroeconomic stability in a RBC model In this section we develop a basic one-sector RBC model augmented with a government sector. We assume an infinite-lived representative consumer who seeks to maximize (1) where C is consumption and L is leisure. We specify the utility function to be given by U(C,, L,) = log C, + H log Lt.4 Our agent has access to a constant-returns technology represented by a Cobb-Douglas production function,5 Y, = A&: ~“(X,NJU (2) for t=0,1,2 ,..., where Y is output, K is physical capital, and N is employment. (X,} represents n onstochastic technical progress, and is given by X,=X,?/‘. {A,) is the stochastic component of technology; its logarithm, denoted by .& is assumed to follow the AR( 1) process (1 --pL) A, =F, where 05~5 1, and {Ed} is an i.i.d. sequence with zero mean and variance v2. The government budget constraint is given by B,, 1 = B, [ 1 + (1 -z)v,] + G, + r, -T Y,, for t = 0, 1,2,. . where T denotes lump-sum transfers, B is the outstanding one-period riskless government debt, r is the interest rate on one-period riskless assets, z is the income tax rate, and G denotes govern- ment purchases. In the initial period the policy maker chooses a tax rate 7 as well as the steady-state share of government purchases in output. That share is denoted by sp. Two alternative rules for government purchases are considered. Under the first rule, G, = sg Y,, all t, i.e. government purchases are a constant fraction of output. Under the second rule, G,=gX,, i.e. govern- ment purchases grow at a constant (gross) growth rate y. Henceforth we refer to those rules as ‘constant share’ and ‘constant growth’ rules, respectively. %uch for U consistent by a constant and constant growth for Y and K, as argued in and Rebel0 (1988). Experimentation with did not affect any the qualitative results we could as selling labor and capital services in to firm to the specified technology. As well known, two market arrangements 120 J. Gali, Gowxment size and macroeconomic stability Admittedly, our assumptions on government behavior are highly stylized and somewhat unrealistic, but they allow us to focus on the impact of the government’s relative size. From this viewpoint, our analysis can be seen as complementary to certain RBC literature that analyzes the effects of stochas- tic variations in the tax rate and/or the share of government purchases in GDP [e.g., Christian0 and Eichenbaum (1992), Braun (1989), McGrattan (1991), Chari et al. (1993)]. Our consumer-producer maximizes (1) subject to K t+l+B,+,=(l-~)K,+(l-z)Y,+B,[I+(l-~)rt]+T~-Cr, (3) L,$N,=l, (4) for t=0,1,2 ,..., as well as a non-negativity condition for K and C, and the transversality condition lim,, 3. E,{n,T=,[l +(I -z)r,]}-’ BTzO. 6 in (3) is the rate of depreciation of physical capital. In solving this problem he takes z, the stochastic sequences {G,), {T,), {r,} and the initial level of capital and government debt as given. It is useful to transform the model above into a stationary one by dividing all variables (except N and L) by X. The transformed variables are denoted by lower case letters: c, r(C’,/X,), k, z(K,/X,), etc. A competitive equilibrium in our economy can then be defined as a sequence {cL,k,, yt, Nt,gt)zO satisfying the following conditions: ec,=(l-N,)r(l-z)(y,lN,), E, {(c&t + i )C(l-~)+(~-~)(l-a)(y,+llk,+,)l)=yB-’, y, = A,k; -‘N’ t> (5) (6) (7) g, = ssy, (‘constant share’), or g, =g (‘constant growth’), $,+I = (I-W,+J+-_g,-c,> for t =O, 1,2,. . . , and the transversality condition lim E,flT(l/c,)k,=O. T-m (10) and initial capital stock, eqs. fully characterize equilibrium k,, yt, N,,g,izo indepen- of the transfer/debt sequence jtr, b,}zO chosen by the government among infinite number of yb, + , = b,[ 1 + ( 1 - T)r, J + g, + t, - T_Yt, for t=O,1,2 )...) and the transversality condition lim T+ n E, p’( l/c,) h, = 0. In other words, a version of Ricardian equivalence holds in the model above. To further characterize the economy’s competitive equilibrium we use the log-linear, certainty equivalence methods described in King et al. (1988). The J. Gali, Government size and macroeconomic stability 121 idea is to solve for the perfect-foresight equilibrium path and to linearize the latter around the economy’s balanced-growth path, replacing future ,? values in that solution with their expected values. Further details can be found in the above reference. That approach can be used to generate univariate processes describing the fluctuations of different variables around their stationary values. Given the purpose of the exercise we focus on the univariate representation for output implied by the model, which takes the form where j, -log(y,/y) is the percent deviation of output from its steady state value y. ,n, &0 and 4i are complicated functions of the model’s parameters.6 Given values for the parameters characterizing technology (y, CI, p, v), prefer- ences (/I, O), and fiscal policy (r, s,, and a rule for government purchases) we can compute the coefficients of the above ARMA(2,l) process and determine the implied unconditional standard deviation of j. In order to do that, we calibrate the model in a way consistent with some stylized features of the postwar U.S. economy. We let y= 1.02, roughly the average (gross) growth rate in per capita income in the postwar period. Under perfect competition, a corresponds to the labor income share, which we set at 0.75. Following King et al. (1988) we use the average return to equity for the postwar period (6.5%) as r’s empirical counterpart. As in other studies a value of 0.10 is assumed for 6, the annual depreciation rate. A benchmark value for r is 0.3, roughly the average ratio of government revenues to GDP, sg is assigned a benchmark value of 0.2, the average share of government purchases in GDP. Given our assumptions on y, r, and Y, and exploiting the fact that /I= y/ [1 + (1 - z)r] must hold in the steady state, we set fl=O.975. We experiment with two alternative values for parameter 8: (i) 8= 1.6014, which, under our benchmark fiscal settings, is consistent with a steady-state value for N equal to l/3, the latter being the average fraction of time allocated to production [Prescott (1986)], and (ii) 8=0, which corresponds to an inelastic labor supply (N, = 1). Finally, v and p are chosen in each case so that, under the benchmark values for the policy parameters, the standard deviation and the first-order autocorrelation of j predicted by the model equal 3.57 and 0.74, their respective sample counterparts our U.S. data set7 Table 1 reports the implied standard deviations of j, for alternative r and sg settings, under the assumptions of a ‘constant growth’ rule for government purchases and transitory shocks (lpi< 1). The model predicts that, holding sg ‘p corresponds to the stable eigenvalue of the linearized dynamical system for k, and ?,. Again, we refer the reader to the King et al. (1988) paper for details. ‘The statement does not apply to the simulation reported in table 4, for which p is no longer a free parameter. See discussion below. 122 .I. Gali, Government size and macroeconomic stability Table 1 Transitory shocks, constant growth rule.” SK = 0.00 s,=O.lO s, = 0.40 “9=0.20 A= 0.30 5 = 0.00 3.33 3.20 3.06 2.92 2.16 r=0.15 3.58 3.44 3.29 3.13 2.96 s=o.30 3.88 3.73 3.51 3.40 3.22 r = 0.45 4.21 4.11 3.95 3.76 3.56 s=O.60 4.78 4.62 4.45 4.26 4.04 “For each pair (r. sJ the table shows the standard deviation of lOOF (i.e. the percent deviations of output from trend) predicted by the RBC model under a ‘constant growth’ rule for government purchases and the following parameter settings: p =0.62, v =0.0132, ;a = 1.02, ~~0.75, 6=0.10, 0= 1.601, and fi=O.975. Table 2 Transitory shocks, constant share rule.” s, = 0.00 s,=O.lO Sg = 0.20 s =0.30 g s,=o.40 s=o.OO 3.36 3.21 3.05 2.87 2.69 s=O.lS 3.61 3.45 3.27 3.09 2.89 r=0.30 3.91 3.74 3.57 3.36 3.14 r = 0.45 4.30 4.13 3.94 3.73 3.49 z=O.60 4.81 4.64 4.45 4.24 3.99 “For each pair (r,sJ the table shows the standard deviation of 1009 (i.e. the percent deviations of output from trend) predicted by the RBC model under a ‘constant share’ rule for government purchases and the following parameter settings: ~~0.63, v=O.O132. y=1.02, x=0.75, 6=0.10, 0=1.601, and p=O.975. constant, variations in z generate changes in the same direction in our measure of output variability. Thus, and given s,=O.2, a reduction in the tax rate from 30 percent to zero lowers the standard deviation of j by 51 basis points. Doubling the tax rate to 60 percent actually increases the variability of j by 88 basis points. In other words, higher taxes appear to be destabilizing in the model, though the effect is relatively small. That basic result carries over to a version of the model with a ‘constant share’ for government purchases (table 2), as well as a version with inelastic labor supply (table 3). We also consider the case of permanent technology shocks, in which the technology parameter is specified to follow a random walk (p= 1). In that case, equilibrium (log) output is a nonstationary process, SO we use the standard deviation of output growth rute as a variability measure.8 We calibrate v so that the implied standard deviation of output growth under benchmark values for r and sp matches its U.S. sample counterpart (2.26). The corresponding results, reported in table 4, are ‘The latter specification excludes the possibility of a ‘constant growth’ rule for government purchases consistent with a constant steady state share, so the analysis in that case was restricted to the ‘constant share’ rule case. J. Gali, Government size and macroeconomic stability 123 Table 3 Transitory shocks, constant growth rule, inelastic labor supply.” sg = 0.00 s,=O.lO sg = 0.20 s = 0.30 sg = 0.40 T=o.m 3.40 3.42 3.44 3.47 3.51 T=0.15 3.45 3.47 3.49 3.53 3.51 rz0.30 3.51 3.53 3.51 3.60 3.65 T = 0.45 3.59 3.62 3.66 3.70 3.76 T = 0.60 3.70 3.74 3.79 3.84 3.92 “For each pair (T,sJ the table shows the standard deviation of 1OOj (i.e. the percent deviations of output from trend) predicted by the RBC model under a ‘constant growth’ rule for government purchases and the following parameter settings: p = 0.59, r= 0.0239 , ~=1.02, a=0.75, 6=0.10, O=O.OO. and fi=O.975. Table 4 Permanent shocks, constant share rule.” s,=o.oo s&.=0.10 sg = 0.20 sg = 0.30 ss = 0.40 T=o.oo 2.19 2.15 2.11 2.07 2.02 T=0.15 2.26 2.22 2.17 2.12 2.07 T =0.30 2.35 2.30 2.26 2.19 2.14 T = 0.45 2.47 2.42 2.36 2.30 2.23 1~0.60 2.64 2.58 2.52 2.45 2.38 “For each pair (r,sJ the table shows the-standard deviation of 10043, (i.e. the percent output growth rates) predicted by the RBC model under a ‘constant share’ rule for government purchases and the following parameter settings: p= 1.00, v=O.o086, y= 1.02, ~~0.75, 6=0.10, O= 1.601, and /I=O.975. qualitatively very similar to the ones discussed above: higher tax rates tend to increase the variability of output growth rates in this case, i.e. they behave as automatic destabilizers. Somewhat related results can be found in other authors’ work. Danthine and Donaldson (1985) introduce a capital income tax in a RBC model with inelastic labor supply, i.i.d. shocks, and one-hundred percent depreciation of physical capital. Under the previous assumptions the equilibrium dynamics can be solved for explicitly. Using the Danthine-Donaldson formulas for steady state y and var(y,) it is straightforward to show that, under their assumptions, var(j) [g (l/y’) var(y,)] is independent of the tax rate level. Our analysis thus shows that the Danthine-Donaldson result depends heavily on their strong assumptions, and does not carry over to a more general RBC setup. Greenwood and Huffman (1991) introduce capital and labor income taxes in a somewhat nonstandard RBC model.9 When evaluating the effects of tax changes on macroeconomic stability, their findings are similar to ours: lower taxes lead to a reduction in volatility. ‘Their model has a variable rate of capital utilization, and technology shocks affect only the ‘efhciency of new additions to the capital stock (but not that of the capital stock in place). 124 .I. Guli, Gol;ernment size and stability What is the economic mechanism underlying the destabilizing effects of taxes? Much of the observed effect results from the increase in the labor supply elasticity brought about by a higher tax rate, through its negative impact on steady state employment. lo The higher labor supply elasticity enhances the response of employment to a given technology shock, leading in turn to a larger response of both output and investment (and thus future output). Yet, the previous mechanism cannot be the only one at work, for (qualitatively) similar results are observed when we assume an inelastic labor supply (see table 3). In that case, the magnitude of the output reponse to a given technology shock ii, depends on the size of the multiplier (~?k^,+,/G;i,), but the latter is positively related to the steady-state output/capital ratio which is, in turn, increasing in z.l’ It follows that higher income tax rate will imply a larger multiplier (8j f+ l/~At), thus explaining the output variability effects observed in table 3. Tables I to 4 can also be used to examine the effects on output variability of changes in the steady-state share of government purchases sg, holding the tax rate constant. The sign of those effects is no longer robust across specifications. In all the specifications considered with Q>O - see tables 1, 2, and 4 - the government spending share sg appears to be inversely related to the standard deviation of output. Thus, for instance, under the calibration corresponding to table 1, a reduction of sg from 20 percent to zero leads to an increase of 31 basis points in our output variability statistic, given r=0.30. Raising s, from 20 to 40 percent reduces the same statistic by 35 basis points. In other words, government spending behaves as an automatic stabilizer under those specifications. That result carries over to the calib- rations corresponding to tables 2 and 4. The basic mechanism responsible for those results is essentially the same as in the tax rate case, though now it works in the opposite direction: an increase in ss leads to a higher steady-state employment and, consequently, a lower labor supply elasticity, and a smaller reponse of employment and output to a technology shock.” On the other hand, when we set H=O ~ thus making labor supply perfectly ‘% our model a higher income tax rate reduces steady-state employment through its negative effect on (after tax) labor productivity, which more than offsets the wealth effect working in the opposite direction. The reduction in N leads to a higher labor supply elasticity, which can be shown to be given by (1~ N)/N. “Some intuition for this result can be obtained by noticing that the (absolute) change in output in response to a technology shock is given by yd,. Letting 4, denote the marginal investment share (i.e., the fraction of the change in output that is allocated to investment) it follows that k^ ,+, =qb, (y/k) ii,. Though 4, also depends on the tax rate (as well as other parameters) in a complicated way, the positive effect of higher tax rates on (p/k) is dominant under all the calibrations considered. “In contrast with the case of distortionary taxation, a change in sg does not affect labor productivity and thus, has no substitution effects on employment. The positive relationship between sg and N arises from the negative wealth erect of higher government purchases (which lead to a lower consumption of leisure). J. Gali, Government size and macroeconomic stability 125 inelastic - the opposite results obtains (see table 3): output variability increases with sgr though, quantitatively, the effect is very sma11.r3 Doubling the value of both the income tax rate and the purchases share simultaneously (from 30 to 60 percent and from 20 to 40 percent, respectively) leads to greater output variability in all the cases considered. Alternatively, when, starting from the benchmark fiscal settings, we eliminate the govern- ment altogether, the standard deviation of output goes down. Given the findings discussed above, this is not surprising in the d =0 case. However, in all the other cases the net effects of such a resealing of both z and s, could not be predicted a priori. The results suggest that the stabilizing effect of a higher spending share is more than offset by the destabilizing effects of a proportional increase in the tax rate. Quantitatively, however, the net effect is very small: for the specification corresponding to table 1, a simultaneous increase in r and sp of 20 and 30 percentage points, respectively, increases the standard deviation of output by less than half a percentage point. When shocks are permanent (table 4) a similar resealing of fiscal variables leads to an increase in the standard deviation of output growth of just 0.12 percentage points. In the following section we confront some of these predictions with the data. 3. Government size and macroeconomic stability: The evidence In this section we present some econometric evidence bearing on the role of income taxes and government purchases as automatic stabilizers. Table 5 reports several statistics related to those fiscal variables, as well as measures of output variability, for 22 OECD countries. All the data are annual, covering the 196&1990 sample period, and were obtained from the OECD database. As proxies for z and s, we use the tax revenues/GDP ratio and the government purchases/GDP ratio, respective1y.14 Their average values for each country over the 1960-1990 sample period are reported under Z and S, in table 5. In the same table we also report three indicators of ‘policy variability’ for each country: the standard deviation of both the tax revenues/ GDP and government purchases/GDP ratios ($71 and s[s,]), as well as their correlation with both the (linearly) detrended and first-differenced logarithm “Again with constant employment the initial response of capital to a technology shock is given by t,+, =v4 (y/k) ii, with c$, being the marginal investment share. Even though an increase in sg does not affect (y/k), it causes the marginal investment share to go up in all our calibrated models, thus amplifying the response of capital and output. 14The model in section 2 treats all government purchases symmetrically, as nonproductive expenditures absorbing part of private sector output. This symmetric treatment is extended to our empirical measure of government purchases which includes both public consumption and public investment. In a way also consistent with our theoretical model, our measure of government purchases does not include government transfers. Table 5 Basic statisttcs.” Country U.S.A. Japan Germany France U.K. Italy Canada Austria Belgium Denmark Finland Greece Iceland Ireland Luxembourg Netherlands Norway Portugal Spain Sweden Switzerland Australia Mean Maximum Minimum i .TlTl 29.2 I .97 24.8 5.22 41.4 3.86 41.4 4.14 37.9 3.74 33.5 3.99 34.5 4.92 42.7 4.22 41.7 9.34 45.0 10.6 35.9 3.85 28.2 4.42 32.5 2.41 35.2 6.20 43.5 9.27 49.6 4.24 46.5 7.30 27.3 7.19 25.4 6.20 50.7 9.71 29.6 4.65 28.7 3.73 36.6 5.51 50.7 10.6 24.8 1.97 ~ 0.01 ~ 0.29 -0.31 -0.65 0.11 -0.43 -0.31 - 0.69 -0.17 ~ 0.28 -0.36 -0.41 0.08 -0.45 0.0 I -0.55 -0.01 -0.62 0.02 ~ 0.48 0.01 - 0.42 -0.18 -0.63 0.93 - 0.46 -0.04 -0.23 ~ 0.027 -0.25 -0.61 -0.71 0.28 -0.29 -0.17 - 0.43 ~ 0.67 -0.49 -0.03 ~ 0.64 -0.20 - 0.48 -0.10 -0.29 -0.09 - 0.46 0.93 -0.23 -0.67 -0.71 19.7 0.78 0.07 14.8 1.63 0.12 22.8 2.07 0.43 20.4 1.65 -0.17 23.3 2.04 0.09 18.7 1.90 -0.35 21.5 2.0 I 0.22 22.6 1.92 0.28 18.5 2.87 0.43 25.8 4.25 0.18 20.9 2.52 -0.25 15.2 3.42 -0.29 21.9 3.49 0.09 20.8 2.86 0.49 19.0 3.99 -0.31 21.5 1.19 0.29 21.8 2.60 0.31 17.0 3.64 -0.16 15.5 2.97 -0.75 28.4 4.12 0.19 11.7 1.39 -0.27 18.8 2.40 0.10 20.1 2.53 0.03 28.4 4.25 0.49 11.7 0.78 -0.75 -0.17 3.57 2.26 -0.79 11.9 3.31 -0.58 5.21 2.14 -0.80 6.77 1.80 - 0.40 3.08 2.04 -0.59 5.59 2.37 -0.49 5.15 2.21 - 0.62 6.23 1.97 -0.65 6.54 2.12 ~ 0.053 4.82 2.38 - 0.42 3.72 2.35 - 0.67 11.5 3.59 -0.29 6.55 4.27 -0.31 3.71 2.17 -0.38 4.28 3.05 ~ 0.064 7.77 2.26 -0.26 3.31 1.82 - 0.47 8.43 3.19 - 0.50 10.3 2.94 -0.63 5.03 1.85 -0.53 6.42 2.73 - 0.42 5.32 2.12 -0.51 6.15 2.50 -0.17 Il.9 4.27 -0.80 3.08 I .80 a7 is the government revenues/GDP ratio. .sp is the government purchases/GDP ratio. An upper bar denotes the average over the period 196&1990. j is the percent deviation of per capita GDP from trend. Ay is the growth rate of per capita GDP. s[ ] and p[.] are the standard deviation and the correlation coefficient. .I. Gali, Gowrnment size and macroeconomic stability 127 of GDP (p[r,j], p[~, dy], p[s,, 91 and p[~~,dy]).~’ Finally, the last two columns of table 5 contain two measures of output variability for each country: the standard deviation of linearly detrended log GDP ($91) and the standard deviation of GDP growth (s[dy]). At the bottom of table 5 we report the mean and both the maximum and minimum values taken by each variable in our cross section of countries. We want to stress here the large observed variability in Z and .Q. We take that feature as being the result of different tastes and/or historical and political experiences, and exogenous to the phenomenon of macroeconomic instabi- lity. In particular, we find it reasonable to assume that the degree to which policy makers pursue active ‘countercyclical’ policies (reflected in short term variations in T and sg) is not systematically related to the size of government. In addition to the large differences in fiscal parameters, the substantial variation across countries in our two measures of GDP variability, s[j] and s[dy], stands out as an important feature of the data, as becomes clear by looking at the corresponding columns of table 5. To what extent can those differences be accounted for by the differences in t and Sg? We address this issue by estimating several cross-country regressions, using the data shown in table 5. We discuss the results of that exercise in the remainder of this section. Table 6 reports the estimated coefficients in regressions with s[jj] as a dependent variable and alternative combinations of fiscal policy statistics as regressors. Standard errors are reported in brackets. The estimates in the first two regressions point to the presence of a significant negative correlation between $91 and each of our government size measures, S and Sg. Interpret- ing the coefficient estimates as partial derivatives, the estimated effects appear to be quantitatively large: an increase of 10 percentage points in Z (without controlling for other variables) is associated with a reduction of 1.5 percentage points in the standard deviation of detrended output. The latter value is even higher (3.9 percentage points) in the case of an analogous increase in Sg. Because of the high correlation between Z and S, across countries, when both variables are included as regressors the individual coefftcients estimates become insignificant [see regression (3)], though they remain negative. Neverthless, they are still highly jointly significant, as indicated by the F-statistics. The high collinearity between the two regressors, makes it difficult to evaluate the importance of each factor as a stabilizing force. Yet, estimates of some linear combinations of the coefficients are still precise. Consider, for instance, an increase of 30 percentage points in Z accompanied by a simultaneous increase of 20 percentage points in Tg, a change well within the “Similar results were obtained when we used p[d~,j}, p[dr,dy], p[ds,,j] and p[ds,, y] as indicators of the extent of fiscal policy variability. For the sake of saving space we do not report those additional results. 128 J. Gali, Government size and macroeconomic stability Table 6 Cross-country regressions, dependent variable: .s[j].” s S(T) f(T>j,) sg 4s.J (1) -0.159 _ 0.22 - _ (0.063) (2) - -0.394 _ 0.34 - _ (0.122) (3) -0.026 _ -0.353 - _ 0.35 5.02 -7.865 (0.089) (0.187) (2.505) (4) -0.162 0.150 -2.400 0.35 (0.069) (0.225) (1.528) (5) _ -0.432 0.517 - 0.029 0.38 (0.161) (0.525) (1.843) (6) -0.012 -0.054 ~ -0.412 0.609 _ 0.38 5.18 -8.622 (0.114) (0.351) (0.220) (0.799) (2.684) (7) -0.056 - 1.695 -0.257 0.149 0.38 2.62 - 6.844 (0.099) (1.838) (0.219) (1.965) (3.110) (8) -0.015 -0.453 -3.856 -0.415 1.720 2.527 0.48 3.66 -8.794 (0.111) (0.419) (2.256) (0.237) (1.052) (2.375) (3.259) ‘T is the government revenues/GDP ratio. sg is the government purchases/GDP ratio. An upper bar denotes the average over the sample period, 196&1990. j is the percent deviation of per capita GDP from trend. s[ ] and p[ .] are the standard deviation and the correlation coefficient, respectively. Each row reports the coefftcient estimates of a cross-country regression of s[j] on a constant term and the listed regressors. The F statistic corresponds to the null hypothesis that the coefftcients of i and S, are jointly zero. A is the linear combination 30 (? coefficient) +20 (Sp coej’kient). Standard errors are reported in brackets. An asterisk denotes significance at the 5:” level. range of the observed variability of Z and Sp in our sample of countries. This experiment is conceptually analogous to the comparative dynamics exercise described in the previous section. Here, we estimate the effect of such a change by looking at the statistic A = 30 (&[Q]/&) +20 (a$$]/&,), while using the individual coefficient estimates to approximate the partial deriva- tives. Our results from regression (3) imply that such a ‘resealing’ of fiscal variables is associated with a reduction of 7.8 percentage points (s.e. = 2.5) in $31. This stands in contrast with the increase of less than 0.5 percentage points predicted by our benchmark mode1 (see table 1). The results just discussed, corresponding to regressions (l)-(3) of table 6, embed the key finding of the paper: the estimated relationship between empirical measures of government size and output variability appears to be far stronger than the standard RBC model predicts and, at least in the case of tax effects, it has the opposite sign. The remaining results in tables 6 and 7 confirm the robustness of the previous finding under alternative regression specifications. We briefly discuss them next. Regressions (4) to (8) aim to overcome a potential source of bias in the J. tiali, Gwernment size and macroeconomic stability 129 Table 7 Cross-country regressions, dependent variable: s[dy]. r $5) P(L AJ’) sg sb,) P(.~~>AY) R F A (1) -0.045 0.31 (0.015) (2) _ -o.oso 0.22 (0.033) (3) -0.037 _ PO.022 0.32 4.41 - 1.569 (0.023) (0.048) (0.648) (4) -0.052 0.036 -0.817 0.36 - (0.017) (0.055) (0.818) (5) -0.106 0.320 -0.383 0.46 (0.031) (0.118) (0.677) (6) -0.001 -0.176 -0.099 0.624 0.67 8.25 -2.004 (0.021) (0.064) (0.040) (0.147) (0.493) (7) -0.039 - 2.934 -0.026 2.360 0.45 5.63 - 1.719 (0.023) ( 1.480) (0.048) (1.351) (0.628) (8) -0.010 -0.149 -2.379 -0.089 0.579 1.61 I 0.76 10.5 -2.100 (0.020) (0.061) (1.067) (0.038) (0.138) (0.984) (0.461) Note: T is the government revenues/GDP ratio, sg is the government purchases/GDP ratio. An upper bar denotes the average over the sample period. dy is the percent growth rate of per capita GDP. s[ ] and p[.] are the standard deviation and the correlation coefficient, respectively. Each row reports the coefficient estimates of a cross-country regression of s[Ay] on a constant term and the listed regressors. The F statistic corresponds to the null hypothesis that the coeflicients of S and Sp are jointly zero. A is the linear combination 30 (5 coefficient) + 20 (se coefficient). Standard errors are reported in brackets. An asterisk denotes significance at the 5% level. results just discussed. That bias would arise if, contrary to our exogeneity assumption, government size was systematically related to some feature of the time series variation in r and/or sg which had a dominant influence in determining the variability of output.” Regressions (4)gS) in table 6 attempt to control for that potential bias by including several combinations of the policy variability statistics introduced in table 5 as regressors, in addition to Z and/or S,. Notice that, with the exception of (7), the ? and S, coefficients in the augmented regressions are always either individually or jointly significant and have a negative sign. Furthermore, the A statistic described above still points to a large (and significant) negative effect on output variability of a simultaneous increase in Y and S,. Table 7 reports the results of a similar exercise, now using the standard deviation of GDP growth, s[dy], as a measure of output variability. Qualitatively, the results are similar to those discussed above. Taking 16For instance, one might argue that the estimated negative correlation between T and s[j] could be reflecting, say, the possibility that countries with higher average tax/GDP ratios tend to engage more in (successful) short-term countercyclical fiscal policies. 130 J. Gali, Government size and macroeconomic stability Fig. I regression (3) as a benchmark, we see that the estimated effect of the government resealing described above is a reduction of 1.56 percentage points (s.e.=0.648) in the standard deviation of output growth (as measured by the d statistic), in contrast with the increase of 0.12 percentage points predicted by the model (see table 4). Even stronger results obtain for the other regressions.17 Our basic finding of this section is illustrated graphically by fig. 1, which plots the position of each OECD country in the (Z,s[dy]) plane, together with a fitted regression line. r8 The picture’s message is clear: economies with ‘large governments’ (e.g., Norway, Sweden, Netherlands) have experienced milder economic fluctuations than economies with ‘small governments’ (e.g., Japan, Spain, Portugal). Our regression analysis suggests that such a conclusion is robust to the use of alternative measures of output variability and government size, and is not altered when we control for a possible policy variability effect. “Notice also that some of the proxies for fiscal policy variability are significant in regressions (5)+8) of table 7 in contrast with the results reported in the previous table. This does not affect any of our conclusions. “A similar picture emerges when we use other combinations of output variability and government size measures. J. 4. Summary and concluding comments Our results in section 3 suggest that both taxes and government purchases seem to be effectively working as ‘automatic stabilizers’, a theme often found in some of the Keynesian literature cited above. We do not view any of those findings as having a normative content, even if consumers value macroecono- mic stability, for large tax rates and government purchases are likely to have important welfare-reducing steady-state effects.” Despite that lack of norma- tive value, we find the results above useful to the extent that they point to a dimension of the data - the role of taxes and government purchases as automatic stabilizers - which the canonical RBC model fails to match. In all the specifications of that model analysed in section 2 income taxes effectively behave as ‘automatic destabilizers’; in contrast, all the evidence points to the presence of a negative relationship between output variability and the tax/ GDP ratio. Such a negative relationship is also found in the data when we look instead at the government purchases/GDP ratio as a measure of government size, but that empirical finding appears to be (qualitatively) consistent with at least some specifications of the RBC model. A more striking contrast between theory and evidence emerges when we focus on the magnitude of the effects: the variability effects of government size predicted by the model (regardless of their sign) is far smaller than those detected in the data. We interpret those results as suggesting that some factors responsible for the stabilizing effects of government size observed in the data are not accounted for in standard versions of the RBC model. To the extent that those factors play a significant role in economic fluctuations and, more specifically, in determining the response of the economy to changes in policy variables, many of the conclusions drawn from the RBC model may be misleading. This will generally be true no matter how well calibrated versions of the model lit time-series data. What are some of the possible elements that are missing in the canonical RBC model, and whose absence is responsible for the model’s counterfactual predictions? It is tempting to point to our stylized modelling of the fiscal sector as the main culprit. One can certainly argue that neither taxes nor government purchases show the kind of stable behavior assumed in our model. Yet, we believe little progress would be made if we were to allow for time variation in t or s,, unless we were willing to assume some systematic relationship between the time series properties of those fiscal parameters and “% our model of section 2 government purchases are wasteful, i.e. they absorb part of the economy’s output without increasing consumers’ utility or private sector Droductivitv. Since there are no other market imperfections the optimal government size in ou; model is zero, in which case the model’s equilibrium allocation is efficient, and any policy that manages to limit the fluctuations induced by technology shocks is welfare reducing. This point was forcefully argued in Kydland and Prescott (1980). 132 J. Gali, Government size and macroeconomic stability their average values. Furthermore, none of our empirical findings were significantly affected when we controlled for some policy variability measures, suggesting that auerage values of tax/GDP and government purchases/GDP ratios are themselves correlated with output variability, independently of their variations. A more fruitful avenue of inquiry, though one that falls beyond the scope of this paper, would augment our government sector model by allowing for possible effects of government purchases on utility [e.g., Christian0 and Eichenbaum (1992)] or private sector productivity [e.g., Baxter and King (1990)]. References Baily, Martin N., 1978, Stabilization policy and private economic behavior, Brookings Papers on Economic Activity 78, 1 I-50. Baxter, Marianne and Robert G. King, 1990, Fiscal policy in general equilibrium, Manuscript (University of Rochester, Rochester, NY). Braun, R. Anton, 1989, The dynamic interaction of distortionary taxes and aggregate variables in postwar U.S. data, Manuscript (University of Virginia, Charlottesville, VA). Burns, Arthur F., 1960, Progress towards economic stability, American Economic Review 50, l-19. Chari, V.V., Lawrence J. Christian0 and Patrick Kehoe. 1993, Optimal fiscal policy in a business cycle model, Manuscript (Northwestern University, Evanston, IL). Christiano, Lawrence J. and Martin Eichenbaum, 1992, Current real business cycle theories and aggregate labor market fluctuations, American Economic Review 82, no. 3, 43&450. Danthine, Jean-Pierre and John B. Donaldson, 1985, A note on the effects of capital income taxation on the dynamics of a competitive economy, Journal of Public Economics 28, 2555265. Danthine, Jean-Pierre and John B. Donaldson, 1993, Methodological and empirical issues in real business cycle theory, European Economic Review 37, no. I, l-36. DeLong, J. Bradford and Lawrence H. Summers, 1986, The changing cyclical variability of economic activity in the United States, in: Robert J. Gordon, ed., The American business cycle: Continuity and change (University of Chicago Press, Chicago, IL). Greenwood, Jeremy and Gregory W. Huffman, 1991, Tax analysis in a real business cycle model: On measuring Harberger triangles and Okun gaps, Journal of Monetary Economics 27, 1677190. King, Robert G., Charles I. Plosser and Sergio T. Rebelo, 1988, Production, growth and business cycles: I. The basic neoclassical model, Journal of Monetary Economics 21, 1955232. Kydland, Finn and Edward C. Prescott, 1980, A competitive theory of fluctuations and the feasibility and desirability of stabilization policy, in: Stanley Fischer, ed., Rational expec- tations and economic policy (University of Chicago Press, Chicago, IL). Lucas, Robert E., 1988, On the mechanics of economic development, Journal of Monetary Economics 22, 342. McGrattan, E.R., 1991, The macroeconomic effects of distortionary taxation, Discussion paper no. 37 (Institute for Empirical Macroeconomics, University of Minnesota, Minneapolis, MN). Plosser, Charles I., 1989, Understanding real business cycles, Journal of Economic Perspectives 3, 51-77. Prescott, Edward C., 1986, Theory ahead of business cycle measurement, Federal Reserve Bank of Minneapolis Quarterly Review 10, 9922.
2076	https://brainly.com/question/45367874	[FREE] What is a benchmark and datum in surveying? - brainly.com 4 Search Learning Mode Cancel Log in / Join for free Browser ExtensionTest PrepBrainly App Brainly TutorFor StudentsFor TeachersFor ParentsHonor CodeTextbook Solutions Log in Join for free Tutoring Session +64,4k Smart guidance, rooted in what you’re studying Get Guidance Test Prep +11,5k Ace exams faster, with practice that adapts to you Practice Worksheets +7,4k Guided help for every grade, topic or textbook Complete See more / Social Studies Textbook & Expert-Verified Textbook & Expert-Verified What is a benchmark and datum in surveying? 1 See answer Explain with Learning Companion NEW Asked by Levantine3768 • 01/01/2024 0:00 / -- Read More Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer This answer helped 10624759 people 10M 0.0 0 Upload your school material for a more relevant answer In surveying, a benchmark is a fixed reference point for elevation, while a datum is a standardized elevation reference level, such as sea level. Explanation In surveying, a benchmark is a fixed reference point of known position and elevation against which the heights of other points are measured. It acts as a starting or reference point for vertical measurements to ensure consistency in the elevational data across a survey. On the other hand, a datum in surveying is a theoretical or virtual reference level from which vertical measurements are taken. It provides a standardized base elevation that is applied universally across a region or country, such as sea level, against which heights and depths are measured. The importance of benchmarks and datums lies in their ability to provide a consistent framework for surveyors to precisely measure and compare elevational changes, ensuring accurate and reliable data that is crucial for various engineering projects, land assessments, and construction planning. Answered by TaylorSmith474 •36.7K answers•10.6M people helped Thanks 0 0.0 (0 votes) Textbook &Expert-Verified⬈(opens in a new tab) This answer helped 10624759 people 10M 0.0 0 Laboratory Manual for Introductory Geology - Bradley Deline Essentials of Geographic Information Systems - Campbell and Shin Geographic Information Systems and Cartography - Adam Dastrup Upload your school material for a more relevant answer In surveying, a benchmark is a fixed reference point for measuring elevation, while a datum is a standardized reference level, such as sea level. Benchmarks provide specific elevations for measurement accuracy, while datums serve as common starting points for various surveying activities. Both are vital for ensuring reliable and consistent data in engineering and construction projects. Explanation In surveying, a benchmark is a fixed reference point used to measure the elevation of other locations. It provides a known elevation point that surveyors can use to ensure their measurements are accurate and consistent. Benchmarks are often marked on maps with symbols like 'BM', 'X', or triangles, and have specific elevations written next to them. On the other hand, a datum is a reference level used as a starting point for measuring elevations or depths. A common type of datum is mean sea level, which serves as a standardized base elevation from which measurements are made across regions. This ensures that all surveyors use the same reference point for their calculations, improving the accuracy of maps and construction designs. Both benchmarks and datums are essential in construction, land surveying, and geographic information systems (GIS) as they contribute to establishing reliable data for various projects. The relationship between a benchmark (specific point) and a datum (the theoretical reference level) is crucial for ensuring that measurements can be understood and compared consistently across different projects and locations. Examples & Evidence For instance, if a surveyor measures a new road's elevation, they might refer to a benchmark labeled 'BM 100' which indicates an elevation of 100 meters. If another point is measured at 105 meters according to the datum of sea level, then the elevation of the new road is 5 meters above the benchmark. Standards for benchmarks and datums are provided by surveying organizations and are critical for engineering projects and mapping, demonstrating their importance in maintaining accurate geographical information. Thanks 0 0.0 (0 votes) Advertisement Levantine3768 has a question! Can you help? Add your answer See Expert-Verified Answer ### Free Social Studies solutions and answers Community Answer 7 You want to restrict the values entered in a cell to a specified set, such as hop, skip, jump. Which type of data validation should you use?. Community Answer 26 Explain four reasons the youth do not want to participate in civic life. Community Answer 4.1 5 as a bystander, which approach may not work when the person you’re trying to stop is likely to become defensive or hostile? Community Answer 6 Outline and analyse the limitations of the traditional medical approach to inclusive education. Community Answer 10 Does the Youth Service Project, or a Civic Organisation, or Community Service Centre pursue the aims of the service centre? YES/NO Elaborate Community Answer In a presentation about voter turnout, you are illustrating various data with charts. Which type of information would you present in a pie chart?. Community Answer 3 Researchers designing online studies should consider the following with respect to participant protections. Community Answer 3 At which event must special care be taken when preparing food for the guests?. Community Answer Additional protections researchers can include in their practice to protect subject privacy and data confidentiality include: Community Answer 4.8 16 You work for caring health, a medicare advantage (ma) plan sponsor. recently, mrs. garcia has completed an enrollment application for a plan offered by caring health, which is waiting for a reply from cms indicating whether or not mrs. garcia’s enrollment has been accepted. once cms replies, how long does caring health have to notify mrs. garcia that her enrollment has been accepted and in what format? New questions in Social Studies Here are several examples of different sources. Sort them based on whether they are primary, secondary, or tertiary sources. Sources: 1. A news article written some time after the event, which cites previous news coverage and summarizes the events. 2. An academic essay analyzing the importance of the event which cites only other academic essays and articles written after the fact. 3. A news article written right after the event, which includes direct quotes and descriptions from attendees. 4. A diary entry written by an attendee's child describing their parent's experiences. 5. A photograph taken at the event/place in question. 6. An academic essay analyzing the importance of the event which cites firsthand accounts. 7. A painting done years later, based on firsthand accounts. 8. A letter written by an attendee describing the event to someone else. 9. An encyclopedia entry summarizing information from various sources. The equal protection clause of the Fourteenth Amendment was most clearly responsible for which of the following? A. the increase in state control of social welfare programs B. the increase in equality promoted by the Civil Rights Act of 1964 C. the Supreme Court's decision in United States v. Lopez D. the desegregation of public schools If the judicial branch declares a law unconstitutional, what option does the legislative branch have? A. override a veto B. propose a constitutional amendment C. issue a pardon D. impeach a Supreme Court justice In a mixed market economy, who makes most of the daily economic decisions? A. an economic expert B. an elected leader C. individual citizens D. the military Analyze why Article II, Section 2 helps the nation provide for the common defense. A. Article II, Section 2 provides for Congress to impeach the president. B. Article II, Section 2 gives the president the role of commander in chief. C. Article II, Section 2 allows for amendments to the Constitution. D. Article II, Section 2 sets up the federal court system. 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2077	https://www.quora.com/How-do-I-show-that-any-palindrome-with-an-even-number-of-digits-is-divisible-by-11	Something went wrong. Wait a moment and try again. Divisibility Tests Even Digits Maths of 11 Number Theory Palindromic Numbers Divisibility Rules 5 How do I show that any palindrome with an even number of digits is divisible by 11? Amitabha Tripathi have more than a working knowledge of Z · Upvoted by David Joyce , Ph.D. Mathematics, University of Pennsylvania (1979) · Author has 4.7K answers and 13.9M answer views · 7y A k-digit number with decimal digits ak−1, ak−2, ak−3, …, a2, a1, a0 read from left to right is a palindrome if it reads the same whether read left to right or right to left. Thus, ak−i=ai−1 for i∈{1,2,3,…,k}. Further, if k=2ℓ, a palindrome is of the form N=2ℓ−1∑i=0ai⋅10i=ℓ−1∑i=0ai(10i+102ℓ−1−i). Since i+(2ℓ−1−i) is odd, i, 2ℓ−1−i are of opposite parity. Hence (−1)i+(−1)2ℓ−1−i=0 for each i∈{0,1,2,…,ℓ−1}. Since 10≡−1(mod11), each term 10^i+10^ A k-digit number with decimal digits ak−1, ak−2, ak−3, …, a2, a1, a0 read from left to right is a palindrome if it reads the same whether read left to right or right to left. Thus, ak−i=ai−1 for i∈{1,2,3,…,k}. Further, if k=2ℓ, a palindrome is of the form N=2ℓ−1∑i=0ai⋅10i=ℓ−1∑i=0ai(10i+102ℓ−1−i). Since i+(2ℓ−1−i) is odd, i, 2ℓ−1−i are of opposite parity. Hence (−1)i+(−1)2ℓ−1−i=0 for each i∈{0,1,2,…,ℓ−1}. Since 10≡−1(mod11), each term 10i+102ℓ−1−i≡0(mod11). Thus 11∣N. ■ Ankit Verma Reshaping product programs @ Meta (formerly called facebook) · Upvoted by David Joyce , Ph.D. Mathematics, University of Pennsylvania (1979) · 11y Originally Answered: how do I show that any palindrome with an even number of digits is divisible by 11? · According to the divisibility rule of 11... Difference of the Sum of digits at even places and sum of digits at odd no of places should be zero or divisible by 11. For any palandron with even no of digits this difference will always be zero. Hence divisible by 11. For example, 1991= (9+1)-(9+1)= 0 123321= (1+3+2)- (2+3+1)=0 You can take any example. Theodore Rice Data Scientist at Samaritan Ministries (2016–present) · Author has 346 answers and 320.1K answer views · 3y Originally Answered: A palindrome is a number that reads the same backward as forward (for instance, 373 and 521125 are palindromes). How can you prove that any Palindrome with an even number of digits is divisible by 11? · A number (expressed in base 10) is divisible by 11 if and only if the sum of the digits in the odd places and the sum of the digits in the even places differ by a multiple of 11. In the case of an even digit palindrome, the difference is 0, which is, of course 110. The reason this is true is that odd powers of 10 are equivalent to -1 mod 11 and even powers of 10 are equivalent to 1 mod 11. So the number abcd is equivalent to a-b+c-d mod 11. Related questions What is the six-digit palindromic number that is divisible by 11? What is the least 7-digit palindromic number divisible by 11 without a single digit 0? How many palindromic numbers have 11 digits and are divisible by 101? What is a palindrome number and give one 5-digit example? Why odd-digit palindromic numbers not divisible by 11? Doug Dillon Ph.D. Mathematics · Author has 12.4K answers and 11.4M answer views · 3y Originally Answered: A palindrome is a number that reads the same backward as forward (for instance, 373 and 521125 are palindromes). How can you prove that any Palindrome with an even number of digits is divisible by 11? · A number with an even number of digits, say, astuffb is divisible by 11 if stuff+a-b is divisible by 11. We use this in a proof by induction and using careless notation. Clearly any 2-digit number of the form aa is divisible by 11. Suppose the palindrome of an even number of digits, called, stuff is divisible by 11. Then the number astuffa is a palindrome of an even number of digits. Since stuff+a-a=stuff divisibility by 11 is for sure. 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. Assistant Bot · 1y To show that any palindrome with an even number of digits is divisible by 11, we can use the properties of palindromes and the divisibility rule for 11. Step-by-step Explanation Definition of a Palindrome: A palindrome is a number that reads the same forwards and backwards. For an even-digit palindrome, we can express it in the following form: N=a1a2a3…akak…a3a2a1 where N has 2k digits. Structure of the Palindrome: The palindrome consists of k digits followed by the same k digits in reverse order. For example, if k=3, the palindrome could be 123321. Divisibilit To show that any palindrome with an even number of digits is divisible by 11, we can use the properties of palindromes and the divisibility rule for 11. Step-by-step Explanation Definition of a Palindrome: A palindrome is a number that reads the same forwards and backwards. For an even-digit palindrome, we can express it in the following form: N=a1a2a3…akak…a3a2a1 where N has 2k digits. Structure of the Palindrome: The palindrome consists of k digits followed by the same k digits in reverse order. For example, if k=3, the palindrome could be 123321. Divisibility Rule for 11: A number is divisible by 11 if the difference between the sum of its digits in odd positions and the sum of its digits in even positions is either 0 or divisible by 11. Identifying Odd and Even Positioned Digits: For the palindrome N=a1a2a3…akak…a3a2a1: Odd-positioned digits: a1,a3,…,ak,ak,ak−1,…,a1 Even-positioned digits: a2,a4,…,ak−1,ak−1,a4,…,a2 Calculating the Sums: The sum of the digits in odd positions: Sodd=a1+a3+…+ak+ak+ak−1+…+a1=2(a1+a3+…+ak) The sum of the digits in even positions: Seven=a2+a4+…+ak−1+ak−1+a4+…+a2=2(a2+a4+…+ak−1) Difference Calculation: The difference Sodd−Seven becomes: Sodd−Seven=2(a1+a3+…+ak)−2(a2+a4+…+ak−1)=2((a1+a3+…+ak)−(a2+a4+…+ak−1)) Conclusion: Since Sodd−Seven is multiplied by 2, the overall difference is clearly divisible by 2. However, for divisibility by 11, we need to check if the difference (a1+a3+…+ak)−(a2+a4+…+ak−1) results in a number that is either 0 or divisible by 11. But since the sums of the digits are structured symmetrically, the sums of the odd and even positioned digits will always balance out, leading us to conclude that the difference will indeed be divisible by 11. Final Statement Thus, we can conclude that any palindrome with an even number of digits is divisible by 11. David Vanderschel PhD in Mathematics & Physics, Rice (Houston neighborhood) (Graduated 1970) · Author has 37.6K answers and 50.1M answer views · 3y Originally Answered: A palindrome is a number that reads the same backward as forward (for instance, 373 and 521125 are palindromes). How can you prove that any Palindrome with an even number of digits is divisible by 11? · A2A: See the following divisibility by 11 rule. Each digit appearing in an odd position on the right also occurs in an even position on the left. Similarly for even/odd. So the odd and even sums cancel out. Divisibility by Eleven It is easy to tell that the following are multiples of 11: 22, 33, 44, 55, etc. But how about: 2728, or 31415? Are they divisible by 11? Here an easy way to test for divisibility by 11. Take the alternating sum of the digits in the number, read from left to right. If that is divisible by 11, so is the original number. So, for instance, 2728 has alternating sum of digits 2 – 7 + 2 – 8 = -11. Since -11 is divisible by 11, so is 2728. Similarly, for 31415, the alternating sum of digits is 3 – 1 + 4 – 1 + 5 = 10. This is not divisible by 11, so neither is 31415. Presentation Suggestions: Students may enjoy thinking about how this divisibility test is related to the Fun Fact Divisibility by Seven. The Math Behind the Fact: This curious fact can be easily shown using modular arithmetic. Since 10 n is congruent to (-1) n mod 11, we see that 1, 100, 10000, 1000000, etc. have remainders 1 when divided by 11, and 10, 1000, 10000, etc. have remainders (-1) when divided by 11. Thus 2728 = 2 1000 + 7 100 + 2 10 + 8, so its remainder when divided by 11 is just 2(-1) + 7(1) + 2(-1) + 8(1), the alternating sum of the digits. (It’s sum is the negative of what we found above because the alternation here begins with a -1.) But either way, if this alternating sum is divisible by 11, then so is the original number. In fact, our observation shows more: that in fact when we take the alternating sum of the digits read from right to left (so that the sign of the units digit is always positive), then we obtain N mod 11. How to Cite this Page: Su, Francis E., et al. “Divisibility by Eleven.” Math Fun Facts . Fun Fact suggested by: Francis Su Related questions How many 7-digit palindromes are there? How many palindromes can you find between 500 and 599? What is the greatest 5-digit palindrome number divisible by 45? How do I find the number of 6 digit palindrome numbers which are divisible by 5? What is the greatest three-digit number divisible by 11 and 12? Wes Browning Studied Mathematics at Cornell University · Author has 8.9K answers and 6.1M answer views · 3y Originally Answered: A palindrome is a number that reads the same backward as forward (for instance, 373 and 521125 are palindromes). 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So,all the four digit palindromes are divisible by 11. Calculate total no of 4-digit no divided by 11. 4-digit no divided by 11 are in AP:- 1001,1012,1023,………,9999 Tn=a+(n-1)d 9999=1001+(n-1)11 9999=1001+11n-11 =990+11n I.e 11n=9009 So,n=819 So,there are 819 4-digit no divided by 11. now, find the total no of 4-digit palindroms:- 4-digit palindromes are of the form “abba”….where a can't be zero,otherwise, it will become a 3-digit no.But there is no problem in “b” being zero. All the four digit palindromes are of the form:- abba Check the divisibility of the no abba by 11 a-b+b-a=0 Hence it is divisible by 11. So,all the four digit palindromes are divisible by 11. Calculate total no of 4-digit no divided by 11. 4-digit no divided by 11 are in AP:- 1001,1012,1023,………,9999 Tn=a+(n-1)d 9999=1001+(n-1)11 9999=1001+11n-11 =990+11n I.e 11n=9009 So,n=819 So,there are 819 4-digit no divided by 11. now, find the total no of 4-digit palindroms:- 4-digit palindromes are of the form “abba”….where a can't be zero,otherwise, it will become a 3-digit no.But there is no problem in “b” being zero. Now,pair of a in Abba can be gotten in 9 ways:-11,22,33,44..,99 And pair of b in abba can be gotten in 10 ways:-00,11,22……,99(00 is also allowed) So total no of 4-digit palindromes=910=90 Finally,total no of 4-digit no divided by11 but not palindromes=819–90=729. Ellis Cave 48 U.S. Patents · Author has 7.9K answers and 4.3M answer views · Feb 13 Related How many palindromic numbers have 11 digits and are divisible by 101? Brute force solution, using the J programming language: 'a b c d e f'=.\|:odo 9,5#10 a=.a+1 m=.n#~0=101\|n=.10#.a,.b,.c,.d,.e,.f,.e,.d,.c,.b,.a 36000 The answer is there are 36,000 palindromic 11-digit integers that are divisible by 101. List the first & last few: m 10000000001 10001010001 10002020001 10003030001 10004040001 10005050001 10006060001 10007070001 10008080001 10009090001 10010201001 10011211001 10012221001 10013231001 10014241001 10015251001 10016261001 10017271001 10018281001 10019291001 10020402001 10021412001 10022422001 10023432001 10024442001 10025452001 10026462001 10027472001 100284 Brute force solution, using the J programming language: 'a b c d e f'=.\|:odo 9,5#10 a=.a+1 m=.n#~0=101\|n=.10#.a,.b,.c,.d,.e,.f,.e,.d,.c,.b,.a 36000 The answer is there are 36,000 palindromic 11-digit integers that are divisible by 101. List the first & last few: m 10000000001 10001010001 10002020001 10003030001 10004040001 10005050001 10006060001 10007070001 10008080001 10009090001 10010201001 10011211001 10012221001 10013231001 10014241001 10015251001 10016261001 10017271001 10018281001 10019291001 10020402001 10021412001 10022422001 10023432001 10024442001 10025452001 10026462001 10027472001 10028482001 10029492001 10030603001 10031613001 10032623001 10033633001 10034643001 10035653001 10036663001 10037673001 10038683001 10039693001 10040804001 10041814001 10042824001 10043834001 10044844001 10045854001 10046864001 10047874001 …………… 99592029599 99593039599 99594049599 99595059599 99596069599 99597079599 99598089599 99599099599 99690009699 99691019699 99692029699 99693039699 99694049699 99695059699 99696069699 99697079699 99698089699 99699099699 99790009799 99791019799 99792029799 99793039799 99794049799 99795059799 99796069799 99797079799 99798089799 99799099799 99890009899 99891019899 99892029899 99893039899 99894049899 99895059899 99896069899 99897079899 99898089899 99899099899 99990009999 99991019999 99992029999 99993039999 99994049999 99995059999 99996069999 99997079999 99998089999 99999099999 Promoted by Webflow Metis Chan Works at Webflow · Aug 11 What are the best AI website builders now? When it comes to AI website builders, there are a growing number of options, but a few stand out for their power, flexibility, and ability to grow with your needs. Webflow’s AI Site Builder is a top choice for small businesses, in-house teams, and agencies who want the speed of AI and the freedom to fully customize every part of their site. With Webflow, you can: Describe your business or idea and instantly generate a unique, production-ready website—no coding required. Edit visually in a powerful no-code canvas, customize layouts, and add advanced interactions. Collaborate with your team in real When it comes to AI website builders, there are a growing number of options, but a few stand out for their power, flexibility, and ability to grow with your needs. Webflow’s AI Site Builder is a top choice for small businesses, in-house teams, and agencies who want the speed of AI and the freedom to fully customize every part of their site. With Webflow, you can: Describe your business or idea and instantly generate a unique, production-ready website—no coding required. Edit visually in a powerful no-code canvas, customize layouts, and add advanced interactions. Collaborate with your team in real time, streamline feedback, and manage all your content in one place. Publish instantly on enterprise-grade hosting with built-in SEO, security, and the flexibility to scale as you grow. Many other tools offer AI-powered templates or quick site launches, but Webflow stands out by letting you take control—so your site never feels generic, and you can easily update, expand, or redesign as your needs change. Want to see what AI-powered site building can really do? Try Webflow AI Site Builder for free today. Mario Skrtic IT at Public Libraries (2003–present) · Author has 4.5K answers and 1.1M answer views · Feb 13 Related How many palindromic numbers have 11 digits and are divisible by 101? The answer is 36000. Brute force in PariGP: { s=0; for(i=10^5,10^6-1, d=digits(i); a=i10^5+d+d10+d100+d1000+d10000; if (a%101==0, s+=1; if (s<=10 \|\| s>35990, print(s,". solution ",a)); ); ); } solution 10000000001 solution 10001010001 solution 10002020001 solution 10003030001 solution 10004040001 solution 10005050001 solution 10006060001 solution 10007070001 solution 10008080001 solution 10009090001 solution 99990009999 solution 99991019999 solution 99992029999 solution 99993039999 solution 99994049999 solution 99995059999 35997. The answer is 36000. Brute force in PariGP: { s=0; for(i=10^5,10^6-1, d=digits(i); a=i10^5+d+d10+d100+d1000+d10000; if (a%101==0, s+=1; if (s<=10 \|\| s>35990, print(s,". solution ",a)); ); ); } solution 10000000001 solution 10001010001 solution 10002020001 solution 10003030001 solution 10004040001 solution 10005050001 solution 10006060001 solution 10007070001 solution 10008080001 solution 10009090001 solution 99990009999 solution 99991019999 solution 99992029999 solution 99993039999 solution 99994049999 solution 99995059999 solution 99996069999 solution 99997079999 solution 99998089999 solution 99999099999 Pablo Emanuel MSc in Math and IMO medalist · Upvoted by Jeremy Collins , M.A. Mathematics, Trinity College, Cambridge and Ermal Feleqi , PhD Mathematics, University of Padua (2010) · Author has 847 answers and 5.5M answer views · Updated 4y Related How do I prove that among any 39 consecutive natural numbers, it is always possible to find one whose sum of digits is divisible by 11? Expanding on Roman Andronov ’s answer, let’s try and answer a much more general question: For each natural number n , find the smallest natural number μ ( n ) with the property that among any μ ( n ) consecutive natural numbers there exists at least one whose sum of (decimal) digits is divisible by n . Your question is equivalent to μ ( 19 ) ≤ 39 . We’ll prove that actually μ ( 19 ) = 39 , and will derive a general explicit formula for μ ( n ) for every n . First of all, it isn’t obvious that μ ( n ) is well defined for every n . That is, we could potentially have, for some n , arbitrarily long sequences of Expanding on Roman Andronov’s answer, let’s try and answer a much more general question: For each natural number n, find the smallest natural number μ(n) with the property that among any μ(n) consecutive natural numbers there exists at least one whose sum of (decimal) digits is divisible by n. Your question is equivalent to μ(19)≤39. We’ll prove that actually μ(19)=39, and will derive a general explicit formula for μ(n) for every n. First of all, it isn’t obvious that μ(n) is well defined for every n. That is, we could potentially have, for some n, arbitrarily long sequences of consecutive natural numbers none of which having its sum of digits divisible by n. I claim this is not possible, and that μ(n)<2⋅10n for all n. Indeed, any set of 2⋅10n natural numbers contains a whole block going from A⋅10n to A⋅10n+999…9=(A+1)⋅10n−1. That means, in particular, that it contains the numbers A000⋯001,A000⋯011,A000⋯111,⋯A011⋯111,A111⋯111. The digit sum of those numbers form a set of n consecutive integers, which therefore span every possible residue class modulo n, including 0, that is, one of them needs to be a multiple of n. So now let’s try and explicitly calculate the value of μ(n) for some small values of n. As we move through the natural number, their digit sum keeps getting increased by 1, until we reach a number whose representation ends in 9. At that point, the digit sum will decrease by exactly 9j−1 where j is the number of nines at the end of the decimal representation. That’s all the preparation we need, let the games begin! μ(1) is trivially 1 For n=2, we need to start with a number with an odd digit sum. Then, if this number doesn’t end in 9, we can move no further. If it ends in an odd number of 9s, though, the next sum will still be odd, but then the next number necessarily needs to have an even sum. So μ(2)=3, and 9, 10 is an example of a maximal sequence of lenght 2. For n=3 we also start with s=1(mod3), then the next will be 2(mod3), and we need it to end in 9 to have a next move. But that doesn’t help much, because will be [math]+1 \mod 3[/math], regardless of [math]j[/math], and we’ll get to a multiple of 3 in 3 steps, i.e. [math]\mu(3) = 3[/math]. For [math]n = 4[/math], we start again with [math]s = 1 \pmod 4[/math] and advance with +1 as far as we can, i.e., until [math]3 \mod 4[/math]. Then we need to bring it back as far as possible, so we can continue further, i.e., ideally we want to bring it all the way back to [math]1 \mod 4[/math]. For that, we need to find [math]j[/math] such that [math]1 - 9j = -2 \pmod{4}[/math], which we can always do because 9 and 4 are relatively prime (we could have simply calculated [math]j[/math] explicitly: [math]1–9\times 3 = -26 = -2 \pmod 4[/math], but the general observation will be useful for us moving forward). In summary, our sequence [math]\pmod 4[/math] will be: [math]1,2,3,1,2,3,0[/math] which means [math]\mu(4) = 7[/math]. And, in fact, one example of a counter-sequence of length 6 would be (remember we had [math]j = 3[/math], so the number where the sequence “turns” needs to end in 999): [math]997, 998, 999, 1000, 1001, 1002[/math] Now we’ve found an useful pattern! For [math]n = 5, 7, 8[/math] we can use the exact same strategy than for [math]n = 4[/math], i.e we start on [math]1 \pmod n[/math], advance with +1’s until [math]n-1[/math], pull it back to [math]1[/math] with a suitable number of 9’s (which we can always do since those numbers are relatively prime to 9), and then we advance again until 0. So the sequence [math]\mod n[/math] will be: [math]1,2,\cdot,n-1, 1,2,\cdot, n-1, 0[/math] and [math]\mu(n) = 2n-1[/math]. For example, for [math]n = 7[/math], we need [math]1–9j = -5 \pmod 7[/math], i.e. [math]j=3[/math], and a counter-sequence of length [math]12[/math] would be [math]994, 995, 996, 997, 998, 999, 1000, 1001, 1002, 1003, 1004, 1005[/math] For [math]n = 6[/math], we can’t get the same result, because [math]1–9j[/math] can only be [math]+1[/math] or [math]-2 \pmod 6[/math], so the best we can do is [math]1,2,3,4,5,3,4,5,0[/math] and [math]\mu(6) = 9[/math] For [math]n=9[/math], every move is a [math]+1[/math], and [math]\mu(9)[/math] is trivially [math]9[/math]. Putting it all together so far: If [math]n \leq 9[/math], then: [math]\mu(3) = 3[/math] [math]\mu(6) = \mu(9) = 9[/math] [math]\mu(n) = 2n-1[/math], otherwise For [math]n = 10[/math], we can follow the same strategy (gcd(10,9) = 1): [math]1, 2, 3, \cdots 9, 1, \cdots 9, 0[/math] and [math]\mu(10) = 19[/math]. But, when we get to [math]n = 11[/math], things get a little more interesting! Let’s start the same way: [math]1, 2, \cdots 10, 1, 2, \cdots 10[/math] But now, notice that the last number in our sequence will end in 9, because that second “1” corresponded to a number ending in 0. Which means that the next number won’t be 11 = 0. More than that, since we had to use more than one 9 on the first turn (actually, we needed to use 6, to have [math]1–9\times 6 = -9 \pmod {11}[/math]), that number ended in “00″, and our last number ends in “09” which has exactly one 9. Therefore, the next move is a -8 instead of a +1, and we have [math]1,2,\cdots 10, 1, 2, \cdots 10, 2, 3, \cdots, 10, 0[/math] But that also means that our very first number ends in “90”, and, moving back, we wouldn’t have a -1, which would bring us to 0, but a +8. So we can also extend the sequence back, to [math]1,2,\cdots, 9, 1, 2, \cdots 10, 1,2, \cdots 10, 2, 3, \cdots 10, 0[/math] which has 9 + 10 + 10 + 10 = 39 terms yielding [math]\mu(19) = 39[/math]. Using the information that our middle turning point had to end in 999999, we can explicit create the optimal 38-term sequence [math]999981 - 1000018[/math] Let’s skip [math]n=12[/math] for a moment (gcd(9,12) = 3) and let’s look at [math]n = 13[/math]. The optimal strategy is still the same. Find a suitable power of 10 such that, on one side we have the sum being obviously 1 (10000…00), and on the other side (99….99) having it being [math]-1 \pmod{13}[/math]. Again, we can do it because gcd(9,13) = 1. When we do it, we’ll be able to extend “forward” all the way to 10…039, and the extension backwards will be similar, as we start at -1, and the steps backwards will be either -1 or +8: going backwards is exactly symmetric to going forward. So we’ll go back 38 steps to 99…960. The complete sequence modulo 13 will be: [math]0 ,-12, \cdots, -4; -12, -11 \cdots -3; -11, \cdots, -2; -10 \cdots -1 : 1 \cdots 10; 2 \cdots 11; 3 \cdots 12; 4 \cdots 0[/math] i.e. [math]\mu(13) = 80 - 1 = 79[/math] We can do the same for [math]n = 14, 16, 17[/math]. The turning point will be a number such that the sum of digits of [math]10^j - 1[/math], i.e. [math]9j[/math] is [math]-1 \pmod n[/math]. Then we’ll be able to extend it fowards to [math]10\cdots 0(k-1)9[/math], i.e. [math]10k[/math] steps, where [math]k = n-9[/math] and backwards the exact number of steps, yielding [math]\mu(9+k) = 20k - 1[/math]. Now, if [math]n = 12[/math] or [math]n=15[/math], the best we can do is to have [math]9j = -3 \pmod n[/math], which means that our sequence will start as [math]-12, \cdots ,-3 : 1 \cdots 10[/math] From that point, we still can extend it forward all the way up to [math]10\cdots 0(k-1)9[/math], i.e. [math]10k[/math] steps. But we lose 2 full blocks going backwards, i.e. we can only go [math]10(k-2)[/math] steps. And we end up with [math]\mu(9+k) = 10(2k-2) - 1 = 20(k-1) - 1[/math]. For [math]n = 18[/math], the situation is even worse, as the best we can hope [math]9j[/math] to be is [math]-9 \pmod n[/math]. That means that we still can go 90 steps going forward, but we lose 8 full blocks going backwards, and we only can go 10 steps. Which means that [math]\mu(18) = 99[/math]. And now it’s pretty straightforward to generalize. If [math]n = 9m + k[/math], with [math]k[/math] not divisible by [math]3[/math], starting with a good [math]10^j[/math], such that [math]9j = -1 \pmod n[/math], we’ll be able to move forward all the way up to [math]10 \cdots 0(k-1)99\cdots9[/math], where we have [math]m[/math] nines at the end, i.e. [math]k\cdot 10^m[/math] steps, and the same number of steps backwards, i.e. [math]\mu(9m + k) = 2k\cdot 10^m - 1[/math]. If [math]n = 9m +k[/math] where [math]k = 3[/math] or [math]k = 6[/math], we’ll be able to go [math]k \cdot 10^m[/math] steps forward and math \cdot 10^m[/math] steps backwards, getting [math]\mu(9m +k) = 2(k-1)\cdot 10^m - 1[/math]. If [math]n = 9m[/math], we’ll be able to go [math]9\cdot 10^{m-1}[/math] steps forward and only [math]10^{m-1}[/math] steps backwards, getting [math]\mu(9m) = 10^m - 1[/math]. Pallav Sahu Mah lyf, mah-thematics! · Author has 646 answers and 2.8M answer views · Updated 10y Related What is an eight digit palindrome divisible by 45? First of all, for a number to be divisible by 45 it should be divisible by 5 and 9 both. A number divisible by 5 has unit digits 0 or 5. Since we require a palindromic number, 0 as a unit digit is ruled out. Let the palindromic number be of the form a, b, c are integers less than or equal to 9. Now, as the number is also divisible by 9, the sum of its digits should be an even multip First of all, for a number to be divisible by 45 it should be divisible by 5 and 9 both. A number divisible by 5 has unit digits 0 or 5. Since we require a palindromic number, 0 as a unit digit is ruled out. Let the palindromic number be of the form 5abccba5 where a, b, c are integers less than or equal to 9. Now, as the number is also divisible by 9, the sum of its digits should be an even multiple of nine i.e. 5+a+b+c+c+b+a+5 = even multiple of nine. Consequently, multiple of nine (say 9k)... Tanmay Pandit IITJ CSE '22 · Author has 571 answers and 2.7M answer views · 9y Related How many four digit numbers divisible by 11 are not palindromes? 