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https://www.amosweb.com/cgi-bin/awb_nav.pl?s=wpd&c=dsp&k=cross%20elasticity%20of%20demand
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Tuesday August 6, 2024
AmosWEB means Economics with a Touch of Whimsy!
EFFICIENT SEARCH: Decreasing marginal search benefit and increasing marginal search cost suggest that the most efficient search lies somewhere between zero effort and complete information. The most efficient level of search effort is given by the equality of marginal search benefit and marginal search cost.
CROSS ELASTICITY OF DEMAND:
The relative response of a change in the demand for one good to a change in the price of another good. More specifically the cross elasticity of demand is percentage change in the demand for one good due to a percentage change in the price of another good. This notion of elasticity captures the other prices demand determinant. Three other notable elasticities are the price elasticity of demand, the price elasticity of supply, and the income elasticity of demand.
The cross elasticity of demand quantifies the theoretical relationship between the price of one good and the demand for another good as identified by the other prices demand determinant. A positive cross elasticity indicates a substitute good and a negative cross elasticity exists for a complement good.
Suppose, for example, that the price of a pecan pie increases by 10 percent (say \$1.00 to \$1.10 per slice). This increase in price is likely to cause the demand of hot fudge sundaes to change. The cross elasticity of demand answers the question: Does demand increase or decrease, and if so, by how much? If the demand increases by 10 percent (say from 100 hot fudge sundaes to 110 hot fudge sundaes), then pecan pie is a substitute good for hot fudge sundaes. If the demand decreases by 10 percent (say from 100 hot fudge sundaes to 90 hot fudge sundaes), then pecan pie is a complement good for hot fudge sundaes. If the demand does not change, then pecan pie is an independent good relative to hot fudge sundaes.
### A Summary Formula
The cross elasticity of demand is often summarized by this handy formula:
cross elasticityof demand = percentage changein demand for good 1percentage changein price of good 2
In theory, the cross elasticity of demand is specified in terms of the "percentage change in demand." The reason is that other prices affect demand not quantity demanded.
However, in practice, the cross elasticity of demand is calculated as the percentage change in "quantity" resulting from the percentage change the price of another good. In other words, the calculation is based on the change in quantity from one value to another.
### Substitute and Complement
AlternativeCoefficient (C)
Substitute GoodC > 0
Complement GoodC < 0
Independent GoodC = 0
Other prices affect the demand for a good in one of three ways. An increase in the price of another good causes an increase in demand, a decrease in demand, or no change in demand. The result is either a substitute good, a complement good, or an independent good.
• Substitute Good: A substitute good exists if an increase in the price of one good causes an increase in demand for the other good. This is seen as a positive value for the cross elasticity of demand, or a coefficient of elasticity of C > 0.
• Complement Good: A complement good exists if an increase in the price of one good causes a decrease in demand for the other good. This is seen as a negative value for the cross elasticity of demand, or a coefficient of elasticity of C < 0.
• Independent Good: An independent good exists if a change in the price of one good has no affect on the demand for another good. This is seen as a zero value for the cross elasticity of demand, or a coefficient of elasticity of C = 0.
Although some goods might intuitively appear to be substitute, complement, or independent goods, economists generally let cross elasticity calculations make the actual determination. One of the more important applications of these calculations is in the area of market control',500,400)">market control and antitrust laws. Because government frowns on markets controlled by a single firm, it is very important to know if the buyers of the good produced by one firm have any reasonable substitute alternatives. If there are NO close substitutes, that is, all cross elasticities are approximately zero, then the firm producing the good probably falls into the monopoly category of market structures and is subject to antitrust scrutiny.
### Three Other Elasticities
The cross elasticity of demand is one of four common elasticities used in the analysis of the market. The other three are price elasticity of demand, price elasticity of supply, and income elasticity of demand.
• Price Elasticity of Demand: On one side of the market is the price elasticity of demand. This is the relative response of quantity demanded to changes in the price. It is specified as the percentage change in quantity demanded to a percentage change in price.
• Price Elasticity of Supply: On the other side of the market is the price elasticity of supply. This is the relative response of quantity supplied to changes in the price. It is also analogously specified as the percentage change in quantity supplied to a percentage change in price.
• Income Elasticity of Demand: This is the relative response of demand to changes in income, or the percentage change in demand due to a percentage change in income. This elasticity quantifies the buyers' income demand determinant.
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CROSS ELASTICITY OF DEMAND, AmosWEB Encyclonomic WEB*pedia, http://www.AmosWEB.com, AmosWEB LLC, 2000-2024. [Accessed: August 6, 2024].
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ALEX Lesson Plan Resources
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ALEX Lesson Plans
Subject: English Language Arts (3), or Technology Education (3 - 5)
Title: Fun With Idioms
Description: This lesson is designed to help students become comfortable with idioms. Students will work closely with idioms to discover meanings and present them to the class. Students will use technology to present the information.
Subject: English Language Arts (2 - 3), or Credit Recovery Science (2), or Science (2)
Title: Splish! Splash!
Description: Let's get wet and wild with this exciting unit on the ocean! Through the use of wonderful literature, art projects, research, and much more, the students will expand their understanding of what life is like for the creatures of the ocean. This is a College- and Career-Ready Standards showcase lesson plan.
Subject: English Language Arts (3)
Title: Alike...but yet Different!
Description: In this lesson, students will compare and contrast the themes, settings, and characters in stories written by the same author about the same or similar characters. This is a College- and Career-Ready Standards showcase lesson plan.