4 digit palindromes are of a form abba where a & b are digits. Let's apply divisibility test for 11 to abba. As a+b=b+a, we can conclude that all palindromes are divisible by 11. Now, let's find the total number of 4 digit palindromes. As a= any digit from 1 to 9 ( if we take a=0, abba will be a 3 digit number. ) & b= any digit from 0 to 9, there are 9 & 10 alternatives for values of a & b respectively. No.of 4 digit palindromes = 910=90. Now, let's find the total number of 4 digit multiples of 11. Their series is - 1001, 1012,......, 9988, 9999. This is an A.P. with a=1001 & d=11. Let tn= 9 4 digit palindromes are of a form abba where a & b are digits. Let's apply divisibility test for 11 to abba. As a+b=b+a, we can conclude that all palindromes are divisible by 11. Now, let's find the total number of 4 digit palindromes. As a= any digit from 1 to 9 ( if we take a=0, abba will be a 3 digit number. ) & b= any digit from 0 to 9, there are 9 & 10 alternatives for values of a & b respectively. No.of 4 digit palindromes = 910=90. Now, let's find the total number of 4 digit multiples of 11. Their series is - 1001, 1012,......, 9988, 9999. This is an A.P. with a=1001 & d=11. Let tn= 9999 & n=? tn= a+(n-1) d 9999=1001+ 11n - 11=11n+990 909=n+90 so n=819. Answer=819-90=729. Related questions What is the six-digit palindromic number that is divisible by 11? What is the least 7-digit palindromic number divisible by 11 without a single digit 0? How many palindromic numbers have 11 digits and are divisible by 101? What is a palindrome number and give one 5-digit example? Why odd-digit palindromic numbers not divisible by 11? How many 7-digit palindromes are there? How many palindromes can you find between 500 and 599? What is the greatest 5-digit palindrome number divisible by 45? How do I find the number of 6 digit palindrome numbers which are divisible by 5? What is the greatest three-digit number divisible by 11 and 12? How many 6 digit palindromes are divisible by 495? What is the least 10 digit prime palindromic number? How many four digit numbers divisible by 11 are not palindromes? What is an eight digit palindrome divisible by 45? 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Authors Editorial Board Authors Author Information Author Instructions Author Awards Blog Question Bank CCN CME Comment Contact Us Games Grants Newsletters Statistics Testimonials Home > CNS nontumor > Multiple sclerosis CNS nontumor CNS myelin disorders Multiple sclerosis Authors:Rachel A. Multz, M.D., Jared T. Ahrendsen, M.D., Ph.D. Editorial Board Member:Meaghan Morris, M.D., Ph.D. Deputy Editor-in-Chief:Chunyu Cai, M.D., Ph.D. Last author update: 23 January 2023 Last staff update: 19 March 2025 Copyright: 2023-2025, PathologyOutlines.com, Inc. PubMed Search: Multiple sclerosis Page views in 2024: 8,742 Page views in 2025 to date: 5,954 Table of Contents Definition / general \| Essential features \| Terminology \| ICD coding \| Epidemiology \| Sites \| Pathophysiology \| Etiology \| Clinical features \| Diagnosis \| Laboratory \| Radiology description \| Radiology images \| Prognostic factors \| Case reports \| Treatment \| Gross description \| Gross images \| Frozen section description \| Frozen section images \| Microscopic (histologic) description \| Microscopic (histologic) images \| Cytology images \| Positive stains \| Negative stains \| Electron microscopy description \| Sample pathology report \| Differential diagnosis \| Additional references \| Practice question #1 \| Practice answer #1 \| Practice question #2 \| Practice answer #2 Cite this page: Multz R, Ahrendsen JT. Multiple sclerosis. PathologyOutlines.com website. Accessed September 28th, 2025. Definition / general Chronic, inflammatory demyelinating disease that may involve any part of the central nervous system Lesions, known as plaques, are characterized by loss of myelin with relative axonal preservation, reactive gliosis and the presence of lymphocytes or macrophages Essential features Most common immune mediated demyelinating disorder of the central nervous system and most common cause of nontrauma related neurologic disability in young adults Female predominant neurologic disorder consisting of episodic nervous system dysfunction and characteristic radiologic findings that demonstrate both dissemination in time (multiple episodes) and dissemination in space (multiple lesions seen on imaging) Terminology Multiple sclerosis MS ICD coding ICD-10: G35 - multiple sclerosis Epidemiology Most frequent cause of nontrauma related permanent disability in young adults (Neurol Clin 2011;29:207) Female predominance with F:M = 2.3:1 Mean age of onset: 28 - 31 years Risk factors: Genetics (HLA::DRB1) Presence of coexisting autoimmune disease Decreased sunlight exposure Low vitamin D levels Geographic location, with a higher latitude corresponding with a higher incidence of disease (Nature 2011;476:214) More recently, it has been posited that there may be an association between EBV infection and the development of MS (Science 2022;375:296) Sites Can involve any site in the central nervous system (CNS), typically within the white matter but also involves the gray matter (Brain 2019;142:1858) Most common sites for MS plaques: Optic chiasm and tract (nearly always involved) Periventricular white matter Subcortical white matter Deep gray nuclei Periventricular areas of the brainstem, cerebellum and spinal cord Pathophysiology Since the cause of MS is unknown, a detailed pathophysiologic pathway has not yet been established (see Etiology) Etiology Cause of MS is unknown, though it is favored to be due to an autoimmune activation of lymphocytes (Arch Neurol 2004;61:1613) Clinical features Patients present with one or more distinct episodes of CNS dysfunction and characteristic MRI findings (dissemination in time and space) (Lancet Neurol 2018;17:162, Nat Rev Dis Primers 2018;4:43) Relapsing remitting MS (RRMS) Most common subtype of MS Episodic neurologic deficits that may partially or fully resolve but are followed by additional relapses Primary progressive MS (PPMS) Nonepisodic progression of disease from the initial onset of symptoms Secondary progressive MS (SPMS) Typically follows RRMS, where the disease transitions from episodic to continued progression Acute Marburg MS Fulminant MS that is monophasic and rapidly progressive, usually leading to death within a year of symptom onset Often seen in children and young adults Tumefactive MS: Uncommon subtype of acute MS that presents as a mass-like lesion with a radiologic differential diagnosis that includes demyelination, infection or neoplasm Radiologically characterized by contrast enhancement often with unique open or incomplete ring Surgical biopsy (with intraoperative frozen section consultation) is sometimes performed to help resolve the differential diagnosis Diagnosis MS is a clinical diagnosis, combining clinical history and physical exam findings with imaging findings 2017 McDonald criteria can be applied where there is clinical suspicion for MS (Lancet Neurol 2018;17:162) Evoked potentials and optical coherence tomography (OCT) can also be used to aid in diagnosis (Lancet Neurol 2006;5:853, Mult Scler 2014;20:1342) Laboratory Lumbar puncture is not required for the diagnosis of MS but can test for the presence of oligoclonal IgG bands in the cerebrospinal fluid (CSF), which is supportive of MS E.g., if a patient is symptomatic with classic MRI findings showing dissemination in space but there is no known dissemination in time, oligoclonal bands can support a diagnosis of MS (Brain 2018;141:1075) Majority of cells in the CSF differential count are lymphocytes (specifically T cells) and the protein levels should be within the reference range (Neurology 2004;63:1966) Radiology description MRI is the best imaging modality since CT findings can be nonspecific Best visualized on T2 / FLAIR sequences: Extends perpendicularly on sagittal view, known as Dawson fingers (Cold Spring Harb Perspect Med 2018;8:a028969) Enhancement is a nonspecific marker of disease activity Tumefactive MS will present as an incomplete / open ring enhancing lesion following contrast administration (Neurol Neuroimmunol Neuroinflamm 2019;6:e589) Persistent hypointensities on T1 weighted images are characteristic of chronic plaques and are referred to as black holes (Acta Neurol Scand 2010;122:1) Healing or fully healed plaques will convert from hypointense to isointense on T1 (Acta Neurol Scand 2010;122:1) Radiology images Contributed by Jared T. Ahrendsen, M.D., Ph.D. and Pouya Jamshidi, M.D. Dawson fingers Inactive MS lesion Images hosted on other servers: Plaques in spinal cord Tumefactive MS Prognostic factors There are no consistently reliable prognostic factors for MS and outcomes for MS patients cannot be accurately predicted Generally, patients with relapsing and remitting courses have a better prognosis than those with progressive forms of the disease (Brain 2006;129:584) Rate of cervical cord atrophy, bowel / bladder symptoms at disease onset and incomplete recovery from initial attack are associated with a worse prognosis (Arch Neurol 2006;63:1686) Pregnancy is thought to be protective, though disease can worsen in the initial postpartum period (Neurology 2017;89:563) Extent of MRI abnormalities does not necessarily correlate with disease severity or prognosis (Neurology 2006;66:1384) Case reports 34 year old woman with acute onset left hemiparesis (Front Neurol 2022;13:891113) 36 year old man with blurred vision, vertigo, perioral numbness and confusion (CMAJ 2022;194:E776) 41 year old man with dysphagia, apraxia and ataxia (Neurol India 2022;70:1226) 47 year old woman with fatigue, blurry vision and numbness 3 weeks after recovering from SARS-CoV-2 infection (Cureus 2021;13:e19036) Treatment Acute exacerbation / episode (Mult Scler 2020;26:1352458520924595) High dose glucocorticoids are the mainstay of treatment for an acute exacerbation Plasma exchange can be considered in patients with poor response Relapsing remitting MS (Mult Scler 2020;26:1352458520924595, Neurology 2018;90:777) Monoclonal antibodies (natalizumab, alemtuzumab, ocrelizumab, rituximab) are generally first line therapy Oral therapies including fumarates and S1P receptor modulators (siponimod, fingolimod) can be used in patients who do not wish to have or are unable to tolerate infusions / injectables Interferons and glatiramer acetate Secondary progressive MS Same treatment as for relapsing remitting MS, though generally ineffective Primary progressive MS Ocrelizumab is the only FDA approved drug that has been shown to slow the progression of disease in PPMS (N Engl J Med 2017;376:209) Gross description Chronic plaques tend to be rounded, tan-gray and variably sized with a sharp demarcation from the surrounding brain tissue More recent lesions will be tan-pink or tan-yellow with less sharply defined borders In the spinal cord, plaques will have a fanned out appearance with sharp borders and can potentially involve multiple adjacent signaling pathways (Pract Neurol 2016;16:279) Gross images Contributed by Rachel A. Multz, M.D. and Jared T. Ahrendsen, M.D., Ph.D. Subcortical and periventricular plaques Subcortical plaques Periventricular plaques Frozen section description Typically only performed in cases of acute tumefactive MS, where there is radiologic suspicion for neoplasm or infection Findings are similar for any type of demyelinating disease: Sheets of foamy macrophages in a background of reactive gliosis Can be misdiagnosed as high grade glioma, especially since reactive atypia and mitotic can activity seen (Creutzfeldt cells) Cell populations will be more heterogeneous than neoplastic processes On permanent sections, there will be a sharp border between the area of interest and background brain tissue Reference: Acta Neuropathol 2017;133:13 Frozen section images Contributed by Jared T. Ahrendsen, M.D., Ph.D. Frozen section of demyelination lesion Microscopic (histologic) description Active MS plaque (Brain Pathol 2005;15:217): Prominent perivascular lymphoid infiltrates consisting predominantly of CD8+ T lymphocytes Parenchymal and perivascular macrophages, some with visible myelin globules on Luxol fast blue / periodic acid-Schiff stain (LFB / PAS) More specific for active demyelination than lymphocytes Preferential loss of myelin with relative axonal preservation and the formation of axonal spheroids (swellings) Reactive astrocytosis In the periphery of the plaque will see remyelination with thinner myelin sheaths than background axons Chronic / inactive MS plaque (Brain Pathol 2005;15:217): Relatively acellular lesions with near complete loss of myelin and sharp borders Should not see inflammation (or if present, rare lymphocytes) and will only see oligodendrocytes at the margin Reactive gliosis Baló concentric sclerosis (Neurol Sci 2004;25:S319): Rare histologic subtype of MS with concentric and alternating rings of demyelination and preserved myelin Areas of demyelination will show features similar to those of an active MS plaque Acute multiple sclerosis, Marburg type (Mult Scler 2015;21:485) Multiple, poorly defined plaques All plaques are active, with numerous macrophages, reactive astrogliosis and perivascular lymphocyte cuffs Sometimes edema or necrosis may be seen Microscopic (histologic) images Contributed by Rachel A. Multz, M.D. and Jared T. Ahrendsen, M.D., Ph.D. Chronic demyelination plaque Partial remyelination Active MS lesion Active MS lesion Cytology images Contributed by Jared T. Ahrendsen, M.D., Ph.D. Smear preparation of demyelination lesion Positive stains Luxol fast blue (LFB): In active plaques, will highlight normal background myelin with complete loss of staining in the area of demyelination Border between demyelination and normal appearing white matter will be distinct but not as sharp as that of an inactive plaque In areas where remyelination has occurred, myelin will be present but less dense than the uninvolved white matter (shadow plaque) Highlights myelin globules within macrophages in areas of active demyelination CD68 or CD163 will highlight macrophages in areas of active demyelination Neurofilament or a silver stain will highlight relative axonal preservation and the presence of occasional axonal spheroids GFAP highlights reactive astrogliosis Ki67 may be elevated in active plaques CD3 will highlight perivascular T lymphocytes but is not necessary for diagnosis Reference: Acta Neuropathol 2017;133:13 Negative stains Viral stains (HIV, HSV, SV40, etc.) Markers of high grade glioma (IDH1 [R132H], p53, ATRX, etc.) There should not be a significant CD20+ B cell population Reference: Acta Neuropathol 2017;133:13 Electron microscopy description No longer used in routine clinical practice Active lesions will show lymphocytes entering the endothelium of blood vessels, axonal spheroids filled with mitochondria and neurofilaments, a vesicular network of degenerated myelin surrounding axons and myelin globules within macrophages Chronic lesions will show demyelinated axons or remyelinated axons with thinner myelin sheaths (N Engl J Med 2006;354:942) Sample pathology report Brain lesion, left parietal lobe, biopsy: Reactive brain tissue with numerous macrophages and evidence of demyelination (see comment) Comment: The findings are suggestive of a demyelinating process. Given the radiologic impression of a mass-like lesion in the brain, tumefactive multiple sclerosis should be considered as a diagnostic consideration. Differential diagnosis Cerebral infarction: Sheets of macrophages with engulfed myelin and evidence of demyelination (but primarily distributed along vascular territories) Loss of axons (as opposed to demyelination, in which axons are relatively spared) Other features of hypoxic ischemic injury are generally present Neuromyelitis optica (NMO): Formerly considered a variant of multiple sclerosis Predominantly involves the optic tract and spinal cord, sometimes extensively involving the gray matter Unique pathogenesis with antiaquaporin 4 antibodies Neutrophils present in areas of demyelination, which are absent in MS Prominent axonal loss and destruction of astrocytes Acute disseminated encephalomyelitis (ADEM): Marked T cell perivascular inflammation with local demyelination, however, will lack large plaque formation Often with history of recent viral illness or vaccination Self limited with rapid recovery and low probability of lasting neurologic sequelae Progressive multifocal leukoencephalopathy (PML): Clinical history of immunosuppression Multiple foci of demyelination with sheets of macrophages and minimal lymphoid infiltrates Unique plum colored viral inclusions in oligodendrocytes and bizarre appearing astrocytes Demonstration of JC viral (PCR assay on cerebrospinal fluid; SV40 immunostain on tissue sections) Note that MS patients treated with certain immunomodulatory agents (especially natalizumab) are at increased risk of developing PML (Front Neurol 2020;11:579438) Additional references Gray: Escourolle & Poirier's Manual of Basic Neuropathology, 6th Edition, 2018, Perry: Practical Surgical Neuropathology - A Diagnostic Approach, 2nd Edition, 2017 Practice question #1 The pathologic features seen in this axial section of the spinal cord (H&E with Luxol fast blue [LFB]) taken at autopsy are most consistent with which of the following conditions? Amyotrophic lateral sclerosis Foix-Alajouanine syndrome Multiple sclerosis Subacute combined degeneration Tabes dorsalis Practice answer #1 C. Multiple sclerosis (MS). As shown in the image, there is asymmetric demyelination in random areas of the spinal cord in MS. Answer A is incorrect because amyotrophic lateral sclerosis shows symmetric degeneration of lateral corticospinal tracts. Answer B is incorrect because Foix-Alajouanine syndrome demonstrates myelopathy caused by dural arteriovenous malformation located in the spinal cord. Answer D is incorrect because subacute combined degeneration demonstrates relatively symmetric degeneration of both the lateral corticospinal tracts and the dorsal columns. Answer E is incorrect because tabes dorsalis shows symmetric degeneration of dorsal columns and dorsal nerve roots. Comment Here Reference: Multiple sclerosis Practice question #2 A 48 year old woman with multiple sclerosis treated with natalizumab develops numerous FLAIR signal abnormalities in the right centrum semiovale and left occipital lobe white matter. The patient deteriorated clinically and died 2 weeks after presentation. Brain autopsy was performed, revealing the microscopic features shown in the image above in areas of FLAIR signal abnormality. What is the most likely diagnosis? Active multiple sclerosis plaque Acute disseminated encephalomyeltis Infiltrating astrocytoma Progressive multifocal leukoenceophalopathy Toxoplasma encephalitis Practice answer #2 D. Progressive multifocal leukoencephalopathy (PML). The presence of macrophages, reactive gliosis, bizarre appearing astrocytes and plum colored oligodendroglial inclusions, along with the characteristic history described in the vignette, is most suggestive of PML. Active multiple sclerosis lesion is incorrect: active MS lesions are characterized by numerous macrophages and reactive gliosis; however, viral inclusions would not be seen in oligodendrocytes. Acute disseminated encephalomyelitis is incorrect: this is a monophasic demyelinating illness, distinct from multiple sclerosis and classically associated with recent viral infection or immunization; viral inclusions would not be seen in oligodendrocytes. Infiltrating astrocytoma is incorrect: while the atypical glial cells shown in the image might raise a concern for an infiltrating astrocytoma, the clinical history and imaging findings are not suggestive of a neoplastic process. Toxoplasma encephalitis is incorrect: Toxoplasma infection in the brain is characterized by necrotizing abscesses and the presence of parasites within pseudocysts (bradyzoites) or free parasites (tachyzoites). Comment Here Reference: Multiple sclerosis Back to top Home > CNS nontumor > Multiple sclerosis © Copyright PathologyOutlines.com, Inc. Click here for information on linking to our website or using our content or images. Home About Us Advertise Amazon.com Authors Board Review Books Cases CCN CME (online) Comment Here Conferences / Webinars Contact Us Copyright Directory FAQ Fellowships Games Grants Jobs Libraries Merchandise Newsletters Privacy Policy Statistics Testimonials 30150 Telegraph Road, Suite 119, Bingham Farms, Michigan 48025 (USA) Telephone: (248) 646-0325; Email: Comments@pathologyoutlines.com This website is intended for pathologists and laboratory personnel but not for patients. We welcome suggestions or questions about using the website. However, we cannot answer medical or research questions or give advice. © Copyright PathologyOutlines.com, Inc. Click here for information on linking to our website or using our content or images.
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2080	http://www.360doc.com/content/22/1222/05/80695768_1061064044.shtml	小学数学知识点解读与学习策略49——轴对称搜索我的图书馆查看信箱系统消息官方通知设置开始对话有 11 人和你对话，查看忽略历史对话记录通知设置发文章发文工具撰写网文摘手文档视频思维导图随笔相册原创同步助手其他工具图片转文字文件清理AI助手留言交流搜索分享 QQ空间QQ好友新浪微博微信生成长图转Word打印朗读全屏修改转藏+1 × 微信扫一扫关注作者 × 微信扫一扫将文章发送给好友了解公众号宣传计划【原】小学数学知识点解读与学习策略49——轴对称慢点数学 2022-12-22 发布于江苏来源\|54阅读\|2 转藏大中小转藏全屏朗读打印转Word生成长图分享 QQ空间QQ好友新浪微博微信展开全文说起轴对称，就不得不提轴对称图形，这两个概念学生最容易混淆。轴对称研究的是两个图形之间的关系，当这两个图形沿着某条直线进行对折，在直线两旁的两个图形或两个图形的部分能够完全重合，那么这两个图形就被称为成轴对称关系的两个图形，简称为成轴对称。而轴对称图形指的是一个图形，如果该图形沿某条直线对折后，能够使图形的两部分完全重合，那么这样的图形就被称为轴对称图形。小学数学所研究的轴对称，指的是轴对称图形，是一个图形的两个部分之间的关系。明确这一点，才能有针对性地指导孩子学习数学。轴对称的学习一定要让学生亲自操作、亲身体验，才能建构起轴对称图形的模型。 1、动手折纸可以先让孩子准备长方形、正方形和平行四边形的纸片各一张，再通过对折纸片，让折痕两边的部分完全重合。操作后发现，长方形、正方形是轴对称图形，平行四边形不是轴对称图形。接着通过不同的对折方法，认识长方形有两条对称轴，正方形有四条对称轴，对称轴的条数因图形的不同而不同。此时有必要指出，对称轴并不是那条折痕，而是那条折痕所在的直线。 2、动手画“轴” 在方格纸上画一些平面图形的对称轴，如： ① 只是轴对称图形：角、等腰三角形、等边三角形、等腰梯形等。 ② 只是中心对称图形：平行四边形等。 ③ 既是轴对称图形又是中心对称图形：正方形、长方形、圆等。 ④ 既不是轴对称图形又不是中心对称图形：不等边三角形、非等腰梯形等。经历画“轴”的过程，可以进一步加深对轴对称图形的认识，初步体会轴对称图形的特征。 3、动手画图轴对称图形到底有什么特征，不妨通过下面三个层次的画图操作，实现对其特征的认识。把下面的图形补全，使其成为轴对称图形。 ① 说明补全图形的方法，体会轴对称图形的对称点到对称轴的距离相等。也就是说，对称点连成的线段被对称轴垂直平分，此结论只要体会理解，不作掌握要求。 ② 说明补全图形的方法，加深对轴对称图形特征的理解（对称点到对称轴的距离相等）。 ③ 该题有些难度，目的就是强化对称点到对称轴的距离相等这一特征。刚开始画图时，如果出现错误也是正常的，不妨提醒孩子联系上面两题得出的结论，再进行操作探究，一定会对轴对称图形有一个统一的认识。 4、动手设计我们可以根据轴对称图形的特征，由一个图形得到与它对称的另一个图形，重复这个过程，便可以得到美丽的图案。经历设计图形的过程不但可以巩固对轴对称图形的认识，而且可以从运动的角度加深对图形的认识。 +关注慢点数学立即填写简介（查看如何填写简介）共 1298 篇原创微信公众号：微信扫一扫关注赞赏转藏分享 QQ空间QQ好友新浪微博微信献花（0） +1 来自：慢点数学>《待分类》举报/认领上一篇：同样是1/4为什么不相等？下一篇：从画到算的转变，是思维能力提升的表现猜你喜欢 0 条评论发表请遵守用户评论公约查看更多评论类似文章更多轴对称与轴对称图形轴对称与轴对称图形欣赏大自然的风景观察下面的图形，你能发现它们有什么共同的特征吗？DEF对称点对称轴轴对称的两个图形是全等的对称轴是对称点连线的垂直平分线轴对称图形轴对称图形\5642.swf如果把... 对称的认识数学上是先定义一个点对一条直线（对称轴）的对称点，再定义一个图形对一条直线（对称轴）的对称图形，最后才透过如果一个图形对直线L（... 中心对称优质教学材料课件PPT 选填中心对称 (1)点O是线段AA的中点（2）△ABC≌△A′B′C′想一想中心对称与轴对称有什么区别?又有什么联系?轴对称中心对称有一条对称轴---直线图形沿对称轴对折(翻折1800)后重合对称点的连线被对称轴垂直平分有一... 第27讲轴对称与中心对称把一个图形沿着某一条直线翻折过去，如果它能够和另一个图形重合，那么这两个图形关于直线对称，两个图形关于直线对称也称轴对称．这条直线叫做对称轴．.【答案】略4．(2012·青岛)下列图形中，既... 解答版《火线100天》2015中考数学复习第24讲图形的平移、对称与旋转中心对称中心对称图形定义把一个图形绕着一点旋转后，如果与另一个图形重合，那么这两个图形成中心对称，这个点叫做其对称中心，旋转前后重合的点叫做对称点.把一个图形绕着某点旋转后，能与其自身重合... 七年级数学下册《轴对称与旋转》练习题及答案(湘教版) (2)满足条件的图形有很多，只要画正确一个，都可以得满分.23.解：24.解：(1)如图所示(2 )图形A的最小旋转角是60度，它是中心对称图形.图形B的最小旋转角是72度，它不是中心对称图形.图形C的最小旋转角... 【人教版】2017届中考复习：第26讲《轴对称与中心对称》ppt课件第七章图形的变化第26讲轴对称与中心对称。考点一轴对称图形与轴对称1．轴对称图形如果一个平面图形沿着一条直线折叠后直线两旁的部分能够互相重合那么这个图形就叫做轴对称图形．.如果一个图形绕某点... 【初二数学】画轴对称图形经典例题解析，逢考必错的高频考点个图VIP年卡，限时优惠价168元>>x 慢点数学教育领域优质作者关注对话 TA的最新馆藏结合经验，让除法自然生长什么是合适的问题？初次学习小括号的主要教学目标小括号有什么作用？分步算式变成综合算式的常用方法 “时间”与“时刻”的区别喜欢该文的人也喜欢更多热门阅读换一换复制打印文章发送到手机微信扫码，在手机上查看选中内容全屏阅读朗读全文分享文章QQ空间QQ好友新浪微博微信 AI解释复制打印文章 adsbygoogle.js 发送到手机微信扫码，在手机上查看选中内容全屏阅读朗读全文 AI助手阅读时有疑惑？点击向AI助手提问吧联系客服在线客服： 360doc小助手2 客服QQ： 1732698931 联系电话：4000-999-276 客服工作时间9:00-18:00，晚上非工作时间，请在QQ留言，第二天客服上班后会立即联系您。
2081	https://www.dummies.com/article/academics-the-arts/math/geometry/applying-the-transversal-theorems-187559/	Book & Article Categories Collections Book & Article Categories Book & Article Categories Collections Applying the Transversal Theorems When you cross two lines with a third line, the third line is called a transversal. You can use the transversal theorems to prove that angles are congruent or supplementary. Here’s a problem that lets you take a look at some of the theorems in action: Given that lines m and n are parallel, find the measure of angle 1. Here’s the solution: (Or you can also say that because you’ve got the parallel-lines-plus-transversal diagram and two angles that are both obviously acute, they must be congruent.) Set them equal to each other and solve for x: This equation has two solutions, so take them one at a time and plug them into the x’s in the alternate exterior angles. So 144° and 155° are your possible answers for angle 1. When you get two solutions (such as x = 6 and x = –5) in a problem like this, you do not plug one of them into one of the x’s and the other solution into the other x (like –5 + 30 = 25). You have to plug one of the solutions into allx’s, giving you one result for both angles then you have to separately plug the other solution into all x’s, giving you a second result for both angles Angles and segments can’t have negative measures or lengths. Make sure that each solution for x produces positive answers for all the angles or segments in a problem. If a solution makes any angle or segment in the diagram negative, it must be rejected even if the angles or segments you care about end up being positive. However, do not reject a solution just because x is negative: x can be negative as long as the angles and segments are positive (x = –5, for example, works just fine in this problem). Now here’s a proof that uses some of the transversal theorems: Check out the formal proof: Statement 1: Reason for statement 1: Given. Statement 2: Reason for statement 2: Given. Statement 3: Reason for statement 3: If lines are parallel, then alternate interior angles are congruent. Statement 4: Reason for statement 4: Given. Statement 5: Reason for statement 5: If a segment (segment GJ) is subtracted from two congruent segments, then the differences are congruent. Statement 6: Reason for statement 6:SAS (using lines 1, 3, and 5). Statement 7: Reason for statement 7:CPCTC (Corresponding Parts of Congruent Triangles are Congruent). Statement 8: Reason for statement 8: If alternate exterior angles are congruent, then lines are parallel. Extend the lines in transversal problems. Extending the parallel lines and transversals may help you see how the angles are related. For instance, if you have a hard time seeing that angle K and angle H are indeed alternate interior angles (for step 3 of the proof), After doing that, you’re looking at the familiar parallel-line scheme shown in the following figure. You can do the same thing for angle LJK and angle IGH by extending About This Article This article is from the book: About the book author: Mark Ryan has more than three decades’ experience as a calculus teacher and tutor. He has a gift for mathematics and a gift for explaining it in plain English. He tutors students in all junior high and high school math courses as well as math test prep, and he’s the founder of The Math Center on Chicago’s North Shore. Ryan is the author of Calculus For Dummies, Calculus Essentials For Dummies, Geometry For Dummies, and several other math books. This article can be found in the category: Quick Links Connect About Dummies Dummies has always stood for taking on complex concepts and making them easy to understand. Dummies helps everyone be more knowledgeable and confident in applying what they know. Whether it's to pass that big test, qualify for that big promotion or even master that cooking technique; people who rely on dummies, rely on it to learn the critical skills and relevant information necessary for success. Copyright © 2000-2025 by John Wiley & Sons, Inc., or related companies. All rights reserved, including rights for text and data mining and training of artificial technologies or similar technologies. © 2025 MARVEL
2082	https://bmcmicrobiol.biomedcentral.com/articles/10.1186/s12866-018-1319-0	Efficient transposon mutagenesis mediated by an IPTG-controlled conditional suicide plasmid \| BMC Microbiology \| Full Text Your privacy, your choice We use essential cookies to make sure the site can function. We also use optional cookies for advertising, personalisation of content, usage analysis, and social media. By accepting optional cookies, you consent to the processing of your personal data - including transfers to third parties. Some third parties are outside of the European Economic Area, with varying standards of data protection. See our privacy policy for more information on the use of your personal data. Manage preferences for further information and to change your choices. Accept all cookies Skip to main content Advertisement Search Explore journals Get published About BMC Login Menu Explore journals Get published About BMC Login Search all BMC articles Search BMC Microbiology Home About Articles Submission Guidelines Collections Join the Editorial Board Submit manuscript Efficient transposon mutagenesis mediated by an IPTG-controlled conditional suicide plasmid Download PDF Download PDF Methodology article Open access Published: 24 October 2018 Efficient transposon mutagenesis mediated by an IPTG-controlled conditional suicide plasmid Santa S. Naorem1na1, Jin Han1na1, Stephanie Y. Zhang1, Junyi Zhang1, Lindsey B. Graham1, Angelou Song1, Cameron V. Smith1, Fariha Rashid1& … Huatao GuoORCID: orcid.org/0000-0002-2396-19341 Show authors BMC Microbiologyvolume 18, Article number:158 (2018) Cite this article 10k Accesses 11 Citations 9 Altmetric Metrics details Abstract Background Transposon mutagenesis is highly valuable for bacterial genetic and genomic studies. The transposons are usually delivered into host cells through conjugation or electroporation of a suicide plasmid. However, many bacterial species cannot be efficiently conjugated or transformed for transposon saturation mutagenesis. For this reason, temperature-sensitive (ts) plasmids have also been developed for transposon mutagenesis, but prolonged incubation at high temperatures to induce ts plasmid loss can be harmful to the hosts and lead to enrichment of mutants with adaptive genetic changes. In addition, the ts phenotype of a plasmid is often strain- or species-specific, as it may become non-ts or suicidal in different bacterial species. Results We have engineered several conditional suicide plasmids that have a broad host range and whose loss is IPTG-controlled. One construct, which has the highest stability in the absence of IPTG induction, was then used as a curable vector to deliver hyperactive miniTn5 transposons for insertional mutagenesis. Our analyses show that these new tools can be used for efficient and regulatable transposon mutagenesis in Escherichia coli, Acinetobacter baylyi and Pseudomonas aeruginosa. In P. aeruginosa PAO1, we have used this method to generate a Tn5 insertion library with an estimated diversity of ~ 10 8, which is ~ 2 logs larger than the best transposon insertional library of PAO1 and related Pseudomonas strains previously reported. Conclusion We have developed a number of IPTG-controlled conditional suicide plasmids. By exploiting one of them for transposon delivery, a highly efficient and broadly useful mutagenesis system has been developed. As the assay condition is mild, we believe that our methodology will have broad applications in microbiology research. Background Transposon mutagenesis is a powerful technique for bacterial genetic and genomic studies. One of the most widely used transposons is derived from Tn5. The Tn5 transposon contains two IS50 elements as inverted terminal repeats (Additional file1: Figure S1) [1, 2]. Both IS50 and Tn5 can be mobilized by their encoded transposase (Tnp) protein, which recognizes two 19 base pair (bp) sequences at their ends, namely outside end (OE) and inside end (IE), for transposition . OE and IE differ by 7 bp (Additional file 1: Figure S1). As Tn5 insertion is almost completely random, it can insert into any gene in a bacterium. The native Tn5/IS50 is not very active, thus avoiding overt deleterious effect on their hosts, but hyperactive mutants have been engineered as genetic manipulation tools [2, 3]. The most active one contains a mosaic sequence of OE and IE (mosaic end; ME) at the transposon termini and an engineered tnp gene encoding a highly active transposase enzyme (Tnp H), which together increase Tn5 transposition by more than 1000-fold. Transposons for insertional mutagenesis are usually delivered into bacteria through conjugation of a suicide plasmid [4,5,6]. Insertion mutants are then selected as the transposons are tagged with an antibiotic-resistance gene. The success of a transposon mutagenesis assay, especially a saturation mutagenesis assay, requires generation of an insertion library with high diversity, which requires efficient plasmid conjugation and transposon transposition. However, conjugation is inefficient in many bacterial species. Occasionally, electroporation has also been used to deliver transposon-containing suicide plasmids for mutagenesis, but low library diversities were often achieved using such approaches [7,8,9]. To perform efficient transposon mutagenesis in these organisms, temperature-sensitive (ts) plasmids are sometimes used for transposon delivery [10,11,12,13,14,15,16]. However, many organisms do not have a ts and easily manipulatable plasmid, and sometimes a ts plasmid in one organism is either non-ts or suicidal in a different organism [10, 14, 17]. In addition, a high temperature is often required to cure the ts plasmids after mutagenesis, which can be inhibitory to cell growth and may result in selection of mutants with adaptive genetic changes [10, 11, 14]. In this study, we have developed an efficient and regulatable transposon mutagenesis tool that exploits an IPTG-controlled conditional suicide plasmid. It contains an RSF1010 replicon, an IncQ-type replication origin that allows plasmid replication in most Gram-negative bacteria, as well as a few Gram-positive bacteria . It is relatively small, so it can be easily modified. To control plasmid replication by IPTG, a second copy of the plasmid-encoded repF repressor gene is cloned downstream of the Escherichia coli tac promoter. For efficient and regulatable transposon mutagenesis, we used miniTn5 (mTn5) transposons and cloned the hyperactive transposase gene downstream of a lac promoter. We show that the resulting constructs can be used for efficient insertional mutagenesis in three different bacterial species. In Pseudomonas aeruginosa PAO1, we show that our system is able to generate a Tn5 insertion library that is almost 2 logs larger than the best library of PAO1 and related Pseudomonas strains previously reported, demonstrating that we have developed a powerful mutagenesis tool that is highly useful for microbiology studies. Results Construction of IPTG-controlled suicide plasmids To develop a method for efficient transposon mutagenesis in bacterial species that are difficult to transform and conjugate, we created multiple IPTG-controlled suicide plasmids that have a broad host range (Fig.1a). The plasmids were derived from pMMB208, which is a conjugatable plasmid containing an RSF1010 oriV (an IncQ-type origin of replication) that can replicate in most Gram-negative bacteria and a few Gram-positive bacteria . Plasmid replication requires three proteins, RepA, MobA/RepB and RepC, which are a helicase, a primase and an oriV-binding protein, respectively. repF encodes a small repressor protein that binds the P4 promoter and controls the repF-repA-repC operon through feedback inhibition [19, 20]. pMMB208 also contains a tac promoter (P tac), a lacI Q gene and a chloramphenicol resistance marker (Cam R). To create a conditional suicide plasmid (pMMB-repF), a second copy of the repF gene was inserted downstream of P tac. Upon IPTG induction, efficient plasmid loss from transformed E. coli DH10B cells was observed (99.97%; Fig. 1b). As an alternative strategy, we inserted two repA helicase dominant negative mutants, K42A and D139A, downstream of P tac . Similarly, IPTG was able to induce efficient plasmid loss from the transformed DH10B cells. In fact, plasmid retention rates of the dominant negative mutants (K42A, 4.9 × 10− 7; D139A, 1.5 × 10− 5) were much lower than that of pMMB-repF (3.5 × 10− 4) (Fig. 1b). However, the two repA dominant negative mutants showed significantly lower plasmid stability in the absence of IPTG induction (Fig. 1b), suggesting that plasmid replication is strongly inhibited by leaky expression of the dominant negative mutants, or by spontaneous recombination of the wild-type and the dominant negative repA genes (~ 861 bp direct repeats). Consistent with that, there were ~ 20–30-fold less plasmid isolated from the same amount of cells for the two mutant constructs (Fig. 1c). Therefore, we decided to choose pMMB-repF for further experiments. To kill the cells that still retain the plasmid after IPTG induction, we inserted a sacB counter selection marker into the vector , resulting in pMMB-repF/sacB. Indeed, insertion of sacB allows efficient killing of plasmid-containing cells by sucrose (data not shown; also see below). Fig. 1 IPTG-controlled conditional suicide plasmids. a Plasmid pMMB208 and its conditional-suicide derivatives. pMMB208 contains an RSF1010 oriV for replication and an oriT for conjugation. Genes repA, mobA/repB and repC encode proteins required for plasmid replication, and repF encodes a transcription repressor that binds promoter P4. pMMB208 also has Cam R and lacI Q genes and a P tac promoter. Plasmid pMMB-repF is a derivative of pMMB208 that has a second copy of the repF gene inserted downstream of P tac. Plasmids pMMB-repA K42A and pMMB-repA D139A have a dominant-negative repA mutant gene, either K42A or D139A, inserted downstream of P tac. b Amount of E. coli DH10B cells retaining the indicated plasmids after 24 h growth in the absence of antibiotics, either with or without IPTG induction. Results were average of three independent experiments, and bars represent mean ± SD (standard deviation). p< 0.0001, p< 0.0001, and p< 0.0001 by unpaired Student’s t-test for IPTG induced cultures. c pMMB208 and its derivatives are digested with Hin dIII (H) and Pst I (P). Comparing to pMMB208 and pMMB-repF, the repA K42A and D139A mutants showed reduced yields in plasmid minipreps (no IPTG induction; 3.0% and 4.7% of that of pMMB208, respectively). Hin dIII and Pst I digestion generates two fragments for each plasmid. The ~ 9 kb fragment is seen on the gel, while the shorter ones, ranging from 18 bp for pMMB208 to 861 bp for the repA mutants, are not visible. Another large band (~ 9 kb) is also seen in restriction digestion of pMMB208 and its derivatives, even after complete digestion, and the cause is unknown Full size image IPTG-controlled mutagenesis of E. coli by a highly-active mTn5 transposon A Kan R-tagged mTn5 was then inserted in pMMB-repF/sacB for transposon mutagenesis (Fig.2a) . The mTn5 contains an OE and an IE at the termini. In addition, it contains an uncoupled, lac promoter (P lac)-controlled tnp H gene encoding the hyperactive transposase (Tnp H) , thus allowing inducible expression of Tnp H. E. coli cells transformed with this plasmid, pSNC-mTn5, were cultured in LB media with and without IPTG induction for 24 h. Cells were then analyzed for efficiencies of plasmid loss, sucrose counter selection and transposon insertion (See Methods). The plasmid is stable without IPTG induction, as ~ 91.4% of cells retained the plasmid (Cam R) after 24 h culture in the absence of antibiotics (Fig. 2b). In contrast, ~ 2.6 × 10− 3 of the cells retained the plasmid post IPTG induction, suggesting that overexpression of the RepF repressor caused efficient plasmid loss. Sucrose counter selection further reduced plasmid-bearing cells (~ 1.6 × 10− 6 are Cam R; ~ 1600-fold reduction). In comparison, the percentage of Suc R Kan R cells after IPTG induction was found to be ~ 2.3 × 10− 4 (Tn5-containing), significantly higher than that of Suc R Cam R cells (~ 1.6 × 10− 6, plasmid-containing), suggesting that Tn5 transposition had occurred efficiently (Suc R Kan R Cam S: ~ 2.3 × 10− 4). Colony restreaking showed that 150/150 Suc R Kan R colonies were Kan R Cam S (Fig. 2c). Colony PCRs, which used two sets of primers (P1 + P2 for detection of Kan R, or mTn5, and P3 + P4 for detection of tac-repF, or plasmid), confirmed plasmid loss in 10 out of 10 colonies (10/10) (Fig. 2d). Sequence analysis showed that all 13 Suc R Kan R colonies analyzed had different Tn5 insertion sites (Fig.3a). Fig. 2 mTn5 transposon mutagenesis using an IPTG-controlled conditional suicide plasmid. a Diagram of plasmid pSNC-mTn5. pSNC-mTn5 is a derivative of pMMB-repF that contains a Kan R-tagged mTn5, a lac promoter-controlled hyperactive transposase gene (tnp H), and a sacB counter selection marker (with its own promoter). OE and IE are outside and inside ends of the mTn5. b Plasmid and transposon retention frequencies in E. coli DH10B. A “+” symbol for IPTG indicates that the inducer was added to the liquid culture, and a “+” symbol for Suc, Cam, and Kan indicates that the chemicals were added to the plates. Black columns represent plasmid retention frequencies, and the blue column represents Tn5 retention frequency. Results were average of three independent experiments, and bars represent mean ± SD (p < 0.0001 and p = 0.0054 by unpaired t-test). (see Methods for details) (c) Colony restreaking. 150/150 Suc R Kan R colonies of DH10B were found to be Kan R Cam S and 50 are shown here. d Colony PCR of 10 restreaked clones in (c). Primer sets P1&P2 and P3&P4 detect Kan R and repF, respectively. All were mTn5-positive and plasmid-negative. Primers P3 and P4 are a functional pair for PCR-amplification of the plasmid sequence (data not shown). e Plasmid and transposon retention frequencies in A. baylyi. Results were average of three independent experiments, and bars represent mean ± SD (p = 0.024 and p < 0.0001 by unpaired t-test). f Plasmid and transposon retention frequencies in P. aeruginosa. Results were average of three independent experiments, and bars represent mean ± SD (p = 0.0013 and p = 0.0038 by unpaired t-test). Colony restreaking and PCR analysis are shown in Additional file 2: Figure S2 Full size image Fig. 3 mTn5 insertion sites in different bacteria. a mTn5 insertion sites in E. coli DH10B. b mTn5 insertion sites in A. baylyi 33,305. c mTn5 insertion sites in P. aeruginosa PAO1. Only the chromosomal sequences next to the OE are shown. The 9 bp duplicated sequences are shown in capital letters. Identical clones are shown only once, with numbers indicated in parenthesis. Either gene names or locus tags are given as genetic locations Full size image Efficient mutagenesis of Acinetobacter baylyi and P. aeruginosa by a highly-active mTn5 transposon Construct pSNC-mTn5 was then tested in two Gram-negative, capsule-bearing bacteria, A. baylyi 33,305 and P. aeruginosa PAO1 [23, 24]. Comparing to E. coli DH10B, transformed A. baylyi 33,305 and P. aeruginosa PAO1 appeared to lose the plasmid more easily in the absence of IPTG, with ~ 56.3% of A. baylyi and ~ 59.6% of P. aeruginosa retaining the plasmid after 24 h culture in LB media without antibiotics (Fig. 2e, f). Following IPTG induction, ~ 14.7% of A. baylyi and ~ 3.2% of P. aeruginosa retained the plasmid, suggesting that IPTG induced additional plasmid loss from these organisms, although their efficiencies were lower than that in DH10B cells. With sucrose counter selection, ~ 1.6 × 10− 7 of A. baylyi remained Cam R, indicating that they contained the plasmid (Fig. 2e). Similarly, ~ 5.4 × 10− 7 of P. aeruginosa cells were found to be Suc R Cam R (Fig. 2f). These results suggest that IPTG and sucrose both contributed in reducing plasmid-bearing cells. In comparison, the percentages of Suc R Kan R cells were 8.4 × 10− 6 for A. baylyi and 3.4 × 10− 6 for P. aeruginosa, suggesting that Tn5 transposition occurred in both organisms prior to plasmid loss. Colony restreaking showed that 100/100 Suc R Kan R colonies are Suc R Cam S, suggesting that efficient plasmid loss had occurred following Tn5 transposition (~ 100% for both; Additional file2: Figure S2a, c). Loss of plasmids was further confirmed by PCR tests (Additional file 2: Figure S2b, d). As observed in DH10B cells, Tn5 insertion also seemed to be random, as 9/9 A. baylyi and 13/15 P. aeruginosa mutants had different Tn5 insertion sites (Fig. 3b, c). The detection of identical mutants suggests that cell growth ensued following transposon transposition (Fig. 3c), which is common in different transposon mutagenesis assays [5, 6, 25]. Construction of a Tn5 insertion library of P. aeruginosa using the highly-active mTn5 transposon To determine whether we can construct a transposon insertion library of P. aeruginosa PAO1 with high diversity, ten pSNC-mTn5 transformants of the bacterium were cultured independently and then combined and induced with IPTG to initiate transposon mutagenesis. Following 24 h culture in LB media containing IPTG, ~ 6.4% of cells retained the plasmid (Additional file3: Figure S3a). The frequencies of Suc R Cam R and Suc R Kan R cells in the IPTG-induced culture were found to be ~ 6.5 × 10− 7 and ~ 3.5 × 10− 6, respectively. Based on the total number of cells cultured and the frequency of Suc R Kan R Cam S cells, the total diversity of the mTn5 insertion library was estimated to be ~ 1.3 × 10 7, which covers the entire gene repertoire (5697) of P. aeruginosa PAO1 by ~ 2238 times . To our knowledge, the diversity of this transposon insertion library is bigger than the best transposon insertion library of PAO1 and related strains previously reported (Table1) [5, 6, 9, 26,27,28,29,30,31,32]. Colony restreaking and PCR tests confirmed plasmid loss in the mutants (Additional file 3: Figure S3b, c), and 28/37 clones analyzed had different Tn5 insertion sites (Additional file 3: Figure S3d). Based on the percentage of independent clones in the library, its diversity is re-estimated to be ~ 1.0 × 10 7. Table 1 Comparison of transposon insertion libraries of P. aeruginosa strains Full size table An mTn5 with MEs enables generation of a P. aeruginosa mutant library with even higher diversity To determine whether the efficiency of mTn5 transposition can be further improved, we replaced both OE and IE of the mTn5 with MEs (Fig.4a). The new plasmid, pSNC-mTn5ME, was transformed into DH10B cells. Cell growth (or colony sizes) appeared to be normal, suggesting that basal-level transposition, if any, did not lead to obvious cellular toxicity, which was our initial concern. The behavior of the plasmid and Tn5 transposition efficiency were determined under the same conditions described above. Without IPTG induction, the plasmid remained relatively stable, as ~ 100% of the cells retained the plasmid (Cam R). After IPTG induction for 24 h, 1.1 × 10− 3 of the cells retained the plasmid, suggesting that RepF overexpression caused efficient plasmid loss. With sucrose counter selection, ~ 2.4 × 10− 5 cells remained Suc R Cam R. Interestingly, the frequency of Suc R Kan R cells was found to be very high (~ 28.0%), indicating that mTn5ME is much more active than the non-ME version (~ 1200 folds). Restreaking of Suc R Kan R colonies showed that they were all Kan R Cam S (100/100) (Additional file4: Figure S4a), and colony PCRs confirmed plasmid loss (10/10) (Additional file 4: Figure S4b). Sequence analysis of the transposon insertion junctions showed that all 13 Suc R Kan R colonies analyzed had different Tn5 integration sites (Additional file 4: Figure S4c). Fig. 4 Generation of a P. aeruginosa insertion library with pSNC-mTn5ME. a Diagram of pSNC-mTn5ME, a derivative of pSNC-mTn5 that has MEs instead of OE and IE at the termini of mTn5. b Plasmid and transposon retention frequencies in E. coli DH10B. Results were average of three independent experiments, and bars represent mean ± SD (p< 0.0001 and p = 0.0004 by unpaired t-test). Colony restreaking and PCR assays are shown in Additional file 4: Figure S4. c Plasmid and transposon retention frequencies in A. baylyi. Results were average of three independent experiments, and bars represent mean ± SD (p = 0.0029 and p = 0.0006 by unpaired t-test). Colony restreaking and PCR assays are shown in Additional file 5: Figure S5. d Plasmid and transposon retention frequencies in P. aeruginosa PAO1. Results were average of three independent experiments, and bars represent mean ± SD (p< 0.0001 and p = 0.0065 by unpaired t-test). Colony restreaking and PCR assays are shown in Additional file 6: Figure S6. e Plasmid and transposon retention frequencies in the P. aeruginosa PAO1 mutant library generated with pSNC-mTn5ME. f Colony restreaking. 100/100 Suc R Kan R colonies of the mTn5ME library of P. aeruginosa were found to be Kan R Cam S. 50 are shown here. g Colony PCR of ten restreaked clones in (f) with the indicated primers. All were mTn5-positive and plasmid-negative. h Transposon insertion sites of 46 mutant clones from the mTn5ME insertion library of P. aeruginosa. Identical clones are shown only once, with their duplication numbers indicated in parenthesis Full size image We then determined whether construct pSNC-mTn5ME would also be more active in A. baylyi 33,305 and in P. aeruginosa PAO1. For A. baylyi, the efficiencies of plasmid loss were higher for pSNC-mTn5ME than for pSNC-mTn5, both in the absence and presence of IPTG induction (Figs. 2e and 4c). Sucrose counter selection was highly effective for both constructs (Figs. 2e and 4c). As in E. coli, mTn5ME was found to be much more active than the non-ME version (~ 140 fold higher; Suc R Kan R cells: ~ 1.2 × 10− 3 for mTn5ME vs. ~ 8.4 × 10− 6 for mTn5). Similarly, colony restreaking of Suc R Kan R cells showed that 100/100 colonies are Kan R Cam S (Additional file5: Figure S5a), and plasmid loss was further verified by PCR (Additional file 5: Figure S5b). 9/14 colonies were found to have different Tn5 insertion sites (Additional file 5: Figure S5c). For PAO1, efficiencies of plasmid loss (±IPTG) were found to be similar for both pSNC-mTn5 and pSNC-mTn5ME (Figs. 2f and 4d), and sucrose counter selection was also effective for pSNC-mTn5ME (Fig. 4d). As in E. coli and in A. baylyi, mTn5ME was found to be more active than mTn5 in PAO1 (~ 11 fold higher; Suc R Kan R cells: ~ 3.9 × 10− 5 for mTn5ME vs. ~ 3.4 × 10− 6 for mTn5) (Figs. 2f and 4d). In the colony restreaking assay, 100/100 Suc R Kan R colonies were found to be Kan R Cam S (Additional file6: Figure S6a), and PCR assays further confirmed plasmid loss (10/10) (Additional file 6: Figure S6b). Sequence determination showed that 7/10 colonies tested had different Tn5 insertion sites (Additional file 6: Figure S6c). We then determined whether pSNC-mTn5ME would be a better construct than pSNC-mTn5 for transposon saturation mutagenesis in P. aeruginosa. Ten transformants were randomly picked for Tn5 insertion library construction using the protocol described above. About 0.87% of cells retained the plasmid after IPTG induction, and the frequency of Suc R Cam R cells was found to be 6.8 × 10− 8. In comparison, the frequency of Suc R Kan R cells was found to be ~ 9.5 × 10− 5, suggesting that efficient mTn5ME transposition has occurred. Based on the total amount of cells cultured and the mTn5ME transposition efficiency (~ 9.5 × 10− 5), the diversity of the mTn5ME insertion library was estimated to be 1.02 × 10 8, which is ~ 3 logs larger than the best PAO1 transposon insertion library previously reported and ~ 2 logs larger than a Tn5 insertion library of P. aeruginosa MPAO1, a derivative of PAO1 with ~ 0.2% genetic variation [6, 33]. This new library is by far the biggest transposon insertion library of PAO1 and related species ever reported (Table 1). The size of our new library is enough to cover the entire gene repertoire of PAO1 by ~ 18,000 times. Colony restreaking (100) and PCR tests (10) confirmed plasmid loss in all the Suc R Kan R clones analyzed (Fig. 4f, g), and 34/46 clones tested had different Tn5 insertion sites (Fig. 4h). Thus, the independent clones in the library (library diversity) are estimated to be ~ 7.5 × 10 7. Discussion We have developed a new transposon mutagenesis system that is efficient, regulatable, easy-to-use, and broadly useful. We believe it will be especially useful for functional genomics studies of Gram-negative bacteria that are difficult to transform and conjugate, such as certain capsule-containing bacteria, obligate anaerobes, and possibly obligate intracellular pathogens as well. The advantage of this method relies on the following features: (i) A broadly-functional plasmid replicon; (ii) Replication of the plasmid is regulated by IPTG; (iii) The inclusion of the sacB gene for counter selection; (iv) A highly-active/hyperactive transposon; (v) Regulatable expression of the hyperactive transposase gene; (vi) mTn5 and mTn5ME transposons insert almost completely randomly in different bacteria (Additional file7: Figure S7) [34, 35]. In addition, the relatively small sizes of the RSF1010-based plasmids also facilitate their transformation and conjugation. Similar to transposon mutagenesis using ts plasmids, our system does not depend on efficient plasmid transformation and conjugation, and requires as few as one transformant or conjugated cell for transposon saturation mutagenesis. Using this new tool, we have generated a Tn5 transposon insertion library of P. aeruginosa PAO1 with a diversity of ~ 10 8, which is ~ 2 logs larger than the best transposon insertion library of PAO1 and related Pseudomonas strains ever generated (Table 1). P. aeruginosa is an important opportunistic pathogen that frequently causes nosocomial infections and many of the strains are multidrug-resistant. The mutant PAO1 library we generated should also be valuable for P. aeruginosa pathogenesis studies. To our knowledge, our plasmids are the only non-ts, conditional suicide plasmids used for transposon mutagenesis, and they replicate in a wide range of bacterial species . In contrast, many ts mutant plasmids seem to have limited host ranges, either due to the limited host ranges of the parental plasmids, or due to the species-specificity of their ts phenotypes [10, 11, 14,15,16,17, 36]. In addition, ts plasmids often require prolonged incubation at high temperatures for plasmid curing, which can be harsh conditions for bacterial growth and survival, thus may lead to accumulation of adaptive genetic changes. Additional file8: Table S1 is a detailed comparison of our systems (pSNC-mTn5 and pSNC-mTn5ME) with various ts plasmid-based platforms that have been used for transposon mutagenesis in Gram-negative bacterial species, which clearly shows that our systems will be more broadly useful. In addition to their utilities in transposon mutagenesis, the IPTG-controlled conditional suicide plasmids that we developed should have many other applications, such as for allelic exchange or as curable vectors for delivering gene targeting systems, e.g., TargeTrons, λ Red, RecET, etc [37, 38]. Conclusion In this work, we have developed a number of IPTG-controlled conditional suicide plasmids that contain the broad-host-range RSF1010 origin. Using one of the constructs to deliver a hyperactive mTn5 transposon, we showed that this system can be used for efficient mutagenesis of different bacterial species. As the assay condition is mild and the host range of the RSF1010 plasmid is extremely wide, we believe that our methodology will have broad applications in microbiology research. Methods Bacterial strains and growth conditions E. coli DH10B was purchased from Invitrogen. A. baylyi (ATCC 33305) and P. aeruginosa PAO1 (ATCC BAA-47) were purchased from ATCC. Unless stated otherwise, all the strains were grown at 37°C in Luria Broth (LB) liquid media with agitation at 200 rpm or on LB plates with 1.5% agar. For sacB counter selection, we used LBNS plates (LB n o s alt: 1% Tryptone, 0.5% yeast extract and 1.5% agar) supplemented with 10% sucrose (Fisher Scientific). Appropriate antibiotics and concentrations were used to select for bacterial cells that are antibiotic resistant. E. coli DH10B: chloramphenicol (Cam; Gold Biotechnology), 25 μg/ml; kanamycin (Kan; Fisher Scientific), 50 μg/ml. A. baylyi: Cam, 10 μg/ml; Kan, 10 μg/ml. P. aeruginosa PAO1: Cam, 250 μg/ml; Kan, 500 μg/ml. Plasmid construction To construct plamsid pMMB-repF, we PCR-amplified the repF gene from pMMB208 . The PCR fragment was digested with Hin dIII and Pst I, and inserted at the corresponding sites of pMMB208, downstream of the tac promoter (P tac). To construct plasmid pMMB-repF/sacB, the sacB gene and its promoter were PCR amplified from plasmid pRE112 and inserted between the unique Sac I and Kpn I sites of pMMB-repF. To construct plasmids pMMB-repA K42A and pMMB-repA D139A, repA genes containing K42A and D139A mutations were generated in two-step PCRs from plasmid pMMB208 . The mutant genes were cloned between the Hin dIII and Pst I sites of pMMB208. Plasmid pSNC-mTn5 was constructed in multiple steps. First, plasmid pUT-mTn5Km/lacEZ was constructed from plasmid pUT-mTn5Km . It contains a lac promoter-driven hyperactive transposase gene (tnp H) that has E54K, M56A and L372P mutations . In addition, inside the mTn5 transposon, the inverted repeats flanking the kanamycin resistance marker (Kan R) were deleted . The entire mTn5 cassette of pUT-mTn5Km/lacEZ, which contains the Kan R-mTn5 transposon and P lac-tnp H, was then PCR amplified and cloned at the Xba I site of pMMB-repF/sacB, resulting in plasmid pSNC-mTn5. It has an OE and an IE at the termini of the mTn5. Plasmid pSNC-mTn5ME was derived from pSNC-mTn5 by replacing both OE and IE with MEs. Characterization of IPTG-induced plasmid loss and transposon mutagenesis To test IPTG-induced plasmid loss of pMMB-repF, pMMB-repA K42A and pMMB-repA D139A, single colonies of E. coli DH10B cells transformed with the plasmids were inoculated into 5 ml LB + Cam media and cultured at 37°C for ~ 14 h (h). After measuring OD 600, 1 ml of each culture was pelleted by centrifugation and washed with 500 μl of fresh LB to remove antibiotics. Cells were then resuspended in 1 ml LB. An aliquot was added to 5 ml LB (final OD 600 = 0.001) with and without 1 mM IPTG and cultured at 37°C for 24 h. 1 ml of the IPTG-induced samples was then pelleted, washed with 500 μl LB, and resuspended in 1 ml LB. Serial dilutions of the samples (±IPTG) were plated on LB and LB + Cam plates to evaluate plasmid loss. Plasmid retention frequencies were calculated as ratios of cfu (colony forming units) on LB + Cam plates and those on LB plates. To perform transposon mutagenesis in E. coli DH10B, single colonies of pSNC-mTn5 and pSNC-mTn5ME transformants were cultured in 5 ml LB + Cam + Kan media overnight at 37°C. Cells were then pelleted and washed as above to remove antibiotics, and an aliquot was inoculated to 5 ml LB (final OD 600 = 0.001) in a 14 ml culture tube and grown at 37°C for 24 h with and without 1 mM IPTG induction. A 1 ml aliquot of the IPTG-induced samples was then pelleted, washed with 500 μl LBNS, and resuspended in 1 ml LBNS. Serial dilutions of the samples (±IPTG) were plated on LB and LB + Cam plates to evaluate plasmid loss. The IPTG induced samples were also plated on LBNS+ 10% sucrose and LBNS+ 10% sucrose+Cam plates to estimate percentage of plasmid-retaining cells in the presence of sucrose counter selection; and LBNS+ 10% sucrose+Kan plates to select for transposition events. Plasmid retention frequencies (PRF) were calculated as the following: (1) -IPTG: (cfu on LB + Cam)/(cfu on LB); (2) + IPTG: (cfu on LB + Cam)/(cfu on LB); (3) + IPTG+Suc: (cfu on LBNS+Suc + Cam)/(cfu on LBNS+Suc). Transposon retention frequencies (TRF) were calculated as the following: +IPTG+Suc: (cfu on LBNS+Suc + Kan)/(cfu on LBNS+Suc). mTn5 (or mTn5ME) transposition frequencies were calculated as TRF+IPTG + Suc – PRF+IPTG + Suc, which essentially equals to TRF+IPTG + Suc if the background (PRF+IPTG + Suc) is low. The same protocol, except for the concentrations of antibiotics (indicated above) and IPTG (10 mM for PAO1), was followed to perform transposon mutagenesis in P. aeruginosa PAO1. Similarly, to perform transposon mutagenesis in A. baylyi 33,305, single colonies of pSNC-mTn5 and pSNC-mTn5ME transformants were cultured in 5 ml LB + Cam + Kan media overnight at 37°C. Cells were pelleted and washed as for E. coli and P. aeruginosa. Then, an aliquot was inoculated to 100 ml LB (final OD 600 = 0.001) in a baffled flask. The cultures were shaken vigorously (~ 250 rpm) at 37°C for 24 h with and without 10 mM IPTG induction. A 1 ml aliquot of the IPTG-induced samples was then pelleted, washed with 500 μl LBNS, and resuspended in 1 ml LBNS. Serial dilutions of the samples (±IPTG) were then plated on appropriate plates to evaluate plasmid loss and mTn5 (or mTn5ME) transposition as in the assays for E. coli and for P. aeruginosa. To verify plasmid loss in cells with potential transposition events, 100–150 Suc R Kan R colonies in each assay were then restreaked on LB + Kan and LB + Cam plates. In addition, presence of the transposon and the plasmid was determined by colony PCRs in a 25 μl reaction containing 25 mM TAPS-HCl (pH 9.3), 50 mM KCl, 2 mM MgCl 2, 1 mM β-mercaptoethanol, 1x GC enhancer, 0.2 mM dNTPs, 0.1 μl of Q5 polymerase (2 u/μl; NEB), 1 μl of resuspended cells, and 150 ng each of the primers (final concentration = ~ 0.5 μM; see figure legends and in Additional file9: Table S2 form oligos used). PCRs were performed using the following condition: 1x (94°C, 2 min); 25x (94°C, 30 s; 50°C, 30 s; 72°C, 1 min); 1x (72°C, 10 min); 1x (4°C, hold). Determination of transposon insertion sites Transposon insertion sites in bacterial chromosomes were determined by arbitrarily primed PCR, in which transposon junctions were amplified in two steps [5, 39]. Bacterial cells were resuspended in 10–20 μl of deionized water and 1 μl was used directly as the PCR template. In the first PCR step, the reaction was performed using a specific primer annealing to the transposon region (Tn5Km1) and a semi-degenerate primer (BDC1) that anneals to many sites on the bacterial chromosome. In the second step, aliquots of the first-round PCR products were amplified using a primer annealing to the transposon region (Tn5Km2), slightly closer to the insertion junction, and a non-degenerate primer (BDC2) that anneals to the constant region of the BDC1-derived sequence. PCRs were carried out under the conditions described above. PCR products from Step 2 were resolved in a 2% agarose gel and major products were gel-purified for sequencing to determine Tn5 insertion sites. Construction of transposon insertion libraries of P. aeruginosa PAO1 To construct an mTn5 (or mTn5ME) insertion library of P. aeruginosa PAO1, plasmid pSNC-mTn5 (or pSNC-mTn5ME) was first electroporated into the bacterial cells. Ten transformants were cultured independently in 5 ml LB + Cam + Kan media at 37°C for ~ 14 h. Equal amount of each sample (equivalent to 0.5 OD 600 × 1 ml) was then combined, pelleted, washed with 500 μl LB, and the pellet was resuspended in 1 ml LB. An aliquot of the mixture was then inoculated into 500 ml LB supplemented with 10 mM IPTG in a baffled flask (final OD 600 = 0.01) and shaken vigorously (300 rpm) at 37°C for 24 h to perform transposon mutagenesis. The cells were then pelleted by centrifugation and washed with 250 ml LBNS medium. The pellet was resuspended in 50 ml LBNS medium and serial dilutions were plated on LB, LB + Cam, LBNS+ 10% sucrose, LBNS+ 10% sucrose+Cam, and LBNS+ 10% sucrose+Kan plates to determine plasmid loss, mTn5 (mTn5ME) transposition, and total library diversity. Arbitrary PCR and DNA sequencing were then performed to determine Tn5 insertion sites. Determination of mTn5 and mTn5ME target site preferences in P. aeruginosa PAO1 To determine if mTn5 and mTn5ME have any target site preferences in P. aeruginosa PAO1, we generated sequence logos of their insertion sites in the bacterium using the WebLogo server ( In total, 40 mTn5 insertion sites and 41 mTn5ME insertion sites were used for the analysis. Abbreviations bp: Base pair Cam: Chloramphenicol IE: Inside end IPTG: Isopropyl β-D-1-thiogalactopyranoside Kan: Kanamycin LB: Luria Broth LBNS: LB no salt ME: Mosaic end OD: Optical density OE: Outside end ts : Temperature-sensitive References Phadnis SH, Berg DE. Identification of base pairs in the outside end of insertion sequence IS50 that are needed for IS50 and Tn5 transposition. Proc Natl Acad Sci U S A. 1987;84(24):9118–22. ArticleCASGoogle Scholar Reznikoff WS. Transposon Tn5. Annu Rev Genet. 2008;42:269–86. ArticleCASGoogle Scholar Goryshin IY, Reznikoff WS. Tn5 in vitro transposition. J Biol Chem. 1998;273(13):7367–74. ArticleCASGoogle Scholar de Lorenzo V, Herrero M, Jakubzik U, Timmis KN. Mini-Tn5 transposon derivatives for insertion mutagenesis, promoter probing, and chromosomal insertion of cloned DNA in gram-negative eubacteria. J Bacteriol. 1990;172(11):6568–72. ArticleCASGoogle Scholar Jacobs MA, Alwood A, Thaipisuttikul I, Spencer D, Haugen E, Ernst S, Will O, Kaul R, Raymond C, Levy R, et al. Comprehensive transposon mutant library of Pseudomonas aeruginosa. Proc Natl Acad Sci U S A. 2003;100(24):14339–44. ArticleCASGoogle Scholar Lee SA, Gallagher LA, Thongdee M, Staudinger BJ, Lippman S, Singh PK, Manoil C. General and condition-specific essential functions of Pseudomonas aeruginosa. Proc Natl Acad Sci U S A. 2015;112(16):5189–94. ArticleCASGoogle Scholar Metzger M, Bellemann P, Schwartz T, Geider K. Site-directed and transposon-mediated mutagenesis with pfd-plasmids by electroporation of Erwinia amylovora and Escherichia coli cells. Nucleic Acids Res. 1992;20(9):2265–70. ArticleCASGoogle Scholar Leahy JG, Jonesmeehan JM, Colwell RR. Transformation of Acinetobacter-Calcoaceticus Rag-1 by electroporation. Can J Microbiol. 1994;40(3):233–6. ArticleCASGoogle Scholar Shan Z, Xu H, Shi X, Yu Y, Yao H, Zhang X, Bai Y, Gao C, Saris PE, Qiao M. Identification of two new genes involved in twitching motility in Pseudomonas aeruginosa. Microbiology. 2004;150(Pt 8):2653–61. ArticleCASGoogle Scholar Sasakawa C, Yoshikawa M. A series of Tn5 variants with various drug-resistance markers and suicide vector for transposon mutagenesis. Gene. 1987;56(2–3):283–8. CASPubMedGoogle Scholar Harayama S, Tsuda M, Iino T. Tn1 insertion mutagenesis in Escherichia coli K-12 using a temperature-sensitive mutant of plasmid RP4. Mol Gen Genet. 1981;184(1):52–5. ArticleCASGoogle Scholar Le Breton Y, Mohapatra NP, Haldenwang WG. In vivo random mutagenesis of Bacillus subtilis by use of TnYLB-1, a mariner-based transposon. Appl Environ Microbiol. 2006;72(1):327–33. ArticleCASGoogle Scholar Stubbendieck RM, Straight PD. Linearmycins activate a two-component signaling system involved in bacterial competition and biofilm morphology. J Bacteriol. 2017;199(18). Maier TM, Pechous R, Casey M, Zahrt TC, Frank DW. In vivo Himar1-based transposon mutagenesis of Francisella tularensis. Appl Environ Microbiol. 2006;72(3):1878–85. ArticleCASGoogle Scholar Rholl DA, Trunck LA, Schweizer HP. In vivo Himar1 transposon mutagenesis of Burkholderia pseudomallei. Appl Environ Microbiol. 2008;74(24):7529–35. ArticleCASGoogle Scholar Pelicic V, Jackson M, Reyrat JM, Jacobs WR Jr, Gicquel B, Guilhot C. Efficient allelic exchange and transposon mutagenesis in Mycobacterium tuberculosis. Proc Natl Acad Sci U S A. 1997;94(20):10955–60. ArticleCASGoogle Scholar Choi KH, Mima T, Casart Y, Rholl D, Kumar A, Beacham IR, Schweizer HP. Genetic tools for select-agent-compliant manipulation of Burkholderia pseudomallei. Appl Environ Microbiol. 2008;74(4):1064–75. ArticleCASGoogle Scholar Morales VM, Backman A, Bagdasarian M. A series of wide-host-range low-copy-number vectors that allow direct screening for recombinants. Gene. 1991;97(1):39–47. ArticleCASGoogle Scholar Meyer R. Replication and conjugative mobilization of broad host-range IncQ plasmids. Plasmid. 2009;62(2):57–70. ArticleCASGoogle Scholar Maeser S, Scholz P, Otto S, Scherzinger E. Gene F of plasmid RSF1010 codes for a low-molecular-weight repressor protein that autoregulates expression of the repAC operon. Nucleic Acids Res. 1990;18(21):6215–22. ArticleCASGoogle Scholar Ziegelin G, Niedenzu T, Lurz R, Saenger W, Lanka E. Hexameric RSF1010 helicase RepA: the structural and functional importance of single amino acid residues. Nucleic Acids Res. 2003;31(20):5917–29. ArticleCASGoogle Scholar Edwards RA, Keller LH, Schifferli DM. Improved allelic exchange vectors and their use to analyze 987P fimbria gene expression. Gene. 1998;207(2):149–57. ArticleCASGoogle Scholar Barbe V, Vallenet D, Fonknechten N, Kreimeyer A, Oztas S, Labarre L, Cruveiller S, Robert C, Duprat S, Wincker P, et al. Unique features revealed by the genome sequence of Acinetobacter sp. ADP1, a versatile and naturally transformation competent bacterium. Nucleic Acids Res. 2004;32(19):5766–79. ArticleCASGoogle Scholar Stover CK, Pham XQ, Erwin AL, Mizoguchi SD, Warrener P, Hickey MJ, Brinkman FS, Hufnagle WO, Kowalik DJ, Lagrou M, et al. Complete genome sequence of Pseudomonas aeruginosa PAO1, an opportunistic pathogen. Nature. 2000;406(6799):959–64. ArticleCASGoogle Scholar Wang N, Ozer EA, Mandel MJ, Hauser AR. Genome-wide identification of Acinetobacter baumannii genes necessary for persistence in the lung. MBio. 2014;5(3):e01163–14. ArticleGoogle Scholar Withers TR, Yin Y. Yu HD: identification of novel genes associated with alginate production in Pseudomonas aeruginosa using mini-himar1 mariner transposon-mediated mutagenesis. J Vis Exp. 2014;85:51346. Google Scholar Lewenza S, Falsafi RK, Winsor G, Gooderham WJ, McPhee JB, Brinkman FS, Hancock RE. Construction of a mini-Tn5-luxCDABE mutant library in Pseudomonas aeruginosa PAO1: a tool for identifying differentially regulated genes. Genome Res. 2005;15(4):583–9. ArticleCASGoogle Scholar Gallagher LA, Shendure J, Manoil C. Genome-scale identification of resistance functions in Pseudomonas aeruginosa using Tn-seq. MBio. 2011;2(1):e00315–0. ArticleGoogle Scholar Wong SM, Mekalanos JJ. Genetic footprinting with mariner-based transposition in Pseudomonas aeruginosa. Proc Natl Acad Sci U S A. 2000;97(18):10191–6. ArticleCASGoogle Scholar Skurnik D, Roux D, Aschard H, Cattoir V, Yoder-Himes D, Lory S, Pier GB. A comprehensive analysis of in vitro and in vivo genetic fitness of Pseudomonas aeruginosa using high-throughput sequencing of transposon libraries. PLoS Pathog. 2013;9(9):e1003582. ArticleCASGoogle Scholar Liberati NT, Urbach JM, Miyata S, Lee DG, Drenkard E, Wu G, Villanueva J, Wei T, Ausubel FM. An ordered, nonredundant library of Pseudomonas aeruginosa strain PA14 transposon insertion mutants. Proc Natl Acad Sci U S A. 2006;103(8):2833–8. ArticleCASGoogle Scholar Seet Q, Zhang LH. Anti-activator QslA defines the quorum sensing threshold and response in Pseudomonas aeruginosa. Mol Microbiol. 2011;80(4):951–65. ArticleCASGoogle Scholar Klockgether J, Munder A, Neugebauer J, Davenport CF, Stanke F, Larbig KD, Heeb S, Schock U, Pohl TM, Wiehlmann L, et al. Genome diversity of Pseudomonas aeruginosa PAO1 laboratory strains. J Bacteriol. 2010;192(4):1113–21. ArticleCASGoogle Scholar Green B, Bouchier C, Fairhead C, Craig NL, Cormack BP. Insertion site preference of mu, Tn5, and Tn7 transposons. Mob DNA. 2012;3(1):3. ArticleCASGoogle Scholar Goryshin IY, Miller JA, Kil YV, Lanzov VA, Reznikoff WS. Tn5/IS50 target recognition. Proc Natl Acad Sci U S A. 1998;95(18):10716–21. ArticleCASGoogle Scholar Maier TM, Havig A, Casey M, Nano FE, Frank DW, Zahrt TC. Construction and characterization of a highly efficient Francisella shuttle plasmid. Appl Environ Microbiol. 2004;70(12):7511–9. ArticleCASGoogle Scholar Enyeart PJ, Mohr G, Ellington AD, Lambowitz AM. Biotechnological applications of mobile group II introns and their reverse transcriptases: gene targeting, RNA-seq, and non-coding RNA analysis. Mob DNA. 2014;5(1):2. ArticleGoogle Scholar Court DL, Sawitzke JA, Thomason LC. Genetic engineering using homologous recombination. Annu Rev Genet. 2002;36:361–88. ArticleCASGoogle Scholar Saavedra JT, Schwartzman JA, Gilmore MS. Mapping transposon insertions in bacterial genomes by arbitrarily primed PCR. Curr Protoc Mol Biol. 2017;118:15.15.11–5. Google Scholar Download references Acknowledgements We thank Drs. Mark McIntosh, David Pintel and Donald H. Burke for helpful discussions. Funding This work was supported by the University of Missouri startup fund to H.G. The funding body had no role in the design of the study and collection, analysis, and interpretation of data and in writing the manuscript. Availability of data and materials The dataset supporting the conclusion of this article are available from the corresponding author on reasonable request. Author information Author notes 1. Santa S. Naorem and Jin Han contributed equally to this work. Authors and Affiliations Department of Molecular Microbiology and Immunology, University of Missouri School of Medicine, Columbia, MO, 65212, USA Santa S. Naorem,Jin Han,Stephanie Y. Zhang,Junyi Zhang,Lindsey B. Graham,Angelou Song,Cameron V. Smith,Fariha Rashid&Huatao Guo Authors 1. Santa S. NaoremView author publications Search author on:PubMedGoogle Scholar 2. Jin HanView author publications Search author on:PubMedGoogle Scholar 3. Stephanie Y. ZhangView author publications Search author on:PubMedGoogle Scholar 4. Junyi ZhangView author publications Search author on:PubMedGoogle Scholar 5. Lindsey B. GrahamView author publications Search author on:PubMedGoogle Scholar 6. Angelou SongView author publications Search author on:PubMedGoogle Scholar 7. Cameron V. SmithView author publications Search author on:PubMedGoogle Scholar 8. Fariha RashidView author publications Search author on:PubMedGoogle Scholar 9. Huatao GuoView author publications Search author on:PubMedGoogle Scholar Contributions SSN, JH and HG conceived the study and designed the experiments; SSN, JH, SYZ, JZ, LBG, AS, CVS and FR performed the experiments; SSN and HG analyzed and wrote the manuscript. All authors have read and approved the manuscript. Corresponding author Correspondence to Huatao Guo. Ethics declarations Ethics approval and consent to participate Not applicable. Consent for publication Not applicable. Competing interests The authors declare that they have no competing interests. Publisher’s Note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. Additional files Additional file 1: Figure S1. Tn5 transposons. (a) Full-length Tn5. The full-length Tn5 contains two inverted IS50 elements at its ends. Only one of them encodes an active Tnp and an Inh (Inhibitor of Tnp). Kan R, kanamycin-resistance gene; Str R, streptomycin-resistance gene; and Ble R, bleomycin-resistance gene. (b) mTn5s. Top, an mTn5 with an OE and an IE at the termini. Bottom, an mTn5 with MEs at the ends. (c) Comparison of OE, IE and ME, with their polymorphisms highlighted in red. (TIF 4125 kb) Additional file 2: Figure S2. Confirmation of mTn5 transposition events in A. baylyi and in P. aeruginosa. (a) Colony restreaking assay of A. baylyi. 100 Suc R Kan R colonies of A. baylyi were restreaked on LB + Kan and LB + Cam plates, and all were found to be Kan R Cam S. 50 are shown here. (b) Colony PCR of 10 restreaked A. baylyi clones with the indicated primers. All were mTn5-positive and plasmid-negative. (c) Colony restreaking assay of P. aeruginosa. 100 Suc R Kan R colonies of P. aeruginosa were restreaked on LB + Kan and LB + Cam plates, and all were found to be Kan R Cam S. 50 are shown here. (d) Colony PCR of ten restreaked P. aeruginosa clones with primers indicated in the diagram. All were mTn5-positive and plasmid-negative. (TIF 6384 kb) Additional file 3: Figure S3. A transposon insertion library of P. aeruginosa PAO1 generated with pSNC-mTn5. (a) Plasmid and transposon retention frequencies of the mTn5 insertion library of P. aeruginosa PAO1. (b) Colony restreaking assay. 100 random Suc R Kan R colonies were restreaked on LB + Kan and LB + Cam plates.100/100 were found to be Kan R Cam S and 50 are shown here. (c) Colony PCR of ten restreaked clones in (b). All were found to be mTn5-positive and plasmid-negative. (d) mTn5 insertion sites of 37 mutant clones from the transposon insertion library of P. aeruginosa. Identical clones are only shown once, and their numbers are indicated in parenthesis. (TIF 1363 kb) Additional file 4: Figure S4. Confirmation of mTn5ME transposition events in E. coli DH10B. (a) Colony restreaking. 100 random Suc R Kan R colonies of E. coli were restreaked on LB + Kan and LB + Cam plates.100/100 were found to be Kan R Cam S and 50 restreaked colonies are shown here. (b) Colony PCR of ten restreaked clones in (a). All were found to be Tn5-positive and plasmid-negative. (c) Tn5 insertion sites of 13 independent DH10B clones. (TIF 7051 kb) Additional file 5: Figure S5. Confirmation of mTn5ME transposition events in A. baylyi 33,305. (a) Colony restreaking. 100 random Suc R Kan R colonies of A. baylyi were restreaked on LB + Kan and LB + Cam plates.100/100 were found to be Kan R Cam S and 50 restreaked colonies are shown here. (b) Colony PCR of ten restreaked clones in (a). All were found to be Tn5-positive and plasmid-negative. (c) Sequence analysis shows that 9/14 A. baylyi clones had different Tn5 insertion sites. (TIF 6938 kb) Additional file 6: Figure S6. Confirmation of mTn5ME transposition events in P. aeruginosa PAO1. (a) Colony restreaking. 100 random Suc R Kan R colonies of P. aeruginosa PAO1 were restreaked on LB + Kan and LB + Cam plates.100/100 were found to be Kan R Cam S and 50 restreaked colonies are shown here. (b) Colony PCR of ten restreaked clones in (a). All were found to be Tn5-positive and plasmid-negative. (c) Sequence analysis shows that 7/10 P. aeruginosa clones had different Tn5 insertion sites. (TIF 6280 kb) Additional file 7: Figure S7. Target site preferences of mTn5 and mTn5ME in P. aeruginosa. (a) Sequence logo of mTn5 insertion sites generated with WebLogo. 40 target sequences were analyzed. The 9 bp duplicated sequences adjacent to the OE are shown. There is a slight preference for certain nucleotides at several positions. (b) Sequence logo of mTn5ME insertion sites. 41 target sequences were analyzed. The 9 bp duplicated sequences adjacent to an ME are shown. It appears that mTn5ME has less nucleotide preference at the duplicated target sequence than mTn5 in Pseudomonas. (TIF 3834 kb) Additional file 8: Table S1. Comparison of conditional suicide vector-based transposon mutagenesis strategies used in Gram-negative bacteria. (TIF 7782 kb) Additional file 9: Table S2. List of primers used. (TIF 2042 kb) Rights and permissions Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License ( which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made. The Creative Commons Public Domain Dedication waiver ( applies to the data made available in this article, unless otherwise stated. Reprints and permissions About this article Cite this article Naorem, S.S., Han, J., Zhang, S.Y. et al. Efficient transposon mutagenesis mediated by an IPTG-controlled conditional suicide plasmid. BMC Microbiol18, 158 (2018). Download citation Received: 12 July 2018 Accepted: 16 October 2018 Published: 24 October 2018 DOI: Share this article Anyone you share the following link with will be able to read this content: Get shareable link Sorry, a shareable link is not currently available for this article. Copy shareable link to clipboard Provided by the Springer Nature SharedIt content-sharing initiative Keywords Transposon mutagenesis IPTG Conditional suicide plasmid Escherichia coli Pseudomonas Acinetobacter Download PDF Sections Figures References Abstract Background Results Discussion Conclusion Methods Abbreviations References Acknowledgements Author information Ethics declarations Additional files Rights and permissions About this article Advertisement Fig. 1 View in articleFull size image Fig. 2 View in articleFull size image Fig. 3 View in articleFull size image Fig. 4 View in articleFull size image Phadnis SH, Berg DE. Identification of base pairs in the outside end of insertion sequence IS50 that are needed for IS50 and Tn5 transposition. Proc Natl Acad Sci U S A. 1987;84(24):9118–22. ArticleCASGoogle Scholar Reznikoff WS. Transposon Tn5. Annu Rev Genet. 2008;42:269–86. ArticleCASGoogle Scholar Goryshin IY, Reznikoff WS. Tn5 in vitro transposition. J Biol Chem. 1998;273(13):7367–74. ArticleCASGoogle Scholar de Lorenzo V, Herrero M, Jakubzik U, Timmis KN. Mini-Tn5 transposon derivatives for insertion mutagenesis, promoter probing, and chromosomal insertion of cloned DNA in gram-negative eubacteria. J Bacteriol. 1990;172(11):6568–72. ArticleCASGoogle Scholar Jacobs MA, Alwood A, Thaipisuttikul I, Spencer D, Haugen E, Ernst S, Will O, Kaul R, Raymond C, Levy R, et al. Comprehensive transposon mutant library of Pseudomonas aeruginosa. Proc Natl Acad Sci U S A. 2003;100(24):14339–44. ArticleCASGoogle Scholar Lee SA, Gallagher LA, Thongdee M, Staudinger BJ, Lippman S, Singh PK, Manoil C. General and condition-specific essential functions of Pseudomonas aeruginosa. Proc Natl Acad Sci U S A. 2015;112(16):5189–94. ArticleCASGoogle Scholar Metzger M, Bellemann P, Schwartz T, Geider K. Site-directed and transposon-mediated mutagenesis with pfd-plasmids by electroporation of Erwinia amylovora and Escherichia coli cells. Nucleic Acids Res. 1992;20(9):2265–70. ArticleCASGoogle Scholar Leahy JG, Jonesmeehan JM, Colwell RR. Transformation of Acinetobacter-Calcoaceticus Rag-1 by electroporation. Can J Microbiol. 1994;40(3):233–6. ArticleCASGoogle Scholar Shan Z, Xu H, Shi X, Yu Y, Yao H, Zhang X, Bai Y, Gao C, Saris PE, Qiao M. Identification of two new genes involved in twitching motility in Pseudomonas aeruginosa. Microbiology. 2004;150(Pt 8):2653–61. ArticleCASGoogle Scholar Sasakawa C, Yoshikawa M. A series of Tn5 variants with various drug-resistance markers and suicide vector for transposon mutagenesis. Gene. 1987;56(2–3):283–8. CASPubMedGoogle Scholar Harayama S, Tsuda M, Iino T. Tn1 insertion mutagenesis in Escherichia coli K-12 using a temperature-sensitive mutant of plasmid RP4. Mol Gen Genet. 1981;184(1):52–5. ArticleCASGoogle Scholar Le Breton Y, Mohapatra NP, Haldenwang WG. In vivo random mutagenesis of Bacillus subtilis by use of TnYLB-1, a mariner-based transposon. Appl Environ Microbiol. 2006;72(1):327–33. ArticleCASGoogle Scholar Stubbendieck RM, Straight PD. Linearmycins activate a two-component signaling system involved in bacterial competition and biofilm morphology. J Bacteriol. 2017;199(18). Maier TM, Pechous R, Casey M, Zahrt TC, Frank DW. In vivo Himar1-based transposon mutagenesis of Francisella tularensis. Appl Environ Microbiol. 2006;72(3):1878–85. ArticleCASGoogle Scholar Rholl DA, Trunck LA, Schweizer HP. In vivo Himar1 transposon mutagenesis of Burkholderia pseudomallei. Appl Environ Microbiol. 2008;74(24):7529–35. ArticleCASGoogle Scholar Pelicic V, Jackson M, Reyrat JM, Jacobs WR Jr, Gicquel B, Guilhot C. Efficient allelic exchange and transposon mutagenesis in Mycobacterium tuberculosis. Proc Natl Acad Sci U S A. 1997;94(20):10955–60. ArticleCASGoogle Scholar Choi KH, Mima T, Casart Y, Rholl D, Kumar A, Beacham IR, Schweizer HP. Genetic tools for select-agent-compliant manipulation of Burkholderia pseudomallei. Appl Environ Microbiol. 2008;74(4):1064–75. ArticleCASGoogle Scholar Morales VM, Backman A, Bagdasarian M. A series of wide-host-range low-copy-number vectors that allow direct screening for recombinants. Gene. 1991;97(1):39–47. ArticleCASGoogle Scholar Meyer R. Replication and conjugative mobilization of broad host-range IncQ plasmids. Plasmid. 2009;62(2):57–70. ArticleCASGoogle Scholar Maeser S, Scholz P, Otto S, Scherzinger E. Gene F of plasmid RSF1010 codes for a low-molecular-weight repressor protein that autoregulates expression of the repAC operon. Nucleic Acids Res. 1990;18(21):6215–22. ArticleCASGoogle Scholar Ziegelin G, Niedenzu T, Lurz R, Saenger W, Lanka E. Hexameric RSF1010 helicase RepA: the structural and functional importance of single amino acid residues. Nucleic Acids Res. 2003;31(20):5917–29. ArticleCASGoogle Scholar Edwards RA, Keller LH, Schifferli DM. Improved allelic exchange vectors and their use to analyze 987P fimbria gene expression. Gene. 1998;207(2):149–57. ArticleCASGoogle Scholar Barbe V, Vallenet D, Fonknechten N, Kreimeyer A, Oztas S, Labarre L, Cruveiller S, Robert C, Duprat S, Wincker P, et al. Unique features revealed by the genome sequence of Acinetobacter sp. ADP1, a versatile and naturally transformation competent bacterium. Nucleic Acids Res. 2004;32(19):5766–79. ArticleCASGoogle Scholar Stover CK, Pham XQ, Erwin AL, Mizoguchi SD, Warrener P, Hickey MJ, Brinkman FS, Hufnagle WO, Kowalik DJ, Lagrou M, et al. Complete genome sequence of Pseudomonas aeruginosa PAO1, an opportunistic pathogen. Nature. 2000;406(6799):959–64. ArticleCASGoogle Scholar Wang N, Ozer EA, Mandel MJ, Hauser AR. Genome-wide identification of Acinetobacter baumannii genes necessary for persistence in the lung. MBio. 2014;5(3):e01163–14. ArticleGoogle Scholar Withers TR, Yin Y. Yu HD: identification of novel genes associated with alginate production in Pseudomonas aeruginosa using mini-himar1 mariner transposon-mediated mutagenesis. J Vis Exp. 2014;85:51346. Google Scholar Lewenza S, Falsafi RK, Winsor G, Gooderham WJ, McPhee JB, Brinkman FS, Hancock RE. Construction of a mini-Tn5-luxCDABE mutant library in Pseudomonas aeruginosa PAO1: a tool for identifying differentially regulated genes. Genome Res. 2005;15(4):583–9. ArticleCASGoogle Scholar Gallagher LA, Shendure J, Manoil C. Genome-scale identification of resistance functions in Pseudomonas aeruginosa using Tn-seq. MBio. 2011;2(1):e00315–0. ArticleGoogle Scholar Wong SM, Mekalanos JJ. Genetic footprinting with mariner-based transposition in Pseudomonas aeruginosa. Proc Natl Acad Sci U S A. 2000;97(18):10191–6. ArticleCASGoogle Scholar Skurnik D, Roux D, Aschard H, Cattoir V, Yoder-Himes D, Lory S, Pier GB. A comprehensive analysis of in vitro and in vivo genetic fitness of Pseudomonas aeruginosa using high-throughput sequencing of transposon libraries. PLoS Pathog. 2013;9(9):e1003582. ArticleCASGoogle Scholar Liberati NT, Urbach JM, Miyata S, Lee DG, Drenkard E, Wu G, Villanueva J, Wei T, Ausubel FM. An ordered, nonredundant library of Pseudomonas aeruginosa strain PA14 transposon insertion mutants. Proc Natl Acad Sci U S A. 2006;103(8):2833–8. ArticleCASGoogle Scholar Seet Q, Zhang LH. Anti-activator QslA defines the quorum sensing threshold and response in Pseudomonas aeruginosa. Mol Microbiol. 2011;80(4):951–65. ArticleCASGoogle Scholar Klockgether J, Munder A, Neugebauer J, Davenport CF, Stanke F, Larbig KD, Heeb S, Schock U, Pohl TM, Wiehlmann L, et al. Genome diversity of Pseudomonas aeruginosa PAO1 laboratory strains. J Bacteriol. 2010;192(4):1113–21. ArticleCASGoogle Scholar Green B, Bouchier C, Fairhead C, Craig NL, Cormack BP. Insertion site preference of mu, Tn5, and Tn7 transposons. Mob DNA. 2012;3(1):3. ArticleCASGoogle Scholar Goryshin IY, Miller JA, Kil YV, Lanzov VA, Reznikoff WS. Tn5/IS50 target recognition. Proc Natl Acad Sci U S A. 1998;95(18):10716–21. ArticleCASGoogle Scholar Maier TM, Havig A, Casey M, Nano FE, Frank DW, Zahrt TC. Construction and characterization of a highly efficient Francisella shuttle plasmid. Appl Environ Microbiol. 2004;70(12):7511–9. ArticleCASGoogle Scholar Enyeart PJ, Mohr G, Ellington AD, Lambowitz AM. Biotechnological applications of mobile group II introns and their reverse transcriptases: gene targeting, RNA-seq, and non-coding RNA analysis. Mob DNA. 2014;5(1):2. ArticleGoogle Scholar Court DL, Sawitzke JA, Thomason LC. Genetic engineering using homologous recombination. Annu Rev Genet. 2002;36:361–88. ArticleCASGoogle Scholar Saavedra JT, Schwartzman JA, Gilmore MS. Mapping transposon insertions in bacterial genomes by arbitrarily primed PCR. Curr Protoc Mol Biol. 2017;118:15.15.11–5. 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2083	https://asq.org/quality-resources/quality-tools?srsltid=AfmBOopMA_df8tnGsgq7deQ1JHlVheq0W6qddV0_m5sb3h-dzbd3ftJJ	Quality Tools & Templates - List of Quality Tools \| ASQ Hello, Guest Login Register Contact Us Become a Member Quincy AI Login Register Contact Us Become a Member Search Term Search \| Cart Total: Checkout Dismiss Menuclose menu ASQ - Excellence Through Quality Home Membership Individual Organizations Plan Your Career - In Beta Learning Corporate Training Specialized Credentials Learning Catalog ASQConnEx Find an Expert Certification Certification Preparation Recertification Quality Resources Conferences & Events Books & Standards Communities Quality Progress Magazine ASQ TV Find Quality Jobs About ASQ About ASQE Learn About Quality About Quality Tools & Templates Resources Related Topics Quality Tools & Templates Resources Articles Books Case Studies Jobs Training Quality Tools & Templates Related Topics Cause Analysis Tools Data Collection & Analysis Tools Decision Making Tools Idea Creation Tools New Management Planning Tools Process Analysis Tools Project Planning Tools Seven Basic Quality Tools Six Sigma Tools Home / Quality Resources / Quality Tools Quality Tools Quality Glossary Definition: Quality tools Quality tools are defined as an instrument or technique to support and improve the activities of quality management and improvement. 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Quality tools A to Z Download quality templates and Excel tools Quality tools resources Quality glossary Quality Tools on ASQTV™ Quality Tools A to Z A A3 report A3 report template (DOC) Affinity diagram Arrow diagram B Balanced scorecard Benchmarking Box and whisker plot Box and whisker plot template (XLS) Brainstorming C Cause-and-effect diagram (also called Ishikawa diagram or fishbone diagram) Cause-and-effect/Ishikawa/fishbone diagram template (XLS) Cause analysis tools Check sheet Check sheet template (XLS) Control chart (also called Shewhart chart) Control chart template (XLS) Critical incident D Data collection and analysis tools Decision matrix Design of experiments (DOE) Design of experiments (DOE) template (XLS) E Eight disciplines (8D) Evaluation and decision-making tools F Failure mode effects analysis (FMEA) Failure mode effects analysis (FMEA) template (XLS) Fishbone diagram (also called Ishikawa diagram or cause-and-effect diagram) Fishbone/Ishikawa/cause-and-effect diagram template (XLS) Five S (5S) Five whys and five hows Flowchart Flowchart template (XLS) Force field analysis G Gage repeatability and reproducibilty (GR&R) Gantt chart Gantt chart template (XLS) H Histogram Histogram template (XLS) House of Quality (HOQ) I Idea creation tools Impact effort matrix Interrelationship diagram (also called relations diagram) K Kano model M Matrix diagram Mistake-proofing (also called poka-yoke) Multivoting N Nine windows Nominal group technique P Pareto chart Pareto chart template (XLS) Plan-do-check-act (PDCA) cycle or plan-do-study-act (PDSA) cycle Problem concentration diagram Process analysis tools Process decision program chart (PDPC) Project planning and implementation tools Q Quality function deployment (QFD) Quality plans R Relations diagram (also called interrelationship diagram) Relations diagram checklist (DOC) Relations diagram template (DOC) Relations diagram instructions (PDF) S Scatter diagram (also called scatter plot or X-Y graph) Scatter diagram template (XLS) Seven basic quality tools (7BQT) Seven new management and planning tools SIPOC+CM diagram SMART matrix Spaghetti diagram Stratification Stratification template (XLS) Success and effect diagram Survey T Tree diagram V Value stream mapping Voice of the customer table (VOCT) X X-Y graph (also called scatter diagram or scatter plot) Download Quality Templates and Excel Tools Box and whisker plot (Excel) This graphical plotting tool goes beyond the traditional histogramby providing you with easy-to-read displays of variation data from multiple sources, for more effective decision making. Check sheet (Excel) Use this simple, flexible tool to collect data and analyze it with histogramand Pareto charts. Control chart (Excel) See how a control chart tracks process change over time, and create your own. Design of experiments (DOE)(Excel) This powerful tool helps you see the effect multiple input factors can have on a desired output (response), exposing important interactions that may be missed when experimenting with one factor at a time. Employee instruction sheet (Excel) Use this employee instruction sheet to capture the components of process documentation on one comprehensive worksheet. The downloadable spreadsheet includes separate tabs with instructions, a template, and an example from Heartland Regional Medical Center, St. Joseph, MO. Failure Mode and Effects Analysis (FMEA)(Excel) Use this template to evaluate the potential failure of a product or process and its effects, and then identify actions that could eliminate or reduce the occurrence of the potential failure. Fishbone (cause-and-effect) diagram (Excel) Analyze process dispersion with this simple, visual tool. The resulting diagram illustrates the main causes and subcauses leading to an effect (symptom). Flowchart(Excel) Create a graphical representation of the steps in a process to better understand it and reveal opportunities for improvement. Gantt chart (Excel) This tool can be used in process planning and control to display planned tasks and finished work in relation to time. Histogram(Excel) Analyze the frequency distribution of up to 200 data points using this simple—but powerful—histogram generating tool. Pareto chart(Excel)Use this quick and very basic tool to capture and analyze problem occurrences. Relations diagramchecklist, template, and template instructions (DOCs and PDF) Mainly used to identify logical relationships in a complex and confusing problem situation, the strength of an interrelationship diagram is its ability to visualize such relationships. The process of creating an interrelationship diagram can help groups analyze the natural links between different aspects of a complex situation. Scatter diagram(Excel)This tool shows the relationship between an input, X, and the output, Y. If a relationship exists, the input is correlated to the output. Stratification diagram (Excel) Analyze data collected from various sources to reveal patterns or relationships often missed by other data analysis techniques. By using unique symbols for each source, you can view data sets independently or in correlation to other data sets. Quality Tools Resources You can also search articles, case studies, and publicationsfor quality tool resources. Books The Quality Toolbox The ASQ Quality Improvement Pocket Guide Failure Mode And Effect Analysis Articles Fish(bone) Stories (Quality Progress) Today’s technology makes it easier than ever to communicate complex concepts more clearly, which is why older, "analog" quality methods should be digitized. The authors explore how digitizing one of the seven basic quality tools—the fishbone diagram—using mind mapping can significantly improve the tool. Tools@Work: More New Twists On Traditional Quality Tools And Techniques (Journal for Quality and Participation) Quality tools and techniques have been developed over the years with specific purposes in mind. This column describes tips for using traditional quality tools in nontraditional applications. Get the Whole Picture(Quality Progress) Two healthcare programs used systems thinking, which helps you analyze how multiple processes fit together and work in tandem, to improve service to patients, save money, and cut waste. A Disciplined Approach (Quality Progress) Nothing causes anxiety for a team quite like the release of a corrective action preventive action (CAPA) system and accompanying eight disciplines (8D) model. Reassure your team that it can easily perform 8D—all that must be done is fit routine problem solving into the eight disciplines. Solve Your FMEA Frustrations (Quality Progress) Despite failure mode and effects analysis (FMEA) being a simple concept, in practice, variations in process and competency exist. Six areas of FMEA that warrant further discussion in the quality community include: types of failure modes, cause or effect, risk assessment, types of control, process vs. design, and interaction between structures. To DMAIC Or Not To DMAIC? (Quality Progress) Define, measure, analyze, improve and control (DMAIC) is a structured problem-solving method. In this Back to Basics article, Carl F. Berardinelli explains how each phase of DMAIC builds on the previous one, with the goal of implementing long-term solutions to problems. Case Studies A Dose of DMAIC(Quality Progress) Hospital pharmacy's use ofprocess capabilitystudies as part of a Six Sigma and lean effort to improve wait times, efficiency, and patient flow. Using Control Charts In A Healthcare Setting(PDF)This teaching case study features characters, hospitals, and healthcare data that are all fictional. Upon use of the case study in classrooms or organizations, readers should be able to create a control chart and interpret its results, and identify situations that would be appropriate for control chart analysis. 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2084	https://www.sciencedirect.com/topics/agricultural-and-biological-sciences/restriction-fragment-length-polymorphism	Skip to Main content Sign in Chapters and Articles You might find these chapters and articles relevant to this topic. CROP IMPROVEMENT \| Marker Assisted Selection Restriction fragment length polymorphism; variation between individuals in DNA fragment sizes cut by specific restriction enzymes. Polymorphic sequences that result in RFLPs are used as markers on genetic linkage maps. View chapterExplore book Read full chapter URL: Reference work2003, Encyclopedia of Applied Plant SciencesP. Langridge, ... K.J. Chalmers Chapter Molecular Markers 2014, Animal BiotechnologyAvinash Marwal, ... R.K. Gaur Restriction Fragment Length Polymorphism (RFLP) Restriction fragment length polymorphism (RFLP) analysis is a method used to indirectly collect information on mutations occurring at specific short regions (restriction sites) (Botstein et al., 1980). RFLP enabled the detection of polymorphisms at the DNA sequence level. RFLP are first class genetic markers, allowing the construction of highly saturated linkage maps. Restriction Fragment Length Polymorphism (RFLP) is a technique in which organisms may be differentiated by analysis of patterns derived from cleavage of their DNA (Flow Chart 16.1). If two organisms differ in the distance between sites of cleavage of a particular restriction endonuclease, the length of the fragments produced will differ when the DNA is digested with a restriction enzyme. The similarity of the patterns generated can be used to differentiate species (and even strains) from one another. Steps Involved in RFLP Analysis 1. : The first step in DNA typing is extraction of the DNA from the sample, be it blood, saliva, semen, or some other biological sample. 2. : The purified DNA is then cut into fragments by restriction enzymes. For example take the pattern GCGC and imagine it occurs more than once in the DNA. The number of times it occurs is unique to the individual. The restriction enzyme chops the DNA in two at every place where the GCGC pattern occurs. 3. : The restriction fragments have negative charge and can be separated by a technique called gel electrophoresis, which separates the pieces of DNA based on their size. The samples of DNA that have been treated with restriction enzymes are placed in separate lanes on a slab of electrophoretic gel across which is placed an electric field. The fragments migrate towards the positive electrode, the smaller fragments moving faster than the larger fragments, thus separating the DNA samples into distinct bands. 4. : Put the nylon membrane onto the porous support, then the plastic mask (the mask must be slightly smaller than the gel), being very careful to slide the gel onto the mask. Close the apparatus and start the pump (you should be able to see if a vacuum is formed; if not, look for the problem). 5. : If the membrane to be used is being employed for the first time, then an overnight pre-hybridization is needed, otherwise 4–5 hours should be enough. Incubate in the rotating hybridization oven at 65°C. After pre-hybridization, start hybridization by adding the boiled, labeled probe to the pre-hybridization. 6. : Incubate overnight. The bands can be visualized using luminescent dyes. Applications of RFLP Analysis of RFLP variation in genomes was a vital tool in genome mapping and genetic disease analysis. If a researcher were to try to initially determine the chromosomal location of a particular disease gene, he/she would analyze the DNA of members of a family afflicted by the disease and look for RFLP alleles that show a similar pattern of inheritance as that of the disease. Once a disease gene was localized, RFLP analysis of other families would reveal who was at risk for the disease, or who was likely to be a carrier of the mutant genes. RFLP analysis was also the basis for early methods of genetic fingerprinting, useful in the identification of samples retrieved from crime scenes, in the determination of paternity, and in the characterization of genetic diversity or breeding patterns in animal populations. RFLP assay was carried out for rapid identification of three target species (Siganus canaliculatus, S. corallinus and S. javus) using mitochondrial gene regions to facilitate studies on species-specific spatio-temporal patterns of larval dispersal and population connectivity to aid fishery management (Ravago-Gotanco et al., 2010). View chapterExplore book Read full chapter URL: Book2014, Animal BiotechnologyAvinash Marwal, ... R.K. Gaur Chapter Volume 2 2022, Encyclopedia of Dairy Sciences (Third Edition)Jasna Kovac, ... Keith A. Lampel Restriction Fragment Length Polymorphisms (RFLP) PCR-restriction fragment length polymorphism is used to differentiate strains based on mutations within a gene of interest (Rasmussen, 2012). The analyses involve the amplification of a target region, followed by restriction enzyme digestion. Restriction endonucleases, such as EcoRI and Hind III, cut the amplified DNA products at certain recognition sequences, usually 4–8 bp long. Sequence polymorphisms that may be present in the amplified product, may alter the site of cleavage for the restriction enzymes. Alterations in the positions of cleavage sites result in DNA fragments of different sizes. The fragments of DNA produced by restriction analysis are then separated by size using gel electrophoresis. PCR-RFLP has been used to characterize Staphylococcus aureus strains obtained from raw milk products (Morandi et al., 2010). View chapterExplore book Read full chapter URL: Reference work2022, Encyclopedia of Dairy Sciences (Third Edition)Jasna Kovac, ... Keith A. Lampel Chapter Diagnostic tools and techniques in tree pathology 2022, Forest MicrobiologyEmad Jaber, ... Fred O. Asiegbu 3.1.5 Restriction fragment length polymorphism (RFLP) Restriction fragment length polymorphism (RFLP) is a technique that displays length differences of restriction fragments on polyacrylamide or agarose gels following hybridization (Jeffreys et al., 1985). The approach is often applied to visualize the specific variations in a sequence of double-stranded DNA. It is based on the specificity of restriction endonucleases and the presence of the restriction sites, and cleaves the DNA at those sites. A specific RFLP pattern emerges on electrophoretic separation of digested DNA, generating different lengths of cleavage fragments which are characteristic of a sequence of DNA from particular isolates. For more specific visualization of the polymorphism, the cleaved fragment pattern from the double-stranded DNA fragments is subjected to blotting, hybridization, and exposed to a labeled probe. Specific probes are usually synthetic DNA that are radioactively labeled and generated from the same genomic DNA, and the resultant pattern of marks is visualized using photographic films. PCR products are equally well suitable as starting material to prepare probes. There are also several non-radioactive options available. RFLP markers were the first markers to be developed, and they are co-dominant (both alleles in heterozygous sample will be detected) and highly locus specific. Indeed, RFLP is useful for detecting locus-specific polymorphisms (genetic variation) in populations even across species boundaries (Liu et al., 1994). RFLP can distinguish between organisms or species based on the presence of any mutation that results in alteration of a restriction site and hence a change in the pattern of fragments obtained. RFLP has previously been widely used in genotyping, DNA fingerprinting, mapping of genes (Neale and Williams, 1991), and diagnosis of genetic disorders (Jeffreys et al., 1985). As direct sequencing has become much more precise and informative, these techniques are rarely used. RFLP was utilized to distinguish two fungal pathogen species causing the oaks canker (Diplodia corticola and Diplodia quercivora) to facilitate survey and detection efforts. This was achieved by developing selective primers based on available published phylogenetic studies and sequence information retrieved from the GenBank. The PCR products were digested by the MseI enzyme to differentiate D. corticola from D. quercivora amplicons (Dreaden et al., 2014). Also, RFLP is one of the most common techniques for phytoplasma identification using phytoplasma-specific PCR primers based on sequences of the 16S-23S rRNA spacer region (Wei et al., 2007). A well-established protocol of RFLP phytoplasma detection and identification is presented by Bertaccini et al. (2018). In many hosts or pathogens, suitable probes may not yet be available; hence, the RFLP analysis would depend on creating a similarity matrix based on either the variations in band intensities or the presence or absence of bands. Thereafter, the fingerprints are grouped based on pattern similarities and the results presented as a dendrogram. A dendrogram is a tree that illustrates the similarities of the RFLP patterns included in the analysis by grouping the fingerprints sharing common bands into a subset or cluster. Sequencing of the digested PCR product is required to develop a specific probe or verify the restriction site base alteration. RFLP is a highly reproducible marker to assess polymorphism in co-dominant alleles. However, RFLP marker generation is a laborious and time-consuming process. Additionally, RFLP requires high-quality DNA and sequence knowledge prior to RFLP development (Liu et al., 1994). The PCR-RFLP procedure for phytoplasma strain diagnostic comprises the following major steps: (1) : DNA extraction: high-yield and high-quality DNA is a requisite for RFLP analysis. (2) : PCR amplification of the sequence diversity of the 16S-23S intergenic spacer (ITS) regions of ribosomal DNA (rDNA) using specific primers. (3) : Digestion of the PCR product by the restriction enzyme. (4) : Digested PCR products are then observed in an agarose gel. Band patterns are determined against a standard ladder or against other digested PCR products. View chapterExplore book Read full chapter URL: Book2022, Forest MicrobiologyEmad Jaber, ... Fred O. Asiegbu Chapter Scallops 2016, Developments in Aquaculture and Fisheries ScienceMaureen K. Krause, Elisabeth von Brand Restriction Fragment Length Polymorphism DNA digested by restriction enzymes results in fragments of varying length because of nucleotide variation within the sequence of bases recognised by the enzyme. Originally, this method required relatively large amounts of DNA because the polymorphisms were detected by probing restriction-digested DNA using southern blots. Modern restriction fragment length polymorphism (RFLP) analyses are applied to regions of DNA amplified by PCR, often for a known region, which are then restriction digested and subject to electrophoresis. Ultimately, what is detected is nucleotide variation for a restriction enzyme recognition site. The variation must lie within the site, a single RFLP typically can only distinguish between two variants and the overall level of polymorphism detected is low in comparison to other markers. One perceived strength of this approach is that these markers are co-dominant, meaning that both alleles contributing to a genotype are scorable. View chapterExplore book Read full chapter URL: Book series2016, Developments in Aquaculture and Fisheries ScienceMaureen K. Krause, Elisabeth von Brand Chapter DNA-Based Technique: Polymerase Chain Reaction (PCR) 2018, Modern Techniques for Food Authentication (Second Edition)Robert E. Levin, ... Da-Wen Sun 4.5 Restriction Fragment Length Polymorphism (RFLP) Restriction fragment length polymorphism (RFLP) is a technique in which species may be differentiated by analysis of patterns derived from cleavage of their DNA by restriction nucleases. If two species differ in the distance between sites of cleavage by a particular restriction endonuclease, the length of the fragments produced will differ when the DNA is digested with such a restriction enzyme. The differences in the patterns generated can be used to differentiate species (and even strains) from one another. DNA is first extracted from tissue and purified, followed by PCR amplification of a sequence from a gene such as that of mitochondrial 12S ribosomal RNA (mt12S rRNA). The resulting amplicon is then cleaved with one or more restriction endonucleases and the resulting segments resolved by AGE with visualization of the DNA bands with EB. Species identification based on PCR-RFLP principally depends on the target sequence chosen, the restriction endonucleases employed, and the length of the amplified fragment. PCR-RFLP is popular among researchers for identification of meat species but may not be suitable for examination of meats exposed to destructive DNA processing, because fairly large DNA fragments are needed for enzymatic restriction and amplification may be hindered by thermal DNA degradation (Wang et al., 2010). Thus, the terminal restriction fragment length polymorphism (T-RFLP) approach based on the amplification of a specific gene using fluorescently labeled primers has been developed for the detection of small restriction fragments. The method entails digesting a mixture of PCR-amplified variants of a single gene using one or more restriction enzymes and detecting the size of each of the individual resulting terminal fragments by CE with laser-induced fluorescence detection. Unlike RFLP that can be done directly on genomic DNA or on PCR products, T-RFLP is specifically designed to be used on PCR amplicons of conserved region genes and can modulate the enumeration of endophytes based on variable number of fragments generated; hence, it is more beneficial since it requires one or two labeled primers. View chapterExplore book Read full chapter URL: Book2018, Modern Techniques for Food Authentication (Second Edition)Robert E. Levin, ... Da-Wen Sun Chapter Molecular markers as tools to improve date palms 2020, Advancement in Crop Improvement TechniquesPushpa Kharb, Rakshita Singh 3.1 Restriction fragment length polymorphism (RFLP) The relative ease with which DNA molecules can currently be cloned along with the availability of a wide range of restriction endonucleases have allowed the assay of a much greater length of the plant genome for genetic markers. DNA samples that differ from one another in base sequence or have been rearranged by insertions, deletions, or inversions produce restriction fragments of different sizes on enzyme digestion, and have therefore been termed “restriction fragment length polymorphism” (RFLP) (Grodzicker et al., 1974; Bostein et al., 1980). RFLP requires both large amounts of DNA and the isolation of informative probes that yield differences between the sources of DNA. The most frequent sources of probes for RFLP analysis include cDNA clones and microsatellites. The use of RFLP in the traditional form of hybridization of labeled probes to filter-bound DNA has been replaced by PCR-based techniques. Cornicquel and Mercier (1994) performed RFLP analysis of five elite cultivars of the date palm (cvs. Barhee, Deglet Nour, Khalassa, Khadrawy, and Medjool) using offshoot leaves surrounding the shoot tips used to initiate tissue culture. Total DNA digested by EcoR1 was hybridized with cDNA probes randomly selected from a cDNA library constructed from the highly organogenic calli of cv. Boustammi Noire, and with a heterologous 1.7 kb nuclear rDNA fragment amplified during the polymerase chain reaction (PCR) of jojoba genomic DNA. Discrimination among the fire cultivars was easily made with cDNA probe 1, which was highly polymorphic. View chapterExplore book Read full chapter URL: Book2020, Advancement in Crop Improvement TechniquesPushpa Kharb, Rakshita Singh Chapter Recent advances in fish disease diagnosis, therapeutics, and vaccine development 2023, Frontiers in Aquaculture BiotechnologySudhansus Mishra, ... P. Swain 3.2.2 Restriction fragment length polymorphism/DNA fingerprinting Restriction fragment length polymorphism (RFLP) is a commonly used technique that can be used for genotyping for nearly all organisms, including plants, animals, and humans. RFLP is widely used in genetic and genomic research, such as genome mapping and gene identification . The availability of a variety of restriction endonuclease enzymes that cleave DNA at specific sites has made it possible to identify the presence of polymorphic regions in the isolated fragments. Such RFLP results owing to a variation in the number of tandem repeats (VNTR) of a short DNA segment. These VNTR sequences can uniquely specify an individual and, as such, are used in DNA fingerprinting and in paternity testing . RFLP, as a molecular marker, is specific to a single clone/restriction enzyme combination. RFLP utilizes restriction endonuclease digestion to identify DNA sequence polymorphisms in genes or DNA regions of interest. When investigating families for inheritance of an RFLP, aliquots of genomic DNA from individual family members are digested to completion with the restriction enzyme known to generate the polymorphism of interest. After size fractionation on an agarose gel, DNA is transferred to a membrane by capillary action in a high-salt buffer. The gel is first treated with NaOH to denature DNA, and after neutralization, the gel is placed between buffer-soaked filter paper and a sheet of membrane. Labeled probe is hybridized overnight to the Southern blot. The blot is washed under conditions designed to remove all nonspecifically adherent probe and exposed to X-ray film. Identified fragment sizes differ among individuals and can be traced from generation to generation . An RFLP probe is a labeled DNA sequence that hybridizes with one or more fragments of the digested DNA sample after they were separated by gel electrophoresis, thus revealing a unique blotting pattern characteristic to a specific genotype at a specific locus. Short, single- or low-copy genomic DNA or cDNA clones are typically used as RFLP probes. The RFLP probes are frequently used in genome mapping and in variation analysis (genotyping, forensics, paternity tests, hereditary disease diagnostics, etc.). Certain criteria need to be fulfilled, however, for RFLP to be useful as a genetic disease marker, such as its closeness to the disease gene. Materials and methodology for detecting RFLP are reviewed with the current emphasis on amplification procedures utilizing the PCR . The advantage of RFLP is that it is a highly specific, reliable, and repeatable method but the method is more time-consuming (10–14 days to produce results). View chapterExplore book Read full chapter URL: Book2023, Frontiers in Aquaculture BiotechnologySudhansus Mishra, ... P. Swain Chapter Modern molecular and omics tools for understanding the plant growth-promoting rhizobacteria 2019, Role of Plant Growth Promoting Microorganisms in Sustainable Agriculture and NanotechnologyRam Krishna, ... Major Singh 3.3.5 Restriction fragment-length polymorphism Restriction fragment-length polymorphism (RFLP) is based on polymorphisms of DNA and is used analyze various communities of microbes (Moyer et al., 1996). A very easy and potent technique for bacterial strain's identification of species and below species level. In RFLP techniques, digested DNA is electrophoresed and then blotted onto nitrocellulose or nylon membranes from agarose gels and hybridized with suitable probes prepared from cloned DNA fragments of linked microbes. This technique has proven to be suitable especially in DNA–DNA hybridization and enzyme electrophoresis combinations for discriminating closely related strains and in intraspecies variation determination. However, occasionally the same banding pattern does not denote a close linkage between the microbes to be compared. View chapterExplore book Read full chapter URL: Book2019, Role of Plant Growth Promoting Microorganisms in Sustainable Agriculture and NanotechnologyRam Krishna, ... Major Singh Chapter Plant Molecular Systematics 2019, Plant Systematics (Third Edition)Michael G. Simpson RFLPs AND AFLPs Restriction fragment length polymorphism, or RFLP, refers to differences between taxa in restriction sites, and therefore the lengths of fragments of DNA following cleavage with restriction enzymes. For example, Figure 14.9A illustrates, for two hypothetical species, amplified DNA lengths of 10,000 base pairs that are subjected to (“digested with”) the restriction enzyme EcoRI. Note, after a reaction with the EcoRI enzyme, that the DNA of species A is cleaved into three fragments, corresponding to two EcoRI restriction sites, whereas that of species B is cleaved into four fragments, corresponding to three EcoRI restriction sites. The relative locations of these restriction sites on the DNA can be mapped; one possibility is seen at Figure 14.9B. (Note that there are other possibilities for this map; precise mapping requires additional work.) More than one (typically two) restriction enzymes can be used. Restriction site fragment data can be coded as characters and character states in a phylogenetic analysis, based on the presence of different sized fragments. Restriction site analysis contains far less data than complete DNA sequencing, accounting only for the presence or absence of sites 6–8 base pairs long. It has the advantage, however, of surveying considerably larger segments of DNA. However, with improved and less expensive sequencing techniques, like RADSeq (above), it is much less often used than in the past. Similar to analysis of RFLPs is that of AFLPs, Amplified Fragment Length Polymorphism. This method also uses a restriction enzyme to cut DNA into numerous, smaller pieces, each of which (because of the action of the restriction enzymes) terminates in a characteristic nucleotide sequence. The numerous, cut DNA fragments are then modified by binding to each end a primer adapter. Primers are then constructed that bind to the primer adapters and amplify the DNA fragments using a polymerase chain reaction. Electrophoresis separates the amplified DNA fragments that exhibit length polymorphism (hence, AFLP), enabling the recognition of numerous genetic markers. View chapterExplore book Read full chapter URL: Book2019, Plant Systematics (Third Edition)Michael G. Simpson Related terms: Quantitative Trait Locus RAPD Restriction Enzyme Ascus Amplified Fragment Length Polymorphism Polymerase Chain Reaction Genetic Divergence Allele Single Nucleotide Polymorphism Ascospore View all Topics We use cookies that are necessary to make our site work. We may also use additional cookies to analyze, improve, and personalize our content and your digital experience. You can manage your cookie preferences using the “Cookie Settings” link. For more information, see ourCookie Policy Cookie Preference Center We use cookies which are necessary to make our site work. 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2085	https://en.wikipedia.org/wiki/Two-point_equidistant_projection	Two-point equidistant projection - 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 See also 2 References Two-point equidistant projection [x] 1 language Nederlands Edit 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 Expand all Edit interlanguage links Print/export Download as PDF Printable version In other projects Wikidata item From Wikipedia, the free encyclopedia Two-point equidistant map projection Two-point equidistant projection of Eurasia. All distances are correct from the two points (45°N, 40°E) and (30°N, 110°E). Two-point equidistant projection of the entire world with Tissot's indicatrix of deformation. The two points are Rome, Italy and Luoyang, China. The two-point equidistant projection or doubly equidistant projection is a map projection first described by Hans Maurer in 1919 and Charles Close in 1921. It is a generalization of the much simpler azimuthal equidistant projection. In this two-point form, two locus points are chosen by the mapmaker to configure the projection. Distances from the two loci to any other point on the map are correct: that is, they scale to the distances of the same points on the sphere. The two-point equidistant projection maps a family of confocal spherical conics onto two families of planar ellipses and hyperbolas. The projection has been used for all maps of the Asian continent by the National Geographic Society atlases since 1959, though its purpose in that case was to reduce distortion throughout Asia rather than to measure from the two loci. The projection sometimes appears in maps of air routes. The Chamberlin trimetric projection is a logical extension of the two-point idea to three points, but the three-point case only yields a sort of minimum error for distances from the three loci, rather than yielding correct distances. Tobler extended this idea to arbitrarily large number of loci by using automated root-mean-square minimization techniques rather than using closed-form formulae. The projection can be generalized to an ellipsoid of revolution by using geodesic distance. See also [edit] List of map projections Chamberlin trimetric projection 3D projection References [edit] ^Hans Maurer (1919). „Doppelbüschelstrahlige, orthodromische“ statt „doppelazimutale, gnomonische“ Kartenentwürfe. Doppel-mittabstandstreue Kartogramme. (Bemerkungen zu den Aufsätzen von W. Immler und H. Thorade. Ann. d. Hydr. usw 1919, S. 22 und 35.), Annalen der Hydrographie und Maritimen Meteorologie, 47 (3–4), 75–8. ^Close, Charles (1921). "Note on a Doubly-Equidistant Projection". The Geographical Journal. 57 (6): 446–448. Bibcode:1921GeogJ..57..446C. doi:10.2307/1780793. ISSN0016-7398. JSTOR1780793. ^Cox, J. F. (1946-10-01). "The doubly equidistant projection". Bulletin Géodésique. 2 (1): 74–76. Bibcode:1946BGeod..20...74C. doi:10.1007/BF02521618. ISSN0007-4632. ^Snyder, J.P. (1993). Flattening the Earth: 2,000 years of map projections. University of Chicago Press. pp.234–235. ISBN0226767469. ^"Portrait of Earth's largest continent", National Geographic Magazine, vol.116, no.6, p.751, 1959 ^Tobler, Waldo (April 1986). "Measuring the Similarity of Map Projections". Cartography and Geographic Information Science. 13 (2): 135–139. doi:10.1559/152304086783900103 – via ResearchGate. ^Karney, Charles F. F. (2011-02-07). "Geodesics on an ellipsoid of revolution". arXiv:1102.1215 [physics.geo-ph]. Charles Close (1934). “A doubly equidistant projection of the sphere.” The Geographical Journal 83(2): 144-145. Charles Close (1947). Geographical By-ways: And Some Other Geographical Essays. E. Arnold. Waldo R. Tobler (1966). “Notes on two projections.” The Cartographic Journal 3(2). 87–89. François Reignier (1957). Les systèmes de projection et leurs applications a la géographie, a la cartographie, a la navigation, a la topométrie, etc... Institut géographique national. \| show v t e Map projection \| \| History List Portal \| \| \| show By surface \| \| \| Cylindrical \| \| Mercator-conformal \| Gauss–Krüger Transverse Mercator Oblique Mercator \| \| Equal-area \| Balthasart Behrmann Gall–Peters Hobo–Dyer Lambert Smyth equal-surface Trystan Edwards \| \| Cassini Central Equirectangular Gall stereographic Gall isographic Miller Space-oblique Mercator Web Mercator \| \| \| Pseudocylindrical \| \| Equal-area \| Collignon Eckert II Eckert IV Eckert VI Equal Earth Goode homolosine Mollweide Sinusoidal Tobler hyperelliptical \| \| Kavrayskiy VII Wagner VI Winkel I and II \| \| \| Conical \| Albers Equidistant Lambert conformal \| \| Pseudoconical \| Bonne Bottomley Polyconic American Chinese Werner \| \| Azimuthal (planar) \| \| General perspective \| Gnomonic Orthographic Stereographic \| \| Equidistant Lambert equal-area \| \| \| Pseudoazimuthal \| Aitoff Hammer Wiechel Winkel tripel \| \| \| \| \| show By metric \| \| \| Conformal \| Adams hemisphere-in-a-square Gauss–Krüger Guyou hemisphere-in-a-square Lambert conformal conic Mercator Peirce quincuncial Stereographic Transverse Mercator \| \| Equal-area \| \| Bonne \| Sinusoidal Werner \| \| Bottomley \| Sinusoidal Werner \| \| Cylindrical \| Balthasart Behrmann Gall–Peters Hobo–Dyer Lambert cylindrical equal-area Smyth equal-surface Trystan Edwards \| \| Tobler hyperelliptical \| Collignon Mollweide \| \| Albers Briesemeister Eckert II Eckert IV Eckert VI Equal Earth Goode homolosine Hammer Lambert azimuthal equal-area Quadrilateralized spherical cube Strebe 1995 \| \| \| Equidistant in some aspect \| Conic Equirectangular Sinusoidal Two-point Werner \| \| Gnomonic \| Gnomonic \| \| Loxodromic \| Loximuthal Mercator \| \| Retroazimuthal (Mecca or Qibla) \| Craig Hammer Littrow \| \| \| \| \| show By construction \| \| \| Compromise \| Chamberlin trimetric Kavrayskiy VII Miller cylindrical Natural Earth Robinson Van der Grinten Wagner VI Winkel tripel \| \| Hybrid \| Goode homolosine HEALPix \| \| Perspective \| \| Planar \| Gnomonic Orthographic Stereographic \| \| Central cylindrical \| \| \| Polyhedral \| AuthaGraph Cahill Butterfly Cahill–Keyes M-shape Dymaxion ISEA Quadrilateralized spherical cube Waterman butterfly \| \| \| \| \| show See also \| \| Interruption (map projection) Latitude Longitude Tissot's indicatrix Map projection of the tri-axial ellipsoid \| \| This cartography or mapping term article is a stub. You can help Wikipedia by expanding it. v t e Retrieved from " Categories: Equidistant projections Cartography stubs Hidden categories: Articles with short description Short description is different from Wikidata All stub articles This page was last edited on 7 September 2025, at 17:06(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 Two-point equidistant projection 1 languageAdd topic