Subject: English Language Arts (3), or Mathematics (3 - 5), or Technology Education (3 - 5)
Title: Weight vs. Capacity
Description: This lesson will introduce students to measuring weight, mass, volume, and capacity using metric and cutomary units. This lesson plan was created by exemplary Alabama Math Teachers through the AMSTI project.
Subject: English Language Arts (3), or Mathematics (7), or Technology Education (3 - 5)
Description: This lesson will allow the student to experience a series of informal comparisons, The student will explore the chances of various outcomes of an event. The student will use vocabulary associated with probability.This lesson plan was created as a result of the Girls Engaged in Math and Science University, GEMS-U Project.
Subject: English Language Arts (3), or Technology Education (3 - 5)
Title: Sharing Original Nursery Rhymes (Distance Learning)
Description: Collaboration of creative writing/creative dramatics using Distance Learning. With a partner, students will write an original nursery rhyme and share/present with another class (Distance Learning) through presentations (dramatization, puppet show, illustration/book, PowerPoint, etc.).
Subject: English Language Arts (3), or Information Literacy (K - 12), or Social Studies (3), or Technology Education (3 - 5)
Title: "What do we do? Natural Disasters in the United States"
Description: During the course of this lesson, the students will research various forms of natural disasters that occur in the United States and what someone should do in the event of that disaster. The teacher and library media specialist will collaborate during the planning, instruction, and research phases of this lesson. The students will take the information they collected and place it on a PowerPoint presentation to be presented to the class. This information should include several pictures that will be located on the Internet. The Library Media Specialist will guide the students through the process of obtaining permission for the use of various images.
Subject: English Language Arts (3), or Technology Education (3 - 5)
Title: Savvy Story Structure
Description: During this lesson, students learn to actively engage in reading by becoming more familiar with the elements of a story. Students will be required to think at a higher level and will enhance their understanding of selected passages and stories. Students should also begin to relate stories to their own lives.
Subject: English Language Arts (3), or Social Studies (3), or Technology Education (3 - 5)
Title: Preparing for Natural Disasters
Description: This social studies lesson provides information and safety tips for different natural disasters. It integrates technology, writing, and art activities. The included slideshow may be presented to the whole group, or students may navigate it independently or in small groups.
Subject: Arts Education (3 - 5), or English Language Arts (3), or Technology Education (3 - 5)
Title: Carle's Crazy Critters
Description: This lesson provides students with the opportunity to read books aloud to the class. Students also spend time researching Eric Carle and his practices as a writer and illustrator. Finally, students become writers and illustrators themselves, as they create their own books and present their books to younger students.
Subject: English Language Arts (3), or Technology Education (3 - 5)
Title: The Stupendous Dodgeball Fiasco by Janice Repka
Description: This lesson will lead students in relating literature to their own lives. The book, The Stupendous Dodgeball Fiasco, will be read aloud to students. Afterward, students will create short stories about things which they'd like to change around their school.
Subject: Arts Education (3 - 5), or English Language Arts (3 - 4), or Credit Recovery Science (4), or Science (3), or Technology Education (3 - 5)
Description: In this lesson students will enjoy different activities focusing on how animals adapt to their environments. They will begin by researching two animals, their environment, and the ways the animals adapt to this environment. In small groups students will invent a new animal, including its environment and adaptations. Using presentation software, students will introduce their new animals to the class.
Subject: English Language Arts (3), or Technology Education (3 - 5)
Title: Irish ABCs
Description: During this lesson, students will have the opportunity to learn about St. Patrick's Day while navigating the Internet. Students will get to transform their research into an ABC Book to share with family and friends.
Subject: English Language Arts (3 - 5), or Technology Education (3 - 5)
Title: Newberry Award
Description: This unit will introduce the history of the Newberry Award to students. The selection process will be discussed. Students will also visit websites and the library to select a Newberry book to read. Finally, students will give book talks to accompany slideshows they have created.
Subject: English Language Arts (3), or Mathematics (3), or Technology Education (3 - 5)
Title: Perimeter of Polygons
Description: The purpose of this lesson is to learn how to find the length, or perimeter of a polygon in real world tasks.
Subject: English Language Arts (3)
Title: Is Laura "Sitting" in the "Setting" of "Little House?"
Description: Using Little House In The Big Woods by Laura Ingalls Wilder, this lesson focuses on the concept of time and place, phonics, and comprehension strategies. It is an interactive lesson with classroom dialogue and small group interaction.
Subject: English Language Arts (3)
Title: What a Character!
Description: This lesson opens up the world of writing by introducing students to the point of view in a story. Students will have the opportunity to read a new version of a classic story and see how it changes when it is written from another character's point of view. Students will have the opportunity to create their own story from another character's point of view.
Subject: English Language Arts (3), or Technology Education (3 - 5)
Title: Just Plain Living
Description: Students work through a teacher-prepared Webquest in order to learn about the art, food, shelter, clothing, etc. of the Plains Indians. Internet resources are used to collect information in order to make group presentations to the class. Students have the opportunity to decide among several projects to complete in order to demonstrate what they have learned.
Subject: English Language Arts (3), or Technology Education (3 - 5)
Title: "Moments in Time" Lines
Description: This lesson will give students the opportunity to share important events in their lives while learning to make a timeline.
Subject: Character Education (K - 12), or English Language Arts (3), or Technology Education (3 - 5)
Description: After reading/hearing Junie B. First Grader Cheater Pants by Barbara Park and discussing the concept of cheating, following directions, and poetry, students write cinquains and publish a class booklet.