2086	https://www.lumoslearning.com/llwp/resources/math-and-ela-worksheets/probability-worksheets.html	What We Do For Schools State Test Prep Blended program State Test Prep Online Program Summer Learning Writing Program Oral Reading Fluency Program Reading Skills Improvement Program After School Program High School Program Professional Development Platform Teacher PD Online Course Smart Communication Platform For Teachers State Assessment Prep Classroom Tools State Test Teacher PD Course For Families State Test Prep Blended Program State Test Prep Online Program Summer Learning Oral Reading Fluency Online Program Oral Reading Fluency Workbooks College Readiness – SAT/ACT High School Math & ELA Curious Reader Series For Libraries For High School Students High School Math, Reading and Writing College Readiness – SAT/ACT For Professionals Online Professional Certification Exams Job Interview Practice Tool For Enterprises and Associations HR Software Online Pre-employment Assessment Platform Online Employee Training & Development Platform Enterprise Video Library Association Software Association Learning Management System Online Course Platform After-event Success Platform Self-service FAQ and Help Center Who Are We Resources Free Resources Success Stories Podcasts Program Evaluation Reports Events Login Free & Printable Probability Worksheets for 3rd to 8th Grades Table of Contents What are Probability Worksheets? Try Now: Purpose of Probability Worksheets Benefits of Probability Worksheets What is Probability? Importance of Probability in Mathematics Real-World Applications of Probability in Daily Life Different Types of Probability in Mathematics How to Find Probability? Basic Steps in Calculating Probability of Simple and Compound Events Understanding and Constructing Sample Spaces Introduction to Probability Notation Everyday Examples of Probability: Rolling Dice, Flipping Coins, Drawing Cards Probability Word Problems with Step-by-Step Solutions Examples of Single Events vs. Compound Events in Probability Role of Probability in Making Predictions and Inferences Understanding Predictions and Inferences through Probability Frequently Asked Questions (FAQs) for Probability Worksheets What are Probability Worksheets? Probability worksheets are educational resources designed to help students understand and practice the fundamental concepts of probability. They typically include exercises, word problems, and real-life scenarios that challenge learners to calculate probabilities, interpret data, and apply mathematical reasoning. These worksheets cater to a range of skill levels and grades, providing structured practice for mastering probability. Try Now: Download Free Probability Worksheet! Probability Sample Questions! Purpose of Probability Worksheets Introduce Probability Concepts: Probability worksheets familiarize students with terms like event, outcome, sample space, and likelihood. Reinforce Learning: Worksheets provide repetitive practice to strengthen comprehension of basic and advanced probability concepts. Encourage Analytical Thinking: By solving real-world problems, students develop critical thinking and problem-solving skills. Promote Hands-On Learning: Worksheets often include interactive activities like drawing probability trees, creating sample spaces, and interpreting data from charts or tables. Benefits of Probability Worksheets Enhanced Understanding of Core Concepts: Worksheets break down abstract probability principles into manageable exercises. Practical Problem-Solving Skills: Through real-world examples, students learn to apply probability to everyday situations like making predictions and evaluating risks. Preparation for Tests and Exams: Probability worksheets align with curriculum standards, ensuring students are ready for assessments. Customizable Learning: Printable worksheets allow educators and parents to tailor practice sessions to individual needs. Builds Confidence: Consistent practice through worksheets helps students master challenging topics and boosts confidence. What is Probability? Probability is a branch of mathematics that measures the likelihood or chance of an event occurring. It is a fundamental concept used to make predictions and analyze outcomes in various scenarios, ranging from simple games of chance to complex real-world situations like weather forecasting, stock market analysis, and decision-making processes.At its core, probability quantifies uncertainty by assigning a numerical value to the likelihood of an event. This value typically ranges between 0 and 1, where: 0 indicates that the event is impossible. 1 indicates that the event is certain to occur. Values between 0 and 1 represent varying degrees of likelihood. For instance: The probability of flipping a coin and landing on heads is 0.5 (50%). The probability of rolling a standard die and getting a 6 is 16\frac{1}{6}61(approximately 0.167 or 16.7%). Importance of Probability in Mathematics: Decision-Making: Helps in assessing risks and predicting outcomes in fields such as finance, healthcare, and weather forecasting. Scientific Research: Facilitates experimental design and data interpretation. Daily Applications: Guides choices, from insurance to games of chance. Real-World Applications of Probability in Daily Life Weather Forecasting: Predicting the likelihood of rain or sunshine. Sports: Analyzing team performances and game outcomes. Healthcare: Assessing the effectiveness of treatments or predicting disease outbreaks. Insurance: Determining premiums based on risk analysis. Gaming: Designing fair games and understanding odds in gambling. Different Types of Probability in Mathematics Probability can be categorized into the following types: Theoretical Probability: Based on reasoning and equally likely outcomes. Formula: P(E) = Number of favorable outcomes/Total number of possible outcomes Experimental Probability: Determined through repeated trials of an experiment. Formula: P(E) = Number of times event occurs/Total trials Subjective Probability: Based on personal judgment or experience rather than data. How to Find Probability? The process of finding probability involves the following steps: Identify the event (e.g., rolling a six on a die). Determine the total number of possible outcomes. Calculate favorable outcomes. Use the formula: P(E) = Number of favorable outcomes/Total number of possible outcomes Basic Steps in Calculating Probability of Simple and Compound Events Simple Events: Involve a single outcome, such as rolling a die once. For Example: Probability of rolling a 4: P(E) = 1/6 Compound Events: Involve two or more outcomes, such as rolling two dice. Independent Events: Outcomes do not affect each other. For Example: Rolling a die and flipping a coin. Dependent Events: Outcomes are interconnected. For Example: Drawing cards without replacement. Understanding and Constructing Sample Spaces A sample space is the set of all possible outcomes of an experiment. Examples: Flipping a coin: {H,T} Rolling a die: {1,2,3,4,5,6} Constructing a sample space helps in analyzing probabilities of events accurately. Introduction to Probability Notation P(E): Probability of event E. E^c: Complement of event E (event not happening). P(A∩B): Probability of events A and B occurring together. P(A∪B): Probability of either A or B occurring. Everyday Examples of Probability: Rolling Dice, Flipping Coins, Drawing Cards Rolling Dice: Probability of rolling an even number: 3/6 = 1/2 Flipping Coins: Probability of getting heads: 1/2 Drawing Cards: Probability of drawing a spade from a standard deck: 13/52 = 1/4 Probability Word Problems with Step-by-Step Solutions Problem: What is the probability of drawing a king from a standard deck of cards? Solution: Total cards = 52 Kings = 4 Probability = 4/52 = 1/13 Problem: A bag contains 3 red, 2 blue, and 5 green balls. What is the probability of picking a red ball? Solution: Total balls = 10 Red balls = 3 Probability = 3/10 Examples of Single Events vs. Compound Events in Probability Single Event: Flipping a coin once. Probability of heads = 1/2 Compound Event: Flipping a coin and rolling a die. Probability of heads and rolling a 5 = 1/2 × 1/6 = 1/12. Role of Probability in Making Predictions and Inferences Probability plays a crucial role in making predictions and drawing inferences in various fields. By quantifying uncertainty, probability allows us to analyze data, evaluate risks, and anticipate outcomes in both everyday situations and complex systems. Understanding Predictions and Inferences through Probability Predictions: Using probability to anticipate future events based on existing data or patterns. Example: Forecasting the likelihood of rain based on historical weather patterns. Inferences: Drawing conclusions or making decisions based on probabilistic analysis. Example: Concluding the effectiveness of a new drug based on clinical trial results. Frequently Asked Questions (FAQs) for Probability Worksheets 1. Why Are Probability Worksheets Important? Probability worksheets are essential for a variety of reasons: Foundational Understanding: They introduce students to the basics of probability, such as understanding outcomes, events, and likelihood, which are essential for more advanced studies in mathematics and statistics. Application to Real-World Scenarios: Probability is used in everyday decisions, such as determining risks, predicting outcomes, and understanding trends. Worksheets provide practical examples that connect math to real life. Skill Development: Working through probability problems helps students improve their logical reasoning, critical thinking, and data analysis skills. Test Preparation: Probability worksheets are designed to align with curriculum standards and are often featured on standardized tests, making them a valuable resource for preparation. 2. What Topics Are Covered in Probability Worksheets? Probability worksheets encompass a wide range of topics, including: Basic Probability Concepts: Understanding outcomes, sample spaces, and the probability of single events. Theoretical vs. Experimental Probability: Exploring the differences between calculated probabilities and those determined through experiments. Compound Events: Working with independent and dependent events, and calculating probabilities for multiple events. Probability Using Fractions, Decimals, and Percentages: Representing probabilities in various forms and converting between them. Tree Diagrams and Tables: Visual tools for organizing outcomes and calculating probabilities systematically. Conditional Probability: Understanding how one event affects the probability of another. Probability Distributions: Exploring concepts like uniform, binomial, and normal distributions (for higher grade levels). Word Problems: Real-world scenarios that require applying probability concepts to solve complex problems. 3. What Grade Levels Are Probability Worksheets Appropriate For? Probability worksheets are tailored to suit various grade levels: Elementary School (Grades 3-5): Worksheets focus on basic probability concepts, such as identifying possible outcomes, simple events, and using visual aids like spinners or dice. Middle School (Grades 6-8): At this stage, students delve into compound events, theoretical vs. experimental probability, and basic probability formulas. Worksheets also incorporate word problems and visual tools like tree diagrams. 4. How Do Probability Worksheets Benefit Students? Probability worksheets provide numerous benefits: Hands-On Practice: They offer a structured way for students to practice and reinforce concepts taught in class. Confidence Building: Regular practice boosts students’ confidence in tackling probability problems. Conceptual Clarity: Worksheets often include visual aids and step-by-step problems, helping students better understand abstract concepts. Skill Enhancement: They improve students’ ability to analyze data, think logically, and solve problems systematically. Exam Readiness: Worksheets align with curriculum standards, ensuring students are well-prepared for tests. 5. What Are Some Key Probability Formulas? Here are some essential probability formulas covered in worksheets: Probability of a Single Event: P(A) = Number of favorable outcomes/Total number of outcomes Probability of Complementary Events: P(Not A) = 1−P(A) Probability of Two Independent Events: P(A and B) = P(A) × P(B) Probability of Two Dependent Events: P(A and B) = P(A) × P(B\|A) Probability of Either of Two Events (Mutually Exclusive): P(A or B) = P(A) + P(B) Probability of Either of Two Events (Non-Mutually Exclusive): P(A or B) = P(A) + P(B) − P(A and B) 6. What Are Some Tips for Solving Probability Worksheet Problems? Understand the Question: Carefully read the problem and identify the key elements, such as outcomes and events. Organize Information: Use visual tools like tables or tree diagrams to arrange data systematically. Break Down Problems: Divide complex problems into smaller, manageable steps. Double-Check Calculations: Ensure accuracy, especially when using fractions, decimals, or percentages. Practice Regularly: Frequent practice improves speed and accuracy. 7. Can Probability Worksheets Be Used for Group Activities? Yes, probability worksheets are excellent for group activities. Group work fosters collaboration and helps students learn from each other. Here are some ideas: Experiments: Use worksheets alongside physical tools like dice or spinners to conduct experiments. Problem-Solving Races: Divide students into teams and have them solve worksheet problems competitively. Collaborative Projects: Assign real-world problems that require students to gather and analyze data together. 8. How Can Probability Worksheets Help with Test Preparation? Probability worksheets are highly effective for test preparation as they: Reinforce Key Concepts: Worksheets cover all essential topics likely to appear on exams. Improve Problem-Solving Speed: Timed practice helps students solve problems more efficiently. Expose Students to Various Question Types: From multiple-choice to word problems, worksheets simulate exam formats. 9. What Is the Difference Between Theoretical and Experimental Probability in Worksheets? Theoretical Probability: Based on mathematical calculations and assumes all outcomes are equally likely. Example: Probability of rolling a 4 on a fair die is 1/6. Experimental Probability: Based on actual experiments and observed outcomes. Example: Rolling a die 60 times and observing 10 fours gives a probability of 10/60 = 1/6. 11. How Do I Use Probability Worksheets to Help a Struggling Student? Start with Simple Problems: Focus on basic concepts before introducing complex problems. Use Visual Aids: Help students organize their thoughts using tree diagrams or charts. Provide Step-by-Step Guidance: Walk students through problems systematically. Offer Frequent Feedback: Correct mistakes and reinforce correct approaches. 12. What Is the Best Way to Teach Probability Concepts Using Worksheets? Begin with Hands-On Activities: Use dice, cards, or spinners to demonstrate concepts before transitioning to worksheets. Incorporate Real-Life Scenarios: Create relatable problems, such as calculating the probability of weather forecasts or sports outcomes. Encourage Group Discussions: Let students discuss and solve problems together for better understanding. 13. Are Probability Worksheets Helpful for Understanding Real-World Applications? Absolutely! Probability is used in many real-world scenarios, including: Risk Assessment: Understanding probabilities is essential in insurance, finance, and health. Sports and Games: Probability helps in analyzing strategies and predicting outcomes. Science and Research: Probability is used to analyze data and validate results. Enhance Learning with Other Free Math and ELA Resources Conjunction Worksheets Adverb Worksheets Punctuation Worksheets Noun Worksheets Context Clues Worksheets Cause and Effect Worksheets Probability Worksheets Figurative Language Worksheets Text Structure Worksheets Rounding Worksheets Graphing Worksheets Idiom Worksheets Literature Worksheets Summarizing Worksheets Telling Time Worksheets Percent Worksheets Vocabulary Worksheets Geometry Worksheets Ratio Worksheets Verb Worksheets Fraction Worksheets Place Value Worksheets Perimeter Worksheets Adjective Worksheets Prefix and Suffix Worksheets Sentence Structure Worksheets Multiple Meaning Words Worksheets Mean Median and Mode Worksheets Adages and Proverbs Worksheet Author’s Point of View Worksheets Thesis Statement Worksheet Pronouns Worksheets Drawing Conclusions Worksheets Fill in the Blank Worksheets Reading Comprehension Worksheets Multiplication Worksheets Division Worksheets Theme Worksheets Point of View Worksheets Pythagorean Theorem Worksheets Rational Numbers Worksheet Word Relationships Worksheets Parts of Speech Worksheets Main Idea Worksheets
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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 Find a point where maximum intersection of interval occurs 2 times Ask Question Asked 7 years, 4 months ago Modified7 years, 4 months ago Viewed 440 times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. I recently went to a competition held offline and they asked this question. Given n intervals, you have to find a point where there is maximum intersection of intervals. You can do this 2 times so the sum of intervals should be maximum. Eg. n=6, (1,3) (2,5) (4,8) (4,8) (3, 7), (6,8) If we greedily choose 4 first time, we'll get 4 intervals overlap. 4: (2,5) (4,8) (4,8) (3, 7). Next time we can get only 1 for a total of 5. either (1,3) or (6,8) But if we choose 2 in the first instance and 7 in the second instance we'll get 6 as answer. 2: (1,3) (2,5) 7: (4,8) (4,8) (3, 7), (6,8) Intervals can be in the range [1, 10^9] and 1 <= n <= 10^5 Note: Twice means, You can do that 2 times. You are allowed to choose 2 points such that the sum of intersections in each interval is maximum.Like, you first do that, find a set of ranges which intersect, then remove those and again run on the remaining set. We need to maximise their sum of intersection as such. algorithm data-structures Share Share a link to this question Copy linkCC BY-SA 4.0 Improve this question Follow Follow this question to receive notifications edited May 5, 2018 at 17:05 ThomasThomas asked May 5, 2018 at 16:14 ThomasThomas 11 2 2 bronze badges Add a comment\| 2 Answers 2 Sorted by: Reset to default This answer is useful 1 Save this answer. Show activity on this post. the optimal solution can be seen visually. there are 3 best solution 2+6, 2+7 and 3+8. algorithmically, the best way is to use the brute force method. start with 1 (see what number remains -> 4,5,6,7,8) check each of the remaining number and remember the score for each pair. proceed to 2 (see what number remains -> 6,7,8) check each of the remaining number and remember the score for each pair. continue all the way then get the pairs with the highest score. note that you don't need to check pairs that was already been checked. e.g. for 4, you don't need to compare with 1,2 or 3 because they have already been checked previously (1+4, 2+4, 3+4). 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 5, 2018 at 16:42 answered May 5, 2018 at 16:37 Angel KohAngel Koh 13.6k 9 9 gold badges 72 72 silver badges 98 98 bronze badges 1 Comment Add a comment Thomas ThomasOver a year ago Intervals can be in the range [1, 10^9]. Sorry for not mentioning. 2018-05-05T16:45:31.5Z+00:00 0 Reply Copy link This answer is useful 0 Save this answer. Show activity on this post. Lest visualize the data set you have: ``` ======= (6,8) ============= (3,7) ============= (4,8) ============= (4,8) ========== (2,5) ======= (1,3) ---------- 1 2 3 4 5 6 7 8 Overlaps: 1 2 3 4 4 4 4 3 ``` It becomes clear that assuming that the interval ends are closed and that all the values are integers there are several point where the maximum amount of intervals overlap: 4, 5, 6 and 7. Finding them should be very simple: ``` (pseudocode) intervals=( (6,8), (3,7), (4,8), (4,8), (2,5), (1,3) ) //Find the maximum point because it would be pointless to compute //up to 10^9. You already know that the line starts at 1, otherwise //you should find the minimum too. lastPoint=1 for each interval if interval(1)>lastPoint then lastPoint=interval(1) //Check each point maxPoint=0 matchesAtMax=0 for i=1 to lastPoint matches=0 for each interval if interval(0)<=i and interval(1)>=i then matches++ end for if matches>=matchesAtMax then maxPoint=i matchesAtMax=matches end if end for ``` This will find the last maximum point, stored at maxPoint, which should be 7 with the sample data. Now, to run it again, we should somehow exclude the intervals that include the maximum point, and check only the remaining intervals. We can do it in the same loop. ``` (pseudocode) intervals=( (6,8), (3,7), (4,8), (4,8), (2,5), (1,3) ) //Find the maximum point because it would be pointless to compute //up to 10^9. You already know that the line starts at 1, otherwise //you should find the minimum too. lastPoint=1 for each interval if interval(1)>lastPoint then lastPoint=interval(1) //Check each point (the greatest first) remainingIntervals=() //some kind of dynamic array maxPoint2=0 matchesAtMax2=0 //Lets save both max matches so we can print the final sum for i=1 to lastPoint matches=0 remainingIntervalsTemp=() for each interval if interval(0)<=i and interval(1)>=i then matches++ else push interval into remainingIntervalsTemp end if end for if matches>=matchesAtMax2 then maxPoint2=i matchesAtMax2=matches remainingIntervals=remainingIntervalsTemp end if end for //Check the points again using only the remaining intervals //(Maybe update lastPoint first?) maxPoint=0 matchesAtMax=0 for i=1 to lastPoint matches=0 allMatchedAreRemaining=true for each interval if interval(0)<=i and interval(1)>=i then //If this interval is not in the remaining, ignore the point if interval is not in remainingIntervals then allMatchedAreRemaining=false exit for end if matches++ end if end for if allMatchedAreRemaining and matches>=matchesAtMax then maxPoint=i matchesAtMax=matches end if end for print maxPoint "+" maxPoint2 "=" (matchesAtMax+matchesAtMax2) ``` The answer will be stored at maxPoint and maxPoint2. This should print 2+7=6 with the sample data. It would be wise to optimize it and maybe use functions, it was just an idea, written to be easy to read, not to run perfectly. 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 5, 2018 at 18:27 answered May 5, 2018 at 16:49 GabrielGabriel 2,180 1 1 gold badge 17 17 silver badges 20 20 bronze badges 4 Comments Add a comment Thomas ThomasOver a year ago You can do that 2 times. You are allowed to choose 2 points such that the sum of intersections in each interval is maximum.Like, you first do that find a set of ranges which intersect, then remove those and again run on the remaining set. We need to maximise their sum of intersection as such. 2018-05-05T16:51:31.513Z+00:00 0 Reply Copy link Thomas ThomasOver a year ago Will always finding maximum help us? Say first we find a point which covers maximum sets but others are disjoint. Whereas if we can choose sum other points so that their sum is maximum but individually not maximum. I will come up with a testcase to make it more clear. 2018-05-05T17:41:48.91Z+00:00 0 Reply Copy link Gabriel GabrielOver a year ago The issue this may have is that by completely excluding the matched intervals, the final solution will be 3+7, because (3,7) literally won't exist in the second loop, therefore 2 and 3 are the maximums with the remainder intervals. This can be solved by using the whole data set both times and, in the second time, discarding the points that matches any already matched intervals. I'll update the answer in a moment. 2018-05-05T18:17:50.28Z+00:00 0 Reply Copy link Thomas ThomasOver a year ago Will having a count of overlap an interval has, be of any use to us? 2018-05-06T03:40:52.667Z+00:00 0 Reply Copy link Add a comment Your Answer Thanks for contributing an answer to Stack Overflow! Please be sure to answer the question. 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2088	https://math.libretexts.org/Courses/Angelo_State_University/Mathematical_Computing_with_Python/3%3A_Interpolation_and_Curve_Fitting/3.2%3A_Polynomial_Interpolation	Published Time: 2023-01-26T16:27:55Z 3.2: Polynomial Interpolation - 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 3: Interpolation and Curve Fitting Mathematical Computing with Python { } { "3.1:_Introduction_to_Interpolation_and_Curve_Fitting" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "3.2:_Polynomial_Interpolation" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } { "00:_Front_Matter" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "01:_Introduction_to_Python" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "02:_Stochastic_Simulation" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "3:_Interpolation_and_Curve_Fitting" : "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, 09 Feb 2023 21:53:52 GMT 3.2: Polynomial Interpolation 122729 122729 Dennis Hall { } Anonymous Anonymous 2 false false [ "article:topic" ] [ "article:topic" ] Search site Search Search Go back to previous article Sign in Username Password Sign in Sign in Sign in Forgot password Contents 1. Home 2. Campus Bookshelves 3. Angelo State University 4. Mathematical Computing with Python 5. 3: Interpolation and Curve Fitting 6. 3.2: Polynomial Interpolation Expand/collapse global location Mathematical Computing with Python Front Matter 1: Introduction to Python 2: Monte-Carlo Simulation 3: Interpolation and Curve Fitting Back Matter 3.2: Polynomial Interpolation Last updated Feb 9, 2023 Save as PDF 3.1: Introduction to Interpolation and Curve Fitting Back Matter picture_as_pdf Full Book Page Downloads Full PDF Import into LMS Individual ZIP Buy Print Copy Print Book Files Buy Print CopySubmit Adoption Report Peer ReviewDonate Page ID 122729 ( \newcommand{\kernel}{\mathrm{null}\,}) Table of contents 1. 3.2.1: Lagrange Polynomial 1. THE LAGRANGE POLYNOMIAL 3.2.2: Newton interpolation NEWTON INTERPOLATION 3.2.3: Cubic Splines Given a set of data, polynomial interpolation is a method of finding a polynomial function that fits a set of data points exactly. Though there are several methods for finding this polynomial, the polynomial itself is unique, which we will prove later. 3.2.1: Lagrange Polynomial One of the most common ways to perform polynomial interpolation is by using the Lagrange polynomial. To motivate this method, we begin by constructing a polynomial that goes through 2 data points (x 0,y 0)(x 0,y 0) and x 1,y 1 x 1,y 1. We use two equations from college algebra. y−y 1=m(x−x 1)and m=y 1−y 0 x 1−x 0(3.2.1)(3.2.1)y−y 1=m(x−x 1)and m=y 1−y 0 x 1−x 0 Combining these, we end up with: y=y 1−y 0 x 1−x 0(x−x 1)+y 1.(3.2.2)(3.2.2)y=y 1−y 0 x 1−x 0(x−x 1)+y 1. Now, to derive a formula similar to that used for the Lagrange polynomial, we perform some algebra. We begin by swapping the terms y 1−y 0 y 1−y 0 and x−x 1 x−x 1: y=x−x 1 x 1−x 0(y 1−y 0)+y 1.(3.2.3)(3.2.3)y=x−x 1 x 1−x 0(y 1−y 0)+y 1. Distributing the fraction: y=x−x 1 x 1−x 0 y 1−x−x 1 x 1−x 0 y 0+y 1.(3.2.4)(3.2.4)y=x−x 1 x 1−x 0 y 1−x−x 1 x 1−x 0 y 0+y 1. Multiplying the right-most y 1 y 1 term by x 1−x 0 x 1−x 0=1 x 1−x 0 x 1−x 0=1: y=x−x 1 x 1−x 0 y 1−x−x 1 x 1−x 0 y 0+x 1−x 0 x 1−x 0 y 1.(3.2.5)(3.2.5)y=x−x 1 x 1−x 0 y 1−x−x 1 x 1−x 0 y 0+x 1−x 0 x 1−x 0 y 1. Combining to the y 1 y 1 terms: y=−x−x 1 x 1−x 0 y 0+x−x 0 x 1−x 0 y 1.(3.2.6)(3.2.6)y=−x−x 1 x 1−x 0 y 0+x−x 0 x 1−x 0 y 1. and finally flipping the denominator of the first term to get rid of the negative: y=x−x 1 x 0−x 1 y 0+x−x 0 x 1−x 0 y 1.(3.2.7)(3.2.7)y=x−x 1 x 0−x 1 y 0+x−x 0 x 1−x 0 y 1. In the above, we can observe that, when x=x 1 x=x 1, it follows that the first term cancels out with a zero on top and the second term ends up as 1⋅y 1=y 1 1⋅y 1=y 1, as desired. Similarly, if x=x 0 x=x 0, then the first term ends up as 1⋅y 0=y 0 1⋅y 0=y 0 and the second term cancels out with a zero on top, causing the entire expression to be y 0 y 0, as desired. For three data points (x 0,y 0),(x 1,y 1),and(x 2,y 2),(x 0,y 0),(x 1,y 1),and(x 2,y 2), we can derive a polynomial that behaves similarly: y=(x−x 1)(x−x 2)(x 0−x 1)(x 0−x 2)y 0+(x−x 0)(x−x 2)(x 1−x 0)(x 1−x 2)y 1+(x−x 0)(x−x 1)(x 2−x 0)(x 2−x 1)y 2.(3.2.8)(3.2.8)y=(x−x 1)(x−x 2)(x 0−x 1)(x 0−x 2)y 0+(x−x 0)(x−x 2)(x 1−x 0)(x 1−x 2)y 1+(x−x 0)(x−x 1)(x 2−x 0)(x 2−x 1)y 2. In the above, plugging in, for example, x=x 1 x=x 1 has the first and third terms cancelling out with zero and the middle term turning into 1⋅y 1 1⋅y 1, as desired. We can follow this pattern to arrive at the full Lagrange polynomial. THE LAGRANGE POLYNOMIAL Given n+1 n+1 data points (x 0,y 0),(x 1,y 1),…,(x n,y n)(x 0,y 0),(x 1,y 1),…,(x n,y n) with x 0<x 1<⋯<x n x 0<x 1<⋯<x n, the Lagrange polynomial is the n n th degree polynomial passing through each of these points and written as: y=(x−x 1)(x−x 2)⋯(x−x n)(x 0−x 1)(x 0−x 2)⋯(x 0−x n)y 0+(x−x 0)(x−x 2)⋯(x−x n)(x 1−x 0)(x 1−x 2)⋯(x 1−x n)y 1+⋯+(x−x 0)(x−x 1)⋯(x−x n−1)(x n−x 0)(x n−x 1)⋯(x n−x n−1)y n(3.2.9)(3.2.9)y=(x−x 1)(x−x 2)⋯(x−x n)(x 0−x 1)(x 0−x 2)⋯(x 0−x n)y 0+(x−x 0)(x−x 2)⋯(x−x n)(x 1−x 0)(x 1−x 2)⋯(x 1−x n)y 1+⋯+(x−x 0)(x−x 1)⋯(x−x n−1)(x n−x 0)(x n−x 1)⋯(x n−x n−1)y n Equivalently, we can use product notation: y=∑i=0 n⎛⎝⎜⎜y i∏j=0 j≠i n x−x j x i−x j⎞⎠⎟⎟(3.2.10)(3.2.10)y=∑i=0 n(y i∏j=0 j≠i n x−x j x i−x j) Note that in the above polynomial, each numerator is written such that, for x=x i x=x i, each coefficient vanishes except for the coefficient to y i y i, which evaluates to y i y i. Thus the above polynomial passes through each of the desired data points and, as can be checked, is of degree n n. Now, let's work with Python. To do this, we use function blocks for the first time. We have used functions in the past in Python, such as sin(x) or cos(x). Here, we create our own function using thedefkeyword. Below, we define f(o), where o is the dynamic variable. At the end of a function block, we "return", using thereturnkeyword, the result of the function calculation. In the code below, the variable o represents the variable x in the Lagrange Polynomial definition above. We do this since x is already used for our data points. view sourceprint? 01``import``numpy as np 02``import``matplotlib.pyplot as plot 03 04``#Data goes through the points (1,3) and (5,7) 05``x``=``[``1``,``5``] 06``y``=``[``3``,``7``] 07 08``#Set the number of data points 09``pts``=``len``(x)``-``1 10``prange``=``np.linspace(x[``0``],x[pts],``50``) 11 12``plot.plot(x,y,marker``=``'o'``, color``=``'r'``, ls``=``'',markersize``=``10``) 13 14``def``f(o): 15````z``=``((o``-``x[``1``])``/``(x[``0``]``-``x[``1``]))````y[``0``]``+``((o``-``x[``0``])``/``(x[``1``]``-``x[``0``]))````y[``1``] 16````return``z 17 18``plot.plot(prange,f(prange)) 19``plot.show() output: Now, we do the same as the above except with 3 data points. Notice how the only changes are the data point lists and the function. Below, we again use the "\" symbol to let Python know to continue on the next line. This is done to make the polynomial easier to read. view sourceprint? 01``import``numpy as np 02``import``matplotlib.pyplot as plot 03 04``#data goes through the points (4,1), (6,1) and (8,0) 05``x``=``[``4``,``6``,``8``] 06``y``=``[``53``,``32``,``60``] 07 08``n``=``len``(x)``-``1 09``prange``=``np.linspace(x[``0``],x[n],``50``) 10 11``plot.plot(x,y,marker``=``'o'``, color``=``'r'``, ls``=``'', markersize``=``10``) 12 13``def``f(o): 14````z``=``\ 15````(((o``-``x[``1``])````(o``-``x[``2``]))``/``((x[``0``]``-``x[``1``])````(x[``0``]``-``x[``2``])))````y[``0``]``+``\ 16````(((o``-``x[``0``])````(o``-``x[``2``]))``/``((x[``1``]``-``x[``0``])````(x[``1``]``-``x[``2``])))````y[``1``]``+``\ 17````(((o``-``x[``0``])````(o``-``x[``1``]))``/``((x[``2``]``-``x[``0``])````(x[``2``]``-``x[``1``])))````y[``2``] 18````return``z 19 20``plot.plot(prange,f(prange)) 21``plot.show() output: Using the product notation of Lagrange Polynomials, we can even come up with a program that accepts any number of data points. y=∑i=0 n⎛⎝⎜⎜y i∏j=0 j≠i n x−x j x i−x j⎞⎠⎟⎟(3.2.11)(3.2.11)y=∑i=0 n(y i∏j=0 j≠i n x−x j x i−x j) In the code below, i i and j j represent the i i and j j in the definition above. view sourceprint? 01``import``numpy as np 02``import``matplotlib.pyplot as plot 03 04``x``=``[``1``,``3``,``5``,``7``,``8``,``9``,``10``,``12``,``13``] 05``y``=``[``50``,``-``30``,``-``20``,``20``,``5``,``1``,``30``,``80``,``-``10``] 06 07``n``=``len``(x)``-``1 08``prange``=``np.linspace(``min``(x),``max``(x),``500``) 09 10``plot.plot(x,y,marker``=``'o'``, color``=``'r'``, ls``=``'', markersize``=``10``) 11 12``def``f(o): 13````sum``=``0 14````for``i``in``range``(n``+``1``): 15````prod``=``y[i] 16````for``j``in``range``(n``+``1``): 17````if``i!``=``j: 18````prod``=``prod````(o``-``x[j])``/``(x[i]``-``x[j]) 19````sum``=``sum``+``prod 20````return``sum 21 22``plot.plot(prange,f(prange)) 23``plot.show() output: While Lagrange polynomials are among the easiest methods to understand intuitively and are efficient for calculating a specific y(x)y(x), they fail when when attempting to either find an explicit formula y=a 0+a 1 x+⋯+a n x n y=a 0+a 1 x+⋯+a n x n or when performing incremental interpolation, that is, adding data points after the initial interpolation is performed. For incremental interpolation, we would need to complete re-perform the entire evaluation. 3.2.2: Newton interpolation Newton interpolation is an alternative to the Lagrange polynomial. Though it appears more cryptic, it allows for incremental interpolation and provides an efficient way to find an explicit formula y=a 0+a 1 x+⋯+a n x n y=a 0+a 1 x+⋯+a n x n. Newton interpolation is all about finding coefficients and then using those coefficients to calculate subsequent coefficients.Since an important part of Newton interpolation is that it can be used for incremental interpolation, let's start with a single data point and show the calculations as we add points. With one data point (x 0,y 0)(x 0,y 0), the calculation is simple: b 0=y 0(3.2.12)(3.2.12)b 0=y 0 and the polynomial is y=b 0.(3.2.13)(3.2.13)y=b 0. Let's add a new data point (x 1,y 1)(x 1,y 1). The next coefficient b 1 b 1 is usually denoted by [y 0,y 1][y 0,y 1]: b 1=[y 0,y 1]=y 1−y 0 x 1−x 0(3.2.14)(3.2.14)b 1=[y 0,y 1]=y 1−y 0 x 1−x 0 and the polynomial is y=b 0+b 1(x−x 0).(3.2.15)(3.2.15)y=b 0+b 1(x−x 0). Notice how we did not have to re-calculate the entire polynomial, only the new coefficient to (x−x 0)(x−x 0). With a third data point (x 2,y 2)(x 2,y 2), we find the coefficient b 2=[y 0,y 1,y 2]b 2=[y 0,y 1,y 2]: b 2=[y 0,y 1,y 2]=[y 1,y 2]−[y 0,y 1]x 2−x 0=y 2−y 1 x 2−x 1−y 1−y 0 x 1−x 0 x 2−x 0(3.2.16)(3.2.16)b 2=[y 0,y 1,y 2]=[y 1,y 2]−[y 0,y 1]x 2−x 0=y 2−y 1 x 2−x 1−y 1−y 0 x 1−x 0 x 2−x 0 with polynomial y=b 0+b 1(x−x 0)+b 2(x−x 0)(x−x 1).(3.2.17)(3.2.17)y=b 0+b 1(x−x 0)+b 2(x−x 0)(x−x 1). You may have noticed the recursive nature of the previous definitions. This continues for Newton interpolation in general. NEWTON INTERPOLATION Given n+1 n+1 data points (x 0,y 0),(x 1,y 1),…,(x n,y n)(x 0,y 0),(x 1,y 1),…,(x n,y n) with x 0<x 1<⋯<x n x 0<x 1<⋯<x n, the Newton interpolation polynomial is the n n th degree polynomial passing through each of these points and written as: y=b 0+b 1(x−x 0)+b 2(x−x 0)(x−x 1)+⋯+b n(x−x 0)(x−x 1)⋯(x−x n−1)(3.2.18)(3.2.18)y=b 0+b 1(x−x 0)+b 2(x−x 0)(x−x 1)+⋯+b n(x−x 0)(x−x 1)⋯(x−x n−1) where b 0 b 2 b 2 b n=y 0=[y 0,y 1]=y 1−y 0 x 1−x 0=[y 0,y 1,y 2]=y 2−y 1 x 2−x 1−y 1−y 0 x 1−x 0 x 2−x 0⋮=[y 0,y 2,…,y n]=[y 1,y 2,…,y n]−[y 0,y 1,…,y n−1]x n−x 0(3.2.19)(3.2.20)(3.2.21)(3.2.22)(3.2.23)(3.2.19)b 0=y 0(3.2.20)b 2=[y 0,y 1]=y 1−y 0 x 1−x 0(3.2.21)b 2=[y 0,y 1,y 2]=y 2−y 1 x 2−x 1−y 1−y 0 x 1−x 0 x 2−x 0(3.2.22)⋮(3.2.23)b n=[y 0,y 2,…,y n]=[y 1,y 2,…,y n]−[y 0,y 1,…,y n−1]x n−x 0 Let's begin by finding a polynomial with three data points using Newton's Method. Example:Use Newton's method to calculating a polynomial through the points (1,3),(5,7)(1,3),(5,7) and (8,0)(8,0). For this example, we will use a "tableau" to help organize our data. x 0 x 0 y 0 y 0 x 1 x 1 y 1 y 1[y 1,y 0][y 1,y 0] x 2 x 2 y 2 y 2[y 2,y 1][y 2,y 1][y 0,y 1,y 2][y 0,y 1,y 2] We start by filling in our data points: 1 3 5 7[y 1,y 0][y 1,y 0] 8 0[y 2,y 1][y 2,y 1][y 0,y 1,y 2][y 0,y 1,y 2] Then we calculate: [y 1,y 0]=7−3 5−1=1(3.2.24)(3.2.24)[y 1,y 0]=7−3 5−1=1 [y 2,y 1]=0−7 8−5=−7 3(3.2.25)(3.2.25)[y 2,y 1]=0−7 8−5=−7 3 and fill in this data: 1 3 5 7 1 8 0−7 3−7 3[y 0,y 1,y 2][y 0,y 1,y 2] Now, we calculate the remaining item, using the previously-calculated terms: [y 0,y 1,y 2]=[y 1,y 2]−[y 0,y 1]x 2−x 0=−7 3−1 8−1=−10 21(3.2.26)(3.2.26)[y 0,y 1,y 2]=[y 1,y 2]−[y 0,y 1]x 2−x 0=−7 3−1 8−1=−10 21 1 3 5 7 1 8 0−7 3−7 3−10 21−10 21 Note, now, that b 0,b 1,b 0,b 1, and b 2 b 2 are written on the top diagonal. Thus our final polynomial is: y=b 0+b 1(x−x 0)+b 2(x−x 0)(x−x 1)=3+1(x−1)−10 21(x−1)(x−5)(3.2.27)(3.2.27)y=b 0+b 1(x−x 0)+b 2(x−x 0)(x−x 1)=3+1(x−1)−10 21(x−1)(x−5) Now, let's program this for three data points to check our work! view sourceprint? 01``import``numpy as np 02``import``matplotlib.pyplot as plot 03 04``x``=``[``1``,``5``,``8``] 05``y``=``[``3``,``7``,``0``] 06 07``n``=``len``(x)``-``1 08``prange``=``np.linspace(``min``(x),``max``(x),``500``) 09 10``plot.plot(x,y,marker``=``'o'``, color``=``'r'``, ls``=``'', markersize``=``10``) 11 12``def``f(o): 13````b0``=``y[``0``] 14````b1``=``(y[``1``]``-``y[``0``])``/``(x[``1``]``-``x[``0``]) 15````b1p``=``(y[``2``]``-``y[``1``])``/``(x[``2``]``-``x[``1``]) 16````b2``=``(b1p``-``b1)``/``(x[``2``]``-``x[``0``]) 17````poly``=``b0``+``b1````(o``-``x[``0``])``+``b2````(o``-``x[``0``])````(o``-``x[``1``]) 18````return``poly 19 20``plot.plot(prange,f(prange)) 21``plot.show() output: A general program can be made by using a recursive function. view sourceprint? 01``import``numpy as nm 02``import``matplotlib.pyplot as plot 03 04``x``=``[``-``5``,``-``4``,``-``3``,``-``2``,``-``1``,``0``,``1``,``2``,``3``,``4``,``5``] 05``y``=``[``0``,``0``,``0.1``,``0.3``,``0.7``,``1``,``0.7``,``0.3``,``0.1``,``0``,``0``] 06 07``plot.plot(x,y,marker``=``'o'``, color``=``'r'``, ls``=``'',markersize``=``10``) 08 09``def``grad(a,b): 10````if``a``=``=``0``:``return``y[b] 11````return``(grad(a``-``1``,b)``-``grad(a``-``1``,a``-``1``))``/``(x[b]``-``x[a``-``1``]) `12``` 13 14``def``f(o): 15````yres``=``0 16````for``p``in``range``(``len``(x)): 17````prodres``=``grad(p,p) 18````for``q``in``range``(p): 19````prodres````=``(o``-``x[q]) 20````yres``+``=``prodres 21````return``yres 22``prange``=``nm.linspace(x[``0``],x[``-``1``],``200``) 23``plot.plot(prange,f(prange)) 24``plot.show() output: Note that above, we have something strange happening near the end points. This is known asRunge's phenomenonand mainly occurs at the endges of an interval when using polynomial interpolation with polynomials of high degree over a set of points that are equally spaced. Because of Runge's phenomenon, it is often the case that going to higher degrees does not always improve accuracy. To combat this, we can use a method such as cubic splines, which we discuss here. 3.2.3: Cubic Splines Asplineis a function defined piecewise by polynomials. Instead of having a single polynomial that goes through each data point, as we have done so far, we instead define several polynomials between each of the points. While defining several polynomials might take more work, it is usually preferred since it gives similar results while avoiding Runge's phenomenon. A linear spline is created by simply drawing lines between each of the data points. Using our data points from earlier, we can create a reasonable approximation of our underlying function. While simplistic, this approximation is likely better than that generated when using the same data points and Newton's method. This is due to the absence of Runge's phenomenon. To make the picture look even better, we can replace the lines by cubic functions. When doing so, we will end up with a number of cubic functions equal to one less than the number of data points - that is, one for every interval between the data points. We call this function s(x)s(x), which can be defined as follows. s(x)=\left{\begin{array}{lr}s_0(x)=a_0x^3+b_0x^2+c_0x+d_0&\text{if}x_0\leq x\leq x_1\s_1(x)=a_1x^3+b_1x^2+c_1x+d_1&\text{if}x_1\leq x\leq x_2\\vdots\s_{n-1}(x)=a_{n-1}x^3+b_{n-1}x^2+c_{n-1}x+d_{n-1}&\text{if}x_{n-1}\leq x\leq x_n\\end{array}\right s(x)=\left{\begin{array}{lr}s_0(x)=a_0x^3+b_0x^2+c_0x+d_0&\text{if}x_0\leq x\leq x_1\s_1(x)=a_1x^3+b_1x^2+c_1x+d_1&\text{if}x_1\leq x\leq x_2\\vdots\s_{n-1}(x)=a_{n-1}x^3+b_{n-1}x^2+c_{n-1}x+d_{n-1}&\text{if}x_{n-1}\leq x\leq x_n\\end{array}\right 3.2: Polynomial Interpolation is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts. Back to top 3.1: Introduction to Interpolation and Curve Fitting Back Matter Was this article helpful? 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2089	https://www.reddit.com/r/German/comments/cc9dhm/b1_self_study_guide/	B1 Self Study Guide : r/German Skip to main contentB1 Self Study Guide : r/German Open menu Open navigationGo to Reddit Home r/German A chip A close button Get App Get the Reddit app Log InLog in to Reddit Expand user menu Open settings menu Go to German r/German r/German /r/German is a community focused on discussion related to learning the German language. It is also a place to discuss the language at large. New visitors, please read the FAQ: /r/German/wiki/faq 410K Members 67 Online •6 yr. ago Nerrroo B1 Self Study Guide Resource I've seen a lot of requests and questions regarding the B1 level, how to reach it, B1 Exams (Goethe Zertifikat B1), etc. and I was wondering if there's any way you can reach B1 on your own. So as soon as I finished my Goethe A2 course, I started my B1 journey. I researched, I asked my teachers about the books/resources before I used them and I studied every day, at least an hour a day and sometimes even three hours/day. So I thought I'd share some of my experiences NOTE: Be aware that some of the things mentioned here might or might not work for you; this is just my personal self-study experience These are the resources that I've used and also how I've used them: 1) BOOKS Get some good books. I can't speak for all the books out there, probably there are better/worse ones, but these are the ones I've used: Menschen B1 Kursbuch, Hueber Menschen B1 Arbeitsbuch (mit CD), Hueber Aspekte Neu B1+ (Teil 1) Lehr- und Arbeitsbuch, Klett I picked these ones because I got them at a very good price and also a friend of mine who was B1 at the time, told me that that's what they use in class at Goethe Institut. I worked through the books every day, doing one chapter from the KB and then one chapter the next day from the AB, so that I could apply what I learned from the previous day. Don't move on to the next chapter if there's something you didn't understand (especially Grammar), because you will encounter it again and again, and you'll soon find yourself unable to understand anything. USE THE CDs. Very important: once you finished a chapter, or a module, go back and do all the listening exercises (especially those in the Arbeitsbuch) because they use the vocab and grammar learned in those chapters. Note: Unfortunately, the Arbeitsbuch has no answers at the back of the book, so if you're not sure about how you did the questions, here are the links to the official answers from Hueber Verlag Menschen Loesungsschluessel Arbeitsbuch B1.1 Menschen Loesungsschluessel Arbeitsbuch B1.2 2) GRAMMAR Grammar is essential and B1 is where it becomes essential. And you MUST know it. (Personal experience: I thought I had a well-consolidated vocabulary because I was able to understand a lot of words, but once I started to encounter structures like "je...desto or entweder...oder" things might start to get a bit tricky and no matter how much vocabulary you know, it'll be really hard to tell the meaning of a sentence. So Learn the GRAMMAR. German level B1 has a lot of grammar topics. In every chapter at least 3–4 grammar topics are present. List of Grammar topics in B1 is as follows. VERB Praeteritum formen: Ich suchte, du suchtest, etc. using “te” instead of partizip II. Vergangenes berichten vergangenheit, vorvergangenheit, plusquamperfekt Futur I Bildung des passiv; werden+partizip II, wurde+partizip II, sein+partizip II Passiv mit Modalverb: Modalverb+Partizip II+werden im infinitiv Konjuktiv II der Modalverben Irreale Bedingungssaetze mit Konjuktiv II Verb mit Pareposition nicht/kein+brauchen+zu, nur+brauchen+zu SUBSTANTIVE Genetiv: “des” n-Deklination Adjektive als Substantive ADJEKTIVE Komparitiv und Superlativ vor Substativen Adjektiv nach dem bestimmten und unbestimmten Artikel: Genitiv Adjektivdeklination ohne Artikel Partizip als Adjektive PRONOMEN Reflexivpronomen im Akkusativ und Dativ Pronomen mit Praeposition und Pronominaladverbien Artikelwoerter als Pronomen Reflexivpronomen was und wo PRAEPOSITIONEN Wegen und Trotz Innerhalb und Ausserhalb Temporale praepositionen Vor, Nach, waehrend aus+material WORTSTELLUNG Stellung von nicht im satz Temporale nebensaetze: bevor, nachdem, seit/seitdem, waehrend, bis Folgen ausdruecken: deshalb, darum, deswegen sodass, so….dass Gruende und Gegenguende ausdruecken: weil/da, obwohl Infinitiv mit zu Relativsaetze: Relativpronomen im Dativ Relativsaetze: Relativpronomen mit Praeposition Verben mit praeposition und Nebensatz Zweiteilige Konnektoren: Sowohl, als auch nicht nur, sondern auch entweder, oder weder, noch zwar, aber einerseits, andererseits Saetze mit je….desto… I recommend using this website: Longua All the Grammar Topics are listed there, from Adjective endings to Irregular Verbs and it also has downloadable PDF files with all essential Grammar (A1-C1) I also suggest getting this book: Deutsch üben Taschentrainer - Fit in Grammatik B1 (Hueber) It has good and really simple to understand questions and exercises and answers as well Note: Some of these (if not all of them) are explained in Textbooks, but some textbooks just go over them briefly. That's why I suggest if you didn't understand something or if it isn't very well explained in the book, look it up 3) VOCABULARY (+Listening) Vocabulary is probably the most important part of learning German, and especially B1 Vocab is the starting point for all the daily and basic conversations that you might have in a German-speaking country. Official Goethe B1 Wortliste: The official B1 vocabulary issued by the Goethe Institut Memrise: It's a great tool honestly, use it every day as much as you can. (I've linked the B1 deck for the exact same B1 Goethe Word List) Search for B1 decks and relevant vocab Duolingo - You can still use Duolingo, but at some point, it's not effective anymore (I use it only for vocabulary) Read as much as you can - Fiction books, magazines, newspapers, articles, etc. This way you'll start encountering the words you've been learning and also understanding them because if you see them in a context they're easier to remember Arte.tv ( - One of the best resources out there if you enjoy watching documentaries, movies and pretty much anything. Arte is a free, and on-demand European (French & German) culture TV channel and most of the content is in German and has German subtitles DW Deutsch - Lots of resources for all levels: For Listening: German Radio - Deutschland.fm (I recommend these: SWR2, NDR1, NDR Info, Bayern 2, WDR 5) ARD Audiothek App (Radio broadcasts/podcasts in German) 50 States - Through the USA with Dirk Rohrbach - Dirk Rohrbach is the first European that goes on America's longest river, Missouri. An amazing & very exciting podcast. Easy to understand as well (Also available in ARD Audiothek App) Herr Professor Podcast (I just found these podcasts, but they are very useful and easy to understand) Watch Youtube Channels in German - there are so many good channels out there and with good content. Here are a few examples: Easy German 24h Deutsch Y-Kollektiv (interesting videos with subtitles in German) WDR Doku (documentaries) - probably a bit more advanced, but good content and clear spoken German Tagesschau in 100 Sekunden - News in 100 seconds, but you can also watch the whole thing (~15 min) Do this before you go to bed if you can every night 4) SPEAKING This is where it gets tricky. If you're like me (you don't live in a German-speaking country/don't have any German friends) then you know what are the odds of running into a Native/Fluent speaker of German. It gets even trickier if you live in an English speaking country because the odds are even thinner. But there are still a couple of things that you might try in order to practice speaking. Apps - it's true, it's not the same thing as speaking with someone in person, but at least it challenges and forces your brain to come up with stuff that you might say in a daily conversation. Tandem - This is a really nice app, where you can connect with people that are native speakers of different languages, and also people that are interested in learning your language. HelloTalk - More people have recommended me this one, but for some reason, I used a lot more Tandem. I thought I'd just link them both here, so you can pick whichever you like Some of you might have friends or at least know people that can speak German on a decent level. Talk to them and tell them to correct you. If not, don't worry, you'll get the chance at some point. Apps are totally fine for B1. Bonus: This is something you don't hear a lot of people do, but I did it a couple of times and it's helpful. Try having mental conversations with yourself and see how long you can keep talking. For example, you can say "Wie war dein Tag?" and go on from there. And you'll see that if you avoid answers like "Gut" or "Toll", it actually gets pretty hard to say everything that you did in one day because you don't have the vocab yet. It's a bit strange, but helpful because it shows you instantly what you know and what you don't know. Other resources for B1: (Goethe Zertifikat B1 Practice) (B1 Exam guide) (More listening) (More German Radio) (Conjugates verbs - Very useful!) (More B1 Practice) I hope this helps! :) Read more Share Share etrade_official•Promoted Focus on your future. Open an account today and get up to $1,000. Sign Up etrade.com Sort by: Best Open comment sort options Best Top New Controversial Old Q&A melina_gamgee •6y ago Definitely ask a mod to include a link to this in the sticky post, or to sticky it! I feel like such a well researched guide would be very helpful for a lot of people. Reply reply Share Share 5 more replies 5 more replies More replies Bhima •6y ago Hi folks. I have copied all this info into the wiki under the heading "Self Study Guides". The wiki can be found here: r/German/wiki/index If anyone has more info to add, contributions are always welcome! Reply reply Share Share no_memes_no_me •6y ago I also have mental convos with myself in German! It does help A LOT Reply reply Share Share B-vinread •6y ago Thank you so much! Would you say that the B1 wortliste is made up of "only B1 words" or "ALL words you must know when you reach B1" ? Reply reply Share Share Nerrroo •6y ago There are also the A1 and A2 words included in there, so the answer would be "ALL words you must know when you reach B1" Reply reply Share Share More replies cstrovn •6y ago Vielen Danke! I just started but like you I've been studying at least 1h a day completely on my own, I know this is going to be extremely helpful and (hopefully) soon! I saved the post for future reference pls never delete it ;) Reply reply Share Share 1 more reply 1 more reply More replies [deleted] •6y ago Thank you, wonderful person! These links are really useful! Reply reply Share Share 1 more reply 1 more reply More replies JeremyDataAnnotation•Promoted Coders: Work wherever you want, whenever you want at DataAnnotation. Apply Now dataannotation.tech melismelismel •6y ago Thank you! Reply reply Share Share ExplosiveGrotto •6y ago Great info, Thanks! Reply reply Share Share aloysiussnuffleupagu •6y ago Thanks for this very detailed post! Reply reply Share Share 1 more reply 1 more reply More replies New to Reddit? Create your account and connect with a world of communities. Continue with Email Continue With Phone Number By continuing, you agree to ourUser Agreementand acknowledge that you understand thePrivacy Policy. More posts you may like I passed my B1 exam after a year of self-study! r/German•1 mo. ago r/German /r/German is a community focused on discussion related to learning the German language. It is also a place to discuss the language at large. New visitors, please read the FAQ: /r/German/wiki/faq 410K Members 67 Online ### I passed my B1 exam after a year of self-study! 758 upvotes ·65 comments German B1 Grammar Guide r/German•3 yr. ago r/German /r/German is a community focused on discussion related to learning the German language. It is also a place to discuss the language at large. New visitors, please read the FAQ: /r/German/wiki/faq 410K Members 67 Online ### German B1 Grammar Guide 119 upvotes ·46 comments How to study B2 by yourself r/German•4 mo. ago r/German /r/German is a community focused on discussion related to learning the German language. It is also a place to discuss the language at large. New visitors, please read the FAQ: /r/German/wiki/faq 410K Members 67 Online ### How to study B2 by yourself 22 upvotes ·34 comments Promoted What's the best book for self studying B1? r/German•4 yr. ago r/German /r/German is a community focused on discussion related to learning the German language. It is also a place to discuss the language at large. New visitors, please read the FAQ: /r/German/wiki/faq 410K Members 67 Online ### What's the best book for self studying B1? 3 upvotes ·3 comments Self learners from B1 up r/German•1 mo. ago r/German /r/German is a community focused on discussion related to learning the German language. It is also a place to discuss the language at large. New visitors, please read the FAQ: /r/German/wiki/faq 410K Members 67 Online ### Self learners from B1 up 30 upvotes ·28 comments What does it feel like to actually be at a B1 level? r/German•1 yr. ago r/German /r/German is a community focused on discussion related to learning the German language. It is also a place to discuss the language at large. New visitors, please read the FAQ: /r/German/wiki/faq 410K Members 67 Online ### What does it feel like to actually be at a B1 level? 97 upvotes ·61 comments Easiest B1 test for citizenship? r/German•1 yr. ago r/German /r/German is a community focused on discussion related to learning the German language. It is also a place to discuss the language at large. New visitors, please read the FAQ: /r/German/wiki/faq 410K Members 67 Online ### Easiest B1 test for citizenship? 9 upvotes ·31 comments B1 in 6 months possible? r/German•5 yr. ago r/German /r/German is a community focused on discussion related to learning the German language. It is also a place to discuss the language at large. New visitors, please read the FAQ: /r/German/wiki/faq 410K Members 67 Online ### B1 in 6 months possible? 16 upvotes ·39 comments Promoted So I just took my Goethe B1 German Exam 😵‍💫 r/German•16 days ago r/German /r/German is a community focused on discussion related to learning the German language. It is also a place to discuss the language at large. New visitors, please read the FAQ: /r/German/wiki/faq 410K Members 67 Online ### So I just took my Goethe B1 German Exam 😵‍💫 337 upvotes ·71 comments 😩🙌🏾🥳🥹 I passed my Goethe B1 Exam r/German•5 days ago r/German /r/German is a community focused on discussion related to learning the German language. It is also a place to discuss the language at large. New visitors, please read the FAQ: /r/German/wiki/faq 410K Members 67 Online ### 😩🙌🏾🥳🥹 I passed my Goethe B1 Exam 361 upvotes ·52 comments Self-Studying German r/German•3 mo. ago r/German /r/German is a community focused on discussion related to learning the German language. It is also a place to discuss the language at large. New visitors, please read the FAQ: /r/German/wiki/faq 410K Members 67 Online ### Self-Studying German 57 upvotes ·34 comments Is there like a free past paper bank for Goethe A2/B1/B2 exams? r/German•4 yr. ago r/German /r/German is a community focused on discussion related to learning the German language. It is also a place to discuss the language at large. New visitors, please read the FAQ: /r/German/wiki/faq 410K Members 67 Online ### Is there like a free past paper bank for Goethe A2/B1/B2 exams? 20 upvotes ·15 comments How can I prepare for the Goethe B1 Exam r/German•2 yr. ago r/German /r/German is a community focused on discussion related to learning the German language. It is also a place to discuss the language at large. New visitors, please read the FAQ: /r/German/wiki/faq 410K Members 67 Online ### How can I prepare for the Goethe B1 Exam 2 upvotes ·1 comment B1 yet struggle to from basic sentences on the spot. r/German•2 mo. ago r/German /r/German is a community focused on discussion related to learning the German language. It is also a place to discuss the language at large. New visitors, please read the FAQ: /r/German/wiki/faq 410K Members 67 Online ### B1 yet struggle to from basic sentences on the spot. 80 upvotes ·32 comments Got B2 test next month, a bit lost, confused, and... scared r/German•2 mo. ago r/German /r/German is a community focused on discussion related to learning the German language. It is also a place to discuss the language at large. New visitors, please read the FAQ: /r/German/wiki/faq 410K Members 67 Online ### Got B2 test next month, a bit lost, confused, and... scared 31 upvotes ·12 comments How to pass B1 in two months r/German•9 days ago r/German /r/German is a community focused on discussion related to learning the German language. It is also a place to discuss the language at large. New visitors, please read the FAQ: /r/German/wiki/faq 410K Members 67 Online ### How to pass B1 in two months 13 upvotes ·16 comments Need help to get to B1 faster without being full time student r/German•6 yr. ago r/German /r/German is a community focused on discussion related to learning the German language. It is also a place to discuss the language at large. New visitors, please read the FAQ: /r/German/wiki/faq 410K Members 67 Online ### Need help to get to B1 faster without being full time student 9 upvotes ·6 comments Feeling like I'm studying for nothing r/German•6 mo. ago r/German /r/German is a community focused on discussion related to learning the German language. It is also a place to discuss the language at large. New visitors, please read the FAQ: /r/German/wiki/faq 410K Members 67 Online ### Feeling like I'm studying for nothing 74 upvotes ·85 comments Just get a good coursebook??? r/German•3 mo. ago r/German /r/German is a community focused on discussion related to learning the German language. It is also a place to discuss the language at large. New visitors, please read the FAQ: /r/German/wiki/faq 410K Members 67 Online ### Just get a good coursebook??? 83 upvotes ·43 comments Any recommendations for online sites that has german exercises? r/German•4 yr. ago r/German /r/German is a community focused on discussion related to learning the German language. It is also a place to discuss the language at large. New visitors, please read the FAQ: /r/German/wiki/faq 410K Members 67 Online ### Any recommendations for online sites that has german exercises? 5 upvotes ·6 comments Passed B1 Goethe with 2 weeks of studying r/German•1 mo. ago r/German /r/German is a community focused on discussion related to learning the German language. It is also a place to discuss the language at large. New visitors, please read the FAQ: /r/German/wiki/faq 410K Members 67 Online ### Passed B1 Goethe with 2 weeks of studying 6 comments B1 exam experience tips r/German•1 yr. ago r/German /r/German is a community focused on discussion related to learning the German language. It is also a place to discuss the language at large. New visitors, please read the FAQ: /r/German/wiki/faq 410K Members 67 Online ### B1 exam experience tips 41 upvotes ·12 comments Does anyone have a list (online) of the grammar topics for each level A1, A2, B1.... in German? r/German•1 yr. ago r/German /r/German is a community focused on discussion related to learning the German language. It is also a place to discuss the language at large. New visitors, please read the FAQ: /r/German/wiki/faq 410K Members 67 Online ### Does anyone have a list (online) of the grammar topics for each level A1, A2, B1.... in German? 4 upvotes ·1 comment This is the single most definite .pdf document I have found on german irregular verbs. r/German•3 yr. ago r/German /r/German is a community focused on discussion related to learning the German language. It is also a place to discuss the language at large. New visitors, please read the FAQ: /r/German/wiki/faq 410K Members 67 Online ### This is the single most definite .pdf document I have found on german irregular verbs. 266 upvotes ·13 comments Can an average person get to B1 level German in about 1 year? r/German•4 yr. ago r/German /r/German is a community focused on discussion related to learning the German language. It is also a place to discuss the language at large. New visitors, please read the FAQ: /r/German/wiki/faq 410K Members 67 Online ### Can an average person get to B1 level German in about 1 year? 14 upvotes ·27 comments Related discussion Best GRE Prep Course Top 1%Rank by size Public Anyone can view, post, and comment to this community Top Posts RedditreReddit: Top posts of July 12, 2019 RedditreReddit: Top posts of July 2019 RedditreReddit: Top posts of 2019 Reddit RulesPrivacy PolicyUser AgreementReddit, Inc. © 2025. All rights reserved. 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2090	https://prase.cz/kalva/putnam/psoln/psol715.html	Putnam 1971/A5 solution 32nd Putnam 1971 Problem A5 A player scores either A or B at each turn, where A and B are unequal positive integers. He notices that his cumulative score can take any positive integer value except for those in a finite set S, where \|S\| =35, and 58 ∈ S. Find A and B. Solution Answer: A, B = 8, 11. Let us call a number green if it can be expressed as a non-negative multiple of A plus a non-negative multiple of B. Assume A > B. A and B must be coprime. Otherwise there would be infinitely many non-green numbers. The set of integers 0, A, 2A, ... , (B-1)A are all incongruent mod B (since A and B are coprime), so they form a complete set of residues mod B. Hence any integer ≥ (B-1)A can be expressed as the sum of a multiple of B and one of the members of the set and is thus green. In fact, the none of the numbers (B-1)A - B + 1, (B-1)A - B + 2, ... , (B-A)A - 1is a multiple of A and they are are incongruent to (B-1)A mod B, so they must equal one of the numbers 0, A, 2A, ... , (B-2)A plus a multiple of B. Thus any integer greater than (B-1)A - B is green. On the other hand, (B-1)A - B itself cannot because it is (B-1)A mod B, so it is incongruent to kA mod B for k < (B-1) (and it cannot be (B-1)A plus a multiple of B because it is too small). So we need to look at the numbers ≤ AB - A - B. Exactly [A/B] of them are congruent to A mod B, but are not green. For if A = [A/B] B + r (with 0 < r < B), then they are r, r + B, r + 2B, ... , r + ([A/B] - 1)B. Similarly, [2A/B] are congruent to 2A mod B, but not green. So the total number of non-green numbers is: [A/B] + [2A/B] + ... + [(B-1)A/B] (). But kA/B cannot be integral for k = 0, 1, ... , (B-1) since A and B are coprime, so [A/B] + [(B-1)A/B] = A-1, [2A/B] + [(B-2)A/B] = A-1, ... . If B - 1 is even, then this establishes that () is 1/2 (A-1)(B-1). If B - 1 is odd, then the central term is [A/2]. B is even, so A must be odd, so [A/2] = 1/2 (A - 1). So in this case also () is 1/2 (A-1)(B-1). So we have (A-1)(B-1) = 2·35 = 70 = 2·5·7. So A = 11, B = 8, or A = 15, B = 6, or A = 36, B = 3. But the last two cases are ruled out because they do not have A, B coprime. So we have A = 11, B = 8. We can easily check that 58 is not green in this case. 32nd Putnam 1971 © John Scholes jscholes@kalva.demon.co.uk 7 Feb 2001
2091	https://artofproblemsolving.com/wiki/index.php/AM-GM_Inequality?srsltid=AfmBOoofjFUbh9fzHLey1-yiU1alBuZDaizfNpyd2CN2zKzaFh6SbMIR	Art of Problem Solving AM-GM 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 AM-GM Inequality Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search AM-GM Inequality In algebra, the AM-GM Inequality, also known formally as the Inequality of Arithmetic and Geometric Means or informally as AM-GM, is an inequality that states that any list of nonnegative reals' arithmetic mean is greater than or equal to its geometric mean. Furthermore, the two means are equal if and only if every number in the list is the same. In symbols, the inequality states that for any real numbers , with equality if and only if . The AM-GM Inequality is among the most famous inequalities in algebra and has cemented itself as ubiquitous across almost all competitions. Applications exist at introductory, intermediate, and olympiad level problems, with AM-GM being particularly crucial in proof-based contests. Contents 1 Proofs 2 Generalizations 2.1 Weighted AM-GM Inequality 2.2 Mean Inequality Chain 2.3 Power Mean Inequality 3 Problems 3.1 Introductory 3.2 Intermediate 3.3 Olympiad 4 See Also Proofs Main article: Proofs of AM-GM All known proofs of AM-GM use induction or other, more advanced inequalities. Furthermore, they are all more complex than their usage in introductory and most intermediate competitions. AM-GM's most elementary proof utilizes Cauchy Induction, a variant of induction where one proves a result for , uses induction to extend this to all powers of , and then shows that assuming the result for implies it holds for . Generalizations The AM-GM Inequality has been generalized into several other inequalities. In addition to those listed, the Minkowski Inequality and Muirhead's Inequality are also generalizations of AM-GM. Weighted AM-GM Inequality The Weighted AM-GM Inequality relates the weighted arithmetic and geometric means. It states that for any list of weights such that , with equality if and only if . When , the weighted form is reduced to the AM-GM Inequality. Several proofs of the Weighted AM-GM Inequality can be found in the proofs of AM-GM article. Mean Inequality Chain Main article: Mean Inequality Chain The Mean Inequality Chain, also called the RMS-AM-GM-HM Inequality, relates the root mean square, arithmetic mean, geometric mean, and harmonic mean of a list of nonnegative reals. In particular, it states that with equality if and only if . As with AM-GM, there also exists a weighted version of the Mean Inequality Chain. Power Mean Inequality Main article: Power Mean Inequality The Power Mean Inequality relates all the different power means of a list of nonnegative reals. The power mean is defined as follows: The Power Mean inequality then states that if , then , with equality holding if and only if Plugging into this inequality reduces it to AM-GM, and gives the Mean Inequality Chain. As with AM-GM, there also exists a weighted version of the Power Mean Inequality. Problems Introductory For nonnegative real numbers , demonstrate that if then . (Solution) Find the maximum of for all positive . (Solution) Intermediate Find the minimum value of for . (Source) Olympiad Let , , and be positive real numbers. Prove that (Source) See Also Proofs of AM-GM Mean Inequality Chain Power Mean Inequality Cauchy-Schwarz Inequality Inequality Retrieved from " Categories: Algebra Inequalities 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.
2092	https://community.cadence.com/cadence_blogs_8/b/cic/posts/skill-for-the-skilled-many-ways-to-sum-a-list-part-1	SKILL for the Skilled: Part 1, Many Ways to Sum a List - Analog/Custom Design - Cadence Blogs - Cadence Community Skip to main content Skip to search Skip to footer Login New User Community Search Search This search text may be transcribed, used, stored, or accessed by our third-party service providers per our Cookie Policy and Privacy Policy. Blogs Analog/Custom Design SKILL for the Skilled: Part 1, Many Ways to Sum a List SKILL for the Skilled: Part 1, Many Ways to Sum a List 5 Sep 2012 • 3 minute read A while back I presented a one day SKILL++ seminar to a group of beginner and advanced SKILL programmers. One example I showed was Variations on how to sum a list of numbers. This is a good example because the problem itself is easy to understand,so the audience can concentrate on the solution techniques rather than on the problem itself. I want to show a few of these examples in this blog post (and a few upcoming posts) in hopes you may find some of the techniques useful. Summing Straightforward One straightforward way to sum a given list of numbers is to use simple iteration and mimic a primitive microprocessor.That is,establish an accumulator initialized to zero, and continue to increment the accumulator by successive numbers from the list until the list is exhausted, and finally return the accumulated value. (defun sumlist_1a (numbers) (let ((sum 0)) (foreach number numbers sum = sum + number) sum)) I imagine the code in this function would look very similar in many different languages such as C or Java -- of course with syntax variations. Summing with apply In SKILL++ (as well as in traditional SKILL) this is not a particularly efficient implementation because the SKILL virtual machine compiler, unlike a native compiler,is not able to make particularly good use of the processor registers. The way to force SKILL to use the processor registers is to organize your code, when possible, to call built-in compiled functions. (defun sumlist_1b (numbers) (apply plus numbers)) Remember -- to force this function to be defined as SKILL++ rather than traditional SKILL, you'll either need to load it from a file with a .ils extension, or you'll need two wrap the definition in (inScheme ...) when you copy and paste it into the CIWindow. A quick test with measureTime shows that sumlist_1b is roughly 10 times faster than sumlist_1a on a 10,000 element list of numbers. How does it work? The plus function returns the sum in its argument list. For example: (plus 1 2 3 4 5) evaluates to 15. The apply function, when called with two arguments, calls the function designated by its first argument. In particular apply calls that function argument list given as its second argument. For example: (apply plus '(1 2 3 4 5)) is a call to plus with two arguments. The first argument plus is the function which apply should call. The second argument, '(1 2 3 4 5) is the list of arguments to call plus with. Thus this use of apply is equivalent to the following. (plus 1 2 3 4 5) Note that plus is used without a leading quote because in SKILL++ all global functions are available in a SKILL++ global variable of the same name. The variable plus evaluates to the plus function. Since sumlist_1b is so much faster (execution-wise) and simpler (line-of-code-wise), it is very tempting to use, but be careful! It suffers from some limitations that do not exist withsumlist_1a. For example, sumlist_1b cannot sum the elements of the nil list. The sumlist_1a, however, claims the sum of the list nil is 0, which indeed sounds like a reasonable answer. Furthermore, sumlist_1a, claims that the sum of a singleton list such as (3.4) is the first (and only) element of the list, which again seems like a reasonable result. Apply in other languages Actually the apply function is not an invention of the SKILL (or SKILL++) language. It is in fact central to programming languages derived from lambda calculus. A quick search on Wikipedia finds an article explaining its use in several languages. More to come In the next posting, I'll show some ways to get the speed and conciseness of sumlist_1b without the caveats. See Also Jim Newton PreviousMixed Signal Design IP Embraces Metric-Driven Verification Using RNM NextThings You Didn't Know About Virtuoso: The (Setup) State of Things CDNS - RequestDemo Try Cadence Software for your next design! Free Trials Team SKILL Community Member Blog Activity Options Subscribe by email CDNS - RequestDemo Have a question? Need more information? Contact Us Jim Newton sum a list summing Virtuoso apply software development SKILL++ sumlist SKILL Related Posts SKILL for the Skilled: Part 9, Many Ways to Sum a List In the previous postings of SKILL for the Skilled , we've looked at different ways to sum the elements of a list of numbers... Team SKILL22 May 2013 • 9 min read Team SKILL, programming, Jim Newton, sum a list, IC615 SKILL for the Skilled: Part 6, Many Ways to Sum a List In a previous post I presented sumlist_2b as a function that would sum lists of length 0, 1, or more. (defun sumlist_2b ... Team SKILL10 Jan 2013 • 5 min read Team SKILL, programming, Jim Newton, sum a list, IC615 SKILL for the Skilled: Part 5, Many Ways to Sum a List In the most recent posts of SKILL for the Skilled (see previous post here ) we looked at different ways to sum a given list... Team SKILL26 Nov 2012 • 5 min read Team SKILL, Jim Newton, summing, Virtuoso, Lisp © 2025 Cadence Design Systems, Inc. All Rights Reserved. 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2093	https://cs.stackexchange.com/questions/108914/prime-implicants-in-boolean-function	digital circuits - Prime Implicants in Boolean Function - Computer Science Stack Exchange Join Computer Science 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 Prime Implicants in Boolean Function Ask Question Asked 6 years, 5 months ago Modified6 years, 4 months ago Viewed 1k times This question shows research effort; it is useful and clear 2 Save this question. Show activity on this post. A Boolean function F(X 1,X 2,X 3,X 4,X 5,X 6)F(X 1,X 2,X 3,X 4,X 5,X 6) of six variables is defined as F=1 F=1, when three or more input variables are at logic 1. otherwise 0. How many essential prime implicants does F have? How to visualize this problem. A Six Variable k-map is very difficult to manage. Any hints? boolean-algebra digital-circuits Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Improve this question Follow Follow this question to receive notifications asked May 3, 2019 at 14:31 VineetVineet 123 5 5 bronze badges 1 see Majoritylox –lox 2019-05-03 15:20:57 +00:00 Commented May 3, 2019 at 15:20 Add a comment\| 2 Answers 2 Sorted by: Reset to default This answer is useful 2 Save this answer. Show activity on this post. Hints Here are two independent hints. Try an easier case such as exercise 1 or exercise 2 below. Can you find one implicant? Can you expand that implicant to a prime implicant? Can you find more? Answer X 1 X 2 X 3 X 1 X 2 X 3 is an implicant since F F becomes 1 when X 1,X 2,X 3 X 1,X 2,X 3 are 1. Since none of X 1 X 2 X 1 X 2, X 1 X 3 X 1 X 3 and X 2 X 3 X 2 X 3 is an implicant, X 1 X 2 X 3 X 1 X 2 X 3 is a prime implicant. Similarly, all X i X j X k X i X j X k are prime implicant where triple i,j,k i,j,k are different from each other. There are (6 3)=20(6 3)=20 of them. Can there be other prime implicant? No. Here is how to see it. Suppose A 1 A 2⋯A m A 1 A 2⋯A m is a prime implicant, where each A i A i is one of X 1,⋯,X 6,X¯¯¯¯1,⋯,X¯¯¯¯6 X 1,⋯,X 6,X¯1,⋯,X¯6. None of A i A i s is X¯¯¯¯1,⋯,X¯¯¯¯6 X¯1,⋯,X¯6. Otherwise, we can remove them to get a larger implicant. The number of A i A i s that are one of X 1,⋯,X 6 X 1,⋯,X 6 must be at least 3. Otherwise, F F can become 1 when only those X i X i s are 1, i.e., less than 3 of input variables are 1. The number of A i A i s that are one of X 1,⋯,X 6 X 1,⋯,X 6 must be at most 3. Otherwise, we can select A i 1,A i 2,A i 3 A i 1,A i 2,A i 3 that are X 1,⋯,X 6 X 1,⋯,X 6. F F can be expanded to the implicant A i 1 A i 2 A i 3 A i 1 A i 2 A i 3 An essential prime implicant is an prime implicant that covers an output of the function that no combination of other prime implicants is able to cover. Based on symmetry, we should expect that every prime implicant is an essential prime implicant, which is indeed the case. Can you find the output that is covered only by X 1 X 2 X 3 X 1 X 2 X 3? In summary, there are 20 essential prime implicants of F F, which are X 1 X 2 X 3,X 1 X 2 X 4,X 1 X 2 X 5,X 1 X 2 X 6,X 1 X 3 X 4,X 1 X 3 X 5,X 1 X 3 X 6,X 1 X 4 X 5,X 1 X 4 X 6,X 1 X 5 X 6,X 1 X 2 X 3,X 1 X 2 X 4,X 1 X 2 X 5,X 1 X 2 X 6,X 1 X 3 X 4,X 1 X 3 X 5,X 1 X 3 X 6,X 1 X 4 X 5,X 1 X 4 X 6,X 1 X 5 X 6, X 2 X 3 X 4,X 2 X 3 X 5,X 2 X 3 X 6,X 2 X 4 X 5,X 2 X 4 X 6,X 2 X 5 X 6,X 2 X 3 X 4,X 2 X 3 X 5,X 2 X 3 X 6,X 2 X 4 X 5,X 2 X 4 X 6,X 2 X 5 X 6, X 3 X 4 X 5,X 3 X 4 X 6,X 3 X 5 X 6,X 3 X 4 X 5,X 3 X 4 X 6,X 3 X 5 X 6, X 4 X 5 X 6 X 4 X 5 X 6. It is not necessary to draw or visualize the Karnaugh map. Exercises Exercise 1. A Boolean function B(X 1,X 2)B(X 1,X 2) of two variables is defined as B=1 B=1, when one or more input variables are at logic 1. Otherwise 0. How many essential prime implicants does B have? Exercise 2. A Boolean function C(X 1,X 2,X 3)C(X 1,X 2,X 3) of three variables is defined as C=1 C=1, when two or more input variables are at logic 1. Otherwise 0. How many essential prime implicants does C have? Exercise 3. A Boolean function F 2(X 1,X 2,X 3,X 4,X 5,X 6)F 2(X 1,X 2,X 3,X 4,X 5,X 6) of six variables is defined as F 2=1 F 2=1, when two or more input variables are at logic 1. Otherwise 0. How many essential prime implicants does F 2 F 2 have? Exercise 4. A Boolean function P(X 1,X 2,X 3,X 4,X 5,X 6)P(X 1,X 2,X 3,X 4,X 5,X 6) of six variables is defined as P=1 P=1, when one or more input variables among X 1,X 2,X 3 X 1,X 2,X 3 are at logic 1 and one or more input variables among X 4,X 5,X 6 X 4,X 5,X 6 are at logic 1. Otherwise 0. How many essential prime implicants does P P have? Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications edited Jun 16, 2020 at 10:30 CommunityBot 1 answered May 4, 2019 at 16:17 喜欢算法和数学喜欢算法和数学 39.2k 4 4 gold badges 35 35 silver badges 96 96 bronze badges Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. An implicant of F F is a partial assignment to the variables of F F which guarantees that F F is satisfied. A prime implicant is a minimal implicant, that is, an implicant that is no longer an implicant if we remove any single variable. Let us start by finding all the implicants of F F. Let α α be any partial assignment to the variables of F F, and suppose that α α assigns k k variables to true. We can complete α α to an assignment by assigning the rest of the variables to false, in which case we get an assignment containing k k true variables. Hence α α can only be an implicant if k≥3 k≥3. Conversely, it is not hard to check that if α α assigns at least 3 variables to true then it is an implicant of F F. When is an implicant a prime implicant? It must be a partial assignment with at least 3 true variables, such that if we remove any single variable, there are at most 2 true variables. Can you figure out how all such implicants look like? Give it a try. As an aside, your function is monotone, and so its prime implicants are also known as minterms. Minterms have a special structure – they cannot be arbitrary partial assignments. I'll let you figure out the special property that minterms have. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications answered May 3, 2019 at 23:55 Yuval FilmusYuval Filmus 281k 27 27 gold badges 317 317 silver badges 515 515 bronze badges 2 As per my understanding I think all such prime implicants must have at most 3 true values. eg. X 2=1,X 4=1,X 6=1 X 2=1,X 4=1,X 6=1 all others X i=0 X i=0. correct me if I am wrong.Vineet –Vineet 2019-05-04 16:07:41 +00:00 Commented May 4, 2019 at 16:07 Implicants are partial assignments. 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2094	https://wayground.com/admin/quiz/67963a423bf870addaaef0b7/solving-two-step-inequalities-with-positive-coefficients	Solving Two-Step Inequalities with Positive Coefficients 6th Grade Quiz \| Wayground (formerly Quizizz) Enter code Login/Signup Find similar resources solving and graphing two step inequalities [x] two step inequalities word problems [x] writing two step inequalities [x] Solving Two-Step Inequalities with Positive Coefficients 6th Grade • 21 Qs Try it as a student Similar activities Solving Inequalities 6th Grade • 25 Qs Test 4.1 6th Grade • 20 Qs Solving & Graphing Inequalities 6th - 7th Grade • 18 Qs Operations with Inequalities 6th Grade • 20 Qs One Step Inequalities Adding and Subtracting 6th Grade - University • 19 Qs One Step Inequalities Adding and Subtracting 6th Grade - University • 19 Qs Variable and Inequalities 6th Grade • 20 Qs WMMMS Integers #2B Plotting Inequalities 6th Grade • 16 Qs View more activities Solving Two-Step Inequalities with Positive Coefficients Quiz • Mathematics • 6th Grade • Easy • CCSS 7.EE.B.4B, 6.EE.B.8 Edit Worksheet Share Save Preview Use this activity Standards-aligned Barbara White Used 2+ times FREE Resource 21 questions Show all answers [x] 1. MULTIPLE CHOICE QUESTION 3 mins • 1 pt Preview Edit -4x + 14 ≤ 54 x ≥ - 10 x ≤ -10 x ≥ 10 x ≤ 10 Tags CCSS.7.EE.B.4B 2. MULTIPLE CHOICE QUESTION 15 mins • 1 pt Preview Edit x > 3 x ≤ ⅕ x ≥ 3 x ≤ ⅕ Tags CCSS.7.EE.B.4B 3. MULTIPLE CHOICE QUESTION 5 mins • 1 pt Preview Edit Solve for k. 2k - 9≤ -19 k≤ -5 k≤ 5 k< -5 k< 5 Tags CCSS.7.EE.B.4B 4. MULTIPLE CHOICE QUESTION 3 mins • 1 pt Preview Edit -3x − 4 < 5 x > -3 x < -3 x < 3 x > 3 Tags CCSS.7.EE.B.4B 5. MULTIPLE CHOICE QUESTION 5 mins • 1 pt Preview Edit Solve for k. 7k–2 ≥ -9 k ≤ -1 k≤1 k≥1 k≥ -1 Tags CCSS.7.EE.B.4B 6. MULTIPLE CHOICE QUESTION 5 mins • 1 pt Preview Edit Solve for t. 6t+4 ≤ -2 t≤ -1 t≤ 1 t=1 1=t Tags CCSS.7.EE.B.4B 7. MULTIPLE CHOICE QUESTION 30 sec • 1 pt Preview Edit Do I need to flip my sign? -3x ≥ 12 Yes, because I'll be dividing by a negative number No, because you never flip the sign. Tags CCSS.6.EE.B.8 8. MULTIPLE CHOICE QUESTION 3 mins • 1 pt Preview Edit Solve and graph the inequality. 7n - 1 > 76 A B C D Tags CCSS.7.EE.B.4B 9. MULTIPLE CHOICE QUESTION 5 mins • 1 pt Preview Edit Match the graph with its inequality. b > –2 b < –2 b ≥ –2 b ≤ –2 Tags CCSS.6.EE.B.8 10. MULTIPLE CHOICE QUESTION 2 mins • 1 pt Preview Edit When graphing b<6 b<6 b<6 , would you use an open circle or a filled-in/closed dot? open circle filled-in/closed dot Create a free account and access millions of resources Create resources Host any resource Get auto-graded reports Continue with GoogleContinue with EmailContinue with ClasslinkContinue with Clever or continue with Microsoft Apple Others By signing up, you agree to our Terms of Service&Privacy PolicyAlready have an account?Log in Let me read it first Similar Resources on Wayground View all 20 questions Inequalities Quiz • 6th - 9th Grade 18 questions Write Inequalities Quiz • 6th - 8th Grade 18 questions Equations and Inequalities Review Quiz • 6th Grade 20 questions Word Problem Test Review (Inequalities and Equations) Quiz • 6th - 8th Grade 20 questions Solving 1-Step Inequalities Quiz • 6th - 8th Grade 20 questions Solving Inequalities Quiz • 6th - 7th Grade 20 questions Math 6 - Lesson 8.3B: Solving Simple Inequalities Quiz • 6th Grade 20 questions Equations and Tables Quiz • 6th Grade Popular Resources on Wayground 10 questions Video Games Quiz • 6th - 12th Grade 10 questions Lab Safety Procedures and Guidelines Interactive video • 6th - 10th Grade 25 questions Multiplication Facts Quiz • 5th Grade 10 questions UPDATED FOREST Kindness 9-22 Lesson • 9th - 12th Grade 22 questions Adding Integers Quiz • 6th Grade 15 questions Subtracting Integers Quiz • 7th Grade 20 questions US Constitution Quiz Quiz • 11th Grade 10 questions Exploring Digital Citizenship Essentials Interactive video • 6th - 10th Grade Discover more resources for Mathematics 22 questions Adding Integers Quiz • 6th Grade 20 questions Adding and Subtracting Integers Quiz • 6th Grade 21 questions Convert Fractions, Decimals, and Percents Quiz • 6th Grade 20 questions Integers, Opposites and Absolute Value Quiz • 6th Grade 15 questions Equivalent Ratios Quiz • 6th Grade 20 questions Order of Operations Quiz • 6th Grade 20 questions Ratios/Rates and Unit Rates Quiz • 6th Grade 18 questions Adding Integers Quiz • 6th - 7th Grade Features School & District Wayground for Business Create a quiz Create a lesson Subjects Mathematics Social Studies Science Physics Chemistry Biology About Our Story Wayground Blog Media Kit Careers Support F.A.Q. Help & Support Privacy Policy Terms of Service Teacher Resources © 2025 Quizizz Inc. (DBA Wayground) Sitemap Get our app
2095	https://alpha.chem.umb.edu/chemistry/ch313/Exp%205%20vinegar.pdf	Experiment 5 Chem 313 Preparation of Standard Sodium Hydroxide Solution and Titration of a Commercial Vinegar Product In this experiment we prepare a standard solution of NaOH, and then use it to determine the concentration of acetic acid in vinegar. We will require knowledge of the exact concentration of the NaOH solution, but it is not convenient either weigh solid NaOH, because it is hydroscopoic. Therefore, we must use a primary standard to determine the sodium hydroxide concentration to four significant figures. The primary standard that is most often used for standardizing sodium hydroxide solutions if potassium hydrogen phthalate, KHP. KHP of very high purity can be purchased, dried it in an oven, cooled it in a desiccator, and a sample can be accurately weighed on an analytical balance. After dissolving a known amount of KHP in distilled water, one can titrate the resulting solution with a sodium hydroxide solution. The volume of the sodium hydroxide solution required to reach the equivalence point and the mass of KHP is used to calculate the concentration of the sodium hydroxide solution to 4 or 5 significant figures. Once the NaOH solution is standardized it can be used to analyze acidic solutions, such as vinegar. The presence of carbon dioxide is another reason why NaOH can not be used as a primary standard. Water can readily absorb carbon dioxide. Dissolved carbon dioxide acts as a weak acid in aqueous solutions. CO2(g) ↔ CO2(aq) CO2(aq) + H2O  H2CO3  HCO3 - + H+ If basic solutions are prepared using water containing dissolved carbon dioxide, a portion of the base reacts with HCO3 - to yield CO3 2-. Sodium hydroxide solutions are often prepared by diluting a 50 % aqueous solution of sodium hydroxide to the desired concentration, followed immediately by standardization against a primary standard. This removes possible sources of error due to varying amounts of carbon dioxide, because the dissolved CO2 precipitates out as sodium carbonate in a 50 % NaOH solution. We will also be using water that has recently been boiled to prepare all of the solution in this experiment. Boiling drives the above equilibriums to the left, driving off CO2(g) and dramatically reducing the bicarbonate concentration. KHP is a weak acid. It is the intermediate species of the phthalic acid system Ka2 = 3.9010-6 pKa2 = 5.408 KHP titration C8H5O4 - + OH-  H2O + C8H4O4 2- Prelab Assignment A 0.5123-g sample of KHP (FW 204.23) was dissolved in about 25 mL of distilled water, and titrated to the phenolphthalein end point with 28.75 mL of a sodium hydroxide solution. Calculate the molar (M) concentration of the hydroxide solution. Apparatus 250 mL Erlenmeyer Flasks 50 mL burette 10 mL graduated cylinder 100 mL graduated cylinder 1000 mL boiling flasks 1 L polyethylene bottles Weighing bottle for drying KHP Chemicals Vinegar Phenolphthalein solution Potassium hydrogen phthalate solid 50 % sodium hydroxide solution PROCEDURE PART A: 1. Dry the KHP in a 105 C oven for at least one hour. Use your weighing bottle, without lid, to hold the solid. Label a 150-mL beaker, put the weighing bottle in the beaker and cover the beaker with a watch glass. 2. Clean the following glassware thoroughly; 25 mL pipet, 250 mL volumetric, and a buret. 3. Add about 6 mL of the 50 % NaOH solution using a 10-mL graduated cylinder to a 1-L bottle. Fill the bottle almost to the top with the boiled water, screw on the cap, and mix the solution thoroughly. (Leaving little head space and screwing on the cap will prevent CO2 in the atmosphere from dissolving into your NaOH). Label it! Handle the 50% NaOH with care!! Avoid contact with your skin or clothes. If you do spill some on yourself wash it off at once with a large amount of water. 4. Obtain about 30 mL of vinegar. There will be two different brands available. Write down the brand of the vinegar that you sample. Pipet a 25 mL of this sample into a 250 mL volumetric flask. Dilute to the mark with the de-ionized water and mix thoroughly. Label it! 5. Pipet 25-mL aliquots of the diluted vinegar solution to five 250 mL Erlenmeyer flask. Add 50 mL of de-ionized water and 3 drops of phenolphthalein indicator and mix thoroughly. Label the flasks. 6. Rinse a clean 50-mL burette with 25 mL of your sodium hydroxide solution. Fill it with approximately 45 mL of the sodium hydroxide solution. Record the level of the of the NaOH solution to the nearest ±0.01 mL. Take your time with this to really understand how to read and work the burette. 7. Take the KHP out of the oven. Place it in a desiccator to cool for 15-20 minutes (until cooled to room temperature). 8. Titrate the diluted vinegar in the five labeled Erlenmeyer flasks. The endpoint is reached when the solution changes from colorless to faint pink. Be careful not to overshoot the endpoint. One drop takes it from clear to faint pink. Be sure to refill burette with NaOH in between titrations. Also, make sure that you recap your 1 L-bottle in between fills. 9. Weigh five portions of 0.4-0.7 g to the nearest 0.1 mg. Place each portion in a labeled 250-mL Erlenmeyer flask. Add about 30 mL of the boiled water, 3 drops of phenolphthalein indicator and mix thoroughly. 10. Titrate each of the five samples to the same phenolphthalein end point as with the vinegar samples. 11. Carefully label your NaOH solution. 12. Trade data with a classmate that sampled from the other brand of vinegar PART 2 CLEAN-UP Rinse the buret, pipet and volumetric flask with several (10) portions of de-ionized water and put them away. (THIS IS CRITICALLY IMPORTANT). If it is not done the pipet and buret will essentially be destroyed. Use a brush and soapy water to wash the Erlenmeyer flasks. (SCRUB!) Data Analysis and Discussion Points: 1. With the data from each of you each of your six KHP samples calculate values for the concentration of your NaOH solution. (5 pts) 2. Calculate the mean, se, sm, and 95% CL of the concentration of your NaOH solution. (5 pts) 3. Calculate the mean, se, sm, and 95% CL of the volume of NaOH used to titrate the diluted vinegar samples (5 pts) 4. Using the mean [NaOH] and the mean V for the diluted vinegar titration, calculate the percentage of acetic acid in the vinegar (w/v). (5 pts) 5. Propagate the uncertainties in this calculation to estimate the uncertainty in the % acetic acid (use se values for [NaOH] and V and the uncertainties in the pipet and volumetric flask for the propagation). (5 pts) 6. Repeat 1-6 with your classmate’s data. (25 pts) 7. Using your estimate of the propagated uncertainties obtained from Question 5 as estimates of standard deviations, se, determine whether the percentage of acetic acid in the different brands of vinegar is statistically different to the 95 % CL. (5 pts) 8. Produce a Master Table that summarizes the results from this experiment and incorporate it into your report. (5 pts) Lab Report Abstract (10 pts) Brief procedure (less detail then in the write-up) (10 pts) Data (10) Data Analysis and Questions (60 pts) Discussion (10 pts)
2096	https://hal.science/hal-01378829/document	Published Time: Sat, 02 Aug 2025 00:02:27 GMT HAL Id: hal-01378829 Submitted on 10 Oct 2016 HAL is a multi-disciplinary open access archive for the deposit and dissemination of sci-entific research documents, whether they are pub-lished or not. The documents may come from teaching and research institutions in France or abroad, or from public or private research centers. L’archive ouverte pluridisciplinaire HAL , est destinée au dépôt et à la diffusion de documents scientifiques de niveau recherche, publiés ou non, émanant des établissements d’enseignement et de recherche français ou étrangers, des laboratoires publics ou privés. Distributed under a Creative Commons Attribution 4.0 International License Circulant Matrices and Their Application to Vibration Analysis Brian Olson, Steven Shaw, Chengzhi Shi, Christophe Pierre, Robert Parker To cite this version: Brian Olson, Steven Shaw, Chengzhi Shi, Christophe Pierre, Robert Parker. Circulant Matri-ces and Their Application to Vibration Analysis. Applied Mechanics Reviews, 2014, 66 (4), 10.1115/1.4027722. hal-01378829 Brian J. Olson Applied Physics Laboratory, Air and Missile Defense Department, The Johns Hopkins University, Laurel, MD 20723-6099 e-mail: brian.olson@jhuapl.edu Steven W. Shaw University Distinguished Professor Department of Mechanical Engineering, Michigan State University, East Lansing, MI 48824-1226 e-mail: shawsw@egr.msu.edu Chengzhi Shi Department of Mechanical Engineering, University of California, Berkeley, Berkeley, CA 94720 e-mail: chengzhi.shi@berkeley.edu Christophe Pierre Professor and Vice President for Academic Affairs Department of Mechanical Science and Engineering, University of Illinois at Urbana-Champaign, Urbana, IL 61801 e-mail: chpierre@uillinois.edu Robert G. Parker L.S. Randolph Professor and Department Head Department of Mechanical Engineering, Virginia Polytechnic Institute and State University, Blacksburg, VA 24061 e-mail: r.parker@vt.edu Circulant Matrices and Their Application to Vibration Analysis This paper provides a tutorial and summary of the theory of circulant matrices and their application to the modeling and analysis of the free and forced vibration of mechanical structures with cyclic symmetry. Our presentation of the basic theory is distilled from the classic book of Davis (1979, Circulant Matrices , 2nd ed., Wiley, New York) with results, proofs, and examples geared specifically to vibration applications. Our aim is to collect the most relevant results of the existing theory in a single paper, couch the mathematics in a form that is accessible to the vibrations analyst, and provide examples to highlight key concepts. A nonexhaustive survey of the relevant literature is also included, which can be used for further examples and to point the reader to important extensions, applica-tions, and generalizations of the theory. 1 Introduction “The theory of matrices exhibits much that is visually attractive. Thus, diagonal matrices, symmetric matrices, (0, 1) matrices, and the like are attractive independently of their applications. In the same category are the circulants.” Philip J. Davis The modeling and analysis of structural vibration is amature field, especially when vibration amplitudes are small and linear models apply. For such models, the powerful tools of modal analysis and superposition allow one to decompose the system and its response into a set of uncoupled single degree-of-freedom (DOF) systems, each of which captures the motion of the overall system in a given normal mode. For geometrically simple continua, the natural frequencies, vibra-tion modes, and response to known excitations can be analyti-cally determined. When discrete models are developed, matrix methods are readily applied for both natural frequency and response analyses. For general system models with no special properties, which occur in the majority of applications, large-scale computational models must be developed, typically using finite element methods. However, certain classes of systems possess special properties, such as symmetries, which aid in the analysis by enabling significant reduction of the fi-nite element models (Fig. 1( a)). This is particularly true for systems with cyclic symmetry. The main goal of this tutorial is to consider the vibrations of structural systems with cyclic symmetry, also known as rotation-ally periodic systems. A useful geometric view of these systems is that of a circular disk (a pie) split into N equal sectors (i.e., equally sized pieces of the pie), each of which contains an identical mechanical structure with identical coupling to forward-nearest-neighbors and to ground (Fig. 1( b)). The theory of circulants also applies to more general forms of coupling with non-nearest-neighbors, for example through a base substructure, as long as the rotational symmetry is preserved. These structural arrangements arise naturally in certain types of rotating machines. Turbomachinery examples include bladed disks, such as fans, compressors, turbines, and impeller stages of aircraft, helicopter engines, power plants, as well as propellers, pumps, and the like. Rotational perio-dicity also arises in some stationary structures such as satel-lite antennae. In some systems, such as planetary gears and rotors with pendulum vibration absorbers, the overall system is not necessarily cyclic, but subcomponents of it may be. When perfect cyclic symmetry of a model is assumed, special modal properties exist that significantly facilitate its vibration analysis. 1The nature of rotationally periodic systems imposes a cyclic structure on their mass and stiffness matrices, which are block cir-culant for systems with many DOFs per sector (Fig. 1( a)) and cir-culant for the special case of a single DOF per sector (Fig. 1( b)). By denoting the stiffness of the internal elements of each sector by K0 and the coupling stiffness between sectors as – K1, the stiff-ness matrices of rotationally periodic structures with nearest-neighbor coupling have the general form K ¼ K0 K1 0 … 0 K1 K1 K0 K1 … 0 00 K1 K0 … 0 0 ... ... ... . .. ... ... 0 0 0 … K0 K1 K1 0 0 … K1 K0 266666664377777775 where K0 and K1 are themselves matrices for the complex model shown in Fig. 1( a) and scalars for the simplest prototypical model shown in Fig. 1( b). A key property of K is that the elements of each row are obtained from the previous row by cyclically per-muting its entries. That is, for j ¼ 2; 3; …; N, row j is obtained from row j – 1 by shifting the elements of row j – 1 to the right by one position and wrapping the right-end element of row j – 1 into the first position. This is precisely the form of a circulant matrix, which is formally defined in Sec. 2.2. The mass matrix of a rota-tionally periodic structure with nearest-neighbor coupling is block diagonal and also shares this cyclic property. The size of the ele-ments of K is equal to the number of DOFs per sector, and is denoted by M. Thus, a system with N sectors and M DOFs per sec-tor has a total of NM DOFs. The most important utility of the theory of circulants in analyzing rotationally periodic systems is that they enable a NM -DOF system to be decomposed to a set of NM -DOF uncoupled systems using the appropriate coordinate transformation. Admittedly, the same can be accomplished using brute-force methods to uncouple the entire system using modal analysis, but such an approach overlooks fundamental properties that are crucial to understanding the free and forced response of these systems and requires significantly more computational power. This is the central motivation for understanding and utiliz-ing circulants to analyze cyclic systems. The vibration modes of rotationally periodic systems consist of multiple pairs of repeated natural frequencies (eigenvalues) that lead to pairs of degenerate normal modes (eigenvectors). The number and nature of such pairs depend on whether N is even or odd. Each mode pair is characterized as a pair of standing waves (SWs) with different spatial phases, or a pair of traveling waves, labeled as a forward traveling wave (FTW) and backward travel-ing wave (BTW) when following the terminology used in applica-tions to rotating machinery. The choice of formulation is based on convenience for a given application, which depends on the nature of the system excitation. For example, the excitation frequency is proportional to the engine speed for many cyclic rotating systems, which leads to the so-called engine order (e.o.) excitation , and often the spatial nature of the excitation (in the rotating frame of reference) is in the form of a traveling wave. When such excita-tion is applied to systems with cyclic symmetry, the response also has special properties that can be easily uncovered by making use of the system traveling wave vibration modes. The strength of the intersector coupling is an important parame-ter in rotationally periodic systems. When the intersector coupling is strong, the frequencies of the mode pairs are well separated. In contrast, weak intersector coupling yields closely spaced frequen-cies, high modal density, and large sensitivity to cyclic-symme-try-breaking imperfections. A wave representation of the response shows that the strength of the coupling determines frequency passband widths, wherein unattenuated propagation of waves takes place. Weak intersector coupling leads to narrow passbands, and the passbands widen as the coupling strength increases. Another important parameter for cyclically symmetric structures is the total number of sectors. The modal density is larger for large N, which corresponds to more natural frequencies within each fre-quency passband. In all cases, the modes are spatially distributed, or extended, for models of cyclic systems. That is, the pattern of displacements in a modal response is uniformly spread around the circumference of the structure. Systems with cyclic symmetry have been studied in the context of vibration analysis for over 40 yr. Early work considered proper-ties of the vibration modes [3,4] and the steady-state response to harmonic excitation [5–7] of tuned and mistuned turbomachinery rotors. Many of these contributions were motivated by vibration studies of general rotationally periodic systems [8–20], bladed disks [1,3,21–30], planetary gear systems [31–43], rings [44,45], circular plates [46–48], disk spindle systems [49–51], centrifugal pendulum vibration absorbers [52–56], space antennae , and microelectromechanical system frequency filters . Implicit in these investigations is the assumption of perfect symmetry which, of course, is an idealization. Perfect symmetry gives rise to well-structured vibration modes [9,31,33–37,39,53,56], which are char-acterized by certain phase indices that define specific phase rela-tionships between cyclic components in each vibration mode . This vibration mode structure is critical in the investigation of dynamic response of cyclic systems using modal analysis . These special properties of rotationally periodic structures save tremendous calculation effort in the analysis of the system dynamics [59–61]. The properties of cyclic symmetry are not only used in the study of mechanical vibrations, they are also important Fig. 1 (a) Finite element model of a bladed disk assembly and ( b) general cyclic system with Nidentical sectors and nearest-neighbor coupling 2 to the analysis of elastic stress [62,63] and coupled cell networks . The special properties of systems with cyclic symmetry extend to nonlinear systems, where they are expressed quite naturally in terms of symmetry groups: the cyclic group, in particular [65–68]. The group theoretic formulation can also be applied to the linear vibration problems considered in this paper [69,70], but the approach presented here is more approachable to readers with a standard engineering background in linear algebra. The extension to systems with small imperfections that perturb the cyclic symmetry has led to important results related to mode localization , which arises in systems with high modal density caused by weak intersector coupling or a large number of sectors. In particular parameter regimes, the mode shapes are highly sensi-tive to small, symmetry-breaking imperfections among the nomi-nally identical sectors, and the spatial nature of the vibration modes can become highly localized. For these cases, the vibration energy is focused in a small number of sectors, and sometimes even a single sector. This behavior, which stems from the seminal work of Anderson on lattices , was originally recognized to be relevant to structural vibrations by Hodges and Woodhouse [72,73] and Pierre and Dowell , and has been extensively studied from both fundamental and applied [76–78] points of view. The phenomenon of mode localization is also observed in the forced response and has practical implications for the fatigue life of bladed disks in turbomachinery [79,80]. It is interesting to note that localization also arises in nonlinear systems with perfect symmetry, where the dependence of the system natural frequen-cies on the amplitudes of vibration naturally leads to the possibil-ity of mistuning of frequencies between sectors if their amplitudes are different [81–88]. Another topic central to vibration analysis that relies on the theory of circulants is the discrete Fourier transform (DFT) [89,90]. The DFT was known to Gauss , and is the most com-mon tool used to process vibration signals from experimental measurements and numerical simulations. The DFT and inverse DFT (IDFT) provide a computationally convenient means of determining the frequency content of a given signal. Because the mathematics of circulants is at the heart of the computation of the DFT, we include a brief introduction to the relationship between the DFT and IDFT, and its connection to the theory of circulants. The goal of this paper is to provide a detailed theory of circu-lant matrices as it applies to the analysis of free and forced struc-tural vibrations. Much of the material was developed as part of the Ph.D. research of the lead author [21,22,92–95]. References to other relevant work are included throughout this paper, but we do not claim to provide an exhaustive survey of the relevant literature. The remainder of the paper is organized as follows. Section 2 gives a quite exhaustive and self-contained treatment of the theory of circulants, which is distilled from the seminal work by Davis . We adopt a presentation style similar to that of Ottarsson , one that should be familiar to an analyst in the vibrations engineering community. This section is meant to act simultaneously as a detailed reference and tutorial, including proofs of the main results and simple illustrative examples. Section 3 provides three examples that make use of the theory, including ordinary circulants and the more general block circulant matrices. Particular attention is given to cyclic systems under trav-eling wave engine order excitation because this type of system forcing appears naturally in many relevant applications of rotating machinery. The apprised reader, or the reader who wishes to learn by example, can skip directly to Sec. 3, depending on their back-ground, and revisit Sec. 2 as warranted. The paper closes with a brief summary in Sec. 4. 2 The Theory of Circulants This section details the theory and mathematics of circulant matrices that are relevant to vibration analysis of mechanical structures with cyclic symmetry. The basic theory is distilled from the seminal work by Davis and is presented using mathemat-ics and notation that should be familiar to the vibrations engineer. Selected topics from linear algebra are reviewed in Sec. 2.1 to introduce relevant notion and support the theoretical development of circulant matrices in Secs. 2.2–2.8. This material is included for completeness; the apprised reader can skip directly to Secs. 2.2 and 2.3, where circulant and block circulant matrices (also referred to as circulants and block circulants ) are defined. Representations of circulants are discussed in Sec. 2.4. Diagonal-ization of circulants and block circulants is discussed at length in Sec. 2.5, which begins with a treatment of the Nth roots of unity in Sec. 2.5.1 and the Fourier matrix in Sec. 2.5.2. It is subse-quently shown how to diagonalize the cyclic forward shift matrix in Sec. 2.5.3 a circulant in Sec. 2.5.4, and a block circulant in Sec. 2.5.5. Some generalizations of the theory are discussed in Sec. 2.6, including the diagonalization of block circulants with circulant blocks. Relevant mathematics of the DFT and IDFT are summarized in Sec. 2.7. Finally, the circulant eigenvalue problem (cEVP) is discussed in Sec. 2.8, including the eigenvalues and eigenvectors of circulants and block circulants, their symmetry characteristics, and connection to the DFT process. 2.1 Mathematical Preliminaries. Definitions and relevant properties of special operators and matrices are discussed in Secs. 2.1.1 and 2.1.2, respectively, including the direct (Kro-necker) product, and Hermitian, unitary, cyclic forward shift, and flip matrices. This is followed in Sec. 2.1.3 with a treatment of matrix diagonalizability. 2.1.1 Special Operators. Let C denote the set of complex numbers and Zþ be the set of positive integers. D EFINITION 1 (Direct Sum). For each i ¼ 1; 2; …; N and p i 2 Zþ, let Ai 2 Cpi pi . Then the direct sum of Ai is denoted by Ni¼1 Ai ¼ A1 A2 … ANand results in the block diagonal square matrix A ¼ A1 0 … 00 A2 … 0 ... ... . .. ... 0 0 … AN 2666437775 of order p 1 þ p2 þ þ p N , where each zero matrix 0 has the appropriate dimension. It is convenient to define the operator diag ðÞ that takes as its argument the ordered set of matrices A1; A2; …; AN and results in the block diagonal matrix given in Definition 1, that is, A ¼ diag ðA1; A2; …; AN Þ ¼ diag i¼1;…;N ðAiÞ For the case when each Ai ¼ a i is a scalar (1 1), the direct sum of a i is denoted by the diagonal matrix diag ða1; a2; …; aN Þ ¼ diag i¼1;…;N ðaiÞ D EFINITION 2 (Direct Product). Let a; b 2 Cn . Then the direct prod-uct (or Kronecker product) of a and bT is the square matrix a bT ¼ a1b1 a1b2 a1bna2b1 a2b2 a2bn ... ... . .. ... an b1 an b2 an bn 2666437775 where ðÞ T denotes transposition. If A 2 Cmn and B 2 Cpq are matrices, then the direct product of A and B is the matrix 3A B ¼ a11 B a12 B a1nB a21 B a22 B a2nB ... ... . .. ... am1B am2B amn B 2666437775 of dimension mp nq. Example 1 . Consider the matrices A ¼ 1 2 3½ and B ¼ 1 23 4 Then the direct product of A and B is given by A B ¼ 1 2 3½ 1 23 4 ¼ 1 1 23 4 ; 2 1 23 4 ; 3 1 23 4 ¼ 1 23 4 ; 2 46 8 ; 3 69 12 ¼ 1 2 2 4 3 63 4 6 8 9 12 Because A is 1 3 and B is 2 2, the direct product A B has dimension 1 2 3 2, or 2 6. Some important properties of the direct product are as follows: (1) The direct product is a bilinear operator. If A and B are square matrices and a is a scalar, then aðA BÞ ¼ ð aAÞ B ¼ A ðaBÞ (1) (2) The direct product distributes over addition. If A, B, and C are square matrices with the same dimension, then ðA þ BÞ C ¼ A C þ B C (2 a) A ðB þ CÞ ¼ A B þ A C (2 b)(3) The direct product is associative. If A, B, and C are square matrices, then A ðB CÞ ¼ ð A BÞ C (3) (4) The product of two direct products yields another direct product. If A, B, C, and D are square matrices such that AC and BD exist, then ðA BÞð C DÞ ¼ ð AC Þ ðBD Þ (4) (5) The inverse of a direct product yields the direct product of two matrix inverses. If A and B are invertible matrices, then ðA BÞ1 ¼ A1 B1 (5) where ðÞ 1 denotes the matrix inverse. (6) The transpose or conjugate transpose of a direct product yields the direct product of two transposes or conjugate transposes. If A and B are square matrices, then ðA BÞT ¼ AT BT (6 a) ðA BÞH ¼ AH BH (6 b)where ðÞ H ¼ ðÞ T is the conjugate transpose and ðÞ denotes complex conjugation. (7) If A and B are square matrices with dimensions n and m,respectively, then det ðA BÞ ¼ ð det AÞmðdet BÞn (7 a)tr ðA BÞ ¼ tr ðAÞtr ðBÞ (7 b)where det ðÞ and tr ðÞ denote the matrix determinant and trace. 2.1.2 Special Matrices. The definitions and relevant proper-ties of selected special matrices are summarized. Hermitian and unitary matrices are defined first (see Table 1), followed by a brief treatment of two important permutation matrices: the cyclic for-ward shift matrix and the flip matrix. The details of circulant mat-rices and the Fourier matrix, which are employed extensively throughout this work, are deferred to Secs. 2.2, 2.3, and 2.5.2. DEFINITION 3 (Hermitian Matrix). A matrix H 2 CNN is Hermi-tian if H ¼ HH. The elements of a Hermitian matrix H satisfy hik ¼ h ki for all i; k ¼ 1; 2; …; N. Thus, the diagonal elements h ii of a Hermitian matrix must be real, while the off-diagonal elements may be com-plex. If H ¼ HT then H is said to be symmetric .D EFINITION 4 (Unitary Matrix). A matrix U 2 CNN is unitary if UHU ¼ I, where I is the N N identity matrix. Real unitary matrices are orthogonal matrices. If a matrix U is unitary, then so too is UH. To see this, consider ðUH ÞH ðUH Þ¼ UU H ¼ I, from which it follows that UHU ¼ UU H ¼ I (8) Finally, if U is unitary and nonsingular, then UH ¼ U1 .A general permutation matrix is formed from the identity matrix by reordering its columns or rows. Here, we introduce two such matrices: the cyclic forward shift matrix and the flip matrix. D EFINITION 5 (Cyclic Forward Shift Matrix). The N N cyclic forward shift matrix is given by rN ¼ 0 1 0 0 00 0 1 0 0 ... ... ... . .. ... ... 0 0 0 1 00 0 0 0 11 0 0 0 0 2666666437777775 NN which is populated with ones along the superdiagonal and in the (N, 1) position, and zeros otherwise. The cyclic forward shift matrix plays a key role in the represen-tation and diagonalization of circulant matrices, which are dis-cussed in Secs. 2.4 and 2.5. Example 2. Let a ¼ (a, b, c) be a three-vector. Then the operation ar3 ¼ a b c½ 0 1 00 0 11 0 0 264375 ¼ ð c; a; bÞ Table 1 Selected special matrices Type Condition Symmetric A¼ATHermitian A¼AH Orthogonal ATA¼I(or) AT¼A1Unitary AHA¼I(or) AH¼A14 cyclically shifts the entries of a by one entry to the right. That is, the ith entry of a is shifted to entry i þ 1, except for entry N ¼ 3, which is placed in position 1 of a. D EFINITION 6 (Flip Matrix). The N N flip matrix is given by jN ¼ 1 0 0 0 00 0 0 0 10 0 0 1 0 ... ... ... . .. ... ... 0 0 1 0 00 1 0 0 0 2666666437777775 NN which is populated with ones in the (1, 1) position and along the subantidiagonal, and zeros otherwise . C OROLLARY 1. Let jN be the flip matrix. Then j2 N ¼ IN jH N ¼ jTN ¼ jN ¼ j1 N ) where IN is the N N identity matrix . 2.1.3 Matrix Diagonalizability. Matrix diagonalization is the process of taking a square matrix and transforming it into a diago-nal matrix that shares the same fundamental properties of the underlying matrix, such as its characteristic polynomial, trace, and determinant. This section defines matrix diagonalizability in terms of similarity, provides necessary conditions for a matrix to be diagonalizable, and summarizes relevant properties of diagonaliz-able matrices. Diagonalization of circulant matrices is deferred to Sec. 2.5. DEFINITION 7 (Similarity Transformation). Let Q be an arbitrary nonsingular matrix. Then B ¼ Q1AQ is a similarity transforma-tion and B is said to be similar to A. If B is similar to A, then A ¼ Q1 1B Q 1 is similar to B.It therefore suffices to say that A and B are similar . A summary of selected additional linear transformations is provided in Table 2. If B is orthogonally (resp. unitarily) similar to A, then we say that A and B are orthogonally (resp. unitarily) similar matrices. T HEOREM 1. If A and B are similar matrices, then they have the same characteristic equation and hence the same eigenvalues. Theorem 1 guarantees that the eigenvalues of a matrix are preserved under a similarity transformation. A proof can be found in any standard textbook on linear algebra [98,99]. Because QT ¼ Q1 for orthogonal Q and QH ¼ Q1 for unitary Q, the eigenvalues are also preserved under orthogonal and unitary transformations. T HEOREM 2. Let the matrices A and B be similar. Then if pðtÞ ¼ XNk¼0 c k tkdenotes a finite polynomial in t with arbitrary coefficients c k ðk ¼ 1; 2; …; NÞ, the matrix polynomials p (A) and p (B) are similar. Proof . Let Q be an arbitrary nonsingular matrix. Then pðBÞ ¼ pðQ1AQ Þ¼ XNk¼0 c kðQ1AQ Þk ¼ c0I þ c1Q1AQ þ c2Q1AQQ 1AQ þ þ c N Q1AQ Q1AQ ¼ c0I þ c1Q1AQ þ c2Q1A2Q þ þ cN Q1AN Q ¼ Q1 c0I þ c1A þ c2A2 þ þ c N AN Q ¼ Q1pðAÞQ which completes the proof. If p(t) ¼ t k with k > 0 in Theorem 2, then we have the following result. C OROLLARY 2. If B ¼ Q1 AQ , then Bk ¼ Q1AkQ for any k 2 Zþ. DEFINITION 8 (Diagonalizable Matrix). A square matrix A is diagonalizable if there exists a nonsigular matrix Q and a diago-nal matrix D such that Q1AQ ¼ D. Thus, a matrix is diagonalizable if it is similar to a diagonal ma-trix. If A is diagonalizable by Q, we say that Q diagonalizes A and that Q is the diagonalizing matrix .T HEOREM 3. An N N matrix A is diagonalizable if it has N lin-early independent eigenvectors . Proof . Suppose A has N linearly independent eigenvectors and denote them by q1; q2; …; qN . Let ki be the eigenvalue of A corre-sponding to qi for each i ¼ 1; …; N. Then if Q is the matrix that has as its ith column the vector qi, it follows that AQ ¼ Aq 1; Aq 2; …; Aq Nð Þ¼ q1k1; q2k2; …; qN kNð Þ¼ q1; q2; …; qNð Þ diag i¼1;…;N ðkiÞ QD Because Q is nonsingular by hypothesis, D ¼ Q1AQ . 2.2 Circulant Matrices DEFINITION 9 (Circulant Matrix). A N N circulant matrix (or circulant, or ordinary circulant) is generated from the N-vector fc1; c2; …; cN g by cyclically permuting its entries, and is of the form C ¼ c1 c2 c Nc N c1 cN1 ... ... . .. ... c2 c3 c1 2666437775 D D EFINITION 10 (Generating Elements). Let the N N circulant matrix C be given by Definition 9. Then the elements of the N-vector fc1; c2; …; c N g are said to be the generating elements of C. Thus, a circulant matrix is defined completely by the generating elements in its first row, which are cyclically shifted to the right by one position per row to form the subsequent rows. The set of all such matrices of order N is denoted by CN . A matrix contained in CN is said to be a circulant of type N.It is convenient to define the circulant operator circ ðÞ that takes as its argument the generating elements c1; c2; …; cN and results in the array given in Definition 9, that is, Table 2 Selected types of linear transformations Type Condition Transformation Equivalence Pand Qare nonsingular B¼PAQ Congruence Qis nonsingular B¼QTAQ Similarity Qis nonsingular B¼Q1AQ Orthogonal Qis nonsingular and orthogonal B¼QTAQ ¼Q1AQ Unitary Qis nonsingular and unitary B¼QHAQ ¼Q1AQ 5 C ¼ circ ðc1; c2; …; c N Þ (9) An N N circulant is also characterized in terms of its ( i, k) entry by ðCÞik ¼ ckiþ1ðmod NÞ for i; k ¼ 1; 2; …; N. Example 3 . The circulant array formed by the generating ele-ments a, b, c, d can be written as circ ða; b; c; dÞ ¼ a b c dd a b cc d a bb c d a 26643775 2 C4which is a circulant matrix of type 4. If a matrix is both circulant and symmetric, its generating ele-ments are c1; …; cN 2 ; cNþ22 ; cN 2 ; …; c3; c2; N even c1; …; cN12 ; cNþ12 ; cNþ12 ; cN12 ; …; c3; c2; N odd ( (10) which are necessarily repeated. Only ( N þ 2)/2 generating ele-ments are distinct if N is even and ( N þ 1)/2 are distinct if N is odd. The set of all N N symmetric circulants is denoted by SC N . A matrix contained in SC N is said to be a symmetric circu-lant of type N. Example 4 . The 5 5 matrix circ ða; b; c; c; bÞ ¼ a b c c bb a b c cc b a b cc c b a bb c c b a 266664377775 2 SC 5is both symmetric and circulant. Because N ¼ 5 is odd, it has (N þ 1)/2 ¼ 3 distinct elements. The matrix defined in Example 3 is not a symmetric circulant because its generating elements are distinct. Next, we give a nec-essary and sufficient condition for a square matrix to be circulant. T HEOREM 4. Let rN be the cyclic forward shift matrix. Then a N N matrix C is circulant if and only if CrN ¼ rN C. Proof. Let C be an N N matrix with arbitrary elements cik for i; k ¼ 1; 2; …; N. Then CrN ¼ c1N c11 c12 c1ðN1Þ c2N c21 c22 c2ðN1Þ ... ... ... . .. ... cNN c N1 c N2 cNðN1Þ 2666437775 and rN C ¼ c21 c22 c23 c2Nc31 c32 c33 c3N ... ... ... . .. ... c11 c12 c13 c1N 2666437775 These matrices are equal if and only if the equalities c1N ¼ c21 ; c11 ¼ c22 ; c1ðN1Þ ¼ c2Nc2N ¼ c31 ; c21 ¼ c32 ; c2ðN1Þ ¼ c3N ... ... . .. ... c NN ¼ c11 ; c N1 ¼ c12 ; c NðN1Þ ¼ c1N are satisfied. Then C can be written as C ¼ c11 c12 c1Nc21 c22 c2N ... ... . .. ... cN1 c N2 c NN 2666437775 ¼ c11 c12 c1Nc1N c11 c1ðN1Þ ... ... . .. ... c12 c13 c11 2666437775 which is a N N circulant matrix with generating elements c11 ; c12 ; …; c1N . Any matrix that commutes with the cyclic forward shift matrix is, therefore, a circulant. Theorem 4 also says that circulant matrices are invariant under similarity transformations involving the cyclic forward shift matrix. That is, C is similar to itself for a similarity transformation using rN . Example 5 . Consider the 3 3 matrix A ¼ a b cc a bb c a 2435 Then a b cc a bb c a 264375 0 1 00 0 11 0 0 264375 ¼ c a bb c aa b c 264375 ¼ 0 1 00 0 11 0 0 264375 a b cc a bb c a 264375 which implies that Ar3 ¼ r3A. Thus, A ¼ circ ða; b; cÞ 2 C3 is a circulant matrix of type N ¼ 3. Next we introduce block circulant matrices, which are natural generalizations of ordinary circulants. 