Subject: English Language Arts (3)
Title: Wordly Wise
Description: Students will learn how to use better word choices by completing a cloze activity, listening to A House for Hermit Crab by Eric Carle, and completing a word web.
Thinkfinity Lesson Plans
Subject: Language Arts
Title: Alaska Native Stories: Using Narrative to Introduce Expository Text Add Bookmark
Description: Tradition and technology come together in this lesson in which students learn about Alaskan animals through Native American tales and their own online research.
Subject: Language Arts
Title: Comics in the Classroom as an Introduction to Narrative Structure Add Bookmark
Description: This lesson uses comic strip frames to define plot and reinforce the structure that underlies a narrative. Students finish by writing their own original narratives.
Subject: Language Arts,Social Studies
Title: What Makes a Hero? Add Bookmark
Description: In this unit of eight lessons, from EDSITEment, students explore heroes and the traits that make them heroic. Students begin by thinking about their own heroes and list the character traits their heroes possess. Students then explore kid heroes, adults heroes, local heroes, and heroes from history, before completing one of several suggested culminating activities.
Thinkfinity Partner: EDSITEment
Subject: Language Arts
Title: Cyberspace Explorer: Getting to Know Christopher Columbus Add Bookmark
Description: Students explore multiple online sources to gather information about the life of Christopher Columbus, complete a cyber scavenger hunt, and use their notes to prepare a timeline and summary report.
Subject: Language Arts
Title: Find favorite book picks in the Children's Choices. Add Bookmark
Description: Students create lists of their favorite books and then the class creates a '' Top Picks'' class booklist. Students can use the Book Cover Creator to create a book jacket for their favorite book.
Subject: Language Arts
Description: Students celebrate a novel they have read and get hands-on experience with technical writing by creating a board game based on the novel and writing the instructions for it.
Subject: Language Arts
Title: Artistic Elements: Exploring Art Through Descriptive Writing Add Bookmark
Description: Paint a vivid picture in your reader's mind with good descriptive writing! Artwork provides the perfect starting point for practicing descriptive writing that conveys color, shape, line, and mood.
Subject: Language Arts
Title: Developing Inferential Comprehension Through DL-TA and Discussion Webs Add Bookmark
Subject: Language Arts
Title: Q is for Duck: Using Alphabet Books With Struggling Writers Add Bookmark
Description: A is for zoo? Q is for duck? The alphabet as students know it is transformed when students create a class book that contains clever associations for each letter of the alphabet.
Subject: Language Arts
Title: Using Science Texts to Teach the Organizational Features of Nonfiction Add Bookmark
Description: Students explore organizational features of nonfiction science. Students then work together to create a two-page spread using those features to present information about their local environment.
Subject: Language Arts
Description: Students develop scripts, perform, and use their voices to depict characters from texts, giving them the opportunity to develop fluency and further enhance comprehension of what they are reading.
Subject: Language Arts,Social Studies
Title: Where I Come From Add Bookmark
Description: In this lesson, from EDSITEment, students take research into their heritage a step beyond the construction of a family tree, traveling through cyberspace to find out what's happening in their ancestral homelands today and explore their sense of connection to these places in their past.
Thinkfinity Partner: EDSITEment
Subject: Language Arts
Title: Beatrix Potter's Naughty Animal Tales Add Bookmark
Description: In this unit of four lessons, from EDSITEment, students gain insight into the unusual, solitary world of Victorian childhood and compare and contrast it with their own world to understand why Beatrix Potter wrote simple stories and why she wrote about animals rather than people. Students can also learn the difference between an author and an illustrator and practice some of the same artistic techniques used by Potter to create masterpieces of their own.
Thinkfinity Partner: EDSITEment
Subject: Language Arts - Writing (composition) - Social Studies - Anthropology - Social Studies - Geography
Title: A Nepalese Village and Your Town: What's the Difference? Add Bookmark
Description: This lesson, from Xpeditions, introduces students to some of the cultural customs of rural Nepal and asks them to consider the differences between Nepalese culture and their own. They write letters as if they were Nepalese young people, describing their culture and lifestyle. They then write additional letters that they might send to Nepalese pen pals, describing the similarities and differences between North American and rural Nepalese cultures.
Thinkfinity Partner: National Geographic Education
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https://questions.examside.com/past-years/jee/question/consider-the-lr-circuit-shown-in-the-figure-if-the-switch-s-jee-main-physics-current-electricity-mem2mwkobt2fwjcu
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1
JEE Main 2019 (Online) 12th April Evening Slot
+4
-1
Consider the LR circuit shown in the figure. If the switch S is closed at t = 0 then the amount of charge that passes through the battery between t = 0 and t = $${L \over R}$$ is :
A
$${{2.7EL} \over {{R^2}}}$$
B
$${{EL} \over {2.7{R^2}}}$$
C
$${{7.3EL} \over {{R^2}}}$$
D
$${{EL} \over {7.3{R^2}}}$$
2
JEE Main 2019 (Online) 10th April Evening Slot
+4
-1
A coil of self inductance 10 mH and resistance 0.1 $$\Omega$$ is connected through a switch to a battery of internal resistance 0.9 $$\Omega$$. After the switch is closed, the time taken for the current to attain 80% of the saturation value is: [take ln 5 = 1.6]
A
0.324 s
B
0.002 s
C
0.103 s
D
0.016 s
3
JEE Main 2019 (Online) 10th April Morning Slot
+4
-1
A transformer consisting of 300 turns in the primary and 150 turns in the secondary gives output power of 2.2 kW. If the current in the secondary coil is 10A, then the input voltage and current in the primary coil are :
A
220 V and 20 A
B
220 V and 10A
C
440 V and 5A
D
440 V and 20 A
4
JEE Main 2019 (Online) 8th April Evening Slot
+4
-1
A circuit connected to an ac source of emf e = e0sin(100t) with t in seconds, gives a phase difference of $$\pi$$/4 between the emf e and current i. Which of the following circuits will exhibit this ?