2.3 Block Circulant Matrices. A block circulant matrix is obtained from a circulant matrix by replacing each entry ck in Definition 9 by the M M matrix Ci for i ¼ 1; 2; …; N.DEFINITION 11 (Block Circulant Matrix). Let Ci be a M Mmatrix for each i ¼ 1; 2; …; N. Then a NM NM block circulant matrix (or block circulant) is generated from the ordered set fC1; C2; …; CN g, and is of the form C ¼ C1 C2 CN CN C1 CN1 ... ... . .. ... C2 C3 C1 2666437775 D D EFINITION 12 (Generating Matrices). Let the NM NM block circulant C be given by Definition 11. Then the elements of the ordered set fC1; C2; ; CN g are said to be the generating matrices of C. A block circulant is therefore defined completely by its generat-ing matrices. The matrix array given by Definition 11 is said to be a block circulant of type ( M, N). The set of all such matrices is denoted by BC M;N . A matrix C 2 BC M;N is not necessarily a cir-culant, as the following example demonstrates. Example 6 . Let A ¼ 2 1 1 2 and B ¼ 1 00 1 Then 6C ¼ A B 0 BB A B 00 B A BB 0 B A 2666437775 ¼ 2 1 1 0 0 0 1 0 1 2 0 1 0 0 0 1 1 0 2 1 1 0 0 00 1 1 2 0 1 0 00 0 1 0 2 1 1 00 0 0 1 1 2 0 1 1 0 0 0 1 0 2 10 1 0 0 0 1 1 2 377777777777775266666666666664 is a block circulant of type (2, 4), but it is not a circulant. Next we give a necessary and sufficient condition for a matrix to be block circulant. T HEOREM 5. Let rN be the cyclic forward shift matrix of dimen-sion N and IM be the identity matrix of dimension M. Then a NM NM matrix C is a block circulant of type (M, N) if and only if CðrN IM Þ ¼ ð rN IM ÞC. The proof of Theorem 5 follows similarly to that of Theorem 4 by replacing the scalar elements c ik with M M matrices Cik for i; k ¼ 1; 2; …; N. The reader can verify that the matrix C in Example 6 satisfies the condition in Theorem 5, but not that in Theorem 4. A block circulant, block symmetric matrix of type ( M, N) has generating matrices of the same form as Eq. (10), and is obtained by replacing each entry ck by the M M matrix Ck for k ¼ 1; 2; …; N. The set of all such matrices is denoted by BCBS M;N . The matrix C in Example 6 is recognized to be a block symmetric, block circulant matrix of type (2, 4), that is, it is contained in BCBS 2;4 , which is a subset of BC 2;4 . 2.4 Representations of Circulants. It is clear from Defini-tion 5 that the N N cyclic forward shift matrix is a circulant with N generating elements 0 ; 1; 0; …; 0; 0. The integer powers of rN can be written as r0 N ¼ circ ð1; 0; 0; 0; 0; …; 0; 0Þ ¼ IN r1 N ¼ rN ¼ circ ð0; 1; 0; 0; 0; …; 0; 0Þ r2 N ¼ circ ð0; 0; 1; 0; 0; …; 0; 0Þ ... rN1 N ¼ circ ð0; 0; 0; 0; 0; …; 0; 1Þ rNN ¼ circ ð1; 0; 0; 0; 0; …; 0; 0Þ ¼ r0 N ¼ IN 9>>>>>>>>>>>=>>>>>>>>>>>; (11) where each successive power cyclically permutes the generating elements. This enables the representation of a general circulant in terms of a finite matrix polynomial involving the cyclic forward shift matrix and its powers. C OROLLARY 3. Let C 2 CN be a circulant matrix of type N with generating elements c 1; c2; …; c N . Then C can be represented by the matrix sum C ¼ c1r0 N þ c2r1 N þ c3r2 N þ þ c N rN1 N where rN is the N N cyclic forward shift matrix . Example 7 . The matrix A ¼ circ ða; b; cÞ from Example 5 can be represented by the matrix sum A ¼ ar03 þ br13 þ cr23 ¼ aI3 þ br3 þ cr23 ¼ a 1 0 00 1 00 0 1 264375 þ b 0 1 00 0 11 0 0 264375 þ c 0 0 11 0 00 1 0 264375 ¼ a b cc a bb c a 264375 where I3 and r3 are the 3 3 identity and cyclic forward shift matrices. Corollary 3 is exploited in Sec. 2.5.4 to diagonalize a general circulant matrix, and can be generalized to represent a general block circulant matrix in terms of the cyclic forward shift matrix and its powers. C OROLLARY 4. Let C 2 BC M;N be a block circulant matrix of type (M, N) with generating matrices C1; C2; …; CN . Then C can be represented by the matrix sum C ¼ r0 N C1 þ r1 N C2 þ þ rN1 N CNwhere rN is the cyclic forward shift matrix. Corollaries 3 and 4 motivate the following result, which cap-tures the representation of circulant and block circulant matrices in terms of the cyclic forward shift matrix, and facilitates their diagonalization in Secs. 2.5.4 and 2.5.5. DEFINITION 13. Let t and s be arbitrary square matrices. Then the function qðt; sÞ ¼ XNk¼1 tk1 s is a finite sum of direct products . C OROLLARY 5. Let rN be the N N cyclic forward shift matrix. Then a circulant matrix with generating elements c 1; c2; …; cNand a block circulant matrix with generating elements C1; C2; …; CN can be represented by the matrix sums qðrN ; ckÞ ¼ XNk¼1 rk1 N c k ¼ circ ðc1; c2; …; c N Þ qðrN ; CkÞ ¼ XNk¼1 rk1 N Ck ¼ circ ðC1; C2; …; CN Þ where the function qðÞ is given by Definition 13 . What is meant by the notation qðrN ; ckÞ, for example, is to sub-stitute t with rN and s with c k in Definition 13 and then perform the summation observing any indices k introduced by the substitution . 2.5 Diagonalization of Circulants. Any circulant or block circulant matrix can be represented in terms of the cyclic forward shift matrix according to Corollary 5. The diagonalization of a general circulant begins, therefore, by finding a matrix that diago-nalizes rN . Together with some basic results from linear algebra (these are summarized in Sec. 2.1), this leads naturally to the diagonalization of an arbitrary circulant. Regarding a suitable diagonalizing matrix, there are a number of candidates [10–12,16,17,100], but all feature powers of the Nth roots of unity or their real/imaginary parts. In this work, we employ an array composed of the distinct Nth roots of unity (Sec. 2.5.1) and their integer powers in the form of the complex Fourier matrix (Sec. 2.5.2). A unitary transformation involving the Fourier matrix is used to diagonalize the cyclic forward shift matrix (Sec. 2.5.3), 7general circulant matrices (Sec. 2.5.4), and general block circulant matrices (Sec. 2.5.5). 2.5.1 Nth Roots of Unity. A root of unity is any complex num-ber that results in 1 when raised to some integer N 2 Zþ . More generally, the Nth roots of a complex number z o ¼ r o e jho are given by a nonzero number z ¼ re jh such that zN ¼ zo or r N e jN h ¼ ro e jho (12) where j ¼ ffiffiffiffiffiffiffi 1 p . Equation (12) holds if and only if r N ¼ ro and Nh ¼ ho þ 2pk with k 2 Z. Therefore, r ¼ ffiffiffiffi r oN p h ¼ ho þ 2pkN 9=;; k 2 Z (13) and the Nth roots are z ¼ ffiffiffiffi r oN p exp j ho þ 2pkN ; k 2 Z (14) Equation (14) shows that the roots all lie on a circle of radius ffiffiffiffi r oN p centered at the origin in the complex plane, and that they are equally distributed every 2 p=N radians. Thus, all of the distinct roots correspond to k ¼ 0; 1; 2; …; N 1. DEFINITION 14 (Distinct Nth Roots of Unity). The distinct N th roots of unity follow from Eq. (14) by setting r o ¼ 1 and ho ¼ 0and are denoted by wðkÞ N ¼ e j 2pkNfor integers k ¼ 0; 1; 2; …; N 1. DEFINITION 15 (Primitive Nth Root of Unity). The primitive N th root of unity is denoted by w N ¼ e j 2p N which corresponds to k ¼ 1 in Definition 14 . C OROLLARY 6. The integer powers w kN of the primitive Nth root of unity are equivalent to the distinct N th roots of unity w ðkÞ N for k ¼ 1; 2; …; N. Proof . Consider wkN ¼ ð e j 2p N Þk ¼ e j 2pkN ¼ wðkÞ N , which follows from Definition 14. Thus, the powers 1; w1 N ; w2 N ; …; w N1 N are equivalent to the distinct Nth roots of unity 1; wð1Þ N ; wð2Þ N ; …; wðN1Þ N where w0 N ¼ wð0Þ N ¼ 1. Example plots of the distinct Nth roots of unity are shown in Fig. 2, where wðkÞ N are arranged on the unit circle in the complex plane (centered at the origin) for N ¼ 1; 2; …; 9. Note that wð0Þ N ¼ 1 is real, as is wðN=2Þ N ¼ 1 if N is even. The remaining roots appear in complex conjugate pairs. Thus, the distinct Nth roots of unity are symmetric about the real axis in the complex plane. Example 8 . Let N ¼ 4. Then the distinct Nth roots of unity are given by the set fw04; w14; w24; w34g ¼ f e j 2p 40 ; e j 2p 41 ; e j 2p 42 ; e j 2p 43 g¼ f e0; e j p 2 ; e j p; e j 3p 2 g¼ f 1; j; 1; jg These four distinct roots of unity can be visualized in Fig. 2 for the case of N ¼ 4. D EFINITION 16. Let w N be the primitive N th root of unity. Then XN ¼ 1 0 w Nw2 N . .. 0 wN1 N 266666664377777775 NN ¼ diag ð1; wN ; w2 N ; …; wN1 N Þ is the N N diagonal matrix formed by placing the distinct N th roots of unity 1; w N ; w2 N ; …; w N1 N along its diagonal . The matrix XN appears naturally in the diagonalization of cir-culants, which is discussed in Secs. 2.5.3–2.5.5. 2.5.2 The Fourier Matrix. This section introduces the com-plex Fourier matrix and its relevant properties, including the sym-metric structure of the N-vectors that compose its columns. A key result is that the Fourier matrix is unitary, which is systematically developed and proved. DEFINITION 17 (Fourier Matrix). The N N Fourier matrix is defined as EN ¼ 1ffiffiffiffi N p 1 1 1 11 w N w2 N w N1 N 1 w2 N w4 N w2ðN1Þ N ... ... ... . .. ... 1 w N1 N w2ðN1Þ N wðN1Þð N1Þ N 266666664377777775 NN where w N is the primitive N th root of unity and N 2 Zþ. Fig. 2 Example plots of the distinct Nth roots of unity 8Example 9 . For the special case of N ¼ 4, the Fourier matrix is given by E4 ¼ 1ffiffiffi 4 p 1 1 1 11 j 1 j 1 1 1 11 j 1 j 2666437775 Clearly the Fourier matrix is symmetric, but generally it is not Hermitian. It can be written element wise as ðEN Þik ¼ 1ffiffiffiffi N p wði1Þð k1Þ N ¼ 1ffiffiffiffi N p e jði1Þuk ¼ 1ffiffiffiffi N p e jðk1Þui ; i; k ¼ 1; 2; …; N (15) where ui ¼ 2p N ði 1Þ (16) is the angle subtended from the positive real axis in the complex plane to the ith power of w N, which is also the ði þ 1Þth of the N roots of unity according to Definition 14 and the numbering scheme in Fig. 2. C OROLLARY 7. The matrix EH N is obtained from EN by changing the signs of the powers of each element . Proof . The Fourier matrix is unaffected by transposition because it is symmetric. Thus, the ( i, k) element of EH N is EH N ik ¼ ð EN Þik ¼ 1ffiffiffiffi N p wði1Þð k1Þ N ¼ 1ffiffiffiffi N p wð i1Þð k1Þ N where the identity w kN ¼ e j 2p Nk ¼ ej 2p Nk ¼ e j 2p N k ¼ wkN ; k 2 Z is employed. It follows that EH N can be obtained from EN by changing the sign of the powers of the Nth roots of unity. It is shown in Sec. 2.8 that all circulant matrices contained in CN share the same linearly independent eigenvectors, the ele-ments of which compose the N columns (or rows) of EN.DEFINITION 18. Let w N be the primitive Nth root of unity. Then for i ¼ 1; 2; …; N the columns of the Fourier matrix EN are defined by ei ¼ 1ffiffiffiffi N p 1; wði1Þ N ; w2ði1Þ N ; …; wðN1Þð i1Þ N T ¼ 1ffiffiffiffi N p 1; ejui ; ej2ui ; …; e jðN1Þui T where ui is given by Eq. (16) . Example 10 . The third column of the Fourier matrix E4 is given by e3 ¼ 1ffiffiffi 4 p 1; wð31Þ 4 ; w2ð31Þ 4 ; w3ð31Þ 4 T ¼ 12 1; w24; w44; w64 T ¼ 12 1; e j 2p 42 ; e j 2p 44 ; e j 2p 46 T ¼ 12 1; ejp; ej2p; ej3p T ¼ 12 ð1; 1; 1; 1ÞTfor the special case of N ¼ 4. The columns ei of the Fourier matrix EN ¼ ð e1; …; eN Þ exhibit a symmetric structure with respect to the index i ¼ (N þ 2)/2 for even N. In Example 9, for instance, the vectors e1 and eðNþ2Þ=2 ¼ e3 are real and distinct, and the vectors e2 and e4appear in complex conjugate pairs. This same structure generally holds for any EN with even N. To see this, consider eNþ22 6q ¼ 1ffiffiffiffi N p 1; wðN 26qÞ N 1 ; …; wðN 26qÞ N N1 T ¼ 1ffiffiffiffi N p 1; w6qN 1; …; w6qN N1 T(17) where the integers 6q correspond to vector pairs relative to the index i ¼ ð N þ 2Þ=2 and the identity w N 26qN ¼ w N 2 N w6qN ¼ w6qN is employed. The case of q ¼ 0 corresponds to i ¼ ð N þ 2Þ=2 and yields the real vector eNþ22 ¼ 1ffiffiffiffi N p ð1; 1; 1; …; 1; 1ÞT (18) which has the same value for each element with alternating signs from element to element. For q 6 ¼ 0 the terms w6qN are complex conjugates according to the proof of Corollary 7 such that e ðNþ2Þ=2ð Þþ q and e ðNþ2Þ=2ð Þ q are complex conjugate pairs. There are (N 2)/2 such pairs corresponding to q ¼ f 1; 2; …; ðN 2Þ=2g in Eq. (17). Finally, the case of i ¼ 1 always yields the real vector e1 ¼ 1ffiffiffiffi N p ð1; 1; 1; …; 1; 1ÞT (19) for even and odd N. A similar formulation for odd N shows that e1is real and distinct and the remaining ( N – 1)/2 vectors appear in complex conjugate pairs. This is shown by example in Sec. 3.3 in the context of vibration modes for a cyclic structure with a single DOF per sector. A key feature of the Fourier matrix is that it is unitary. This is essentially a statement of orthogonality of each column of EN and is captured by the following lemmas. L EMMA 1 (Finite Geometric Series Identity). Let N 2 Zþ and q 2 C. Then XsþN1 r¼s qr ¼ qsð1 qN Þ 1 qfor any s 2 Z and q 6 ¼ 1. Proof. Consider the finite geometric series XsþN1 r¼s qr ¼ qs þ q sþ1 þ qsþ2 þ qsþN1 ¼ qsð1 þ q þ q2 þ þ qN1Þ Multiplying from the left by q yields q XsþN1 r¼s qr ¼ qsðq þ q2 þ q3 þ þ q N Þ Subtraction of the second equation from the first results in ð1 qÞ XsþN1 r¼s qr ¼ qsð1 qN Þ9from which the proof is established by division of the term (1 – q)because q 6 ¼ 1 by restriction. Lemma 1 is used to establish the following result, which is required to show that the Fourier matrix is unitary. The orthogon-ality condition is fundamental to the diagonalization of circulants in Secs. 2.5.3–2.5.5, and the relationship between the DFT and IDFT in Sec. 2.7. L EMMA 2. Let w N be the primitive Nth root of unity with N 2 Zþ. Then XsþN1 r¼s wrðikÞ N ¼ N; i k ¼ mN 0; otherwise for i ; k 2 Z and any s ; m 2 Z. Proof. Let q ¼ wðikÞ N ¼ e j 2p NðikÞ and note that qN ¼ 1. If ik ¼ mN , then q ¼ e j2pm ¼ 1 for any integer m, and it follows that XsþN1 r¼s w rðikÞ N ¼ XsþN1 r¼s qr ¼ ð 1Þs þ ð 1Þsþ1 þ þ ð 1ÞsþN1 \|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} Nterms ¼ N For the case of i k 6 ¼ mN it follows from Lemma 1 that XN1 r¼0 w rðikÞ N ¼ XN1 r¼0 qr ¼ 1 qN 1 q ¼ 1 11 q ¼ 0(Lem. 1) which completes the proof. Example 11 . Consider the orthogonality condition given by Lemma 2 for s ¼ 0. Then if i k ¼ N ¼ 5, X51 r¼0 wr55 ¼ X4 r¼0 e j 2p 5 5r ¼ X4 r¼0 e j 2p 55r ¼ e0 þ e j2p þ e j4p þ e j6p þ e j8p ¼ 1 þ 1 þ 1 þ 1 þ 1 ¼ 5which is numerically equal to N, as expected. If instead we set i k ¼ 5 but N ¼ 4, then X41 r¼0 w r54 ¼ X3 r¼0 e j 2p 4 5r ¼ X3 r¼0 e j 2p 45r ¼ X3 r¼0 e j 5p 2r ¼ e5p 20 þ e5p 21 þ e5p 22 þ e5p 23 ¼ e0 þ e5p 2 þ e5p þ e15 p 2 ¼ 1 þ j 1 j ¼ 0sums to zero. Lemma 2 allows for representations of the N N identity, flip, and cyclic forward shift matrices in terms of certain conditions on their indices relative to N.C OROLLARY 8. For i ; k ¼ 1; …; N and any integer m, the (i, k) elements of the N N identity, flip, and cyclic forward shift matri-ces can be represented by the summations dik ¼ ð IN Þik ¼ 1 N XN1 r¼0 w rðikÞ N ¼ 1; i k ¼ mN 0; otherwise ( ðjN Þik ¼ 1 N XN1 r¼0 wrðiþk2Þ N ¼ 1; i þ k 2 ¼ mN 0; otherwise ( ðrN Þik ¼ 1 N XN1 r¼0 wrðikþ1Þ N ¼ 1; i k þ 1 ¼ mN 0; otherwise ( where dik is the Kronecker delta . The reader can verify Corollary 8 for the special case of N ¼ 3by inspection of the arrays in Fig. 3. We are now ready to state the key result required to diagonalize a general circulant matrix. Corollaries 7 and 8 are used to show that the Fourier matrix is unitary. T HEOREM 6 (Unitary Fourier Matrix). The Fourier matrix EN is unitary. Proof. For 1 i; k N, the ( i, k) entry of EH N EN is given by EH N EN ik ¼ XNr¼1 ðEH N Þir ðEN Þrk ¼ XNr¼11ffiffiffiffi N p wð i1Þð r1Þ N 1ffiffiffiffi N p wðr1Þð k1Þ N ¼ 1 N XNr¼1 wðr1Þð ikÞ N ¼ 1 N XN1 r¼0 w rðikÞ N (Eq. 15 and Cor. 7) ¼ INð Þik (Cor. 8) from which it follows that EH N EN ¼ IN . Example 12 . Consider the matrix E4 from Example 9. Because the Fourier matrix is unitary, it follows that Fig. 3 Arrays showing the ( i, k) elements of the ( a) identity, ( b)flip, and ( c) cyclic forward shift matrices of dimension N 5 3 for i, k 5 1, 2, 3 10 EH 4 E4 ¼ 1ffiffiffi 4 p 1 1 1 11 j 1 j 1 1 1 11 j 1 j 2666666437777775 1ffiffiffi 4 p 1 1 1 11 j 1 j 1 1 1 11 j 1 j 2666666437777775 ¼ 144 0 0 00 4 0 00 0 4 00 0 0 4 2666666437777775 ¼ I4where I4 is the 4 4 identity matrix. The following corollaries follow from Theorem 6. C OROLLARY 9. If EN is the unitary Fourier matrix, then EN is also unitary . Proof. Note that EH N ¼ ETN ¼ EN because EN ¼ ETN is symmet-ric. It follows that EH N EN ¼ ETN EN ¼ EN EN ¼ EN EH N ¼ EH N EN (Eq. 8) ¼ IN (Thm. 6) which implies that EN is unitary. Example 13 . Consider the unitary matrix E4 from Example 12. Corollary 9 guarantees that E4 is also unitary. Thus, EH 4 E4 ¼ 1ffiffiffi 4 p 1 1 1 11 j 1 j 1 1 1 11 j 1 j 2666666437777775 1ffiffiffi 4 p 1 1 1 11 j 1 j 1 1 1 11 j 1 j 2666666437777775 ¼ 144 0 0 00 4 0 00 0 4 00 0 0 4 2666666437777775 ¼ I4as expected. C OROLLARY 10. Let ei denote the ith column of the Fourier matrix EN and dik be the Kronecker delta. Then eH i ek ¼ dik for i ; k ¼ 1; 2; …; N. Proof. For i; k ¼ 1; 2; …; N, the ( i, k) entry of EH N EN can be written as eH i ek ik ¼ EH N EN ik ¼ INð Þik (Thm. 6) where IN is the N N identity matrix. Thus, eH i ek ¼ dik . Corollary 10 shows that the columns (and rows) of the Fourier matrix are mutually orthogonal. The result can also be proved by expanding the product eH i ek according to Eq. (15) and invoking Lemma 2, as is done in the proof of Theorem 6. The same result also follows by expanding EH N EN ¼ eH 1 ... eH i ... eH N 2666666666666437777777777775 e1 ek eN½ ¼ eH 1 e1 eH 1 ek … eH 1 eN ... . .. ... ... eH i e1 eH i ek eH i eN ... ... . .. ... eH N e1 … eH N ek … eH N eN 2666666666666437777777777775 ¼ IN (20) which is a statement of unitary EN. The matrix equality holds only if each eH i ek ¼ dik for i; k ¼ 1; 2; …N. Example 14 . Consider the vector e3 ¼ 12 ð1; 1; 1; 1ÞT from Example 10. Then the product eH 3 e3 ¼ 12 1 1 1 1½ 121 11 1 2666666437777775 ¼ 14 ð1 þ 1 þ 1 þ 1Þ¼ 1corresponds to d33 ¼ 1 in Corollary 10. However, the product eH 3 e2 ¼ 12 1 1 1 1½ 121 j 1 j 2666666437777775 ¼ 14 ð1 j 1 þ jÞ¼ 0vanishes because e2 and e3 are mutually orthogonal. C OROLLARY 11. If EN is the N N Fourier matrix and IM is the identity matrix of dimension M, then the NM NM matrix EN IM is unitary . Proof. Consider the matrix product ðEN IM ÞH ðEN IM Þ¼ ð EH N IH M Þð EN IM Þ (Eq. 6 b) ¼ ð EH N EN Þ ðIH M IMÞ (Eq. 4) 11 ¼ IN IM ¼ INM (Thm. 6) where IN and INM are identity matrices of dimension N and NM ,respectively. C OROLLARY 12. If ei denotes the ith column of the Fourier matrix EN , IM is the identity matrix of dimension M, and dik is the Kro-necker delta, then the NM M matrices ei IM are such that ðei IMÞH ðek IMÞ ¼ dik IMfor i ; k ¼ 1; 2; …; N. Proof. Consider the matrix product ðei IM ÞH ðek IM Þ ¼ ð eH i IH M Þð ek IM Þ (Eq. 6 b) ¼ ð eH i ekÞ ðIH M IM Þ (Eq. 4) ¼ dik IM (Cor. 10) ¼ dik IM which completes the proof. Corollary 12 can also be obtained directly from Corollary 11 by writing EH N IM ¼ eH 1 eH 2 ... eH N 266666664377777775 IM ¼ eH 1 IM eH 2 IM ... eH N IM 266666664377777775 and EN IM ¼ ð e1; e2; …; eN Þ IM ¼ ð e1 IM; e2 IM; …; eN IM Þ expanding these matrices similarly to Eq. (20), and setting the result equal to INM ¼ diag ðIM; IM; …; IMÞ.Next we derive a relationship between the Fourier and flip matrices. T HEOREM 7. Let EN and jN denote the N N Fourier and flip matrices, respectively. Then E2 N ¼ jN ¼ EH N 2: Proof. We first shown that E2 N ¼ EN EN ¼ jN . For any inte-ger m and i; k ¼ 1; 2; …; N, the ( i, k) entry of ENEN is given by EN ENð Þik ¼ XNr¼1 ðEN Þir ðEN Þrk ¼ XNr¼11ffiffiffiffi N p wði1Þð r1Þ N 1ffiffiffiffi N p wðr1Þð k1Þ N (Eq. 15) ¼ 1 N XN1 r¼0 w rðiþk2Þ N ¼ jNð Þik (Cor. 8) from which it follows that E2 N ¼ jN . The result jN ¼ EH N 2follows from complex conjugation and transposition of jN ¼ EN EN , and by invoking the properties jH N ¼ jN and E2 N H ¼ EH N 2 . A number of properties follow directly from Theorem 7. C OROLLARY 13. Let EN and jN be the N N Fourier and flip matrices. Then (a) EN jN ¼ jN EN ;(b) j2 N ¼ IN or jN ¼ ffiffiffiffiffi IN p ; and (c) E4 N ¼ IN or EN ¼ ffiffiffiffiffi IN 4 p where IN is the N N identity matrix . Property ( a) of Corollary 13 says that the flip and Fourier matri-ces commute or, since EN is unitary, that jN is invariant under a unitary transformation with respect to EN. Thus, jN is not diago-nalizable by EN. Properties ( b) and ( c) give alternative definitions of the flip and Fourier matrices. Moreover, because the power of a diagonal matrix is obtained by raising each diagonal element to the power in question, if follows that the eigenvalues of jN are 61 and those of EN are 61 and 6j, each with the appropriate multiplicities. 2.5.3 Diagonalization of the Cyclic Forward Shift Matrix. In light of Corollary 5, diagonalization of a general circulant or block circulant matrix begins by diagonalizing the cyclic forward shift matrix. T HEOREM 8. Let EN be the N N Fourier matrix and rN be the N N cyclic forward shift matrix. Then EH N rN EN ¼ XNis a diagonal matrix, where XN is given by Definition 16 . Proof. For i; k ¼ 1; 2; …; N, the ( i, k) entry of EN XN EH N is given by EN XN EH N ik ¼ XNr¼1 XNp¼1 ðEN Þip ðXN Þpr ðEH N Þrk ¼ XNr¼1 XNp¼11ffiffiffiffi N p wði1Þð p1Þ N dpr wðr1Þ N 1ffiffiffiffi N p wð r1Þð k1Þ N (Eq. 15) ¼ 1 N XNr¼1 wði1Þð r1Þ N wðr1Þ N wð r1Þð k1Þ¼ 1 N XNr¼1 wðr1Þð ikþ1Þ N ¼ 1 N XN1 r¼0 wrðikþ1Þ N ¼ rNð Þik (Cor. 8) from which it follows that EN XN EH N ¼ rN . The desired result fol-lows by multiplying from the left by EH N , multiplying from the right by EN, and invoking Theorem 6. Theorem 8 implies that rN is unitarily similar to a diagonal ma-trix whose diagonal elements are the distinct Nth root of unity (i.e., Definition 16). Because the eigenvalues of a matrix are pre-served under such a transformation (this is guaranteed by Theo-rem 1), it follows that aðrN Þ ¼ aðXN Þ ¼ f 1; w N ; w2 N ; …; w N1 N g where aðÞ denotes the matrix spectrum. The eigenvectors of the circulant matrix rN are the linearly independent columns of EN ¼ ð e1; e2; …; eN Þ, which are given by Definition 18. In fact, all circulant matrices contained in CN share the same eigenvectors 12 ei, which is shown in Sec. 2.8. In light of Corollary 2, we have the following results. C OROLLARY 14. Let EN and rN be the N N Fourier and cyclic forward shift matrices. Then for any n 2 Zþ, EH N rnN EN ¼ XnNwhere XN given by Definition 16 . C OROLLARY 15. Let ei be the ith column of the Fourier matrix EN and rN be cyclic forward shift matrix. Then for any n 2 Zþ and i ; k ¼ 1; 2; … N, eH i rnN ek ¼ ð w i1 N Þndik ¼ w nði1Þ N dik where dik is the Kronecker delta . Corollary 15 shows that the columns of the Fourier matrix are orthogonal with respect to the cyclic forward shift matrix and its powers. It is shown in Sec. 2.5.4 that they are in fact orthogonal with respect to any circulant matrix. 2.5.4 Diagonalization of a Circulant. Corollary 16. Let the matrix XN be populated with the distinct Nth roots of unity according to Definition 16 and s be an arbitrary square matrix. Then qðXN ; sÞ ¼ diag i¼1;…;N ðqðw i1 N ; sÞÞ where qðÞ is given by Definition 13 . Proof. Consider the representation qðXN ; sÞ ¼ qðdiag ð1; w N ; w2 N ; …; w N1 N Þ; sÞ¼ XNk¼1diag ðw0ð k1Þ N ; w1ð k1Þ N ; …; wðN1Þð k1Þ N Þ s ¼ XNk¼1diag ðsw0ð k1Þ N ; sw1ð k1Þ N ; …; swðN1Þð k1Þ N Þ¼ diag i¼1;…;N XNk¼1 swði1Þð k1Þ N ! ¼ diag i¼1;…;N ðqðwi1 N ; sÞÞ which is diagonal when s is a scalar and block diagonal when s is a matrix. T HEOREM 9 (Diagonalization of a Circulant). Let C 2 CN have generating elements c 1; c2; …; c N . Then if EN is the N N Fourier matrix , EH N CE N ¼ k1 0 k2 . .. 0 kN 26666643777775 is a diagonal matrix. For i ¼ 1; 2; …; N, the diagonal elements are ki ¼ qðw i1 N ; ckÞ ¼ XNk¼1 c k wðk1Þð i1Þ N (21) where w N is the primitive Nth root of unity and the function qðÞ is given by Definition 13 . Proof. Consider the representation C ¼ qðrN ; ckÞ (Cor. 5) ¼ qðEN XN EH N ; ckÞ (Thm. 8) ¼ EN qðXN ; c kÞEH N (Thms. 2 and 6) where the last step follows directly from the proof of Theorem 2 and the polynomial term qðXN ; c kÞ ¼ diag i¼1;…;N XNk¼1 c k wðk1Þð i1Þ N ! ¼ diag i¼1;…;N ðqw i1 N ; ckÞÞ follows from Corollary 16 and Definition 13. The desired result is obtained by multiplying from the left by EH N , multiplying from the right by EN, and invoking Theorem 6. Theorem 9 shows that the Fourier matrix EN diagonalizes any N N circulant matrix. As discussed in Sec. 2.8, the columns e1; e2; …; eN of EN are the eigenvectors of C. The scalars ki in Theorem 9 are the eigenvalues of C. Unlike the eigenvectors, they depend on the elements (i.e., generating elements) of C. In fact, Eq. (21) has the same form as the DFT of a discrete time series, which is clear by comparing it to Definition 20 of Sec. 2.7. In this case, c k ðk ¼ 1; 2; …; NÞ and kiði ¼ 1; 2; …; NÞ are analogous to a discrete signal and its DFT, and are related by k1 k2 ... kN 26666643777775 ¼ ffiffiffiffi N p ENc1 c2 ... cN 26666643777775 (22) which has the same form as the matrix–vector representation of the DFT defined by Eq. (27) of Sec. 2.7. Thus, the eigenvalues ki can be calculated directly using Eq. (21) or Eq. (22), which are equivalent. If the circulant matrix in Theorem 9 is also symmetric, the eigenvalues are real-valued and certain ones are repeated, as shown in Corollary 17. C OROLLARY 17. If C 2 SC N is a symmetric circulant, Eq. (21) reduces to ki ¼ c1 þ 2 XN=2 k¼2 c k cos 2pðk 1Þð i 1Þ N þ ð 1Þi1c Nþ22 ; N even c1 þ 2 XðNþ1Þ=2 k¼2 c k cos 2pðk 1Þð i 1Þ N ; N odd 8>>>>>>>>><>>>>>>>>>: where the generating elements are given by Eq. (10) . Proof. If N is even, the generating elements of C are given by c1; c2; …; cN 2 ; c Nþ22 ; c N 2 ; …; c3; c2and Eq. (21) reduces to 13 ki ¼ XNk¼1 c k wðk1Þð i1Þ N ¼ c1w0ð i1Þ N þ c2w1ð i1Þ N þ c3w2ð i1Þ N þ þ c N 2 w N 21 ð Þði1Þ N þ c Nþ22 w Nþ22 1ð Þði1Þ N þ c N 2 w Nþ42 1ð Þði1Þ N þ þ c3wðN11Þð i1Þ N þ c2wðN1Þð i1Þ N ¼ c1 þ c2 wði1Þ N þ wðN1Þð i1Þ N þ c3 w2ði1Þ N þ wðN2Þð i1Þ N þ þ c N 2 wN22 ði1Þ N þ w NN22ð Þði1Þ N þ c Nþ22 wN 2ði1Þ N ¼ c1 þ 2 XN=2 k¼2 c k cos 2pðk 1Þð i 1Þ N þ c Nþ22 ð 1Þi1where the identity wkði1Þ N þ wðNkÞð i1Þ N ¼ 2 cos 2pkN ði 1Þ is employed. If N is odd, the generating elements of C are given by c1; c2; …; c N12 ; c Nþ12 ; c Nþ12 ; c N12 ; …; c3; c2and Eq. (21) reduces similarly to the case for even N, which is left as an exercise for the reader. The cosinusoidal nature of ki in Corollary 17 implies that k1 is distinct for a symmetric circulant matrix C 2 SC N , but the remaining elements ki ¼ kNþ2i appear in repeated pairs. How-ever, k Nþ2=2ð Þ is also distinct if N is even. Example 15 . Let C ¼ circ ð4; 1; 0; 1Þ 2 SC 4 be a symmetric circulant matrix. Then the product EH 4 CE 4 ¼ 1ffiffiffi 4 p 1 1 1 11 j 1 j 1 1 1 11 j 1 j 2666666437777775 4 1 0 1 1 4 1 00 1 4 1 1 0 1 4 2666666437777775 1ffiffiffi 4 p 1 1 1 11 j 1 j 1 1 1 11 j 1 j 2666666437777775 ¼ 141 1 1 11 j 1 j 1 1 1 11 j 1 j 2666666437777775 2 4 6 42 4j 6 4j 2 4 6 42 4j 6 4j 2666666437777775 ¼ 2 0 0 00 4 0 00 0 6 00 0 0 4 2666666437777775 ¼ diag ð2; 4; 6; 4Þ is a diagonal matrix. The diagonal elements can be computed directly using Eq. (21) for N ¼ 4. For example, k3 ¼ X4 k¼1 c k wðk1Þð 31Þ 4 ¼ 4wð11Þð 31Þ 4 þ ð 1Þwð21Þð 31Þ 4 þ 0 wð31Þð 31Þ 4 þ ð 1Þwð41Þð 31Þ 4 ¼ 4ð1Þ e j 2p 42 þ 0 e j 2p 46 ¼ 4 e jp þ 0 e j3p ¼ 4 ð 1Þ þ 0 ð 1Þ ¼ 6which is recognized to be the third diagonal element of the matrix product EH 4 CE 4 . Because C is a symmetric circulant matrix, Corol-lary 17 can also be used. Observing that N ¼ 4 is even, it follows that k3 ¼ c1 þ 2 X4=2 k¼2 c k cos 2pðk 1Þð 3 1Þ 4 þ ð 1Þ ð31Þc4þ22 ¼ 4 þ 2 ð 1Þ cos 2pð2 1Þð 3 1Þ 4 þ ð 1Þ2 0 ¼ 4 2 cos p þ 0 ¼ 4 2ð 1Þ þ 0 ¼ 6as before. The eigenvalues and eigenvectors of C are discussed in Example 24 of Sec. 2.8. Example 16 . Let C ¼ circ ð4; 1; 0; 1Þ 2 C4 be a nonsymmetric circulant matrix. Then the product EH 4 CE 4 ¼ 1ffiffiffi 4 p 1 1 1 11 j 1 j 1 1 1 11 j 1 j 26666643777775 4 1 0 11 4 1 00 1 4 1 1 0 1 4 26666643777775 1ffiffiffi 4 p 1 1 1 11 j 1 j 1 1 1 11 j 1 j 26666643777775 ¼ 141 1 1 11 j 1 j 1 1 1 11 j 1 j 26666643777775 4 4 2j 4 4 þ 2j 4 2 þ 4j 4 2 4j 4 4 þ 2j 4 4 2j 4 2 4j 4 2 þ 4j 26666643777775 ¼ 4 0 0 00 4 2j 0 00 0 4 00 0 0 4 þ 2j 26666643777775 is a diagonal matrix. The diagonal elements are the eigenvalues of C,which is stated explicitly for general C in the following corollary. They may be computed directly using Eq. (21), as it is done in Example 15, but Corollary 17 cannot be used because C is not symmetric. The eigenvalue magnitudes of the nonsymmetric matrix C in Example 16 are observed to exhibit the same multiplicity (and symmetry) as the eigenvalues in Example 15 for a symmetric cir-culant. It is shown in Sec. 2.8.3 that the eigenvalues of any circu-lant matrix C 2 CN with real-valued generating elements exhibit 14 a certain symmetry about the so-called “Nyquist” component, which is analogous to the DFT of a real-valued sequence. This is because Eq. (22) represents the DFT of the generating elements c1; c2; …; c N .C OROLLARY 18. Let ei be the ith column of the Fourier matrix EN and C be a N N circulant matrix. Then for i ; k ¼ 1; 2; …; N, eH i Ce k ¼ kidik where dik is the Kronecker delta and ki is defined by Eq. (21) for C 2 CN or Corollary 17 if C 2 SC N . Thus, the columns of the Fourier matrix are mutually orthogo-nal (Corollary 10) and orthogonal with respect to any circulant matrix (Corollary 18), not just the cyclic forward shift matrix and its integer powers (Corollary 15). Example 17 . Consider the circulant C ¼ circ ð4; 1; 0; 1Þ from Example 15. Then eH 3 Ce 3 ¼ 1ffiffiffi 4 p 1 1 1 1½ 4 1 0 1 1 4 1 00 1 4 1 1 0 1 4 2666437775 1ffiffiffi 4 p 1 11 1 2666437775 ¼ 14 1 1 1 1½ 6 66 6 2666437775 ¼ 14 24 ¼ 6which is recognized to be the third diagonal element k3 of the ma-trix EH 4 CE 4 in Example 15. However, the scalar eH 3 Ce 1 ¼ 1ffiffiffi 4 p 1 1 1 1½ 4 1 0 1 1 4 1 00 1 4 1 1 0 1 4 26666643777775 1ffiffiffi 4 p 1111 26666643777775 ¼ 14 1 1 1 1½ 2222 26666643777775 ¼ 14 0 ¼ 0vanishes, as expected, because i 6 ¼ k in Corollary 18 such that dik ¼ 0. 2.5.5 Block Diagonalization of a Block Circulant. Theorem 9 is generalized to handle block circulants using the Fourier and identity matrices together with the Kronecker product. The choice of diagonalizing matrix EH N IM is discussed in Sec. 2.6, where generalizations of Theorem 10 are considered. T HEOREM 10 (Block Diagonalization of a Block Circulant). Let C 2 BC M;N and denote its M M generating matrices by C1; C2; …; CN . Then if EN is the N N Fourier matrix and IM is the identity matrix of dimension M, ðEH N IM ÞCðEN IMÞ ¼ K1 0 K2 . .. 0 KN 266664377775 is a NM NM block diagonal matrix. For i ¼ 1; 2; …; N, the M M diagonal blocks are Ki ¼ qðw i1 N ; CkÞ ¼ XNk¼1 Ck wðk1Þð i1Þ N (23) where w N is the primitive Nth root of unity and the function qðÞ is given by Definition 13. Proof. Consider the representation C ¼ XNk¼1 rk1 N Ck (Cor. 5) ¼ XNk¼1 ðEN Xk1 N EH N Þ Ck (Cor. 14) ¼ XNk¼1 ðEN IMÞ Xk1 N Ck ðEH N IMÞ (Eq. 4) ¼ ð EN IMÞqðXN ; CkÞð EH N IMÞ (Def. 13) where qðXN ; CkÞ ¼ diag i¼1;…;N XNk¼1 Ck wðk1Þð i1Þ N ! ¼ diag i¼1;…;N ðqðw i1 N ; CkÞÞ follows from Corollary 16 and Definition 13. The desired result follows by multiplying from the left by ðEH N IM Þ ¼ ð EN IMÞH multiplying from the right by ðEN IM Þ, and invoking Corollary 11. Thus, the unitary matrix EN IM reduces any NM NM block circulant matrix with M M blocks to a block diagonal matrix with M M diagonal blocks. Example 18 . Consider C ¼ circ ðA; B; 0; BÞ 2 BC 2;4 from Example 6. It can be block diagonalized via the transformation ðEH 4 I2ÞCðE4 I2Þ. That is, 1ffiffiffi 4 p 1 0 1 0 1 0 1 00 1 0 1 0 1 0 11 0 j 0 1 0 j 00 1 0 j 0 1 0 j 1 0 1 0 1 0 1 00 1 0 1 0 1 0 11 0 j 0 1 0 j 00 1 0 j 0 1 0 j 377777777777777777775 C 266666666666666666664 1ffiffiffi 4 p 1 0 1 0 1 0 1 00 1 0 1 0 1 0 11 0 j 0 1 0 j 00 1 0 j 0 1 0 j 1 0 1 0 1 0 1 00 1 0 1 0 1 0 11 0 j 0 1 0 j 00 1 0 j 0 1 0 j 377777777777777777775266666666666666666664 ¼ diag 0 1 1 0 " # ; 2 1 1 2 " # ; 4 1 1 4 " # ; 2 1 1 2 " #! 15 which is a block diagonal matrix with 2 2 diagonal blocks. The eigenvalues and eigenvectors of C are discussed in Example 25 of Sec. 2.8. C OROLLARY 19. Let C 2 BC M;N have M M generating matri-ces C1; C2; …; CN and Ki be defined by Eq. (23). Then if each Ciis symmetric, it follows that Ki is symmetric for i ¼ 1; 2; …; N. Proof. If each Ck is symmetric for k ¼ 1; 2; …; N, then so too are the matrices Ck wðk1Þð i1Þ N for each i because wðk1Þð i1Þ N is a scalar. Moreover, the sum and difference of two symmetric matri-ces is again symmetric. If follows that Ki ¼ XNk¼1 Ck wðk1Þð i1Þ N is symmetric for i ¼ 1; 2; …; N. C OROLLARY 20. Let C 2 BC M;N be a NM NM block circulant matrix. Then for i ; k ¼ 1; 2; …; N, ðeH i IM ÞCðek IMÞ ¼ Kidik where Ki is defined by Eq. (23) . Example 19 . Consider C ¼ circ ðA; B; 0; BÞ 2 BC 2;4 from Examples 6 and 18. Then ðeH 3 I2ÞCðe3 I2Þ ¼ 1ffiffiffi 4 p 1 1 1 1½ 1 00 1 C 1ffiffiffi 4 p 1 11 1 2666437775 1 00 1 0BBB@1CCCA ¼ 4 1 1 4 is the third 2 2 block of the block diagonal matrix obtained in Example 18. 2.6 Generalizations. Let ðÞ denote an arbitrary operation that takes a square matrix as its argument and returns another square matrix with the same dimension. For example, the opera-tion could denote a matrix inverse such that ðÞ ¼ ðÞ 1 . Let ðÞ # be another arbitrary matrix operation with the same restrictions (i.e., returns another square matrix with the same dimension). Then if C 2 BC M;N has generating matrices C1; C2; …; CN , it fol-lows that ðA B#ÞCðA BÞ¼ ð A B#Þ XNk¼1 rk1 N Ck ! ðA BÞ (Cor. 5) ¼ XNk¼1 ðð A rk1 N Þ ðB#CkÞÞð A BÞ (Eq. 4) ¼ XNk¼1 ðA rk1 N AÞ ðB#CkBÞ (Eq. 4) for any matrices A 2 CNN and B 2 CMM . The importance of this result is that C can be decomposed into a summation of direct products of two separate equivalence transformations, one that operates on the cyclic forward shift matrix and the other on the generating matrices of C. This decomposition justifies the diago-nalizing matrix used in Sec. 2.5, motivates some generalizations of Theorem 10, and aids in proving orthogonality relationships for the cyclic eigenvalue problems described in Sec. 2.8. In light of Corollary 14, it is clear that the choice of A ¼ EN and ðÞ ¼ ðÞ H accomplishes block diagonalization of a matrix C 2 BC M;N . Then if B ¼ IM , the appropriate diagonalizing matrix to block decouple C without operating on its generating matrices is EN IM (see Theorem 10). However, if B and ðÞ # are kept general, we have the following result. T HEOREM 11. Let C 2 BC M;N have M M generating matrices C1; C2; …; CN and EN be the N N Fourier matrix. Then for an arbitrary matrix B 2 CMM and operator ðÞ #, ðEH N B#ÞCðEN BÞ ¼ W1 0 W2 . .. 0 WN 2666437775 is a block diagonal matrix, where Wi ¼ qðw i1 N ; B#CkBÞ ¼ XNk¼1 B#CkBwðk1Þð i1Þ N (24) is the ith M M diagonal block for i ¼ 1; 2; …; N. C OROLLARY 21. Let C 2 BC M;N be a NM NM block circulant matrix, B 2 CMM be an arbitrary square matrix, and ðÞ # denote an arbitrary operation that takes a square matrix as its argument and returns another square matrix with the same dimension. Then for i ; k ¼ 1; 2; …; N, ðeH i B#ÞCðek BÞ ¼ Widik where Wi is defined by Eq. (24) . Theorem 11 is useful if there exists an equivalence transforma-tion B#CkB that simplifies each of the generating matrices. For example, if each Ck is a circulant of type M, then the additional choice of B ¼ EM and ðÞ # ¼ ðÞ H fully diagonalizes a block cir-culant matrix C 2 BC N;M with circulant blocks. C OROLLARY 22. Let C 2 BC M;N have generating matrices C1; C2; …; CN 2 CM and denote the generating elements of each Ci by c ð1Þ i ; cð2Þ i ; …; cðMÞ i . Then ðEH N EH M ÞCðEN EM Þ ¼ diag i¼1;…;N kð1Þ i 0 kð2Þ i . .. 0 kðMÞ i 2666666437777775 is a NM NM diagonal matrix, where kðpÞ i ¼ XNk¼1 XMl¼1 cðlÞ k wðl1Þð p1Þ M wðk1Þð i1Þ N (25) is the pth diagonal element of the ith M M block for i ¼ 1; 2; …; N and p ¼ 1; 2; …; M. Corollary 22 shows that the NM -dimensional eigenvectors qðpÞ i of C are the columns of EN EM . This is in contrast to rotation-ally periodic structures, where q is partitioned into N M -vectors corresponding to each sector and decomposed into a set of N reduced-order eigenvalue problems described in Sec. 2.8. Example 20 . Consider C ¼ circ ðA; B; 0; BÞ 2 BC 2;4 from Examples 6, 18, and 19. Because each of its generating matrices is a circulant, that is, ðA; B; 0Þ 2 C2 , the block circulant C is diagon-alized via the transformation 16 ðEH 4 EH 2 ÞCðE4 E2Þ ¼ 1ffiffiffi 4 p 1 1 1 11 j 1 j 1 1 1 11 j 1 j 26666643777775 1 11 j " #0BBBBB@1CCCCCA C 1ffiffiffi 4 p 1 1 1 11 j 1 j 1 1 1 11 j 1 j 26666643777775 1 11 j " #0BBBBB@1CCCCCA ¼1 0 0 0 0 0 0 00 1 0 0 0 0 0 00 0 1 0 0 0 0 00 0 0 3 0 0 0 00 0 0 0 3 0 0 00 0 0 0 0 5 0 00 0 0 0 0 0 1 00 0 0 0 0 0 0 3 37777777777777777752666666666666666664 from which is follows that aðCÞ ¼ f 1; 1; 1; 3; 3; 5; 1; 3g. Observ-ing that cð1Þ 1 ¼ 2; cð1Þ 2 ¼ 1; cð1Þ 3 ¼ 0; cð1Þ 4 ¼ 1 cð2Þ 1 ¼ 1; cð2Þ 2 ¼ 0; cð2Þ 3 ¼ 0; cð2Þ 4 ¼ 0 9=; are the generating elements of C, the NM ¼ 4 2 ¼ 8 diagonal elements can be calculated directly using Eq. (25). For example, the second diagonal element ( p ¼ 2) of the third 4 4 diagonal block ( i ¼ 3) is given by kð2Þ 3 ¼ cð1Þ 1 wð11Þð 21Þ 2 wð11Þð 31Þ 4 þ cð2Þ 1 wð21Þð 21Þ 2 wð11Þð 31Þ 4 þ cð1Þ 2 wð11Þð 21Þ 2 wð21Þð 31Þ 4 þ cð2Þ 2 wð21Þð 21Þ 2 wð21Þð 31Þ 4 þ cð1Þ 3 wð11Þð 21Þ 2 wð31Þð 31Þ 4 þ cð2Þ 3 wð21Þð 21Þ 2 wð31Þð 31Þ 4 þ cð1Þ 4 wð11Þð 21Þ 2 wð41Þð 31Þ 4 þ cð2Þ 4 wð21Þð 21Þ 2 wð41Þð 31Þ 4 ¼ ð 2Þð 1Þð 1Þ þ ð 1Þð 1Þð 1Þþ ð 1Þð 1Þð 1Þ þ ð 0Þð 1Þð 1Þþ ð 0Þð 1Þð 1Þ þ ð 0Þð 1Þð 1Þþ ð 1Þð 1Þð 1Þ þ ð 0Þð 1Þð 1Þ¼ 5The reader can compare this matrix decomposition to the results in Example 25 of Sec. 2.8.1. Finally, the eigenvectors qðpÞ i are the columns of E4 E2 and are stated explicitly in Example 25. 2.7 Relationship to the Discrete Fourier Transform. Before turning to the circulant eigenvalue problem, we consider a somewhat tangential but relevant subject on the DFT and its inverse, which are central to the analysis of experimental data in a wide range of fields, including mechanical vibrations. In this detour we show that computation of the DFT is, in fact, a multipli-cation of an N-vector of discrete signal samples by the Fourier matrix and a constant c f. The IDFT is similarly defined using the Hermitian of the Fourier matrix and a constant ci, where c f c i ¼ 1=N. More importantly, it is shown that the DFT computation is exactly analogous to the determination of the eigenvalues of a circulant matrix given its generating elements. We begin by defining the DFT sinusoids, which provide a conven-ient means of representing the DFT of a discretized signal, and then develop the relationships of interest for the DFT and the IDFT. We present only the basic results as they relate to the theory and mathematics of circulants. The reader can find a vast literature on related topics [89–91,102–110]. DEFINITION 19 (DFT Sinusoids). Let w kN denote the distinct Nth roots of unity, where w N ¼ e j 2p N is the primitive root. Then the DFT sinusoids are SkðrÞ ¼ ð w kN Þr ¼ wkr N ¼ e j 2p Nkr for k ; r ¼ 0; 1; …; N 1. C OROLLARY 23. The DFT sinusoids are orthogonal . Proof. Consider the DFT sinusoids SiðrÞ ¼ e j 2p Nir SkðrÞ ¼ e j 2p Nkr ) ; i; k; r ¼ 0; 1; …; N 1The inner product of Si(r) and Sk(r) is given by hSiðrÞ; SkðrÞi ¼ XN1 r¼0 SiðrÞS kðrÞ¼ XN1 r¼0 wir N w kr N (Def. 19) ¼ XN1 r¼0 w ir N wkr N (Cor. 7) ¼ XN1 r¼0 w rðikÞ N ¼ N; i ¼ k 0; otherwise ( (Lem. 2) which shows that the DFT sinusoids are orthogonal. The Fourier matrix in Definition 17 can be written element wise as ðEN Þik ¼ 1ffiffiffiffi N p wðk1Þð i1Þ N ¼ 1ffiffiffiffi N p e j 2p Nðk1Þð i1Þ ¼ 1ffiffiffiffi N p Si1ðk 1Þ ¼ 1ffiffiffiffi N p Sk1ði 1Þ for each i; k ¼ 1; 2; …; N. It follows that EN ¼ 1ffiffiffiffi N p S0ð0Þ S0ð1Þ … S0ðN 1Þ S1ð0Þ S1ð1Þ … S1ðN 1Þ S2ð0Þ S2ð1Þ … S2ðN 1Þ ... ... . .. ... SN1ð0Þ SN1ð1Þ … S N1ðN 1Þ 2666666666437777777775 ¼ 1ffiffiffiffi N p WN (26) is a representation of the Fourier matrix in terms of the DFT sinu-soids, where WN is the DFT sinusoid matrix. The DFT is formally 17 defined next and subsequently reformulated in terms of a matrix multiplication involving WN.D EFINITION 20 (DFT). Let x(i) denote a finite sequence with indi-ces i ¼ 1; 2; …; N. Then for k ¼ 1; 2; …; N, the DFT of x(i) is the sequence XðkÞ ¼ XNi¼1 xðiÞe j 2p Nði1Þð k1Þ ¼ XNi¼1 xðiÞwði1Þð k1Þ N ¼ XNi¼1 xðiÞSi1ðk 1Þ where w N is the primitive Nth root of unity and S iðkÞ ¼ w ik N is a DFT sinusoid . The DFT preserves the units of x(i). That is, if x(i) has engineer-ing units EU , then the units of X(k) are also EU . This is clear from Definition 20, where the exponential function is dimensionless. Expanding each X(k) in Definition 20 yields Xð1Þ ¼ xð1ÞS0ð0Þ þ þ xðNÞS0ðN 1Þ Xð2Þ ¼ xð1ÞS1ð0Þ þ þ xðNÞS1ðN 1Þ ... XðkÞ ¼ xð1ÞSi1ð0Þ þ þ xðNÞSi1ðN 1Þ ... XðNÞ ¼ xð1ÞSN1ð0Þ þ þ xðNÞS N1ðN 1Þ 9>>>>>>>>>>>>=>>>>>>>>>>>>; If the discrete samples x(i) and corresponding sequence of DFTs X(k) are assembled into the configuration vectors xN ¼ ð xð1Þ; xð2Þ; …; xðNÞÞ T XN ¼ ð Xð1Þ; Xð2Þ; …; XðNÞÞ T ) then the DFT of xN can be represented in the matrix–vector form XN ¼ WN xN (27) where WN ¼ ffiffiffiffi N p EN follows from Eq. (26). Thus, the DFT com-putation is simply a multiplication of the configuration vector xN with the DFT sinusoid matrix WN and constant c f ¼ 1. Equation (27) has exactly the same form as Eq. (22), where the generating elements of a circulant matrix are analogous to the sequence of signals x(i) and the resulting eigenvalues are analogous to the sequence of DFTs X(k). Example 21 . Consider the configuration vector of samples x4 ¼ ð 0; 1; 2; 3ÞT . The DFT of x4 follows from Eq. (27) with N ¼ 4 and is given by X4 ¼ W4x4 ¼ S0ð0Þ S0ð1Þ S0ð2Þ S0ð3Þ S1ð0Þ S1ð1Þ S1ð2Þ S1ð3Þ S2ð0Þ S2ð1Þ S2ð2Þ S2ð3Þ S3ð0Þ S3ð1Þ S3ð2Þ S3ð3Þ 266664377775 xð1Þ xð2Þ xð3Þ xð4Þ 266664377775 ¼ 1 1 1 11 j 1 j 1 1 1 11 j 1 j 266664377775 0123 266664377775 ¼ ð 6; 2 þ 2j; 2; 2 2jÞTwhere W4 ¼ ffiffiffi 4 p E4 follows from Example 9 or by elementwise direct computation according to Definition 19. Alternatively, each element X(k) of X4 can be computed using the summation given in Definition 20. If k ¼ 3, for example, Xð3Þ ¼ X4 i¼1 xðiÞe j 2p 4ði1Þð 31Þ ¼ X4 i¼1 xðiÞe jpði1Þ¼ xð1Þejp0 þ xð2Þejp1 þ xð3Þejp2 þ xð4Þejp3 ¼ ð 0Þð 1Þ þ ð 1Þð 1Þ þ ð 2Þð 1Þ þ ð 3Þð 1Þ¼ 0 1 þ 2 3 ¼ 2which is the third element of X4 , as expected. Results for k ¼ 1, 2, 4 follow similarly. If XN is known, the N-vector xN is recovered from Eq. (27) by multiplying from the left by EH N and invoking Theorem 6. Then EH N XN ¼ EH N WN xN (Eq. 27) ¼ EH N ffiffiffiffi N p EN xN (Eq. 26) ¼ ffiffiffiffi N p EH N EN xN ¼ ffiffiffiffi N p IN xN (Thm. 6) ¼ ffiffiffiffi N p xN and it follows that xN ¼ 1ffiffiffiffi N p EH N XN ¼ 1 N WH N XN (28) is a matrix–vector representation of the IDFT of XN. That is, the IDFT computation is simply a matrix multiplication of XN with the Hermitian of WN and constant c i ¼ 1=N such that cf c i ¼ 1=N.In light of Corollary 7, Eq. (28) can be written as xð1Þ xð2Þ ... xðNÞ 2666666437777775 ¼ 1 NS0ð0Þ S0ð1Þ … S0ðN 1Þ S1ð0Þ S1ð1Þ … S1ðN 1Þ ... ... . .. ... SN1ð0Þ SN1ð1Þ … S N1ðN 1Þ 266666664377777775 Xð1Þ Xð2Þ ... XðNÞ 2666666437777775 Expanding each row yields xð1Þ ¼ 1 N ðXð1ÞS0ð0Þ þ þ XðNÞS0ðN 1ÞÞ xð2Þ ¼ 1 N ðXð1ÞS1ð0Þ þ þ XðNÞS1ðN 1ÞÞ ... xðiÞ ¼ 1 N ðXð1ÞSk1ð0Þ þ þ XðNÞSk1ðN 1ÞÞ ... xðNÞ ¼ 1 N ðXð1ÞSN1ð0Þ þ þ XðNÞS N1ðN 1ÞÞ 9>>>>>>>>>>>>>>>>>>=>>>>>>>>>>>>>>>>>>; which provides an alternative representation of the IDFT. 18 D EFINITION 21 (IDFT). Let the sequence X(k) be defined accord-ing to Definition 20. Then for each i ¼ 1; 2; …; N, the IDFT of X(k) is the sequence xðiÞ ¼ 1 N XNk¼1 XðkÞSk1ði 1Þ¼ 1 N XNk¼1 XðkÞwð i1Þð k1Þ N ¼ 1 N XNk¼1 XðkÞej 2p Nði1Þð k1Þ where w N is the primitive Nth root of unity and S iðkÞ ¼ w ik N is a DFT sinusoid . The IDFT preserves the units of X(k). If X(k) has engineering units EU , then the units of x(i) are also EU . This is clear from Def-inition 21, where the exponential function is dimensionless, as is the number N. Example 22 . Reconsider Example 21, where it is shown that the DFT of x4 ¼ ð 0; 1; 2; 3ÞT is the four-vector X4 ¼ ð 6; 2 þ 2j; 2; 2 2jÞT . The IDFT of X4 follows from Eq. (28) with N ¼ 4and is given by x4 ¼ 1 N WH 4 X4 ¼ 1 NS0ð0Þ S0ð1Þ S0ð2Þ S0ð3Þ S1ð0Þ S1ð1Þ S1ð2Þ S1ð3Þ S2ð0Þ S2ð1Þ S2ð2Þ S2ð3Þ S3ð0Þ S3ð1Þ S3ð2Þ S3ð3Þ 2666666437777775 Xð1Þ Xð2Þ Xð3Þ Xð4Þ 2666666437777775 ¼ 141 1 1 11 j 1 j 1 1 1 11 j 1 j 2666666437777775 6 2 þ 2j 2 2 2j 2666666437777775 ¼ 14 ð0; 4; 8; 12 ÞT ¼ ð 0; 1; 2; 3ÞTwhich is the same four-vector from Example 21. Alternatively, each element x(i) of x4 can be computed using the summation given in Definition 21. If i ¼ 3, for example, xð3Þ ¼ 14 X4 k¼1 XðkÞej 2p 4ð31Þð k1Þ ¼ 14 Xð1Þejp0 þ Xð2Þejp1 þ Xð3Þejp2 þ Xð4Þejp3 ¼ 14 ð6Þð 1Þ þ ð 2 2jÞð 1Þ þ ð 2Þð 1Þ þ ð 2 þ 2jÞð 1Þð Þ¼ 2which is the third element of x4 , as expected. Results for k ¼ 1, 2, 4 follow similarly. There are other suitable definitions of the DFT/IDFT pair. For example, the signs of the exponents are irrelevant as long as they are opposite in the DFT and IDFT. To see this, suppose that the sign of the exponential in Definition 20 is negative instead of positive. Then each entry S k(r) in WN is replaced with SkðrÞ to produce WN (see Corollary 7), and the DFT is instead given by XN ¼ WN xN (29) The corresponding IDFT takes the form xN ¼ 1 N WH N XN (30) which follows in the same way as Eq. (28) by replacing EN with EN and invoking Corollary 9. Similarly, the multiplicative con-stants that define the DFT/IDFT pair are arbitrary as long as the product of the constants is equal to 1/ N. This is important for applications involving a transformation of data to the frequency domain for analysis and then back to the time domain for results. However, the multiplicative constant is irrelevant if the analysis goal is only to identify periodicities in a data set (e.g., frequencies that correspond to amplification of a structural response). Thus, Definitions 20 and 21 form a representation of the DFT/ IDFT pair, where Eqs. (27) and (28) are the corresponding matrix–vector forms. The DFT and IDFT pair can be written in the general form XðkÞ ¼ cf XNi¼1 xðiÞe6j 2p Nði1Þð k1Þ (31 a) xðiÞ ¼ c i XNk¼1 XðkÞe j 2p Nði1Þð k1Þ (31 b)where c i and c f are such that cf c i ¼ 1=N but are otherwise arbi-trary. Common choices are fcf ; cig ¼ f 1= ffiffiffiffi N p ; 1= ffiffiffiffi N p g or fcf ; cig ¼ f 1; 1=Ng, where the multiplicative constants in the matrix–vector formulation (i.e., Eqs. (27) and (28) or Eqs. (29) and (30)) are adjusted accordingly. If the generally complex sequences x(i) and X(k) are dissected into their real and imaginary parts, the DFT/IDFT pair is repre-sented by xðiÞ ¼ xRðiÞ þ jx I ðiÞ (32 a) XðkÞ ¼ XRðkÞ þ jX I ðkÞ (32 b)where the subscripts R and I denote the real and imaginary parts and each sequence pair ðxRðiÞ; x I ðiÞÞ and ðX RðkÞ; XI ðkÞÞ is real for i; k ¼ 1; 2; …; N. An alternative representation of the DFT and IDFT is obtained by substituting Eq. (32) in Definition 20, writing the complex exponential in terms of sines and cosines, and col-lecting the real and imaginary terms. The result is XRðkÞ ¼ XNi¼1 x RðiÞ cos 2pði 1Þð k 1Þ N x I ðiÞ sin 2pði 1Þð k 1Þ N (33 a) X I ðkÞ ¼ XNi¼1 x RðiÞ sin 2pði 1Þð k 1Þ N þ x I ðiÞ cos 2pði 1Þð k 1Þ N (33 b)which together form the DFT of x(i) according to Eq. (32 b). A similar relationship for the IDFT is obtained by substituting Eq. (32) into Definition 21 and collecting the real and imaginary terms. Then 19 x RðiÞ ¼ þ 1 N XNi¼1 X RðkÞ cos 2pði 1Þð k 1Þ N þ X I ðkÞ sin 2pði 1Þð k 1Þ N (34 a) x I ðiÞ ¼ 1 N XNi¼1 X RðkÞ sin 2pði 1Þð k 1Þ N XI ðkÞ cos 2pði 1Þð k 1Þ N (34 b)together form the IDFT of X(k) according to Eq. (32 a). The DFT/ IDFT pair defined by Eqs. (32)–(34) are equivalent to Definitions 20 and 21. Example 23 . Reconsider Example 21, where it is shown that the DFT of x4 ¼ ð 0; 1; 2; 3ÞT is the four-vector X4 ¼ ð 6; 2 2j; 2; 2 þ 2jÞT . Because x4 is real, Eq. (33) can be used to com-pute each XðkÞ ¼ X RðkÞ þ jX I ðkÞ for k ¼ 1, 2, 3, 4. If k ¼ 2, for example, the real and imaginary DFT components are given by XRð2Þ ¼ X4 i¼1 x RðiÞ cos 2pði 1Þð 2 1Þ 4 ¼ 0 cos 2pð1 1Þð 1Þ 4 þ 1 cos 2pð2 1Þð 1Þ 4 þ 2 cos 2pð3 1Þð 1Þ 4 þ 3 cos 2pð4 1Þð 1Þ 4 ¼ 0 cos 0p 2 þ 1 cos 1p 2 þ 2 cos 2p 2 þ 3 cos 3p 2 ¼ ð 0Þð 1Þ þ ð 1Þð 0Þ þ ð 2Þð 1Þ þ ð 3Þð 0Þ¼ 2 XI ð2Þ ¼ X4 i¼1 x RðiÞ sin 2pði 1Þð 2 1Þ 4 ¼ 0 sin 0p 2 þ 1 sin 1p 2 þ 2 sin 2p 2 þ 3 sin 3p 2 ¼ ð 0Þð 0Þ þ ð 1Þð 1Þ þ ð 2Þð 0Þ þ ð 3Þð 1Þ¼ 2such that Xð2Þ ¼ 2 2j, as expected. Results for k ¼ 1, 3, 4 fol-low similarly. 2.8 The Circulant Eigenvalue Problem. The eigenvalues and eigenvectors of circulants and block circulants with circulant blocks have thus far been inferred from Theorem 9 (circulant matrix) and Corollary 22 (block circulant matrix with circulant blocks), where these matrices are fully diagonalized. Determination of the eigenvalues and eigenvectors of a block circulant matrix with gener-ally noncirculant and nonsymmetric blocks is systematically described here, which reinforces the results already obtained for the special cases of C 2 CN and C 2 BC M;N with circulant generating matrices. The standard cEVP is discussed in Sec. 2.8.1, where an eigensolution is obtained for a general matrix C 2 BC M;N . Section 2.8.2 generalizes these results to handle ( M,K) systems with system matrices contained in BC M;N , which arise naturally in vibration stud-ies of rotationally periodic structures. The structure of the eigenval-ues and eigenvectors is discussed in Sec. 2.8.3, which builds upon the relationship to the DFT described in Sec. 2.7. Section 2.8.4 intro-duces fundamental orthogonality conditions for the special case of block circulants with symmetric generating matrices. 2.8.1 Standard Circulant Eigenvalue Problem. Consider the block circulant matrix C 2 BC M;N with arbitrary generating mat-rices C1; C2; …; CN . The standard cEVP involves determination of the scalar eigenvalues k and NM 1 eigenvectors q of C such that 0NM ¼ C kINM ð Þq (35) where 0NM is a NM 1 vector of zeros and INM is the identity ma-trix of dimension NM . The problem can be simplified consider-ably by exploiting the block circulant nature of C. To this end, partition q ¼ q1; q2; …; qNð ÞT into M 1 vectors qi ði ¼ 1; …; NÞ and introduce the change of coordinates q ¼ ð EN IMÞu (36) where IM is the M M identity matrix (same dimension as the gener-ating matrices of C) and u ¼ u1; u2; …; uNð ÞT is composed of N M -vectors ui. Substituting Eq. (36) into Eq. (35) and multiplying from the left by the unitary matrix ðEN IM ÞH ¼ ð EH N IM Þ yields 0NM ¼ ð EH N IM Þ C kINM ð Þð EN IM Þu ¼ ðEH N IM ÞCðEN IM Þ kðEH N IMÞINM ðEN IM Þu ¼ K1 0 K2 . .. 0 KN 2666666437777775 k IM 0IM . .. 0 IM 26666664377777750BBBBBB@1CCCCCCA u1 u2 ... uN 2666666437777775 which follows from Theorem 10, and from ðEH N IM ÞINM ðEN IMÞ ¼ ð EN IM ÞH ðEN IM Þ¼ INM ¼ diag ðIM; IM; …; IMÞ \|fflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflffl} Nterms in light of Corollary 11. Thus, the single eigenvalue problem defined by Eq. (35) is decomposed into the N reduced-order stand-ard EVPs 0M ¼ Ki kIMð Þui; i ¼ 1; 2; …; N (37) where 0M is a M 1 vector of zeros and each Ki is defined by Eq. (23) in terms of the generating matrices of C. Because the transforma-tion defined by Eq. (36) is unitary, it preserves the eigenvalues of C.Thus, the NM eigenvalues of C are the eigenvalues of the N M M matrices Ki, which follow from the characteristic polynomials det ðKi kIM Þ ¼ 0; i ¼ 1; 2; …; N (38) If k ¼ kðpÞ i denotes the pth eigenvalue of the ith matrix Ki for p ¼ 1; 2; …; M, then the attendant M 1 eigenvector uðpÞ i is obtained from Eq. (37). The corresponding NM 1 eigenvector of C follows from Eq. (36) and is given by qðpÞ i ¼ ð EN IM Þ 0M; …; uðpÞ i ; …; 0M T ¼ e1; …; ei; …; eNð Þ IMð Þ 0M ; …; uðpÞ i ; …; 0M T ¼ e1 IM; …; ei IM; …eN IMð Þ 0M; …; uðpÞ i ; …; 0M T ¼ ð ei IM ÞuðpÞ i ¼ ei uðpÞ i (39) 20 for i ¼ 1; 2; …; N and p ¼ 1; 2; …; M. If the scalars aðpÞ i are such that each ~uðpÞ i ¼ aðpÞ i uðpÞ i (40) is orthonormal with respect to Ki, then the corresponding ortho-normal eigenvectors of C are given by ~qðpÞ i ¼ ei ~uðpÞ i ¼ ei aðpÞ i uðpÞ i ¼ aðpÞ i ei uðpÞ i ¼ aðpÞ i qðpÞ i (41) for i ¼ 1; 2; …; N and p ¼ 1; 2; …; M.If C is an ordinary (not block) circulant with M ¼ 1, Eq. (41) reduces to ~qð1Þ i ¼ qð1Þ i ¼ ei 1 ¼ ei; ðM ¼ 1Þ (42) which confirms that all circulants contained in CN share the same N 1 eigenvectors e1; e2; …; eN , and each ei is orthonormal with respect to C. The eigenvalues of a matrix C 2 CN with generating elements c1; c2; …; cN are given by Eq. (21), or equivalently by Eq. (22). If C 2 SC N , the eigenvalues follow from Corollary 17. Example 24 . Consider C ¼ circ ð4; 1; 0; 1Þ 2 C4 from Examples 15 and 17, where it is shown that EH 4 CE 4 ¼ diag ð2; 4; 6; 4Þ Thus, the eigenvalues of C are 2, 4, 6, and 4. Because N ¼ 4 is even, k1 ¼ 2 and kðNþ2Þ=2 ¼ k3 ¼ 6 are distinct and k2 ¼ k4 ¼ 4are repeated. The reason for this eigenvalue symmetry is dis-cussed in Sec. 2.8.3. The eigenvalues can be verified using ki ¼ 4 2 cos p 2 ði 1Þ; i ¼ 1; 2; 3; 4which follows from Corollary 17 for the case of even N because C is also contained in SC 4 . Because C is an ordinary circulant, its orthonormal eigenvectors are simply the columns of the Fourier matrix E4 , and are given by Definition 18 according to Eq. (42). For example, e2 ¼ 1ffiffiffi 4 p 1; w14; w24; w34 T ¼ 12 1; e j 2p 41 ; e j 2p 42 ; e j 2p 43 T ¼ 12 1; e j p 2 ; e j p; e j 3p 2 T ¼ 12 ð1; j; 1; jÞTwhich can be visualized in Fig. 2 for the case of N ¼ 4. The com-plete set of eigenvectors is given by e1 ¼ 121111 266664377775; e2 ¼ 121 j 1 j 266664377775; e3 ¼ 121 11 1 266664377775; e4 ¼ 121 j 1 j 266664377775 The eigenvectors e1 and e3 are real and correspond to the distinct eigenvalues k1 ¼ 2 and k3 ¼ 6. The eigenvectors e2 and e4 are complex conjugates according to Eq. (17) and correspond to the repeated eigenvalue k2 ¼ k4 ¼ 4. Example 29 of Sec. 2.8.4 dis-cusses orthogonality of the orthonormal eigenvectors 1=2ð Þei with respect to the matrix C. The same basic results also hold for the nonsymmetric circulant C ¼ circ ð4; 1; 0; 1Þ 2 C4 in Example 16, where it is shown that EH 4 CE 4 ¼ diag ð4; 4 2j; 4; 4 þ 2jÞ. In this case, the eigenvalues are given by aðCÞ ¼ f 4; 4 2j; 4; 4 þ 2jg and the corresponding eigenvectors are the same as those in Example 24 because all circulants contained in CN share the same linearly independent eigenvectors. Example 25 . Consider C ¼ circ ðA; B; 0; BÞ 2 BC 2;4 from Examples 6, 18, 19, and 20, where it is shown that ðEH 4 I2ÞCðE4 I2Þ¼ diag 0 1 1 0 " # ; 2 1 1 2 " # ; 4 1 1 4 " # ; 2 1 1 2 " #! diag ðK1; K2; K3; K4Þ The eigenvalues of C are obtained from the reduced-order matri-ces Ki ði ¼ 1; 2; 3; 4Þ. For example, the eigenvalues of K3 follow from the characteristic polynomial det ðK3 kI2Þ ¼ 4 1 1 4 " # k 1 00 1 " # ¼ k2 8k þ 15 ¼ ð k 3Þð k 5Þ ¼ 0and are given by aðK3Þ ¼ f 3; 5g. Similarly, aðK1Þ ¼ f 1; 1g and aðK2Þ ¼ aðK4Þ ¼ f 1; 3g. Because N ¼ 4 is even, aðK1Þ and aðK3Þ are distinct sets of eigenvalues and aðK2Þ ¼ aðK4Þ are repeated sets. Section 2.8.3 discusses the symmetry characteristics and multiplicities of these groups of eigenvalues. The eigenvectors of C follow from the eigenvectors of each Ki according to Eq. (39). For example, for the eigenvector uð1Þ 3 ¼ ð 1; 1ÞT of K3 with eigen-value kð1Þ 3 ¼ 3, the corresponding eigenvector of C is qð1Þ 3 ¼ e3 uð1Þ 3 ¼ 1ffiffiffiffi N p ð1; w24; w44; w64ÞT 11 " # ¼ 1ffiffiffi 4 p ð1; 1; 1; 1ÞT 11 " # ¼ 1211 " # ; 1 1 " # ; 11 " # ; 1 1 " #! T ¼ 12 ð1; 1; 1; 1; 1; 1; 1; 1ÞTThe entire set of NM ¼ 8 reduced-order eigenvectors and eigen-values is given by uð1Þ 1 ¼ ð 1; 1ÞT; kð1Þ 1 ¼ 1 uð2Þ 1 ¼ ð 1; 1ÞT; kð2Þ 1 ¼ 1 uð1Þ 2 ¼ ð 1; 1ÞT; kð1Þ 2 ¼ 1 uð2Þ 2 ¼ ð 1; 1ÞT; kð2Þ 2 ¼ 3 uð1Þ 3 ¼ ð 1; 1ÞT; kð1Þ 3 ¼ 3 uð2Þ 3 ¼ ð 1; 1ÞT; kð2Þ 3 ¼ 5 uð1Þ 4 ¼ ð 1; 1ÞT; kð1Þ 4 ¼ 1 uð2Þ 4 ¼ ð 1; 1ÞT; kð2Þ 4 ¼ 3 9>>>>>>>>>>>>>>>>>>>=>>>>>>>>>>>>>>>>>>>;21 where each pair satisfies Eq. (37). The corresponding eigenvectors of C are qð1Þ 1 ¼ 12 ð1; 1; 1; 1; 1; 1; 1; 1ÞT qð2Þ 1 ¼ 12 ð 1; 1; 1; 1; 1; 1; 1; 1ÞT qð1Þ 2 ¼ 12 ð 1; 1; j; j; 1; 1; j; jÞT qð2Þ 2 ¼ 12 ð 1; 1; j; j; 1; 1; j; jÞT qð1Þ 3 ¼ 12 ð 1; 1; 1; 1; 1; 1; 1; 1ÞT qð2Þ 3 ¼ 12 ð 1; 1; 1; 1; 1; 1; 1; 1ÞT qð1Þ 4 ¼ 12 ð 1; 1; j; j; 1; 1; j; jÞT qð2Þ 4 ¼ 12 ð 1; 1; j; j; 1; 1; j; jÞT 9>>>>>>>>>>>>>>>>>>>>>>>>>>>>>=>>>>>>>>>>>>>>>>>>>>>>>>>>>>>; The eigenvectors are made orthonormal according to Eqs. (40) and (41) by setting each aðpÞ i ¼ 1= ffiffiffi 2 p for i ¼ 1; 2; …; N and p ¼ 1; 2; …; M. Orthogonality of the eigenvectors ~qðpÞ i with respect to C is discussed in Example 30 of Sec. 2.8.4. The determinant of any matrix is the product of its eigenvalues, which yields the following result. C OROLLARY 24. Let C 2 BC M;N be a block circulant matrix. Then the determinant of C is given by det C ¼ YNM i¼1 kiwhere ki is the ith eigenvalue of C. The determinant of C 2 CN follows from Corollary 24 by set-ting M ¼ 1. In light of Eq. (21), det C ¼ YNi¼1 XNk¼1 c k wðk1Þð i1Þ N ðC 2 CN Þ (43) for an ordinary circulant matrix with generating elements c1; c2; …; c N . For the special case of C 2 SC N , the ith eigenvalue ki in Corollary 24 can be replaced by the result given in Corollary 17. Similarly, ki can be replaced with Eq. (25) if C 2 BC M;N has circulant generating matrices. Example 26 . Consider C ¼ circ ðA; B; 0; BÞ 2 BC 2;4 from Example 25, where it is shown that the eigenvalues are given by the set aðCÞ ¼ f 1; 1; 1; 3; 3; 5; 1; 3g. The determinant of C fol-lows from Corollary (24) and is given by det C ¼ 1 1 1 3 3 5 1 3 ¼ 135 which is simply the product of the eigenvalues of C. 2.8.2 Generalized Circulant Eigenvalue Problem. The gener-alized cEVP involves determination of the scalar eigenvalues k and NM 1 eigenvectors q of a cyclic system ( M,K) such that 0NM ¼ K kMð Þq (44) where M ¼ circ ðM1; M2; …; MN Þ K ¼ circ ðK1; K2; …; KN Þ ) are block circulant matrices contained in BC M;N . This type of problem arises in the study of cyclic vibratory mechanical systems composed of N sectors with M DOFs per sector, where M and K are NM NM block circulant mass and stiffness matrices. Several examples are discussed in Sec. 3, where the eigenvectors q are the system mode shapes and the eigenvalues k correspond to the sys-tem natural frequencies. The generalized cEVP defined by Eq. (44) is handled in the same way as the standard cEVP of Sec. 2.8.1. To this end, partition q ¼ q1; q2; …; qNð ÞT into M 1 vec-tors qi ði ¼ 1; 2; …; NÞ, transform to a new set of coordinates u ¼ u1; u2; …; uNð ÞT by substituting Eq. (36) into Eq. (44), and left-multiply the result by the unitary matrix EH N IM. It follows that 0NM ¼ ð EH N IMÞ K kMð Þð EN IM Þu ¼ ðEH N IMÞKðEN IMÞ kðEH N IM ÞMðEN IM Þu ¼ eK1 0 eK2 . .. 0 eKN 266664377775 k eM1 0 eM2 . .. 0 eMN 2666643777750BBBB@1CCCCA u1 u2 ... uN 266664377775 (45) where the decomposed M M matrices eMi ¼ XNk¼1 Mk wðk1Þð i1Þ N eKi ¼ XNk¼1 Kk wðk1Þð i1Þ N 9>>>>>=>>>>>; ; i ¼ 1; 2; …; N (46) follow from Theorem 10. Equation (45) represents a set of N reduced-order generalized EVPs 0M ¼ eKi k eMi ui; i ¼ 1; 2; …; N (47) which are analogous to the reduced-order standard EVPs defined by Eq. (37). The eigenvalues are determined from the characteris-tic polynomials det ð eKi k eMiÞ ¼ 0; i ¼ 1; 2; …; N (48) and are denoted by k ¼ kðpÞ i for p ¼ 1; 2; …; M. Each kðpÞ i also satisfies Eq. (44) because the transformation to new coordi-nates via Eq. (36) is unitary, and hence preserves the system eigenvalues. The reduced-order eigenvectors ui ¼ uðpÞ i are obtained from Eq. (47) for each kðpÞ i . The eigenvectors of the full system ( M,K) are given by qðpÞ i ¼ ei uðpÞ i , which is the same as Eq. (39). The relationships defined by Eqs. (40) and (41) also hold for the generalized cEVP to make the eigen-vectors orthonormal. If M and K are ordinary circulants (i.e., M ¼ 1), then the system eigenvectors reduce to ~qð1Þ i ¼ qð1Þ i ¼ ei 1 ¼ ei, which is the same as Eq. (42) and shows that e1; e2; …; eN are the orthonormal eigenvectors of all generalized eigensystems defined by M; K 2 CN . If M1; M2; …; MN and K1; K2; …; K N are the generat-ing elements of M and K, respectively, the corresponding eigen-values are ki ¼ XNk¼1 Kk wðk1Þð i1Þ N XNk¼1 Mk wðk1Þð i1Þ N ; i ¼ 1; 2; …; N (49) 22 which follows from Eq. (45) by replacing each eMk with M k and eKk with Kk. For the special case of M ¼ IN with generating ele-ments 1 ; 0; …; 0 (i.e., for the standard cEVP), Eq. (49) reduces to the form shown in Theorem 9, as expected. It is clear from this formulation that the same basic steps are followed to solve the standard and generalized cEVPs. For the standard cEVP, we say that kðpÞ i and qðpÞ i are the eigenvalues and eigenvectors of the matrix C. For the generalized cEVP, the eigenvalues and eigenvectors correspond to the system (M,K), not the individual matrices M and K. Example 27 . Consider the generalized cEVP defined by Eq. (44). Let the generating elements of M 2 C2 be M1 ¼ 2 and M2 ¼ 1 and the generating elements of K 2 C2 be K1 ¼ 5 and K2 ¼ 1 such that M ¼ 2 1 1 2 and K ¼ 5 1 1 5 Then the eigenvalues of the system ( M, K ) follow from Eq. (49) with N ¼ 2 and are given by ki ¼ X2 p¼1 Kp wðp1Þð i1Þ N X2 p¼1 M p wðp1Þð i1Þ N ¼ 5w02 þ ð 1Þw i122w02 þ ð 1Þw i12 ¼ 5 e jpði1Þ 2 e jpði1Þ for i ¼ 1, 2. It follows that k1 ¼ 4 and k2 ¼ 2. The corresponding orthonormal eigenvectors are given by ~qð1Þ 1 ¼ e1 ¼ 1ffiffiffi 2 p 11 " # ~qð1Þ 2 ¼ e2 ¼ 1ffiffiffi 2 p 1 w2 " # ¼ 1ffiffiffi 2 p 1 1 " # which follow from Definition 18 according to Eq. (42). 2.8.3 Eigenvalue and Eigenvector Structure. For real-valued generating elements, such as those that arise in models of physical systems with cyclic symmetry, the eigenvalues of a circulant ma-trix C 2 CN are endowed with certain symmetry characteristics, and the same is true for the eigenvalues of systems M; K 2 CN .However, we do not require the circulants to be symmetric, as it is assumed in Corollary 17. We begin by systematically describing the eigenvalue structure of an ordinary circulant with real generat-ing elements. The results are generalized by inspection to handle block circulants and ( M,K) systems composed of circulant or block circulant matrices. The eigenvalues ki of a circulant matrix C 2 CN with generat-ing elements c1; c2; …; c N are given by Eq. (21), or the equivalent matrix–vector form in Eq. (22). It is convenient to re-index each c k and ki such that yðpÞj p¼0;1;2;…;N1 ¼ c kjk¼1;2;3;…;NYðrÞj r¼0;1;2;…;N1 ¼ kiji¼1;2;3;…;N ) (50) which facilitates the results for real generating elements and clari-fies their interpretation. If the eigenvalues are dissected according to YðrÞ ¼ Y RðrÞ þ jY I ðrÞ, then the real and imaginary components Y RðrÞ ¼ XN1 p¼0 yðpÞ cos 2ppr N (51 a) Y I ðrÞ ¼ XN1 p¼0 yðpÞ sin 2ppr N (51 b)follow from the formulation of Eq. (33) in Sec. 2.7 for a real-valued signal where, recall, the eigenvalue expression defined by Eq. (21) has exactly the same form as the DFT in Definition 20. That is, Eq. (51) also represents the real and imaginary parts of the DFT of a real-valued signal, where the sequence of generating elements y(p) is analogous to a real signal and the eigenvalues Y(r) are analogous to its DFT. It is shown that the symmetry of the DFT about the so-called Nyquist component also exists for the eigenvalues of a circulant matrix with real generating elements. As is customary in signal processing, we restrict the formulation to even N. The case of odd N also yields symmetric eigenvalues, but with multiplicity of the Nyquist component. This is handled by example in Sec. 3 (for instance, see Fig. 9). For even N, the zeroth eigenvalue Y(0) and “Nyquist” eigen-value Y N =2ð Þ are always real because Y0 Y Rð0Þ ¼ XN1 p¼0 yðpÞ cos ð0Þ ¼ XN1 p¼0 yðpÞ YI ð0Þ ¼ XN1 p¼0 yðpÞ sin ð0Þ ¼ 0(52 a) Y N=2 YRN 2 ¼ XN1 p¼0 yðpÞ cos ðppÞ ¼ XN1 p¼0 yðpÞð 1ÞpY IN 2 ¼ XN1 p¼0 yðpÞ sin ðppÞ ¼ 0(52 b)are such that the imaginary parts vanish. 2 The remaining eigenval-ues appear in complex conjugate pairs, as the following corolla-ries show. C OROLLARY 25. Let y(p) be the real-valued generating elements of a circulant matrix for p ¼ 0; 1; 2… ; N 1 and Y(r) denote its eigenvalues according to Eq. (21). Then YðN rÞ ¼ YðrÞ ¼ Yð rÞ fo r r ¼ 0; 1; 2… ; N 1. Proof. The eigenvalues of y(p) are given by YðrÞ ¼ YRðrÞ þ jY I ðrÞ, where Y RðrÞ ¼ XN1 p¼0 yðpÞ cos 2ppr N ¼ XN1 p¼0 yðpÞ cos 2ppð rÞ N ¼ YRð rÞ (Eq. 51 a)and the property cos ð hÞ ¼ cos ðhÞ is employed. Similarly, YI ðrÞ ¼ XN1 p¼0 yðpÞ sin 2ppr N ¼ XN1 p¼0 yðpÞ sin 2ppð rÞ N ¼ Y I ð rÞ (Eq. 51 b)2Equation (52 a) also holds if N is odd, but the Nyquist component is repeated with multiplicity of two. 23 where the property sin ð hÞ ¼ sin ðhÞ is employed. It follows that YðrÞ ¼ Y RðrÞ jY I ðrÞ¼ Y Rð rÞ þ jY I ð rÞ¼ Yð rÞ which completes the right-hand side of the proof. To prove the left-hand side, consider Y RðN rÞ ¼ XN1 p¼0 yðpÞ cos 2ppðN rÞ N ¼ XN1 p¼0 yðpÞ cos 2pp 2ppr N ¼ XN1 p¼0 yðpÞ cos ð2ppÞ cos 2ppr N sin ð2ppÞ sin 2ppr N ¼ XN1 p¼0 yðpÞ 1 cos 2ppr N 0 sin 2ppr N ¼ XN1 p¼0 yðpÞ cos 2ppð rÞ N ¼ YRð rÞ (Eq. 51 a)and a similar expansion shows that Y I ðN rÞ ¼ Y I ðrÞ. It follows that YðN rÞ ¼ Y RðN rÞ jY I ðN rÞ¼ Y RðrÞ jY I ðrÞ¼ YðrÞ which completes the proof. Corollary 25 establishes the following result. C OROLLARY 26. Let y(p) be the real-valued generating elements of a circulant matrix for p ¼ 0; 1; 2… ; N 1 and Y(r) denote its eigenvalues according to Eq. (21). Then jYðN rÞj ¼ j YðrÞj ff YðN rÞ ¼ ff YðrÞ for integers r ¼ 0; 1; 2… ; N 1. The magnitudes jYðrÞj and arguments ff YðrÞ of the eigenvalues Y(r) are listed in Table 3 for the special case of even N ¼ 8, where YðrÞ ¼ YðN rÞ follows from complex conjugation of the left-hand equality given by Corollary 25. The zeroth eigenvalue Y(0) ¼ Y0 is real and generally distinct, as is Y N =2ð Þ ¼ Y N=2 for even N, but the remaining eigenvalues generally appear in com-plex conjugate pairs. It follows that the eigenvalue magnitudes (frequencies) are symmetric about the Nyquist component Y N=2 ,as are the arguments (phase angles) but with the opposite sign for indices r > N=2. The eigenvalue symmetries can be observed in the examples of Sec. 2.5.4. The eigenvalues of the matrix C ¼ circ ð4; 1; 0; 1Þ in Example 15 are aðCÞ ¼ f 2; 4; 6; 4g, where Y(0) ¼ 2 and YN=2 ¼ Yð2Þ ¼ 6 are real and distinct and Y(1) ¼ Y(3) ¼ 4 are real and repeated. Similarly, for the matrix C ¼ circ ð4; 1; 0; 1Þ in Example 16, the eigenvalues are aðCÞ ¼ f 4; 4 2j; 4; 4 þ 2jg,where Yð1Þ ¼ 4 2j and Yð3Þ ¼ 4 þ 2j are complex conjugates and Yð0Þ ¼ Y N=2 ¼ Yð2Þ ¼ 4 are real-valued. As expected, these same symmetries are also observed in Example 21 for the DFT of a real-valued signal. A similar formulation shows that the eigenvalues of real-valued (M,K) systems exhibit the same symmetry characteristics because the numerator and denominator of Eq. (49) have exactly the same form as Eq. (21). If the scalars Mk and K k in Eq. (49) are re-indexed and restricted to be real-valued, then Corollaries 25 and 26 are generalized accordingly. For signal processing applications, the symmetry characteristics summarized in Table 3 have practical significance because the subset of ðN þ 2Þ=2 frequency-domain components 0 r N=2is endowed with the same basic “information” contained in all N time-domain signal samples for 0 p N 1. That is, when transformed to the frequency domain by the DFT process, any real-valued signal has a zero-frequency or dc component ( r ¼ 0), distinct magnitude and phase information for 0 r N=2, and repeated magnitude and phase information for r > N=2. For circu-lant matrices that describe physical systems with cyclic symmetry, Y0 and YN=2 correspond to standing wave vibration modes and the remaining eigenvalues are associated with traveling wave modes. This is discussed in Sec. 3. Equation (42) shows that e1; e2; …; eN are the eigenvectors of any circulant matrix C, and the same is true for ( M,K) systems com-posed of circulant matrices. Each eigenvector ei is associated with an eigenvalue Y(r) (i.e., ki) according to the indices defined by Eq. (50). For the special case of real generating elements, the eigenvalues summarized in Table 3 have exactly the same symmetry characteris-tics as the vectors e1; e2; …; eN , which are discussed in Sec. 2.5.2. For example, if N is even, the real eigenvalues YN/2 (i.e., k(N+2)/2 ) and Y(0) (i.e., k1) correspond to the real eigenvectors e Nþ2=2ð Þ and e1defined by Eqs. (18) and (19), respectively. The remaining eigenval-ues appear in complex conjugate pairs and are associated with the complex conjugate eigenvectors according to Eq. (17). If the scalar sequences y(p) and Y(r) for a circulant matrix are replaced by a sequence of matrices y(p) and Y(r), it is clear that the formulation given above also holds for Eq. (23), which defines the eigenvalues for block circulant matrices. In this case, the groups of eigenvalues associated with each Ki are endowed with the symmetry properties given in Table 3. This is confirmed by Example 25 where, using the indexing scheme of that section, aðK1Þ ¼ f 1; 1g and aðK3Þ ¼ f 3; 5g are distinct sets of eigenval-ues and aðK2Þ ¼ aðK4Þ ¼ f 1; 3g are repeated. 2.8.4 Eigenvector Orthogonality. Here, we consider eigen-vector orthogonality relationships for block circulant matrices C and systems ( M,K) contained in BC M;N for the special case of symmetric generating matrices . However, we do not restrict either C or ( M,K) to be symmetric or block symmetric. We require only that the generating matrices C1; C2; …; CN of C are symmetric for the standard cEVP, which guarantees that each Ki is symmetric according to Corollary 19. Similarly, we require that the generat-ing matrices M1; M2; …; MN and K1; K2; …; KN of ( M,K) are symmetric for the generalized cEVP, which implies that ð eKi; eMiÞ are symmetric. Symmetric generating matrices commonly arise in models of rotating flexible structures, including the ones consid-ered in Sec. 3, where the sector models and intersector coupling are described by symmetric matrices. Then the usual orthogonal-ity relationships hold for the reduced-order eigenvectors defined in Secs. 2.8.1 and 2.8.2. It is first shown that each ~uðpÞ i is Table 3 Magnitudes and arguments of Y(r) for even N58 Index Eigenvalue Magnitude Argument rY(r)YðNrÞjYðrÞj ff YðrÞ 0Yð0Þ ¼ Y0Yð8ÞjYð0Þj ¼ Y0ff Yð0Þ 1Y(1) Yð7ÞjYð1Þj ff Yð1Þ 2Y(2) Yð6ÞjYð2Þj ff Yð2Þ 3Y(3) Yð5ÞjYð3Þj ff Yð3Þ 4¼N=2ðÞYð4Þ ¼ Yð4Þ ¼ YN=2Yð4ÞjYð4Þj ¼ YN=2ff Yð4Þ 5Yð5Þ ¼ Yð3ÞYð3ÞjYð3Þj ff Yð3Þ 6Yð6Þ ¼ Yð2ÞYð2ÞjYð2Þj ff Yð2Þ 7Yð7Þ ¼ Yð1ÞYð1ÞjYð1Þj ff Yð1Þ24 orthogonal with respect to Ki for the standard cEVP. Proofs can be found in any standard textbook on linear algebra [98,99]. To-gether with generic orthogonality conditions on the basic circulant structure of a matrix C 2 BC M;N , this gives rise to an orthogonal-ity condition on the eigenvector ~qðpÞ i ¼ ei ~uðpÞ i with respect to C.Orthogonality of ~qðpÞ i with respect to the system M; K 2 BC M;N is handled similarly. The results of this section, and the requirement of symmetric gener-ating matrices, are meant to highlight how orthogonality of an eigen-vector ~qðpÞ i ¼ ei ~uðpÞ i is essentially dissected into the orthogonality of ei with respect to the circulant structure of C 2 BC M;N and ortho-gonality of ~uðpÞ i with respect to the generating matrices (e.g., Ki), where the latter requires symmetric C1; C2; …; CN . It should be noted that none of the results in Sec. 2, aside from this section, require symmetric generating matrices. In particular, Theorems 9 and 10, upon which block reduction of the cEVPs in Secs. 2.8.1 and 2.8.2 are based, are valid for arbitrary M M generating matrices. COROLLARY 27. Suppose each Ki defined by Eq. (23) is symmetric. Let ~uðpÞ i be the pth orthonormal eigenvector of Ki and kðpÞ i be the corresponding eigenvalue. Then if eUi ¼ ð ~uð1Þ i ; ~uð2Þ i ; …; ~uðMÞ i Þ is the M M orthonormal modal matrix associated with Ki , it follows that eUT i Ki eUi ¼ kð1Þ i 0 kð2Þ i . .. 0 kðMÞ i 2666437775; i ¼ 1; 2; …; Nis a diagonal matrix with eigenvalues kðpÞ i along its diagonal for p ¼ 1; 2; …; M and ~uðpÞT i Ki ~uðqÞ i ¼ kðpÞ i dpq where dpq is the Kronecker delta . Example 28 . Consider the eigensolutions ~uð1Þ 3 ¼ 1ffiffiffi 2 p ð1; 1ÞT; kð1Þ 3 ¼ 3 ~uð2Þ 3 ¼ 1ffiffiffi 2 p ð 1; 1ÞT; kð2Þ 3 ¼ 5 9>>=>>; corresponding to the symmetric matrix K3 ¼ 4 1 1 4 from Example 25, where the reduced-order eigenvectors are in orthonormal form. The corresponding reduced-order modal matrix is denoted by eU3 ¼ ð ~uð1Þ 3 ; ~uð2Þ 3 Þ¼ 1ffiffiffi 2 p 1 11 1 " # It follows from Corollary 27 that the diagonal matrix eUT3 K3 eU3 ¼ 1ffiffiffi 2 p 1 11 1 " #T 4 1 1 4 " # 1ffiffiffi 2 p 1 11 1 " # ¼ 121 1 1 1 " # 3 53 5 " # ¼ 126 00 10 " # ¼ 3 00 5 " # has eigenvalues kð1Þ 3 ¼ 3 and kð2Þ 3 ¼ 5 as its diagonal elements. The reader can verify that ~uð1ÞT3 K3 ~uð1Þ 3 ¼ 3 and ~uð2ÞT3 K3 ~uð2Þ 3 ¼ 5according to Corollary 27. Orthogonality of an eigenvector ~qðpÞ i ¼ ei ~uðpÞ i with respect to C 2 BC M;N is essentially decomposed into orthogonality of ei with respect to the circulant structure of C and orthogonality of each ~uðpÞ i with respect to the symmetric matrices Ki. These indi-vidual orthogonality conditions are captured by Corollaries 18 and 27, which lead to the following fundamental result. C OROLLARY 28. Let C 2 BC M;N be a block circulant matrix with symmetric generating matrices, ei be the ith column of the N NFourier matrix EN , and ~uðpÞ i be the pth reduced-order orthonormal eigenvector corresponding to Ki defined by Eq. (23). Then ðeH i ð~uðpÞ i ÞTÞCðek ~uðqÞ i Þ ¼ kðpÞ i dik dpq for i ; k ¼ 1; 2; …; N and p ; q ¼ 1; 2; …; M, where kðpÞ i is the eigen-value associated with ~uðpÞ i and dik is the Kronecker delta . Proof. Let eUi ¼ ð ~uð1Þ i ; ~uð2Þ i ; …; ~uðMÞ i Þ be the orthonormal modal matrix associated with Ki and C1; C2; …; CN be the symmetric generating matrices of C. Corollary 19 guarantees that Ki is sym-metric because the generating matrices are symmetric. By setting B ¼ eUi and ðÞ # ¼ ðÞ T in Corollary 21, it follows that ðeH i eUT i ÞCðek eUiÞ ¼ Widik ¼ Wi; i ¼ k 0; otherwise ( where Wi ¼ XNn¼1 B#CnBwðn1Þð i1Þ N (Eq. 24) ¼ XNn¼1 eUTi Cn eUi wðn1Þð i1Þ N (by substitution) ¼ eUTi XNn¼1 Cn wðn1Þð i1Þ N !eUi ¼ eUTi Ki eUi (Eq. 23) ¼ kð1Þ i 0 kð2Þ i . .. 0 kðMÞ i 2666437775 (Cor. 27) for i ¼ 1; 2; …; N. Expanding the M M matrix product ðeH i eUTi ÞCðek eUiÞ yields eH i ð~uð1Þ i ÞT ... ð~uðpÞ i ÞT ... ð~uðMÞ i ÞT 2666666666643777777777750BBBBBBBBBB@1CCCCCCCCCCA C e k ð~uð1Þ i ; …; ~uðqÞ i ; …; ~uðMÞ i Þ ¼ eH i ð~uð1Þ i ÞT ... eH i ð~uðpÞ i ÞT ... eH i ð~uðMÞ i ÞT 266666666664377777777775 Cðek ~uð1Þ i ; …; ek ~uðqÞ i ; …; ek ~uðMÞ i Þ which produces an M M array with scalar elements 25 ðeH i ð~uðpÞ i ÞTÞCðek ~uðqÞ i Þ in the ( p, q) position for p; q ¼ 1; 2; …; M. However, in light of the diagonal structure of each Wi, only the diagonal elements sur-vive in this expansion. That is, ðeH i ð~uðpÞ i ÞTÞCðek ~uðqÞ i Þ ¼ kðpÞ i dik ; p ¼ q 0; otherwise ( ¼ kðpÞ i dik dpq which completes the proof. If i ¼ k in Corollary 28, the orthogonality condition can be stated in terms of the eigenvectors ~qðpÞ i ¼ ei ~uðpÞ i . That is, ð~qðpÞ i ÞHC~qðqÞ i ¼ kðpÞ i dpq ; i ¼ 1; 2; …; N (53) For an ordinary circulant, each ~uðpÞ i ¼ 1 and Corollary 28 reduces to eH i Ce k ¼ kidik ; i ¼ 1; 2; …; N (54) which is the same result given by Corollary 18. Example 29 . Consider C ¼ circ ð4; 1; 0; 1Þ 2 C4 from Examples 15, 17, and 24, where it is shown that e2 ¼ 12 ð1; j; 1; jÞT is an eigenvector of C corresponding to the eigenvalue k2 ¼ 4. Thus, the product eH 2 Ce 2 ¼ 12 1 j 1 j½ 4 1 0 1 1 4 1 00 1 4 1 1 0 1 4 2666666437777775 121 j 1 j 2666666437777775 ¼ 14 1 j 1 j½ ð 4; 4j; 4; 4jÞT ¼ 14 ð4 þ 4 þ 4 þ 4Þ¼ 4is numerically equal to k2 according to Eq. (54). However, the product eH 4 Ce 2 ¼ 14 1 j 1 j½ ð 4; 4j; 4; 4jÞT ¼ 14 ð4 4 þ 4 4Þ¼ 0vanishes because i 6 ¼ k. Example 30 . Consider C ¼ circ ðA; B; 0; BÞ 2 BC 2;4 from Examples 6, 18, 19, 20, and 25, where it is shown that ~qð2Þ 3 ¼ 1ffiffiffi 2 p qð2Þ 3 ¼ 12 ffiffiffi 2 p ð 1; 1; 1; 1; 1; 1; 1; 1ÞTis an orthonormal eigenvector of C corresponding to the eigen-value kð2Þ 3 ¼ 5. Because the generating matrices A, B, 0, B are symmetric (see Example 6), Corollary 28 guarantees that each ~qðpÞ i ¼ ei ~uðpÞ i is mutually orthogonal with respect to C. For example, it follows from Eq. (53) that qð2ÞH 3 Cq ð2Þ 3 ¼ 12 ffiffiffi 2 p 1 1 1 1 1 1 1 1½ 2 1 1 0 0 0 1 0 1 2 0 1 0 0 0 1 1 0 2 1 1 0 0 00 1 1 2 0 1 0 00 0 1 0 2 1 1 00 0 0 1 1 2 0 1 1 0 0 0 1 0 2 10 1 0 0 0 1 1 2 377777777777777775266666666666666664 12 ffiffiffi 2 p111 1 111 1 266666666666666664377777777777777775 ¼ 18 1 1 1 1 1 1 1 1½ ð 5; 5; 5; 5; 5; 5; 5; 5ÞT ¼ 5which is numerically equal to the eigenvalue kð2Þ 3 . However, the matrix product ðe3 uð1Þ 3 ÞHCðe3 uð2Þ 3 Þ¼ ð eH 3 ðuð1Þ 3 ÞH ÞCðe3 uð2Þ 3 Þ¼ ð eH 3 ðuð1Þ 3 ÞTÞCðe3 uð2Þ 3 Þ¼ 12 ffiffiffi 2 p 1 1 1 1 1 1 1 1½ C 12 ffiffiffi 2 p ð 1; 1; 1; 1; 1; 1; 1; 1ÞT ¼ 18 1 1 1 1 1 1 1 1½ ð 5; 5; 5; 5; 5; 5; 5; 5ÞT ¼ 0vanishes because dpq ¼ 0 ðp 6 ¼ qÞ in Corollary 28. Similarly, ðe2 uð2Þ 3 ÞHCðe3 uð2Þ 3 Þ¼ 12 ffiffiffi 2 p 1 1 j j 1 1 j j½ C 12 ffiffiffi 2 p ð 1; 1; 1; 1; 1; 1; 1; 1ÞT ¼ 18 1 1 j j 1 1 j j½ ð 5; 5; 5; 5; 5; 5; 5; 5ÞT ¼ 0because dik ¼ 0 ði 6 ¼ kÞ.26 If instead we set A ¼ 4 and B ¼ 1 such that C 2 C4 is an ordi-nary circulant, we recover the orthogonality condition in Example 15 for i ¼ 3. The orthogonality conditions used in Example 30 for the special case of an ordinary circulant does not require that the circulant structure is symmetric (i.e., C need not be contained in SC N ). The reader can verify that Eq. (54) also holds for nonsymmetric matrices C 2 CN by inspection of Example 16. The fundamental orthogonality relationship given by Corollary 28 also holds for M; K 2 BC M;N systems with symmetric generat-ing matrices, as the following corollary shows. Corollary 29. Let M; K 2 BC M;N be block circulant with symmetric generating matrices, ei be the ith column of the N NFourier matrix EN , and ~uðpÞ i be the pth M 1 reduced-order orthonormal eigenvector corresponding to the ith system ð eMi; eKiÞ defined by Eq. (46). Then ðeH i ð~uðpÞ i ÞTÞMðei uðpÞ i Þ ¼ dik dpq ðeH i ð~uðpÞ i ÞTÞKðei uðpÞ i Þ ¼ kðpÞ i dik dpq 9=; for i ; k ¼ 1; 2; …; N and p ; q ¼ 1; 2; …; M, where kðpÞ i is the eigen-value associated with ~uðpÞ i and dik is the Kronecker delta . In practice, of course, the M 1 reduced-order eigenvectors ~uðpÞ i are not known a priori. Instead, Theorem 10 is used in Sec. 3 to decouple the NM -DOF system equations into a set of N reduced-order M-DOF standard vibratory problems, from which the system eigenvalues (natural frequencies) and eigenvectors (normal modes) are extracted. 3 Example Applications In this section we apply the results developed in Sec. 2 to vibra-tion models of systems with cyclic symmetry. For each model considered, we begin by formulating the equations of motion (EOM) and then use the theory of circulants to diagonalize or block diagonalize the governing equations. This is achieved by a coordinate transformation that exploits the special relationship between circulant matrices and the Fourier matrix. The process also shows how external forces are projected on the resulting block diagonal EOM. The special case of traveling wave engine order excitation is also presented in some detail because it appears in many relevant applications of rotating machinery. Three examples are presented. The first example (Sec. 3.1) con-siders the structure of the EOM for a general cyclic system with N sectors, M DOFs per sector, and arbitrary excitation. It is shown how to block diagonalize the system equations via a modal trans-formation involving the Fourier matrix, which results in NM -DOF reduced-order vibratory systems. If engine order excitation is assumed (Sec. 3.2), it is shown that the steady-state forced response of the NM -DOF system can be obtained from a single M -DOF harmonically forced, reduced-order system in modal space. The mathematical and physical details of engine order excitation are discussed, including its temporal and spatial duality. The sec-ond example (Sec. 3.3) considers a cyclic system with one DOF per sector under engine order excitation. This system is fully dia-gonalized by the Fourier matrix. The example is presented in detail, showing the nature of the natural modes and frequencies, and the resonant response to excitation of various engine orders. The third example (Sec. 3.4) has two DOFs per sector and demon-strates the block diagonalization process for a perfectly cyclic sys-tem with specified sector models, as opposed to the general sector models in the first example. In each sector, one DOF is due to flexure, and thus has a constant frequency, while the other DOF is a centrifugally driven pendulum whose frequency is proportional to the rotor speed. The coupling between these DOFs leads to some interesting behavior in both the free and forced vibration of the system, which is discussed in Refs. [21,22] and [92–95]. More importantly, this example shows the process of handling multiple DOFs per sector, which easily extends from two to M DOFs per sector using the theory presented in Sec. 2. 3.1 General Cyclic System 3.1.1 Equations of Motion. Consider the general cyclic vibra-tory system shown schematically in Fig. 4, which consists of N sectors with coupling (elastic and damping) to adjacent neighbors. The topology diagram only indicates nearest-neighbor coupling, but more general coupling is admissible as long as the cyclic sym-metry is preserved. If there are M DOFs per sector, each M 1vector qi describes the dynamics of the ith sector for i 2 f 1; 2; …; Ng N . Then the linear EOM takes the form M€q þ C _q þ Kq ¼ bf (55) where q ¼ ð q1; q2; …; qN ÞT is a NM 1 configuration vector, the system matrices are block circulant with M M blocks, and over-dots denote differentiation with respect to time. If the M 1 vec-tor fi denotes the component of forcing on the ith sector, then bf ¼ ð f1; f2; …; fN ÞT is a NM 1 general forcing vector. The NM NM system mass, damping, and stiffness matrices are of the form M ¼ circ ðM1; M2; …; MN Þ 2 BC M;N C ¼ circ ðC1; C2; …; CN Þ 2 BC M;N K ¼ circ ðK1; K2; …; KN Þ 2 BC M;N 9>=>; (56) where the generating matrices Mi, Ci, and Ki depend on the M M sector mass, damping, and stiffness matrices and the inter-sector coupling (stiffness and damping). Equation (55) is a general model for any linear, lumped-parameter, conservative, nongyro-scipic, cyclic vibratory system with N sectors and M DOFs per sector. For example, a linearized lumped-parameter model of a bladed disk assembly under engine order excitation is captured by Eq. (55), where N is the number of blades, M is the number of DOFs per blade, f has the special properties described in Sec. 3.2, and the system matrices depend on the structural details of each blade (i.e., sector) and its connectivity to adjacent blades and rotating hub. 3.1.2 Modal Transformation. Of course, one can apply stand-ard techniques to investigate the free and forced response of the model given by Eq. (55). However, this requires solving an NM NM eigenvalue problem to determine the modal properties, or inversion of an NM NM impedance matrix to determine the response to harmonic excitation. Such an approach may be pro-hibitive or computationally expensive for practical models with Fig. 4 Topology diagram of a general cyclic system 27 many sectors and many DOFs per sector, nor does it highlight or exploit the underlying features of the cyclic system. It is precisely these special features that are brought to light by the properties of the circulants or block circulants that describe the system. Solving for the system response is significantly facilitated by a modal transformation that exploits the cyclic symmetry among the N sectors. Specifically, it is straightforward to block decouple the EOM into a set of N systems, each with M DOFs. To this end, we introduce the change of coordinates q ¼ ð EN IMÞu (57) where u ¼ ð u1; u2; …; uN ÞT is a NM 1 vector of modal coordi-nates. Each ui is M 1 and describes the sector dynamics in modal space, where the EOM are block decoupled (as described below). In this formulation the physical coordinates q are expressed in terms of the modal coordinates ui and the Fourier elements of length N, which accounts for the overall cyclic nature of the EOM. Substituting Eq. (57) into Eq. (55) and multiplying from the left by ðEN IMÞH ¼ ð EH N IMÞ yields ðEH N IMÞMðEN IMÞ€u þ ð EH N IMÞCðEN IMÞ _u þ ð EH N IMÞKðEN IMÞu ¼ ð EH N IMÞbf or eM1 0 eM2 . .. 0 eMN 2666666437777775 €u1 €u2 ...€uN 26666643777775 þ eC1 0 eC2 . .. 0 eCN 2666666437777775 _u1 _u2 ..._uN 26666643777775 þ eK1 0 eK2 . .. 0 eKN 2666666437777775 u1 u2 ... uN 26666643777775 ¼ðeH 1 IM Þbf ðeH 2 IM Þbf ... ðeH N IM Þbf 2666666437777775 (58) The block diagonal structure on the left-hand side of Eq. (58) fol-lows from Theorem 10, where the M M modal mass, damping, and stiffness matrices associated with the pth mode follow from Eq. (23) and are given by eMp ¼ XNk¼1 Mk wðk1Þð p1Þ N eCp ¼ XNk¼1 Ck wðk1Þð p1Þ N eKp ¼ XNk¼1 Kk wðk1Þð p1Þ N 9>>>>>>>>>=>>>>>>>>>; ; p 2 N (59) The forcing terms on the right-hand side of Eq. (58) follow from the decomposition ðEH N IM Þbf ¼ ðeH 1 ; eH 2 ; …; eH N ÞT IM bf ¼ eH 1 IM ; eH 2 IM ; …; eH N IM Tbf ¼ðeH 1 IMÞbf ðeH 2 IMÞbf ... ðeH N IMÞbf 2666666437777775 (60) and make no assumptions on the nature of the applied forcing. Thus, the solution to the NM -DOF matrix EOM given by Eq. (58) reduces to solving NM -DOF uncoupled systems eMp €up þ eCp _up þ eKpup ¼ ð eH p IMÞbf; p 2 N (61) for the modal solutions up. These N reduced-order EOMs of order M offer a substantial computational savings compared to the full NM -DOF system defined by Eq. (55). The reduced equations can be solved directly or with standard modal analysis by solving a set of N generalized eigenvalue problems, each of order M, from which one can form the global eigenvectors of the system. Specifi-cally, solving the N eigenvalue problems associated with Eq. (61) yields a set of normalized M M modal matrices Up, which define a change to reduced modal coordinates via up ¼ Upsp. The sp are modes that define the behavior of the internal sector dynamics and account for the manner in which these are coupled to each other in a given Fourier mode of the overall system. The EOM for the si form a set of M uncoupled equations, and these are related to the physical coordinates of the original system by q ¼ ð EN IM Þu where u ¼ ð U1s1; U2s2; …; UN sN ÞT . This process is general, and it allows one to decouple the full EOM in two steps, one which accounts for the global cyclic nature of the system, and the other which accounts for the details of the sector model. Equation (61) can be simplified even further if the system is subjected to the so-called engine order excitation . The mathemat-ics and physics of this type of excitation are developed in the next section, which closes with a treatment of the general cyclic system governed by Eq. (61) under engine order excitation. 3.2 Engine Order Excitation. Traveling wave engine order excitation arises in rotating machinery and is a primary source of forced vibration response in bladed disk assemblies [11,111]. A mathematical model of this common form of excitation is devel-oped in Sec. 3.2.1 and its traveling wave characteristics are described in Sec. 3.2.2. While this material is known, the discus-sion that follows is unique because it provides physical insights and a systematic explanation of the temporal and spatial duality of engine order excitation. The general cyclic system of Sec. 3.1 is reconsidered in Sec. 3.2.3 under engine order excitation, where it is shown that the steady-state forced response of the NM -DOF system reduces to that of a single sector in modal space. 3.2.1 Mathematical Model. Ideally, the steady axial gas pres-sure in a jet engine might vary with radius but is otherwise uni-form in the circumferential direction, thus resulting in an identical force field on each blade in a particular fan, compressor, or turbine within the engine. In practice, however, flow entering an engine inlet invariably meets static obstructions, such as struts, stator vanes, etc., in addition to rotating bladed disk assemblies in its path to the exhaust. Even in steady operation, therefore, the flow slightly upstream of these bladed assemblies is nonuniform in pressure, temperature, and so on. This results in a static pressure (effective force) field on the blades that vary circumferentially, a notional example of which is shown in Fig. 5. Consider, for example, an engine in steady operation with n evenly spaced struts slightly upstream (or downstream) of a bladed assembly. As explained in Ref. , these obstructions produce a circumferential variation upon the mean axial gas pressure that is essentially proportional to cos nh, where h is an angular posi-tion. Thus, a blade rotating through this static pressure field expe-riences a force proportional to cos nXt, where X is the constant angular speed of the bladed disk assembly and t is time. An adjacent blade experiences the same force, but at a constant frac-tion of time later. This type of excitation is defined as engine order (e.o.) excitation and n is said to be the order of the excitation. To be more precise, the axial gas pressure of a steady flow through a jet engine may be described by the function pðhÞ ¼ pðh þ 2pÞ, where h is an angular coordinate measured 28 relative to a fixed origin on the machine. That is, the pressure field is rotationally periodic and can therefore be expanded in a Fourier series with terms of the form po cos nh. If the angular position of the ith blade relative to the same origin is defined by hiðtÞ ¼ Xt þ 2p N ði 1Þ; i 2 N where N is the total number of blades and N ¼ f 1; 2; …; Ng is the set of blade, or sector numbers, it follows that the total effective force exerted on blade i due to the nth harmonic of the pressure field pðhÞ is captured by F cos nXt þ 2p nN ði 1Þ ; i 2 N (62) Upon complexifying, FiðtÞ ¼ Fe j/i e jn Xt; i 2 N (63) is a model for the nth predominant component of the excitation. Eq. (63) has period T ¼ 2p=nX, strength F, and is said to have angular speed X. The so-called interblade phase angle is defined by /i ¼ /ðnÞ i ¼ 2p nN ði 1Þ ¼ nui; i 2 N (64) where n 2 Zþ and ui is the angle subtended from blade 1 to blade i and is defined by Eq. (16). Equation (63) is defined as nth engine order, or traveling wave excitation. The traveling wave character-istics of this type of excitation are considered next. 3.2.2 Traveling Wave Characteristics. The function defined by Eq. (63) is continuous in time and discretized in space via the index i. This gives rise to two interpretations of engine order exci-tation relative to the rotating hub, one discrete and the other con-tinuous. These can be visualized in Fig. 6, which shows adissection of the excitation amplitudes along time and sector axes. In the first and usual sense, Eq. (63) is a discrete temporal varia-tion of the dynamic loading applied to individual blades. That is, under an engine order n excitation, each sector is harmonically forced with strength F and frequency nX, but with a fixed phase difference relative to its nearest neighbors. Physically, one can think of this as placing N different observers at the discrete sectors and having the ith observer record the excitation strength applied to sector i as a function of time. Their recorded time traces would resemble those shown in Fig. 6( a). In the second and more general sense, Eq. (63) can be viewed as a continuous spatial variation of the excitation strength relative to the rotating hub (along the sector axis) that evolves with increasing time, i.e., it is a propagating waveform, or traveling wave . If a single observer was placed on the rotating hub and recorded the strength of this traveling wave as a function of i (taken here to be continuous ), it would resemble the curve shown in Fig. 6( b). In this context, the instantaneous loading applied to individual blades is obtained by essentially “sampling” the continuous traveling wave at each sector i 2 N and, as time evolves, these sampled points define N time profiles of the force amplitudes, which is equivalent to the discrete tempo-ral interpretation described above. However, the latter interpreta-tion illuminates some important traveling wave characteristics of the engine order excitation that are otherwise difficult to explain, and in what follows these are systematically described. To explain the traveling wave mathematically, it is convenient to define the function UkðvÞ ¼ cos 2pðk 1Þ N v ¼ cos ðukvÞ (65) which is a harmonic waveform with wavelength 2 p=uk. Then for i 2 N and noting that /i ¼ unþ1ði 1Þ, Eq. (63) can be written in real form as F iðtÞ ¼ F cos ðunþ1ði 1Þ þ nXtÞ¼ F cos unþ1 i 1 þ nX unþ1 t ¼ FUnþ1ði 1 þ Ct Þ (66) which is a harmonic function with a wavelength of 2p=unþ1 ¼ N=n (unþ1 is the wave number) and angular frequency nX. Equation (66) shows that engine order excitation is a TW in the negative i-direction (descending blade number) with speed C ¼ nX=unþ1 ¼ NX=2p, measured in sectors per second . An example plot of this continuous BTW is shown in Fig. 6( b) and, as described above, the applied loads can be obtained from this figure by continuously “sampling” the waveform at the discrete sector numbers as time evolves. Then the engine order excitation applied to the individual blades consists of a wave composed of these N discrete points, examples of which are shown in Figs. 7( a)–7( d). Interestingly, this gives rise to discrete SW or even FTW applied dynamic loads (depending on the value of n relative to N), even though Eq. (66) is strictly a BTW relative to the rotating hub. These additional possibilities arise due to alias-ing of the “sampled points” just as it occurs in elementary signal processing theory [112,113]. Before characterizing the traveling and standing waveforms, it is shown that one need only consider engine orders n 2 N .The traveling wave nature of the discrete applied loads (i.e., SW, BTW, or FTW) depends only on the value of n relative to N.To see this, let n ¼ n mod N 2 N ; n 2 Zþ (67) and assume n ¼ n þ mN for some integer m. Then UnþmN þ1ðvÞ ¼ Unþ1ðvÞ That is, if n ¼ n corresponds to a SW, BTW, or FTW, then so does n þ mN for any m 2 Zþ. In this sense, the traveling wave na-ture of the applied dynamic loads is seen to alias relative to N.These features are characterized for engine orders n 2 N ¼ N O;EBTW [ N O;EFTW [ N O;ESW where it is understood that the results apply to any n > N simply by taking n modulo N, as appropriate. The subsets N O;EBTW , N O;EFTW ,and N O;ESW are defined in Table 4 and discussed below. For the special case of n ¼ N the rotating blades become entrained with the excitation because /ðNÞ i ¼ 2pnði 1Þ with Fig. 5 The axial gas pressure pðhÞ: ideal and (notional) actual conditions 29 i; n 2 Zþ, and hence each is forced with the same strength and phase. As illustrated in Fig. 7( d), this is a SW excitation where each blade is harmonically forced according to F iðtÞ ¼ F cos nXt.Entrainment also occurs when n ¼ N/2 if N is even, in which case /ðN=2Þ i ¼ pnði 1Þ, where ( i 1) is odd (resp. even) for even (resp. odd) sector numbers i 2 N . Accordingly, all blades with odd sector numbers are driven by FiðtÞ ¼ F cos nXt, as are the blades with even sector numbers, but with a 180-deg phase shift. As shown in Fig. 7( b), this amounts to the same standing wave ex-citation as the n ¼ N case, except for a phase reversal in the excita-tion among adjacent blades. The engine orders corresponding to SW excitations for odd and even N are denoted by the sets N O;ESW N , and all other values of n 2 N correspond to traveling waves. Engine orders n 2 N O;EBTW (resp. n 2 N O;EFTW ) correspond to BTW (resp. FTW) excitation, an example of which is shown in Fig. 7( a) (resp. Fig. 7( c)), where N O;EBTW and N O;EFTW are also defined in Table 4. These sets can be visualized in Figs. 7( i) and 7( ii ) for odd and even N, respectively. The manner in which cyclic systems respond to this type of excitation is considered in the examples that follow. In Sec. 3.2.3, we prove the most important result related to the forced response of cyclic systems, namely, that in the case of perfect symmetry each engine order excites only a single mode. This is clear mathe-matically and will be explored with more physical insight in the examples presented in Secs. 3.3 and 3.4. 3.2.3 Forced Response of a General Cyclic System Under Engine Order Excitation. The general cyclic system governed by Eq. (55) is reconsidered here under engine order excitation. Using the notation of Sec. 3.1, a model for the nth engine order excita-tion is fi ¼ fe jn ui e jn Xt; i 2 N (68) where f is a constant M M vector of sector force amplitudes, t is time, ui is the angle subtended from sector 1 to sector i and is defined by Eq. (16), n is the order of the excitation, nX is the angular frequency of the excitation, and X is the angular speed of the system relative to the excitation. Under this type of excitation, the system forcing vector becomes bf ¼ f1 f2 ... fN 26666666643777777775 ¼ fe jn u1 e jn Xt fe jn u2 e jn Xt ... fe jn uN e jn Xt 26666666643777777775 ¼ f0 fe jn Xt where Fig. 6 Example illustration of the discrete temporal and continuous spatial variations of the traveling wave excitation defined by Eq. (63) in real form: ( a) the discrete dynamic loads with amplitude Fand period T52p=nXapplied to each sector; and ( b) the continuous BTW excitation with wavelength N/nand speed C5NX=2prelative to the rotating hub 30 f0 ¼ ð e jn u1 ; ejn u2 ; …; e jn uN ÞT ¼ ð e j/1 ; e j/2 ; …; e j/N ÞT (69) is a vector of constant intersector phase angles /i ¼ nui. Thus, the pth modal forcing term on the right-hand side of Eq. (61) reduces to ðeH p IMÞbf ¼ ð eH p IMÞð f0 fÞe jn Xt ¼ ð eH p f0Þ ðIM fÞe jn Xt ¼ ð eH p f0Þ fe jn Xt; p 2 N (70) which is a direct product of the scalar product eH p f0 with the nth order harmonic excitation fe jn Xt imparted to each sector. This dis-section of the modal forcing term highlights an orthogonality con-dition that is shared by all cyclic systems under engine order excitation. In particular, eH p f0 ¼ 1ffiffiffiffi N p ej0up ; ej1up ; …; ejðN1Þup e jn u1 ; e jn u2 ; …; ejn uN T ¼ 1ffiffiffiffi N p XNk¼1 ejðk1Þup e jn uk ¼ 1ffiffiffiffi N p XNk¼1 ejðk1Þ2p Nðp1Þ e jn 2p Nðk1Þ ¼ 1ffiffiffiffi N p XNk¼1 wð k1Þð p1Þ N wnðk1Þ N ¼ 1ffiffiffiffi N p XNk¼1 wðk1Þð nþ1pÞ N ¼ ffiffiffiffi N p ; p ¼ n þ 10; otherwise ( (71) which follows from Lemma 2 and shows that the force vector f0 is mutually orthogonal to all but one of the modal vectors ep. This result is obtained more directly from Corollary 10 by observing that f0 ¼ e jn u1 ; ejn u2 ; …; e jn uN T ¼ e j0unþ1 ; ej1unþ1 ; …; e jðN1Þunþ1 T ¼ ffiffiffiffi N p enþ1where enþ1 is the ðn þ 1Þth column of the Fourier matrix. It fol-lows that eH p f0 ¼ eH p ffiffiffiffi N p enþ1 ¼ ffiffiffiffi N p eH p enþ1 ¼ ffiffiffiffi N p dpðnþ1Þ (72) which is the same orthogonality condition given by Eq. (71). Thus, for p 2 N , the only excited mode is p ¼ n mod N þ 1 (73) for an engine order n 2 Zþ excitation. Then Eq. (70) becomes ðeH p IMÞbf ¼ ffiffiffiffi N p fejn Xt; p ¼ n þ 10; otherwise ( (74) and the forced response of the NM -DOF matrix EOM given by Eq. (58) reduces to solving a single , M-DOF system eMnþ1 €unþ1 þ eCnþ1 _unþ1 þ eKnþ1unþ1 ¼ ffiffiffiffi N p fejn Xt (75) in modal space. Assuming harmonic motion, the steady-state modal response is given by uss nþ1 ðtÞ ¼ ffiffiffiffi N p eZnþ1fe jn Xt (76) where Fig. 7 Engine orders n mod N corresponding to BTW, FTW and SW applied dynamic loading for ( i) odd N and ( ii ) even N (see also Table 4); example plots of applied dynamic loading (represented by the dots) for a model with N 5 10 sectors and with ( a) n 5 1(BTW), ( b) n 5 5 (SW), ( c) n 5 9 (FTW), and ( d) n 5 10 (SW). The BTW engine order excitation is represented by the solid lines. Table 4 Sets of engine orders nðmod NÞ ‰ N corresponding to BTW, FTW, SW dynamics loads applied to the blades for odd and even N. These can be visualized in Figs. 7( i) and 7( ii ). N Type Set Odd BTW N OBTW ¼ n 2 Zþ : 1 n N 12 FTW N OFTW ¼ n 2 Zþ : N þ 12 n N 1 SW N OSW ¼ Nf g Even BTW N EBTW ¼ n 2 Zþ : 1 n N 22 FTW N EFTW ¼ n 2 Zþ : N þ 22 n N 1 SW N ESW ¼ N 2 ; N 31 eZnþ1 ¼ eKnþ1 þ jn XeCnþ1 ð nXÞ2 eMnþ1is the ðn þ 1Þth modal impedance matrix. All other steady-state modal responses are zero because only mode p ¼ n þ 1 is excited. In light of the decomposition shown in Eq. (39), the forced response in physical coordinates follows from Eq. (57) and is given by qss ðtÞ ¼ enþ1 uss nþ1 ðtÞ (77) where enþ1 is the ðn þ 1Þth column of the Fourier matrix and uss nþ1 ðtÞ is given by Eq. (76). Expanding Eq. (77) into its sector components yields qss 1 ðtÞ qss 2 ðtÞ ... qss i ðtÞ ... qss N ðtÞ 266666666664377777777775 ¼ 1ffiffiffiffi N p e j0unþ1 e j1unþ1 ... ejð i1Þ unþ1 ... e jð N1Þ unþ1 266666666664377777777775 ffiffiffiffi N p eZnþ1fe jn Xt (78) Thus, the steady-state forced response of the ith sector in physical coordinates is given by qss i ðtÞ ¼ 1ffiffiffiffi N p e jð i1Þ unþ1 ffiffiffiffi N p eZnþ1fe jn Xt ¼ eZnþ1fe jn /i e jn Xt; i 2 N (79) where /i ¼ 2pðn=NÞð i 1Þ. The response of each sector is identi-cal but simply shifted in time by a constant phase relative to its nearest neighbors. 3.3 Cyclic System With One DOF Per Sector. This exam-ple considers the simplest prototypical model for vibrations of a bladed disk with only one DOF per sector, nearest-neighbor cou-pling, and perfect symmetry. Despite the simplicity of the model, we show that the resonant response can be quite complicated when the system is subjected to engine order excitation with mul-tiple orders. We begin by formulating the EOM, and then consider a direct (traditional) approach to deriving the response to traveling wave excitation. The forced response is then derived using the modal analysis based on the Fourier matrix, which decouples the EOM. The nature of the natural frequencies and modes is consid-ered next, which sets the stage for examining the resonance behavior of the system when subjected to engine order excitation. This provides a quite general view of the forced response of these models. 3.3.1 Equations of Motion. The undamped cyclic system to be considered is shown in Fig. 8 in dimensionless form. It consists of a cyclic chain of N identical and identically coupled single-DOF oscillators with unit mass, the dynamics of which are cap-tured by the dimensionless transverse displacements qi for i 2 N .The oscillators are uniformly attached around the circumference of a stationary rigid hub via linear elastic elements with unit stiff-ness and unit effective length. Adjacent masses are elastically coupled via linear springs, each with nondimensional stiffness .It is assumed that the elastic elements are unstressed when the oscillators are in a purely radial configuration, that is, when qi ¼ 0for each i 2 N . An individual oscillator, together with the for-ward-nearest-neighbor elastic coupling, forms one fundamental sector and there are N such sectors in the overall system. The os-cillator chain has cyclic boundary conditions such that q0 ¼ qN and qNþ1 ¼ q1. Finally, the system is subjected to engine order ex-citation (Sec. 3.2) according to f iðtÞ ¼ fe j/i e jn rs ; i 2 N (80) where f is the strength of the excitation, the interphase blade angle /i is defined by Eq. (64), n 2 Zþ is the excitation order, r is the angular speed, and s is time (all dimensionless). The linear dynamics of the ith sector are obtained using New-ton’s laws and are governed by €qi þ q i þ t2ð qi1 þ 2qi qiþ1Þ ¼ fe j/i e jn rs ; i 2 N (81) where overdots denote differentiation with respect to dimension-less time s. In Eq. (81), the qi61 terms arise from the left-nearest-neighbor ( i 1) and right-nearest-neighbor ( i þ 1) elastic cou-pling. By stacking the N coordinates qi into the configuration vector q ¼ ð q1; q2; …; qN ÞT , the governing EOM for the overall N-DOF system takes the form €q þ K11 q ¼ f11 e jn rs ; i 2 N (82) where f11 ¼ ð fe j/1 ; fe j/2 ; …; fe j/N ÞT is the system forcing vector, which accounts for the constant phase difference in the dynamic loading from one sector to the next. The N N matrix K11 ¼ 1 þ 2t2 t2 0 … 0 t2 t2 1 þ 2t2 t2 … 0 00 t2 1 þ 2t2 … 0 0 ... ... ... . .. ... ... 0 0 0 … 1 þ 2t2 t2 t2 0 0 … t2 1 þ 2t2 2666666666437777777775 (83) captures the nondimensional stiffness of each sector relative to the hub (additive unity along its diagonal) and the intersector cou-pling ( t2 along the super-diagonal and subdiagonal). The elements t2 appearing in the (1, N) and ( N, 1) positions of K11 are due to the cyclic boundary conditions q0 ¼ q N and q Nþ1 ¼ q1 . In the ab-sence of these cyclic coupling terms, the system represents a finite chain of N oscillators. Thus, in addition to being symmetric, Eq. (83) is also a circulant and can be written as K11 ¼ circ ð1 þ 2t2; t2; 0; …; 0; t2Þ 2 SC N (84) where 1 þ 2t2; t2; 0; …; 0; t2 are the N generating elements. In the absence of coupling (that is, if t ¼ 0) K11 is diagonal and Eq. (82) represents a decoupled set of N harmonically forced, single-DOF oscillators. The forced response of Eq. (82) is considered next with empha-sis on a modal analysis whereby the fully coupled system (that is, one in which t 6 ¼ 0) is reduced to a set of N single-DOF oscilla-tors, only one of which is harmonically excited. The approach taken here, and a generalization in which each sector has multiple DOFs, is applied to the linear system in Sec. 3.4 to block decouple the system matrices as it is done in Sec. 3.1. 