A
RC circuit with R = 1 k$$\Omega$$ and C = 1μF
B
RL circuit with R = 1k$$\Omega$$ and L = 1mH
C
RC circuit with R = 1k$$\Omega$$ and C = 10 μF
D
RL circuit with R = 1 k$$\Omega$$ and L = 10 mH
EXAM MAP
Medical
NEET
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https://www.shanghai-optics.com/lesson-5-intro-to-optical-components/
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## Lesson 5 – Intro to Optical Components
Posted on: Friday, February 18th, 2022 In: Learning Optics with Austin
The properties of light evident in nature are utilized by humans through instruments such as lenses and mirrors. While one might think that glass is the main material used to make lenses, resin has become the most common material due to its lightness yet hardness (resistance to scratching). Lenses refract light while mirrors reflect light.
Glass slab
Other than being made of glass, traditional lenses tend to be spherical lenses which are characterized by having a bump or curve. Thinner aspherical lenses produce a better image than spherical lenses and have thus supplanted spherical lenses as the better choice.
Aspherical (left) vs. Spherical (right) lenses
Another common way of categorizing lenses is convex versus concave. A convex lens has a thicker center compared with its edge, while a concave lens has a thicker edge than its center. Since light converges after passing through convex lenses, these lenses are also called converging lenses. As concave lenses have the opposite effect, they are known as diverging lenses.
Concave (left) vs. Convex (right)
There is however, another pair of names synonymous to convex and concave- positive and negative respectively. The positive and negative labels have to do with focal length. In short, we know that light converges after passing through a convex lens. The point at which light converges is called the focal point and the length from the lens to that point is called the focal length. Convex lenses have a positive focal length because the focal point exists. For concave lenses however, light diverges after passing through. Instead of finding the length to a focal point, the diverging light is traced backwards to a single point prior to entering the lens. This point is called the virtual focal point and the distance to it is the negative focal length.
Group of Concave (left) vs. Convex (right) lenses
The introduction to lenses is the first step in understanding how the concepts of optics relates to the products we make. Stay tuned for the next post as we look at the different applications of focal length!
Next: Lesson 6 – Focal Length and Applications of Lenses
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https://de.maplesoft.com/support/help/errors/view.aspx?path=updates%2FMaple2018%2FUnits
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Units - Maple Help
Units
Introduction Maple 2018 offers enhanced tools for using units throughout your analysis.
Units in Solving, Integration, and Optimization
You can now use use units in numeric equation solving, integration, and optimization. Several functions support units, including fsolve and the optimizers in the Optimization package. In addition, units handling has been improved in the solve command and the int command for symbolic and numeric inputs.
Example: Adiabatic Flame Temperature of Butane
Liquid butane is burnt with 100% theoretical air at an initial temperature of 298.15 K. The combustion reaction is
C4H10 (l) + 6.5 O2 (g)+ 24.44 N2 (g) → 4 CO2 (g) + 5 H2O (g) + 24.44 N2 (g)
Here, we will calculate the adiabatic flame temperature of the combustion products.
> $\mathrm{with}\left(\mathrm{ThermophysicalData}\right):$
Heat of formation of butane
> $\mathrm{h_f_C4H10}≔\mathrm{Chemicals}:-\mathrm{Property}\left("HeatOfFormation","C4H10\left(l\right),n-buta",\mathrm{useunits}\right)$
${-}{150.6640000}{}⟦\frac{{\mathrm{kJ}}}{{\mathrm{mol}}}⟧$ (2.1.1)
Enthalpies of the combustion products at a temperature T
> $\mathrm{h_N2}≔\mathrm{Chemicals}:-\mathrm{Property}\left("Hmolar","N2\left(g\right)","temperature"=\mathrm{T}\right):$$\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}\mathrm{h_O2}≔\mathrm{Chemicals}:-\mathrm{Property}\left("Hmolar","O2\left(g\right)","temperature"=\mathrm{T}\right):\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}$$\mathrm{h_H2O}≔\mathrm{Chemicals}:-\mathrm{Property}\left("Hmolar","H2O\left(g\right)","temperature"=\mathrm{T}\right):\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}$$\mathrm{h_CO2}≔\mathrm{Chemicals}:-\mathrm{Property}\left("Hmolar","CO2\left(g\right)","temperature"=\mathrm{T}\right):$
Enthalpy of the reactants
>
${-}{150.6640000}{}⟦{\mathrm{kJ}}⟧$ (2.1.2)
Total enthalpy of the combustion products
>
${\mathrm{H_products}}{≔}{4}{}{\mathrm{Chemicals}}{:-}{\mathrm{Property}}{}\left({"Hmolar"}{,}{"CO2\left(g\right)"}{,}{"temperature"}{=}{T}\right){}⟦{\mathrm{mol}}⟧{+}{5}{}{\mathrm{Chemicals}}{:-}{\mathrm{Property}}{}\left({"Hmolar"}{,}{"H2O\left(g\right)"}{,}{"temperature"}{=}{T}\right){}⟦{\mathrm{mol}}⟧{+}{24.44}{}{\mathrm{Chemicals}}{:-}{\mathrm{Property}}{}\left({"Hmolar"}{,}{"N2\left(g\right)"}{,}{"temperature"}{=}{T}\right){}⟦{\mathrm{mol}}⟧$ (2.1.3)
Equating the enthalpy of the reactants and the enthalpy of the combustion products gives the adiabatic flame temperature
> $\mathrm{fsolve}\left(\mathrm{H_reactants}=\mathrm{H_products},\mathrm{T}=2000⟦\mathrm{K}⟧\right)$
${2379.853026}{}⟦{K}⟧$ (2.1.4)
Example: Optimizing the Design of a Fuel Pod
You are designing a fuel pod with a hemispherical cap, cylindrical mid-section and conical cap.
What are values of L, H and R that minimize the surface area while maintaining the volume V at 3 m3?
> $\mathrm{restart}:$
Objective function - surface area of pod
> $\mathrm{obj}≔\frac{1}{2}\cdot 4\cdot \mathrm{\pi }\cdot {\mathrm{R}}^{2}+2\cdot \mathrm{\pi }\cdot \mathrm{R}\cdot \mathrm{L}+\mathrm{\pi }\cdot \mathrm{R}\cdot \sqrt{{\mathrm{H}}^{2}+{\mathrm{R}}^{2}}:$
Constraint on the volume area of pod
>
All dimensions must be greater than 0
> $\mathrm{cons2}≔0\le \mathrm{R},0\le \mathrm{L},0\le \mathrm{H}:$
Hence the optimized dimensions are
> $\mathrm{dimensions}≔\mathrm{Optimization}:-\mathrm{Minimize}\left(\mathrm{obj},\left\{\mathrm{cons1},\mathrm{cons2}\right\},\mathrm{initialpoint}=\left\{\mathrm{H}=1⟦\mathrm{m}⟧,\mathrm{L}=1⟦\mathrm{m}⟧,\mathrm{R}=1⟦\mathrm{m}⟧\right\}\right)$
$\left[{10.2533536615869920}{}⟦{{m}}^{{2}}⟧{,}\left[{H}{=}{0.785093823049978}{}⟦{m}⟧{,}{L}{=}{0.392546902492684}{}⟦{m}⟧{,}{R}{=}{0.877761593519080}{}⟦{m}⟧\right]\right]$ (2.2.1)
Example: Work Done in Isothermal Compression of Methane
Methane at 350 K is isothermally compressed from a specific volume of 1.0 m3 kg-1 to 0.5 m3 kg-1. Here, we will calculate the work done.
First, we define an expression that gives the pressure of methane as a function of the specific volume V.
> $P≔\mathrm{ThermophysicalData}:-\mathrm{Property}\left("pressure","methane","temperature"=350⟦K⟧,"density"=\frac{1}{V}\right):$
We then perform the numeric integration to calculate the work done.
>
${125.4345869}{}⟦\frac{{\mathrm{kJ}}}{{\mathrm{kg}}}⟧$ (2.3.1)
Unit Conversion
Maple 2018 moves the units formatting dialog into the new Context Panel, and enhances its functionality.
Changing units takes fewer mouse clicks, and can be used in tandem with numeric formatting to quickly create presentable, readable results. For example, you can reformat this result...
> $\mathrm{with}\left(\mathrm{ThermophysicalData}:-\mathrm{Chemicals}\right):\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}\mathrm{Property}\left("Hmolar","C2H5OH\left(L\right)","temperature"=320⟦\mathrm{K}⟧\right)$
${-}{274957.7153}{}⟦\frac{{J}}{{\mathrm{mol}}}⟧$ (3.1)
...to...
> $\mathrm{Property}\left("Hmolar","C2H5OH\left(L\right)","temperature"=320⟦\mathrm{K}⟧\right)$
${-}{274.9577153}{}⟦\frac{{\mathrm{kJ}}}{{\mathrm{mol}}}⟧$ (3.2)
>
...with a few mouse clicks, and without navigating between different dialogs.
The list of suggested unit conversions now gives you
• typeset Greek letters, e.g. micro Joules are typeset as μJ
• typeset math, e.g. instead of J/mol/K
• sensible suggestions for unit conversions, e.g. selecting a result with units of offers $\frac{\mathrm{J}}{\mathrm{mol}}$ and $\frac{\mathrm{kJ}}{\mathrm{mol}}$ as potential conversions
• a cleaner, more streamlined experience by removing little used unit systems (although all unit systems can still be set via the command line)
Applications
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Mass of a black hole
I know that if a star collapses into a volume with radius less or equal to the Schwarzschild radius $r_s=\frac{2GM}{c^2}$ then a black hole is created and it has the same mass of the star that gave it origin. But is there a way to calculate the mass of a black hole without knowing the volume of the star?
[a black hole] has the same mass of the star that gave it origin
No, not really. A stellar-mass black hole is formed after a star with mass around 20 times that of our sun collapses due to lack of core fusion and by some (as of yet) unknown process, rebounds and explodes. This explosion, a Type II supernova, kicks off something like 90% of its matter. Thus, the black hole would be something around the 5 solar mass range.
is there a way to calculate the mass of a black hole
Well you can't actually observe black holes in free space (because you can't see them), you need to infer it by using something around it (either a star or gas cloud) and computing the mass from Kepler's laws.