3.3.2 Forced Response. The steady-state forced response of Eq. (82) can be obtained using standard techniques and, for nonresonant forcing, is given by Fig. 8 Linear cyclic vibratory system with Nsectors and one DOF per sector 32 qss ðsÞ ¼ ð K11 n2r2IÞ1f11 e jn rs (85) where I is the N N identity matrix. However, this requires inver-sion of the impedance matrix K11 n2r2I, which is computation-ally expensive for a large number of sectors, and it offers little insight into the basic vibration characteristics. In what follows, a transformation based on the cyclic symmetry of the system is exploited to fully decouple the single N-DOF system to a set of N single-DOF oscillators from which the steady-state response is easily obtained. The procedure is similar to the usual modal analy-sis from elementary vibration theory. However, a key difference is that the transformation matrix (and hence the system mode shapes) is known a priori and, because the transformation is uni-tary (thus preserving the system eigenvalues), the natural frequen-cies are obtained after the transformation is carried out. Moreover, due to orthogonality conditions between the normal modes and forcing vector, the steady-state response of the overall system reduces to finding the forced response of a single harmoni-cally forced, single-DOF oscillator in modal space, which offers a clear advantage over the direct computation of Eq. (85). As described in Sec. 3.2.2, engine order excitation can be regarded as traveling wave dynamic loading. It is therefore rea-sonable to expect steady-state solutions of the same type. We begin with a simple way to show the existence of such a response, and then systematically describe it using the modal analysis tech-niques described in Secs. 3.1.2 and 3.2.3. Existence of a Traveling Wave Response. It is natural to search for steady-state solutions of the form qss i ðsÞ ¼ Ae j/i e jn rs ; i 2 N (86) which has the same traveling wave characteristics as the engine order excitation described in Sec. 3.2.2. Equation (86) assumes that each sector responds with the same amplitude A, but with a constant phase difference relative to its nearest neighbors, and to-gether all N such solutions form a traveling wave response among the sectors. By mapping this trial solution into Eq. (81) and divid-ing through by the common term e j/i e jn rs , it follows that ð nrÞ2A þ A þ t2 Ae junþ1 þ 2A Ae junþ1 ¼ f (87) where the identity /i61 /i ¼ 6unþ1 is employed. Solving for the amplitude A yields A ¼ f 1 þ 2t2ð1 cos unþ1Þ ð nrÞ2 (88) from which it follows that xnþ1 ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 þ 2t2ð1 cos unþ1Þ q is one of the N natural frequencies of the coupled system corre-sponding to mode p ¼ n mod N þ 1. Equation (88) shows that mode n þ 1 is excited, but the reason is not clear from this approach. A modal analysis that considers the fully coupled sys-tem is required to systematically describe the response character-istics of the cyclic system under engine order excitation. Modal Analysis. Theorem 9 guarantees that circulant matrices, such as the stiffness matrix defined by Eq. (84), are diagonalizable via a unitary transformation involving the Fourier matrix, and in what follows this property is exploited to fully decouple the matrix EOM defined by Eq. (82). To this end, the change of coordinates qðsÞ ¼ Eu ðsÞ or q iðsÞ ¼ eTi uðsÞ; i 2 N (89) is introduced, where E is the N N complex Fourier matrix (Defini-tion 17), ei is its ith column (Definition 18), and u ¼ ð u1; u2; …; uN ÞTis a vector of modal, or cyclic coordinates. Substituting Eq. (89) in Eq. (82) and multiplying from the left by EH yields EHE€u þ EHK11 Eu ¼ EHf11 e jn rs (90) where EHE ¼ I because E is unitary (Theorem 6). In light of Theorem 9, it follows that €u1 €u2 ...€uN 26666643777775 þ x21 0 x22 . .. 0 x2 N 26666643777775 u1 u2 ... uN 26666643777775 ¼ eH 1 f11 eH 2 f11 ... eH N f11 26666643777775 e jn rs (91) where the pth scalar element of the N 1 modal forcing vector EHf11 is eH p f11 . Decomposition of EHf11 follows from Eq. (60) by replacing the identity matrix IM with unity and the vector bf with f11 . Equation (89) is a unitary (similarity) transformation and hence the system natural frequencies are preserved, which is guar-anteed by Theorem 1. For each p 2 N , the dimensionless natural frequencies follow from Eq. (21) and are given by x2 p ¼ 1 þ 2t2 t2wðp1Þ N þ 0 þ þ 0 t2wðN1Þð p1Þ N ¼ 1 þ 2t2 t2 wðp1Þ N þ wðN1Þð p1Þ N ¼ 1 þ 2t2ð1 cos upÞ (92) where w N is the primitive Nth root of unity and the identity wðp1Þ N þ wðN1Þð p1Þ N ¼ 2 cos up is employed. Equation (91) is a decoupled set of N single-DOF harmonically forced modal oscil-lators of the form €up þ x2 p up ¼ eH p f11 e jn rs ; p 2 N (93) Thus, the single N-DOF system given by Eq. (82) is transformed to a system of N decoupled single-DOF systems defined by Eq. (93). The steady-state, nonresonant modal response of the pth decoupled system follows from Eq. (93) using standard techniques . Assuming harmonic motion, the solution is uss p ðsÞ ¼ eH p f11 x2 p ð nrÞ2 e jn rs ; p 2 N (94) from which the steady-state modal response vector uss ðsÞ ¼ uss 1 ðsÞ; uss 2 ðsÞ; …; uss N ðsÞ T is constructed. In physical coordinates, the steady-state response of sector i follows from Eq. (89) and is given by qss i ðsÞ ¼ eTi uss ðsÞ¼ 1ffiffiffiffi N p e j0ui ; ej1ui ; …; e jðN1Þui uss 1 ðsÞ; uss 2 ðsÞ; …; uss N ðsÞ T ¼ XNp¼11ffiffiffiffi N p e jðp1Þui uss p ðsÞ¼ XNp¼11ffiffiffiffi N p e jup ði1Þ eH p f11 x2 p ð nrÞ2 e jn rs ¼ 1ffiffiffiffi N p XNp¼1 eH p f11 x2 p ð nrÞ2e jði1Þup e jn rs ; i 2 N (95) where the identity ðp 1Þui ¼ upði 1Þ is employed. Equation (95) shows that there are N possible resonances, depending on the details of the modal forcing terms eH p f11 . However, only a single mode survives under an engine order excitation of order n, which 33 is clear from the orthogonality condition described in Sec. 3.2.3. Noting that f11 ¼ f f0 ¼ ffiffiffiffi N p f enþ1 , it follows from Eq. (72) that eH p f11 ¼ ffiffiffiffi N p f eH p enþ1 ¼ ffiffiffiffi N p f dpðnþ1Þ¼ ffiffiffiffi N p f ; p ¼ n þ 10; otherwise ( (96) which shows that the force vector f11 is mutually orthogonal to all but one of the modal vectors ep. That is, for p 2 N , the only excited mode is p ¼ n mod N þ 1 (97) for an engine order n 2 Zþ excitation. Thus, Eq. (95) reduces to qss i ðsÞ ¼ f x2 nþ1 ð nrÞ2 e j/i e jn rs ; i 2 N (98) where the identity ði 1Þunþ1 ¼ /i is employed and x2 nþ1 ¼ 1 þ 2t2ð1 cos unþ1Þ from Eq. (92). Equation (98) is the same result as that obtained from Eq. (88). Indeed, the process described here is significantly more laborious than the direct approach, but many general fea-tures can be gleaned from the analysis. The eigenfrequency char-acteristics (Sec. 3.3.3), normal modes of vibration (Sec. 3.3.4), and resonance structure (Sec. 3.3.5) are systematically described based on the modal decomposition results of this section. 3.3.3 Eigenfrequency Characteristics. The dimensionless nat-ural frequencies follow from Eq. (92) and are given by xp ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 þ 2t2ð1 cos upÞ q ; p 2 N (99) which clearly exhibit the effect of cyclic coupling. For the special case of t ¼ 0, the sectors are dynamically isolated and each has the same natural frequency xp ¼ 1. There are repeated natural fre-quencies for nonzero coupling ðt 6 ¼ 0Þ, a degeneracy that is due to the circulant structure of K. This is captured by the cyclic term cos up ¼ cos 2pðp 1Þ N ¼ Re w p1 N (100) which is obtained by projecting the powers of the Nth roots of unity onto the real axis (see Fig. 2). Multiplicity of the eigenfre-quencies can also be visualized in Fig. 9, which shows the dimen-sionless natural frequencies in terms of the number of nodal diameters (n.d.) in their attendant mode shapes versus: the mode number p for the special case of N ¼ 10 sectors; the wave type (i.e., BTW, FTW, or SW); the number of n.d.; and the sector num-ber i. 3 Results are shown for weak coupling (WC), strong cou-pling (SC), odd N (Fig. 9( a)), and even N (Fig. 9( b)). These cyclic features are described in terms of mode numbers p 2 N ¼ P O;ESW [ P O;EBTW [ P O;EFTW where each subset is defined in Table 5. A description of the BTW, FTW, and SW designations of these sets is deferred to Sec 3.3.4. The natural frequency corresponding to mode p ¼ 1 2 P O;ESW (zero harmonic of Eq. (100)) is distinct, but the remaining natural frequencies appear in repeated pairs, except for the case of even N, in which case the p ¼ ð N þ 2Þ=2 2 P ESW frequency ( N/2 har-monic) is also distinct. There are ( N 1)/2 such pairs if N is odd, and these correspond to mode numbers in POBTW and POFTW ,respectively. For even N there are ( N 2)/2 repeated natural fre-quencies corresponding to mode numbers in PEBTW and PEFTW .Finally, if k 2 P O;EBTW then the mode number of the corresponding repeated eigenfrequency is N þ 2 k 2 P O;EFTW . The normal modes Table 5 Sets of mode numbers p ‰ N corresponding to BTW, FTW, and SW normal modes of free vibration for odd and even N N Type Set Odd BTW POBTW ¼ p 2 Zþ : 2 n N þ 12 FTW POFTW ¼ p 2 Zþ : N þ 32 n N SW POSW ¼ 1f g Even BTW PEBTW ¼ p 2 Zþ : 2 n N 2 FTW PEFTW ¼ p 2 Zþ : N þ 42 n N SW PESW ¼ 1; N þ 22 Fig. 9 Dimensionless natural frequencies xp in terms of the number of n.d. versus mode number p for WC and SC: ( a) N 5 11 (odd) and ( b) N 5 10 (even). Also indicated below each figure is, for general N, the number of n.d. at each value of p and also the mode numbers corresponding to SW, BTW, and FTW. 3A mode shape nodal diameter refers to a line of zero sector responses across which adjacent sectors respond out of phase. For example, in Fig. 11 of Sec. 3.3.4, mode 1 has 0 n.d., modes 2 and 100 have 1 n.d., modes 3 and 99 have 2 n.d., and so on. 34 of vibration are described next, where it is shown that each can be categorized as a SW, BTW, or FTW. 3.3.4 Normal Modes of Vibration. It was shown that Eq. (82) can be decoupled via a unitary transformation involving the Fou-rier matrix E ¼ ð e1; e2; …; eN Þ. As a consequence, ep is the pth normal mode of vibration corresponding to the natural frequency xp. In what follows these mode shapes are characterized by inves-tigating the free response of the system, and it is shown that they are of the SW, BTW, or FTW variety. The free response of the system in its pth mode of vibration can be described by qðpÞ ðsÞ ¼ a pep e j xp s where ap is a modal amplitude and the natural frequency xp is defined by Eq. (99). There is generally a phase angle as well, which is omitted because its presence does not affect the arguments that follow. Noting that element i of ep can be written as wðp1Þð i1Þ N ¼ e jup ði1Þ, the free response of sector i can be written in real form as qðpÞ i ðsÞ ¼ ap cos ðupði 1Þ þ xpsÞ¼ ap cos up i 1 þ xp up s !! ¼ apUpði 1 þ C psÞ; i; p 2 N (101) where Cp ¼ xp=up and the function UpðvÞ is defined by Eq. (65). Equation (101) is a function of continuous time s and it is discre-tized according to the sector number i. In this way, it is endowed with the same discrete temporal and continuous spatial duality that is described in Sec. 3.2.2 in the context of traveling wave engine order excitation. That is, it can be regarded as the time–res-ponse of individual (discrete) sectors, or a continuous spatial vari-ation of displacements among the sectors that evolves with increasing time (i.e., a traveling wave). The propagating wave-form is strictly a BTW in the negative i-direction (descending sec-tor number) with wavelength 2 p=up ¼ N=ðp 1Þ and speed C p,an illustration of which is shown in Fig. 10. However, depending on the value of p, this gives rise to SW, BTW, or FTW mode shapes, a property that follows analogously from the features described in Sec. 3.2.2, where it is seen that Eq. (101) has the same form as Eq. (66). For the special case of p ¼ 1 it is clear from Eq. (101) that each sector behaves identically with the same amplitude and the same phase because u1 ¼ 0. An additional special case occurs when p ¼ (N þ 2)/2 if N is even. Then uðNþ2Þ=2 ¼ p and each sector has the same amplitude but adjacent sectors oscillate with a 180-deg phase difference. In this case, the vibration modes p 2 P O;ESW corre-spond to SW mode shapes whose characteristics can be visualized in Figs. 7( b) and 7( d) by replacing the amplitude F with ap. The remaining mode shapes correspond to repeated natural frequen-cies and are either BTWs or FTWs. In particular, the normal modes p 2 P O;EBTW (resp. p 2 P O;EFTW ) are backward (resp. forward) traveling waves and can be visualized in Fig. 7( a) (resp. Fig 7( c)). If mode k 2 P O;EBTW is a BTW corresponding to a natural frequency xk, then the attendant FTW mode is N þ 2 k 2 P O;EFTW with the same natural frequency xNþ2k ¼ xk.Figure 11 illustrates the normal modes of free vibration for a model with N ¼ 100 sectors. In this figure, the extent of the radial lines represents sector displacements. Those appearing outside the hub are to be interpreted as being positively displaced relative to their zero positions, and the opposite is true for lines inside the hub. Modes 1 and 51 are SWs, modes 2–50 are BTWs, and modes 52–100 are FTWs. Finally, the number of nodal diameters can be clearly identi-fied in Fig. 11. For example, modes 4 and 98 have 3 n.d. 3.3.5 Resonance Structure. In general, there may be a system resonance if the excitation frequency matches a natural frequency, that is, if nr ¼ xp. These possible resonances are conveniently identified in a Campbell diagram , an example of which is shown in Fig. 12( a) for engine orders n 2 N (the general case of n 2 Zþ is considered below), N ¼ 10, and ¼ 0:5. The natural frequen-cies are plotted in terms of the dimensionless rotor speed and sev-eral engine order lines n r are superimposed. Possible resonances correspond to intersections of the order lines and eigenfrequency loci. There are ðN þ 2Þ=2 such possibilities for each engine order if N is odd and ðN þ 1Þ=2 possible resonances if N is even. In light of Eq. (96), however, there is only a single resonance associated with each n under the traveling wave dynamic loading of Sec. 3.2, which corresponds to mode p ¼ n mod N þ 1. The set of N resonances for a system excited by N engine orders ðn ¼ 1; 2; …; NÞ are indicated by the black dots in Fig. 12( a) and the corresponding frequency response curves jqss i ðsÞj (for each n)are shown in Fig. 12( b) for a model with f ¼ 0.01. For example, a 3 e.o. excitation resonates mode 4 ( p ¼ 4), which is a BTW with 3 n.d. Mode 8 ( p ¼ 4) also has 3 n.d. and is excited by a 7 e.o. exci-tation. The TW and n.d. designations can be verified in Fig. 9. The basic resonance structure shown in Fig. 12( a) for n 2 N essentially aliases relative to the total number of sectors, in the sense that the excited modes for n ¼ mN þ 1; …; ðm þ 1ÞN with m 2 Zþ are the same as those for n 2 N . This follows from the orthogonality condition given by Eq. (96) and is manifested in Eq. (97), which gives a relationship for the excited mode in terms of the engine order n and total number of sectors N. Because n > 0 by assumption (see Sec. 3.2) the first mode ( p ¼ 1) is excited when n ¼ mN ¼ 10 ; 20 ; 30 ; …, the second mode ( p ¼ 2) is excited when n ¼ 1 þ mN ¼ 1; 11 ; 21 ; …, and so on. Table 6 summarizes these conditions for a model with N ¼ 10 sectors and the corresponding resonance structure for n ¼ N 1; …; 20 N is shown in Fig. 13( a). Each collection of resonance points n ¼ mN þ 1; …; ðm þ 1ÞN is qualitatively the same in structure. However, for m > 1 the resonances become increasingly clustered, which is shown in Fig. 13( b) for n ¼ N; …; 2N. In terms of the sets defined in Tables 4 and 5, an engine order n mod N 2 N O;ESW excites a SW mode p 2 P O;ESW . Similarly, an engine order n mod N 2 N O;EFTW (resp. n mod N 2 N O;EBTW ) excites a FTW (resp. BTW) mode p 2 P O;EFTW (resp. p 2 P O;EBTW ). While each engine order excites only a single mode, realistic excitation it composed of multiple harmonics (that is, orders), so that many modes can be excited. The nature of the natural fre-quencies and the order excitation lines leads to nontrivial reso-nance behavior even in the case of perfect symmetry. Of course, as noted elsewhere in this paper, imperfections that disturb the symmetry lead to even more complicated responses, in which ev-ery intersection between e.o. and natural frequency lines can lead to a resonance. These are especially important when the intersec-tor coupling is small. Fig. 10 Abackward traveling wave apUpði11CpsÞ 5apcos ðupði21Þ1xpsÞwith amplitude ap, wavelength 2 p=up 5N=ðp21Þ, and speed Cp5xp=up35 3.4 Cyclic System With Two DOFs Per Sector. This exam-ple generalizes the one DOF per sector model of Sec. 3.3 to a sim-ple system with two DOFs per sector, which demonstrates the process of block diagonalizing the EOM when there are multiple DOFs per sector. The mathematics of the decoupling process described here applies equally as well to models with two or N DOFs per sector. Of course, the nature of the natural frequencies and mode shapes depend on the details of each sector model which, for the cyclic system considered here, is discussed in Refs. [92–94]. There is much more to the topic of multiple DOFs per sector; the reader is referred to the works of Ottarsson [97,114] and Bladh [29,115–117] for more details and more complex examples. 3.4.1 Equations of Motion. The nondimensional bladed disk model shown in Fig. 14( a) consists of a rotationally periodic array of N identical, identically coupled sector models (Fig. 14( b)). The disk has radius d and rotates with a fixed speed r about an axis through C. Each blade is modeled by a simple pendulum with unit mass and length, the dynamics of which are captured by the nor-malized angles x i with i 2 N . The blades are attached to the rotat-ing disk via linear torsional springs with unit stiffness, and adjacent blades are elastically coupled by linear springs with stiff-ness . It is assumed that the springs are unstretched when the blades are in a purely radial configuration, that is, when each x i ¼ 0. As shown in the inset of Fig. 14( b), each blade is fitted with a pendulum like, circular-path vibration absorber with radius c and mass l at an effective distance a along the blade length. The absorber dynamics are captured by the normalized pendulum angles yi, which are physically limited to jy ij 1 by stops that represent the rattling space limits imposed by the blade geometry. This feature is included for generality, but in all of what follows it is assumed that jyij < 1, i.e., that impacts do not occur. Linear Fig. 11 Normal modes of free vibration for a model with N5100 sectors. Mode 1 consists of a SW, in which each sector oscillates with the same amplitude and phase. Mode 51 also corresponds to a SW, but neighboring oscillators oscillate exactly 180 deg out of phase. Modes 2–50 (resp. 52–100) consist of BTWs (resp. FTWs). 36 viscous damping is also included at the spring locations, but is not indicated in Fig. 14. Blade and interblade damping is captured by linear torsional and translational dampers with constants nb and nc,respectively, and the absorber damping is captured by a torsional damper with constant na. Finally, the system is subjected to the traveling wave dynamic loading defined by Eq. (80), as shown in Fig. 14( b). Sector Model. The EOM for each two-DOF sector are derived using Lagrange’s method and linearized for small motions of the primary and absorber systems, that is, for small xi and yi. Then for each i 2 N , the dynamics of the ith sector are governed by [92,93] lc 2ð€x i þ €y iÞ þ na _y i þ lcdr 2ðx i þ y iÞþ lca ð€x i þ r2y iÞ ¼ 0 (102 a) €x i þ nb _x i na _y i þ x i þ dr 2x i þ l a2 €x i þc2ð€x i þ €y iÞ þ ac ð€y i þ 2€x iÞþadr 2x i þ cdr 2ðx i þ y iÞ " # þ ncð _xi 1 þ 2 _x i _x iþ1Þþ 2ð xi 1 þ 2x i x iþ1Þ ¼ fe j/i e jn rs (102 b)where Eq. (102 a) describes the absorber dynamics and Eq. (102 b)describes the blade dynamics. The indices i are taken mod N such that x Nþ1 ¼ x1 and x0 ¼ x N , which are cyclic boundary con-ditions implying that the Nth blade is coupled to the first. In matrix–vector form, and for each i 2 N , Eq. (102) becomes M€zi þ C_zi þ Kz iþCcð _zi1 þ 2_zi _ziþ1Þþ Kcð zi1 þ 2zi ziþ1Þ¼ fe j/i e jn rs 9>=>; (103) where zi ¼ x i; yið ÞT captures the sector dynamics, f ¼ f ; 0ð ÞT is a sector forcing vector, and the elements of the sector mass, damp-ing, and stiffness matrices are defined in Table 7. The matrices Cc ¼ nc 00 0 " # ; Kc ¼ 2 00 0 " # (104) capture the interblade coupling and vanish if nc ¼ ¼ 0, in which case Eq. (103) describes the forced motion of N isolated blade/ absorber systems. System Model. By stacking each zi into the configuration vector q ¼ z1; z2; …; zNð ÞT , the governing matrix EOM for the overall 2N-DOF system takes the form bM€q þ bC _q þ bKq ¼ bfe jn rs (105) where bM 2 BCBS 2;N is block diagonal with diagonal blocks M and bK 2 BCBS 2;N has generating matrices K þ 2Kc; Kc; 0; …; 0; Kc. The matrix bC 2 BCBS 2;N is similarly defined by replacing K with C and Kc with Cc in bK. In terms of the circulant Fig. 12 (a) Campbell diagram and ( b) corresponding frequency response curves jqss i ðsÞj for N 5 10, m 5 0:5, f 5 0.01, and each n 5 1; 2; . . . ; N Table 6 Condition on the engine order n ‰ Zþ to excite mode p ‰ N for N 5 10 Excited mode Conditions on engine order n 1 mN ¼ 10 ; 20 ; 30 ; …2 1 þ mN ¼ 1; 11 ; 21 ; …3 2 þ mN ¼ 2; 12 ; 22 ; … ... ... N 1 N 2 þ mN ¼ 8; 18 ; 28 ; … N N 1 þ mN ¼ 9; 19 ; 29 ; … Fig. 13 (a) Campbell diagram for N 5 10, m 5 0:5, f 5 0.01, and n 5 1; . . . ; 20 N and ( b) the corresponding frequency response curves jqss i ðsÞj corresponding to n 5 N; . . . ; 2N. Engine order lines are not shown for n 5 N 1 1; N 1 2; . . . ; 2N 2 1, and so on. 37 operator, the system mass, damping, and stiffness matrices are defined by bM ¼ circ ðM; 0; 0; …; 0; 0Þ bC ¼ circ ðC þ 2Cc; Cc; 0; …; 0; CcÞ bK ¼ circ ðK þ 2Kc; Kc; 0; …; 0; KcÞ 9>>=>>; (106) The 2 N 1 system forcing vector is bf ¼ fej/1 ; fe j/2 ; …; fej/N T ¼ f0 f (107) where the N 1 vector f0 is defined by Eq. (69) and the interblade phase angle /i is given by Eq. (64). 3.4.2 Forced Response. The forced response of the overall system defined by Eq. (105) can be handled directly using stand-ard techniques . Its nonresonant solution in the steady-state follows in the usual way and is given by qss ðsÞ ¼ bZ1bfe jn rs (108) where bZ ¼ bK n2r2 bM þ jn rbC is the system impedance matrix of dimension 2 N 2N. However, Eq. (108) does not offer any insight into the system’s modal characteristics and it requires computation of bZ1 , which can be prohibitive for practical bladed disk models with many sectors and many DOFs per sector. We thus turn to a decoupling strategy that exploits the system symme-try and the theory developed in Sec. 2. The analysis follows simi-larly to that presented in Sec. 3.1, except in this case the single coupled 2 N-DOF system is transformed into a set of N block decoupled two-DOF systems. To this end, we introduce the change of coordinates q ¼ ð E IÞu; or zi ¼ ð eTi IÞu; i 2 N (109) where E is the N N complex Fourier matrix and ei is its ith col-umn, is the Kronecker product, I is the 2 2 identity matrix (the dimension of I corresponds to the number of DOFs per sec-tor), and u ¼ u1; u2; …; uNð ÞT is a vector of modal, or cyclic coor-dinates. Each up is 2 1 and describes the sector dynamics in modal space. Substituting Eq. (109) into Eq. (105), multiplying from the left by the unitary matrix ðE IÞ H ¼ ð EH IÞ, and invoking Theorem 10 yields a system of N block decoupled equa-tions, each with two DOFs. They are eMp €up þ eCp _up þ eKpup ¼ ð eH p IÞbfe jn rs ; p 2 N (110) where ðeH p IÞbf is the pth 2 1 block of ðEH IÞbf. Equation (110) is analogous to the N M -DOF systems given by Eq. (61) for the general formulation in Sec. 3.1, but in this case M ¼ 2 and engine order excitation is assumed from the onset. Figure 15 illus-trates the transformation of the single 2 N-DOF system given by Eq. (105) to a system of N block decoupled two-DOF forced oscil-lators defined by Eq. (110). The 2 2 mass, damping, and stiffness matrices associated with the pth mode follow from Theorem 10 and are given by eMp ¼ M eCp ¼ C þ 2Ccð1 cos upÞ eKp ¼ K þ 2Kcð1 cos upÞ 9>>=>>; ; p 2 N (111) where up is defined by Eq. (16), the elements of M, C, and K are defined in Table 7 and the coupling matrices Cc and Kc are defined by Eq. (104). In light of Eqs. (70) and (71), the pth modal forcing vector takes the form ðeH p IÞbf ¼ ð eH p IÞð f0 fÞ¼ eH p f0 f (112) ¼ ffiffiffiffi N p f; p ¼ n þ 1 0; otherwise ( Fig. 14 (a) Model of bladed disk assembly and ( b) sector model Table 7 Elements of the sector mass, damping, and stiffness matrices M, C, and K Matrix Notation Elements Mass MM11 ¼1þlðaþcÞ2 M12 ¼lc ðaþcÞ M21 ¼M12 M22 ¼lc 2Damping CC11 ¼nbC12 ¼ naC21 ¼0 C22 ¼na Stiffness KK11 ¼1þ1þlðaþcÞðÞdr 2 K12 ¼lcdr 2 K21 ¼K12 K22 ¼lc a þdðÞr238 where 0 ¼ (0,0) T and the scalar product eH p f0 vanishes except for p ¼ n þ 1. Because only mode p ¼ n þ 1 is excited, unþ1ðsÞ is the only nonzero modal response in the steady-state. Assuming harmonic motion, and in light of Eq. (112), the pth steady-state modal response follows easily from Eq. (110) and is given by uss p ðsÞ ¼ ffiffiffiffi N p eZ1 nþ1 fe jn rs ; p ¼ n þ 1 0; otherwise ( (113) where eZp ¼ eKp n2r2 eMp þ jn reCp; p 2 N (114) is the pth modal impedance matrix. The response of sector i (in physical coordinates) follows from the transformation given by Eq. (109) with uss ðsÞ ¼ ð 0; …; 0; uss nþ1 ðsÞ; 0; …; 0ÞTand is given by zss i ðsÞ ¼ eZ1 nþ1 fe j/i e jn rs ; i 2 N (115) where w nði1Þ ¼ e j/i is employed. From Eq. (115) it is clear that each blade/absorber combination behaves identically except for a constant phase shift from one sector to another, which is captured by the interblade phase angle /i. This approach offers a signifi-cant computational advantage over the direct solution to the full 2N-DOF system, as given by Eq. (108). 4 Conclusions The goal of this paper is to provide the mathematical tools for handling circulant matrices as they apply to the free and forced vibration analysis of structures with cyclic symmetry. As demon-strated by past work in this area and the review provided here, the theory of circulants provides a useful description of the fundamen-tal structure of the mode shapes and spectrum of systems with cyclic symmetry, including those of large scale. The theory also provides a convenient means for computing the vibration response of these systems, even when the idealized symmetry is broken by mistuning or by nonlinear effects. As with any mathematical tool, the overhead in learning it must provide appropriate benefit, whether in terms of fundamental understanding, insight, or ease of computation. We trust that the results presented here offer such benefits to readers interested in vibration analysis of cyclic systems. It must be noted that no physical system has perfect symmetry, as assumed herein. This assumption must be examined in light of the system under consideration. One key to the suitability of a cyclically symmetric model is the intersector coupling. If the cou-pling is strong, so that the pairs of modes have well separated fre-quencies, then small imperfections will not alter the picture substantially, and one can consider each mode pair as robust against coupling to other modes. However, if the coupling is weak, so that the system frequencies are clustered near those of the isolated sector model, the possibility of localization is signifi-cantly increased. This topic has been investigated quite thor-oughly, primarily in the context of the vibration of bladed disk assemblies with small blade mistuning; see, for example, [72–88]. Also, nonlinear effects can couple linear modes under certain res-onance conditions, even at small amplitudes . In cyclic sys-tems this can occur for the pairs of modes with equal frequencies , and this possibility expands to groups of modes for the case of weak coupling, leading to extremely complicated behavior [82,84,88]. 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Sound Vib., 145 (1), pp. 95–110. Tseng, J. G., and Wickert, J. A., 1994, “On the Vibration of Bolted Plate and Flange Assemblies,” ASME J. Vib. Acoust., 116 (4), pp. 468–473. Shahab, A. A. S., and Thomas, J., 1987, “Coupling Effects of Disc Flexibility on the Dynamic Behaviour of Multi Disc-Shaft Systems,” J. Sound Vib., 114 (3), pp. 435–452. Kim, H., and Shen, I.-Y., 2009, “Ground-Based Vibration Response of a Spin-ning, Cyclic, Symmetric Rotor With Gyroscopic and Centrifugal Softening Effects,” ASME J. Vib. Acoust., 131 (2), p. 021007. Kim, H., Colonnese, N. T. K., and Shen, I. Y., 2009, “Mode Evolution of Cyclic Symmetric Rotors Assembled to Flexible Bearings and Housing,” ASME J. Vib. Acoust., 131 (5), p. 051008. Shaw, S. W., and Pierre, C., 2006, “The Dynamic Response of Tuned Impact Absorbers for Rotating Flexible Structures,” ASME J. Comput. Nonlinear Dyn., 1(1), pp. 13–24. Shi, C., and Parker, R. 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2097	https://ocw.mit.edu/courses/8-01sc-classical-mechanics-fall-2016/resources/36-1/	36.2 Worked Example - Wheel Rolling Without Slipping Down Inclined Plane - Torque Method \| Classical Mechanics \| Physics \| MIT OpenCourseWare Browse Course Material Syllabus About the Team Online Textbook Readings Assignments Review: Vectors [Lesson 0: Vectors [0.1 - 0.6]]( Week 1: Kinematics Week 1 Introduction [Lesson 1: 1D Kinematics - Position and Velocity [1.1-1.7]]( [Lesson 2: 1D Kinematics - Acceleration [2.1-2.5]]( [Lesson 3: 2D Kinematics [3.1-3.5]]( [Week 1 Worked Examples [PS.1.1-PS.1.5]]( Problem Set 1 Week 2: Newton's Laws Week 2 Introduction [Lesson 4: Newton's Laws of Motion [4.1-4.4]]( [Lesson 5: Gravity [5.1-5.3]]( [Lesson 6: Contact Forces [6.1-6.2]]( [Lesson 7: Tension and Springs [7.1-7.4]]( [Deep Dive: Friction [DD 1.1]]( [Week 2 Worked Examples [PS.2.1-PS.2.3]]( Problem Set 2 Week 3: Circular Motion Week 3 Introduction [Lesson 8: Circular Motion - Position and Velocity [8.1-8.3]]( [Lesson 9: Uniform Circular Motion [9.1-9.3]]( [Lesson 10: Circular Motion – Acceleration [10.1-10.4]]( [Lesson 11: Newton's 2nd Law and Circular Motion [11.1-11.3]]( Week 3 Worked Example Problem Set 3 Week 4: Drag Forces, Constraints and Continuous Systems Week 4 Introduction [Lesson 12: Pulleys and Constraints [12.1-12.5]]( [Lesson 13: Massive Rope [13.1-13.6]]( [Lesson 14: Resistive Forces [14.1-14.3]]( Problem Set 4 Week 5: Momentum and Impulse Week 5 Introduction [Lesson 15: Momentum and Impulse [15.1-15.5]]( [Lesson 16: Conservation of Momentum [16.1-16.2]]( [Lesson 17: Center of Mass and Motion [17.1-17.7]]( Problem Set 5 Week 6: Continuous Mass Transfer Week 6 Introduction [Lesson 18: Relative Velocity and Recoil [18.1-18.4]]( [Lesson 19: Continuous Mass Transfer [19.1-19.7]]( [Week 6 Worked Examples [PS.6.1-PS.6.2]]( Problem Set 6 Week 7: Kinetic Energy and Work Week 7 Introduction [Lesson 20: Kinetic Energy and Work in 1D [20.1-20.6]]( [Lesson 21: Kinetic Energy and Work in 2D and 3D [21.1-21.6]]( [Lesson 22: Conservative and Non-Conservative Forces [22.1-22.5]]( Week 7 Worked Example Problem Set 7 Week 8: Potential Energy and Energy Conservation Week 8 Introduction [Lesson 23: Potential Energy [23.1-23.5]]( [Lesson 24: Conservation of Energy [24.1-24.4]]( [Lesson 25: Potential Energy Diagrams [25.1-25.3]]( Problem Set 8 Week 9: Collision Theory Week 9 Introduction [Lesson 26: Types of Collision [26.1-26.3]]( [Lesson 27: Elastic Collisions [27.1-27.6]]( [Deep Dive: Center of Mass Reference Frame [DD.2.1-DD.2.7]]( Problem Set 9 Week 10: Rotational Motion Week 10 Introduction [Lesson 28: Motion of a Rigid Body [28.1-28.3]]( [Lesson 29: Moment of Inertia [29.1-29.6]]( [Lesson 30: Torque [30.1-30.5]]( [Lesson 31: Rotational Dynamics [31.1-31.7]]( Week 10 Worked Example Problem Set 10 Week 11: Angular Momentum Week 11 Introduction [Lesson 32: Angular Momentum of a Point Particle [32.1-32.4]]( [Lesson 33: Angular Momentum of a Rigid Body [33.1-33.5]]( [Lesson 34: Torque and Angular Impulse [34.1-34.5]]( Problem Set 11 Week 12: Rotations and Translation - Rolling Week 12 Introduction [Lesson 35: Rolling Kinematics [35.1-35.5]]( [Lesson 36: Rolling Dynamics [36.1-36.5]]( [Lesson 37: Rolling Kinetic Energy & Angular Momentum [37.1-37.4]]( [Deep Dive: Gyroscopes [DD.3.1-DD.3.3]]( Problem Set 12 Course Info Instructors Prof. Deepto Chakrabarty Dr. Peter Dourmashkin Dr. Michelle Tomasik Prof. Anna Frebel Prof. Vladan Vuletic Departments Physics As Taught In Fall 2016 Level Undergraduate Topics Science Physics Classical Mechanics Learning Resource Types theaters Lecture Videos assignment Problem Sets menu_book Open Textbooks Download Course menu search Give Now About OCW Help & Faqs Contact Us searchGIVE NOWabout ocwhelp & faqscontact us 8.01SC \| Fall 2016 \| Undergraduate Classical Mechanics Menu More Info Syllabus About the Team Online Textbook Readings Assignments Review: Vectors [Lesson 0: Vectors [0.1 - 0.6]]( Week 1: Kinematics Week 1 Introduction [Lesson 1: 1D Kinematics - Position and Velocity [1.1-1.7]]( [Lesson 2: 1D Kinematics - Acceleration [2.1-2.5]]( [Lesson 3: 2D Kinematics [3.1-3.5]]( [Week 1 Worked Examples [PS.1.1-PS.1.5]]( Problem Set 1 Week 2: Newton's Laws Week 2 Introduction [Lesson 4: Newton's Laws of Motion [4.1-4.4]]( [Lesson 5: Gravity [5.1-5.3]]( [Lesson 6: Contact Forces [6.1-6.2]]( [Lesson 7: Tension and Springs [7.1-7.4]]( [Deep Dive: Friction [DD 1.1]]( [Week 2 Worked Examples [PS.2.1-PS.2.3]]( Problem Set 2 Week 3: Circular Motion Week 3 Introduction [Lesson 8: Circular Motion - Position and Velocity [8.1-8.3]]( [Lesson 9: Uniform Circular Motion [9.1-9.3]]( [Lesson 10: Circular Motion – Acceleration [10.1-10.4]]( [Lesson 11: Newton's 2nd Law and Circular Motion [11.1-11.3]]( Week 3 Worked Example Problem Set 3 Week 4: Drag Forces, Constraints and Continuous Systems Week 4 Introduction [Lesson 12: Pulleys and Constraints [12.1-12.5]]( [Lesson 13: Massive Rope [13.1-13.6]]( [Lesson 14: Resistive Forces [14.1-14.3]]( Problem Set 4 Week 5: Momentum and Impulse Week 5 Introduction [Lesson 15: Momentum and Impulse [15.1-15.5]]( [Lesson 16: Conservation of Momentum [16.1-16.2]]( [Lesson 17: Center of Mass and Motion [17.1-17.7]]( Problem Set 5 Week 6: Continuous Mass Transfer Week 6 Introduction [Lesson 18: Relative Velocity and Recoil [18.1-18.4]]( [Lesson 19: Continuous Mass Transfer [19.1-19.7]]( [Week 6 Worked Examples [PS.6.1-PS.6.2]]( Problem Set 6 Week 7: Kinetic Energy and Work Week 7 Introduction [Lesson 20: Kinetic Energy and Work in 1D [20.1-20.6]]( [Lesson 21: Kinetic Energy and Work in 2D and 3D [21.1-21.6]]( [Lesson 22: Conservative and Non-Conservative Forces [22.1-22.5]]( Week 7 Worked Example Problem Set 7 Week 8: Potential Energy and Energy Conservation Week 8 Introduction [Lesson 23: Potential Energy [23.1-23.5]]( [Lesson 24: Conservation of Energy [24.1-24.4]]( [Lesson 25: Potential Energy Diagrams [25.1-25.3]]( Problem Set 8 Week 9: Collision Theory Week 9 Introduction [Lesson 26: Types of Collision [26.1-26.3]]( [Lesson 27: Elastic Collisions [27.1-27.6]]( [Deep Dive: Center of Mass Reference Frame [DD.2.1-DD.2.7]]( Problem Set 9 Week 10: Rotational Motion Week 10 Introduction [Lesson 28: Motion of a Rigid Body [28.1-28.3]]( [Lesson 29: Moment of Inertia [29.1-29.6]]( [Lesson 30: Torque [30.1-30.5]]( [Lesson 31: Rotational Dynamics [31.1-31.7]]( Week 10 Worked Example Problem Set 10 Week 11: Angular Momentum Week 11 Introduction [Lesson 32: Angular Momentum of a Point Particle [32.1-32.4]]( [Lesson 33: Angular Momentum of a Rigid Body [33.1-33.5]]( [Lesson 34: Torque and Angular Impulse [34.1-34.5]]( Problem Set 11 Week 12: Rotations and Translation - Rolling Week 12 Introduction [Lesson 35: Rolling Kinematics [35.1-35.5]]( [Lesson 36: Rolling Dynamics [36.1-36.5]]( [Lesson 37: Rolling Kinetic Energy & Angular Momentum [37.1-37.4]]( [Deep Dive: Gyroscopes [DD.3.1-DD.3.3]]( Problem Set 12 36.2 Worked Example - Wheel Rolling Without Slipping Down Inclined Plane - Torque Method 36.2 Worked Example - Wheel Rolling Without Slipping Down Inclined Plane - Torque Method Instructor: Dr. Peter Dourmashkin Video Player is loading. 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Transcript Download video Download transcript 0:03 Now let's go back and analyze that same problem 0:05 that we were looking at before of a wheel 0:09 rolling down an incline plane. 0:11 And it's rolling without slipping. 0:15 But instead of using the energy method, 0:18 we're now going to use the torque method. 0:21 So let's consider the center of mass. 0:24 And what we want to do is draw the forces. 0:26 We have a normal force. 0:28 We have gravity. 0:30 And we have the friction force about the center of mass. 0:34 Our wheel has a radius R. 0:37 Let's choose a coordinate system i hat, j hat. 0:42 And so we have a right-handed system k hat, 0:45 and that will correspond to some angle theta. 0:49 Now here we're now going to enlarge how we apply 0:54 both translation and rotation. 0:56 And the beauty of this problem is 0:58 we now can decompose our motion into translational motion 1:04 and rotational motion. 1:05 So for the transitional motion, we'll 1:07 apply Newton's second law. 1:11 Now, if this is the angle phi, then that's 1:13 the angle phi as well. 1:15 And so our forces in the i hat direction, we 1:19 have mg sine phi minus the friction force. 1:25 And that's equal to the x component of the acceleration. 1:31 Now we also can choose the center of mass 1:35 to calculate the torque. 1:38 And so what we're really just studying here 1:40 is simply our old problem in the center of mass frame of fixed 1:44 axis rotation. 1:46 And you can see gravity is acting at the center of mass. 1:49 So it produces no torque about the center of mass. 1:52 The normal force is directed towards the center of mass. 1:56 And so when we take that vector product of R cross n from cm 2:06 to this point down here, the contact point, 2:08 these forces are anti-parallel. 2:10 So the normal force produces no torque. 2:12 And the only torque that we have is from the friction force, 2:16 and that friction torque is going 2:18 to give us a positive angular acceleration in the k hat 2:22 direction. 2:23 It's at right angles with the vector R. 2:26 So we have fs times R equals I center of mass times alpha. 2:33 And these are our two dynamic equations. 2:37 But remember, when the object is rolling without slipping, 2:40 let's just remind ourselves that Vcm equals R omega. 2:45 And if I differentiate, the Acm which is what we're calling ax, 2:50 is equal to R alpha. 2:53 So this ax here is the acceleration 2:57 of the center of mass. 2:59 And that's our third condition. 3:02 And so now I see that I have three equations 3:06 and my unknowns-- 3:09 fs, ax and alpha. 3:12 And so I'm going to solve these equations for a. 3:16 And I'll look at these equations. 3:18 And what I'll do is I'll just substitute for alpha ax over R. 3:24 And then solve this equation for fs, and put it in there. 3:28 And so I get mg sine phi. 3:32 Now my fs is equal to Icm over R times alpha. 3:39 But alpha is ax over R. So that's ax and an R squared. 3:45 Notice dimensionally, I is mr squared. 3:48 So this is just ma, the dimensions 3:51 of force, mg dimensions of force, and that's equal to max. 3:56 And now I can solve for ax. 4:00 And I get mg sine phi divided by m plus Icm over R squared. 4:11 Now ax is a constant. 4:15 And we can, from our kinematic equations, 4:19 if our object is moving a distance s 4:28 as it drops height h, we know from kinematics 4:34 and we can work this out. 4:35 We have that Xcm is one half Acm t squared. 4:42 And we know that the velocity Vcm equals Acm times t. 4:53 And so when we put these two together, 4:56 this is the distance s, we get that the velocity 5:02 cm is equal to the square root of 2s times Acm. 5:12 Now s is equal to h over sine phi. 5:21 S h over s is sine phi. 5:26 And so the Vcm equals the square root of 2h sine phi times Acm-- 5:34 but we've solved for Acm-- 5:36 mg sine phi over m plus Icm over R squared. 5:44 And so we get the square root of 2mgh 5:50 over m plus Icm divided by R squared. 5:56 And this agrees with our energy method. Course Info Instructors Prof. Deepto Chakrabarty Dr. Peter Dourmashkin Dr. Michelle Tomasik Prof. Anna Frebel Prof. Vladan Vuletic Departments Physics As Taught In Fall 2016 Level Undergraduate Topics Science Physics Classical Mechanics Learning Resource Types theaters Lecture Videos assignment Problem Sets menu_book Open Textbooks Download Course Over 2,500 courses & materials Freely sharing knowledge with learners and educators around the world. Learn more © 2001–2025 Massachusetts Institute of Technology Accessibility Creative Commons License Terms and Conditions Proud member of: © 2001–2025 Massachusetts Institute of Technology You are leaving MIT OpenCourseWare close Please be advised that external sites may have terms and conditions, including license rights, that differ from ours. MIT OCW is not responsible for any content on third party sites, nor does a link suggest an endorsement of those sites and/or their content. Stay Here Continue
2098	https://www.chemguide.co.uk/14to16/equations/equations.html	How to write equations for simple reactions Chemguide: Core Chemistry 14 - 16 How to write symbol equations for simple chemical reactions This page explains how to work out equations for simple chemical reactions. You have to get this right, because chemistry will be a nightmare to you if you find equations a mystery - it is an essential tool in understanding chemistry. Learning to write equations can take quite a lot of time when you are starting chemistry. This is one of the major hurdles in the subject. Don't try to rush it - spending time now will save you a lot of problems later on. You will need to be reasonably competent at writing formulae for ionic compounds and there is no point in continuing with this page until you are. How writing equations works You need to know what is actually happening! You can't just make up an equation - it has to be based on exactly what happened in a chemical reaction. You need to know what you started with and what you ended up with. Have a look at this YouTube video which is a snippet from a Royal Society of Chemistry lecture, and shows magnesium metal burning in oxygen in the air. It forms magnesium oxide. The video displayed this as a simple word equation. magnesium + oxygen magnesium oxide The arrow here be translated as "goes to" or "produces" or whatever similar term you want to use. In the very early stages of a chemistry course, you might be asked to write word equations - usually for reactions slightly more complicated than this one. But if you are doing chemistry properly, you may never actually write a word equation down, but it has to be in your mind. You have to know what you started with and what you ended up with. Translating the words into symbols Still using the example of magnesium burning in oxygen . . . The next thing you have to do is to translate the names of the substances involved into their formulae. Mg + O 2 MgO Don't forget that oxygen (in common with things like hydrogen and chlorine) goes around in pairs. But this isn't the finished equation. Note:I know I keep on about this, but if you can't work out that magnesium oxide is MgO, or you didn't know that oxygen goes around in pairs, then you must go back and read the pages about writing formulae. It is pointless going on with this page if you can't do basic things like that. Balancing the equation Each symbol you write counts as one atom (or one ion) of that element. (You will later come across another more mathematical way of looking at this, but that is some distance in the future.) That means that in the equation we have just written, you seem to have started with 2 atoms of oxygen, but only ended up with 1 in the product. You can't just lose an oxygen atom - everything must be accounted for. Here is the unfinished equation again . . . Mg + O 2 MgO So how do you get another oxygen on the right-hand side? What you absolutely must not do is change the formula of the magnesium oxide. You can't just re-write the formula of magnesium oxide as MgO 2 - that doesn't exist! You must never try to balance an equation by changing a formula. What you need is to form twice as much MgO. You can do this by writing a big "2" in front of the formula. A big number in front of a formula multiplies up the whole formula until you get to a break - the end of the equation, a plus sign, or an arrow, for example. So a revised version would be . . . Mg + O 2 2MgO But beware! By writing that 2, you have messed up the balance of the magnesium. If you leave it like that, it seems that you have by magic generated another magnesium atom out of nowhere. That is easily put right, of course, by re-writing the equation to start with 2 atoms of magnesium. 2Mg + O 2 2MgO And that is the proper equation - everything is now balanced on both sides of it. This is a trivial example, of course, and all of this would just be running through your head, but it illustrates the principles. You have to start with a word equation, even if that only exists in your head. Write down the formulae for everything in your word equation. Look for where the equation doesn't balance, and put it right. (I will give you some more help with this in examples to come.) At the end, go back and check everything again in case by altering something you have messed up something you have already checked - as, for example, with the magnesium above. Examples Example 1 Heptane is one of the hydrocarbons in petrol (gasoline). It has the formula C 7 H 16, and burns to form carbon dioxide and water. Write the equation for this. You have all the information you need in the question, so all you need to do is to write the formulae down to start the equation writing process. (I am assuming you know the formulae for carbon dioxide and water.) C 7 H 16 + O 2 CO 2 + H 2 O The best way of balancing most equations is to start from the beginning and work through the elements one at a time. However, it is a good idea to leave until the end anything which occurs more than once on either side of the equation - in this case, oxygen. You will usually find that this will sort itself out easily later on. So start with the carbon. If there are 7 atoms on the left-hand side, there must be 7 on the right-hand side as well. That means you must have produced 7 molecules of carbon dioxide. C 7 H 16 + O 2 7CO 2 + H 2 O Now the hydrogens. There are 16 hydrogens on the left, and so there have to be 16 on the right as well. You must have produced 8 molecules of water. C 7 H 16 + O 2 7CO 2 + 8H 2 O Now count the oxygens. There are a total of 22 on the right-hand side. To get 22 atoms, you must have started with 11 molecules of oxygen. C 7 H 16 + 11O 2 7CO 2 + 8H 2 O And that's all OK. You wouldn't normally write all these stages down, of course - you just write the original formulae down leaving enough space for the balancing numbers, and then write them in as you work them out. Example 2 Calcium carbonate reacts with dilute hydrochloric acid to give a solution of calcium chloride, carbon dioxide gas and water. Write the equation for this. You have to know that dilute hydrochloric acid is a solution of hydrogen chloride in water. For equation purposes, you can just write this as HCl. Obviously, you should know how to work this formula out but, to be honest, you will meet it so many times, that you will just know it. If you work out the other formulae, you will get . . . CaCO 3 + HCl CaCl 2 + CO 2 + H 2 O Working from left to right, the calcium is OK; the carbon is also OK; the oxygen is OK. The hydrogen and chlorine are obviously not OK. You can put the chlorine right by having two lots of HCl - NOT by turning the HCl into the non-existent HCl 2. This also puts the hydrogen right. That gives the final balanced equation . . . CaCO 3 + 2HCl CaCl 2 + CO 2 + H 2 O Example 3 This is to illustrate how you get around a particular problem that comes up now and then. If you heat solid copper(II) nitrate, you get copper(II) oxide, nitrogen dioxide and oxygen. The unbalanced symbol equation is . . . Cu(NO 3)2 CuO + NO 2 + O 2 Working systematically, the copper is OK. On the left-hand side, there are 2 nitrogen atoms. The little 2 after the brackets applies to everything inside the brackets. You therefore need 2 nitrogen dioxides on the right-hand side. Cu(NO 3)2 CuO + 2NO 2 + O 2 That now leaves a tricky problem with the oxygens. There are 6 (2 lots of 3) on the left-hand side, and 7 on the right. It isn't immediately obvious how to sort that out. The trick is to allow yourself to have half an oxygen molecule - half an oxygen molecule is just one oxygen atom. Cu(NO 3)2 CuO + 2NO 2 + ½O 2 Now there are 6 oxygen atoms on both sides. You wouldn't normally leave it like this though. Having halves in equations feels wrong, although if you do chemistry to a more advanced level, they aren't uncommon. You can get rid of the half by doubling everything in the equation. 2Cu(NO 3)2 2CuO + 4NO 2 + O 2 If you count everything up now, it all balances. Note:If you can't get the adding up right, it is probably because you have a problem with the 2Cu(NO 3)2. The 2 at the beginning means that there are two lots of Cu(NO 3)2. That is two times 1 copper, 2 nitrogens and 6 oxygens. The 2 after the bracket multiplies everything inside the bracket. So 2Cu(NO 3)2 has 2 coppers, 4 nitrogens and 12 oxygens. Problems to practise on Do one set of these questions at a time, and then check your answers. Don't go on to the next set until you are sure you understand any mistakes that you made. Set 1 Balance these equations where the formulae are given to you. Mg + HCl MgCl 2 + H 2 Ba + H 2 O Ba(OH)2 + H 2 FeSO 4 + NaOH Fe(OH)2 + Na 2 SO 4 Al + Fe 2 O 3 Al 2 O 3 + Fe Na 2 CO 3 + HNO 3 NaNO 3 + CO 2 + H 2 O set 1 balancing equations answers Set 2 Balance these equations where the formulae are also given to you. Al + H 2 SO 4 Al 2(SO 4)3 + H 2 Fe 2(SO 4)3 + KOH Fe(OH)3 + K 2 SO 4 C 4 H 10 + O 2 CO 2 + H 2 O K + H 2 O KOH + H 2 HNO 3H 2 O + NO 2 + O 2 set 2 balancing equations answers The rest of these questions are likely to take longer because you are being asked to work out formulae as well as balancing the equation. Take your time to make sure the formulae are right - you can't get the equation properly balanced it it contains a wrong formula. There aren't any difficult problems here. If you find an equation impossible to balance it is probably because you have got a formula wrong. Don't be surprised if the occasional equation is actually balanced when you have written the formulae - it isn't uncommon. Set 3 Write balanced equations for the following reactions. sodium carbonate + hydrochloric acid (HCl) sodium chloride + carbon dioxide + water sodium + chlorine sodium chloride magnesium + copper(II) oxide magnesium oxide + copper sodium hydroxide + sulfuric acid (H 2 SO 4) sodium sulfate + water copper(II) oxide + hydrochloric acid (HCl) copper(II) chloride + water set 3 balancing equations answers Set 4 Write balanced equations for the following reactions. potassium + water potassium hydroxide + hydrogen iron(III) oxide + nitric acid (HNO 3) iron(III) nitrate + water barium chloride + potassium sulfate barium sulfate + potassium chloride magnesium + sulfuric acid (H 2 SO 4) magnesium sulfate + hydrogen sodium + oxygen sodium oxide set 4 balancing equations answers State symbols in equations I don't want to make too big a deal of this at the moment, because for now you need to concentrate on learning to write the basic equations that we have covered above. But quite often you will find equations which have state symbols added in brackets after each formula. There are four of these, and they tell you what physical state the substance is in. \| state symbol \| means \| --- \| \| (s) \| The substance is a solid. \| \| (l) \| The substance is a liquid. \| \| (g) \| The substance is a gas. \| \| (aq) \| The substance is in solution in water. "aq" is short for aqueous. An aqueous solution is a solution in water. \| So, for example, if you just wrote H 2 O in an equation, it isn't always clear whether you are talking about water or steam. Liquid water would be written H 2 O(l) and steam would be H 2 O(g). Solid magnesium burns in steam to produce a white powder, magnesium oxide, and hydrogen gas. That would be shown using state symbols as . . . Mg(s) + H 2 O(g) MgO(s) + H 2(g) In this case, the state symbols matter, because magnesium has only a very, very slight reaction with cold water, and produces a different product - magnesium hydroxide rather than magnesium oxide. To take another example, if you write HCl in an equation, it isn't always totally obvious whether you are talking about hydrogen chloride gas, or the solution of hydrogen chloride gas in water, known as hydrochloric acid. The gas would be HCl(g) and its solution (hydrochloric acid) would be HCl(aq). We will talk some more about this in places where it really matters. Note:If you have worked through these formulae and equations pages and still feel you need more examples to practise with, you could explore Philip Sheldon's site. This has numerous examples of writing formulae and equations, as well as a section on basic atomic structure which you might also want to look at. For now, ignore the part which deals with ionic equations. I prefer to introduce these gently a bit at a time when they are needed, and if you start worrying about them at this point, you will just get confused. It is important that you read the short User Guide before you begin because the site is correctly fussy about how you enter subscripts (e.g. H 2) and superscripts (e.g. Cu 2+). You will get reminders about this on the pages where it is relevant. Use the keyboard shortcuts wherever possible - it's a bit tedious otherwise. It is much easier to use the site on a computer with a traditional keyboard rather than a tablet. You can access a necessary Periodic Table from the title bar on each page. This should open in a new tab, which makes it easy to keep referring to. You can set the level of difficulty yourself. If you are coming at this for the first time, choose "beginner". Where would you like to go now? To the formulae and equations menu . . . To the Chemistry 14-16 menu . . . To Chemguide Main Menu . . . © Jim Clark 2019
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