• Thanks. I heard that the mass can be also calculated using GR is that true? Apr 21, 2014 at 20:47
• The Schwarzschild radius is derived from a particular metric for the GR field equations. Outside of that, I'm not sure what you mean. Apr 21, 2014 at 20:50
• I heard that it has something to do with space-time bent but I may be wrong Apr 21, 2014 at 20:54
• @Peterix GR (General Relativity) describes gravity as space-time curvature, so black holes curve space and time around them so it means as you get near to black hole time will pass slower and slower and someone who is watching you will never see you crossing the event horizon. Sample image that shows curvature of space around black hole: encrypted-tbn2.gstatic.com/… , or just google "black hole curvature of space" and open images tab
– G B
Apr 21, 2014 at 21:02
• @Peterix: yes, someone far from a black hole can measure small variations in the gravitational force that the y observe and infer the mass and spin of the black hole. Apr 21, 2014 at 22:10
|[a black hole] has the same mass of the star that gave it origin
as Kyle figured out, the mass of the black hole correlates with the star, but in general a star loses much mass in a supernova.
Furthermore, there are many more ways to create a black hole, like colloding neutron stars in theory, even at the LHC/Cern, colliding nuclei could lead to black holes. So they can be large, or very small.
To calculate the mass of a black hole: well, that's a bit tricky, because you can fully describe a black hole by the parameters: 1) mass, 2) electric charge, 3) rotation momentum.
So the question is: how to calculate the mass from what? What you could do: measure the Schwarzschild radius or mesaure the gravity field of the black hole to derive it's mass. But there is no real way to "calculate" the mass of a black hole. Its like if you want to calculate the amount of water in a bucket. It just depends on the amount of water in the bucket.
About your comment on GR: At the current stage of phsics, the general relativity theory is the only theory, that can describe black holes. They only exist in this theory.
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# acoth
Inverse hyperbolic cotangent
## Syntax
``Y = acoth(X)``
## Description
example
````Y = acoth(X)` returns the inverse hyperbolic cotangent of the elements of `X`. The function accepts both real and complex inputs. All angles are in radians.```
## Examples
collapse all
Find the inverse hyperbolic cotangent of the elements of vector `X`. The `acoth` function acts on `X` element-wise.
```X = [2 -3 1+2i]; Y = acoth(X)```
```Y = 1×3 complex 0.5493 + 0.0000i -0.3466 + 0.0000i 0.1733 - 0.3927i ```
Plot the inverse hyperbolic cotangent function over the intervals $-30\le x<-1$ and $1.
```x1 = -30:0.1:-1.1; x2 = 1.1:0.1:30; plot(x1,acoth(x1),x2,acoth(x2)) grid on xlabel('x') ylabel('acoth(x)')```
## Input Arguments
collapse all
Hyperbolic cotangent of angle, specified as a scalar, vector, matrix, multidimensional array, table, or timetable. The `acoth` operation is element-wise when `X` is nonscalar.
Data Types: `single` | `double` | `table` | `timetable`
Complex Number Support: Yes
collapse all
### Inverse Hyperbolic Cotangent
For real values $x$ in the domain $-\text{\hspace{0.17em}}\infty and $1, the inverse hyperbolic cotangent satisfies
`${\mathrm{coth}}^{-1}\left(x\right)={\mathrm{tanh}}^{-1}\left(\frac{1}{x}\right)=\frac{1}{2}\mathrm{log}\left(\frac{x+1}{x-1}\right).$`
For complex numbers $z=x+iy$ as well as real values in the domain $-1\le z\le 1$, the call `acoth(z)` returns complex results.
## Version History
Introduced before R2006a
expand all
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# How to find range in which a number falls, from given list of numbers?
How can one find the range in which a number falls, from given list of numbers?
f[x_, list_List] := ???
(*
Return {a,b} where
a & b belongs to list
{a,b} forms shortest possible interval which match condition
if a<=x<=b {a,b}
if x <= a {-∞,a}
if x >= b {b,∞}
*)
f Should also consider outer ranges $-\infty$ and $\infty$
-
And what do you want to happen with f if x is an element of list? – J. M. Jan 21 '12 at 10:16
it should return two ranges then – Prashant Bhate Jan 21 '12 at 10:21
And when there is only one element in the list , and if it is x then it should return {{-∞,x},{x,∞}} – Prashant Bhate Jan 21 '12 at 10:26
I propose using Interpolation.
list = Prime~Array~3000;
intf = Interpolation[
{list, Range@Length@list}\[Transpose],
InterpolationOrder -> 0
];
Then, for point x:
x = 12225.4;
Which[
x < First@list , {-\[Infinity], First@list},
x > Last@list , {Last@list, \[Infinity]},
True , list[[#-1 ;; #]]& @ intf @ x
]
{12211, 12227}
This could all be done inside Interpolation as well:
intf2 =
Interpolation[
Join[
{{First@list, {-\[Infinity], 2}}},
Thread[{Rest@list, Partition[list, 2, 1]}],
{{Last@list + 1, {Last@list, \[Infinity]}}}
],
InterpolationOrder -> 0
];
intf2[12225.4]
{12211, 12227}
-
Yet another non-obvious gem +1 – kglr Jan 21 '12 at 13:05
This method requires pre-sorted input data without duplicates. For an arbitrary list you need to pre-process the list with DeleteDuplicates@Sort or Union. Interpolation gives error messages when input data is unsorted or contains duplicates. – kglr Jan 21 '12 at 16:06
@kguler Indeed. Thank you for stating that. I made assumptions on the meaning and structure of a "given list of numbers." – Mr.Wizard Jan 22 '12 at 7:09
You can make use of BinCounts. I think this is a very simple to understand solution because BinCounts does almost exactly what you need already.
f[x_, list_List] :=
Module[{bins},
bins = Join[{-Infinity}, Sort[list], {Infinity}];
First@Pick[Partition[bins, 2, 1], BinCounts[{x}, {bins}], 1]
]
But it won't give you two intervals if the number is part of both. Of course it's very easy to put in an extra check and include this feature if you need it, but I just wanted to show the concept now.
-
Cool.. I was trying to figure out how to use the third argument of Pick, This is a great example.+1 – kglr Jan 21 '12 at 16:23
Assuming the list list is already ordered, the following should answer your question:
f[x_,list_List]:=
Module[{pos=Last@Ordering@Ordering[Append[list,x]]},
Which[pos==1,{-Infinity,First@list},
pos==Length[list]+1,{Last@list,Infinity},
True,list[[{pos-1,pos}]]]]
-
My favorite:
interval[x_,list_List]:=ReplaceList[Append[Prepend[Sort@list, -Infinity],
Infinity], {___, a_, b_, ___} /; (a <= x <= b) :> {a, b}]
EDIT: Much faster if we eliminate the condition checking as follows:
interval2[x_, list_List] := ReplaceList[#, {___, a_, x, b_, ___} :> {a, b} ] &@
Join[{-Infinity}, Sort[Join[list, {x}]], {Infinity}]
-
This is clever, but it is going to be quite inefficient. This is 10,000X slower than InterpolatingFunction on the data in my answer. This may or may not matter, but it should be noted. – Mr.Wizard Jan 21 '12 at 12:30
Mr.Wizard, i agree. Both your InterpolatingFunction method and Pick-based method in my other answer (after removing Sort from the definition) are uncomparably faster than ReplaceList. However, ReplaceList and pattern-based approaches often have this hard-to-resist simplicity. – kglr Jan 21 '12 at 13:02
I yield; +1 for simplicity. – Mr.Wizard Jan 21 '12 at 13:15
intervals[x_, list_List] :=
Cases[
Partition[Flatten[{-Infinity, Union[list], Infinity}], 2, 1]
, {l_, u_} /; l <= x <= u
]
Use cases:
In[53]:= intervals[3, Range[10]]
Out[53]= {{2,3},{3,4}}
In[54]:= intervals[3, 2 * Range[10]]
Out[54]= {{2,4}}
intervals[-3, Range[10]]
Out[55]= {{-∞,2}}
In[56]:= intervals[999, Range[10]]
Out[56]= {{10,∞}}
In[58]:= intervals[37, {13,8,1,28,87,14,61,20,91,92,37,93,76,83,32}]
Out[58]= {{32,37},{37,61}}
-
While working on this question I was surprised to find that out Sort[{4, 3, 2, 1, -∞, ∞}] yielded {1, 2, 3, 4, -∞, ∞} (and Union too). – WReach Jan 21 '12 at 22:54
That's because Sort[] uses OrderedQ[] by default. Try Sort[{4, 3, 2, 1, -∞, ∞}, Less]. – J. M. Jan 21 '12 at 23:49
+1 for clarity of purpose. – Brett Champion Jan 22 '12 at 3:58
As it turns out, Combinatorica has the function BinarySearch[] implemented. The code in the package is attributed to Paul Abbott. What follows is a modification of the routine that gives results in the format desired by the OP:
bisect[k_?NumericQ, l_List] :=
Block[{n = Length[l], lo, mid, hi, el},
{lo, hi} = {1, n};
While[lo <= hi,
If[(el = l[[mid = Quotient[lo + hi, 2]]]) === k,
Which[
mid == 1,
Return[{{-Infinity, First[l]}, Take[l, 2]}],
mid == n,
Return[{Take[l, -2], {Last[l], Infinity}}],
True,
Return[Partition[Take[l, mid + {-1, 1}], 2, 1]]]];
If[el > k, hi = mid - 1, lo = mid + 1]
];
Which[
lo == 1,
Return[{-Infinity, First[l]}],
lo == n + 1,
Return[{Last[l], Infinity}],
True,
Return[l[[{lo - 1, lo}]]]
]
] /; VectorQ[l, NumericQ]
I'll leave to you how to handle the case of the singleton list.
-
Here is another version:
ClearAll[getInterval];
getInterval[ints_List, num_] :=
Position[UnitStep[ints - num], 1, 1, 1] /.
{
{{1}} -> { -Infinity, First@ints},
{} -> {Last@ints, Infinity},
{{n_}} :> ints[[n - 1 ;; n]]
};
-
Among many alternatives you can use something like
glblub1[x_, data_List] := Pick[#, IntervalMemberQ[Interval@#, x] & /@ #]& @
( Partition[#, 2, 1]& @
Append[Prepend[Sort@data, -Infinity], Infinity])
or
glblub2[x_, data_List] := Pick[#, (#1 <= x <= #2) & @@@ #]& @
( Partition[#, 2, 1]& @
Append[Prepend[Sort@data, -Infinity], Infinity])
If x is a member of the list and you wish to return x rather than two intervals containing x use as
Intersection@@glblub1[x,data]
Intersection@@glblub2[x,data]
or just redefine the two functions by prefixing both with Intersection@@.
-
WReach's answer prompted me to write another answer:
intervals[x_?NumericQ, list_List] :=
With[{sl = Sort[Flatten[{-Infinity, list, Infinity}], LessEqual]},
sl[[{#, # + 1}]] & /@
Flatten[Position[
Times @@@ Partition[Sign[x - sl], 2, 1], -1 | 0]]] /;
VectorQ[list, NumericQ]
-
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# QlikView Creating Analytics
Discussion Board for collaboration related to Creating Analytics for QlikView.
Not applicable
## Subtotals for a calculated dimension?
Spent a few days and couldn't find a solution -- your help is much appreciated.
Data fields: Type, Product, Qty
PIVOT TABLE
Dimensions:
=Type
=if(aggr(sum(Qty),Product,Type)>500, '>500','<500') /* labelled as Group
Expressions:
=Sum(Qty)
Calculated dimension is grouping for Products with Sum(Qty)>500 and <500.
Question: How to get subtotals for the calculated dimension to calculate percentage ?
This is the table I have:
This is the table I need to get:
Problem is not that so trivial as it seems -- try to play with the application (attached).
Thanks!
Tags (2)
1 Solution
Accepted Solutions
Not applicable
## Subtotals for a calculated dimension?
OK, here is the answer. Works like a charm with any number of rows in a calculated dimension, and without any additional data structures.
=sum(Qty)/rangesum(top(Sum(Qty),1,NoOfRows()))
8 Replies
Honored Contributor
## Re: How to create subtotals for a calculated dimension?
Hi Dmitry,
In [% of Type] Expression try with this
`=NUM(Sum(Qty) / (IF(ISNULL(AGGR(SUM(Qty),Type)),ABOVE(AGGR(SUM(Qty),Type)),AGGR(SUM(Qty),Type))),'00%')`
Do let me know, if you looking for this one.
Regards,
Sokkorn
Not applicable
## Re: How to create subtotals for a calculated dimension?
Thanks, Sokkorn -- it works exactly as needed.
Not applicable
## Re: How to create subtotals for a calculated dimension?
This solution works well if there are only two values in Group. What about more flexible solution -- when it could be any number of values in group?
Actually what is needed is to aggregate over a calculated dimension.
Message was edited by: Dmitry Gudkov
Contributor
## Re: How to create subtotals for a calculated dimension?
Hi, Dmitry
You can use valuelist('<500','>500') in calculated dimension. And the same in the expression: if(valuelist('<500','>500')='<500',.....)
BR,
Konstantins
Honored Contributor
## Re: How to create subtotals for a calculated dimension?
Hi Dmitry,
Try to load data like this
`[Data1]:LOAD * INLINE [Type,Product,QtyXXX,A,13XXX,A,34XXX,A,123XXX,A,325XXX,B,13XXX,B,34XXX,B,123XXX,B,325XXX,C,34XXX,C,123XXX,C,325XXX,C,453XXX,D,4XXX,D,13XXX,E,325XXX,E,454ZZZ,A,45ZZZ,A,56ZZZ,A,65ZZZ,B,45ZZZ,B,435ZZZ,C,45ZZZ,C,65ZZZ,C,435ZZZ,C,657ZZZ,D,5ZZZ,D,34ZZZ,D,45ZZZ,E,45ZZZ,E,53];[Data2]:LOAD *,IF([TotalQty]>500, '>500', IF([TotalQty]<=500 AND [TotalQty]>=400,'400-500','<400')) AS [Range];LOAD Type &' - '& Product AS [Key], SUM(Qty) AS [TotalQty]RESIDENT [Data1] GROUP BY Type &' - '& Product;`
Then use the second table to do with business need.
This one just an idea to solve your problem. It not reply to your issue yet. So we need to do more on it. Will reply you back when I'm in office.
Hope this help.
Regards,
Sokkorn
Not applicable
## Re: How to create subtotals for a calculated dimension?
Sokkorn,
Thanks for helping me. This sample application is just a representation of the problem. Actual application is much more complex and table data changes dynamically, so additional static data structures barely help here.
Probably I should find a way how to get rid of the calculated dimension, like konstantins advised.
Honored Contributor
## Re: How to create subtotals for a calculated dimension?
Hi Dmitry,
Check my sample attached file.
Regards,
Sokkorn
Not applicable
## Subtotals for a calculated dimension?
OK, here is the answer. Works like a charm with any number of rows in a calculated dimension, and without any additional data structures.
=sum(Qty)/rangesum(top(Sum(Qty),1,NoOfRows()))
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# More Approximations for Trigonometric Functions with the Binomial Series
Posted on August 31, 2015
In this post, we will create a new set of approximations for sine and cosine that utilize the binomial series. Before you go any further, you may want to read this previous post where we walk through the binomial series and comparisons to the Maclaurin series.
## Derivation
We are going to exploit Euler's formula with a twist. To refresh your memory, here's Euler's Formula. $$\cos(\theta)+i\sin(\theta)=e^{i\theta}$$ Now let's employ one of the older tricks in the book, adding and subtracting one from the same expression: $$\cos(\theta)+i\sin(\theta)=\left[(e-1)+1\right]^{i\theta}.$$ Now we have a binomial raised to a power. Once again, we will use known values and the binomial series to generate approximations for sine and cosine.
## Fine Tuning For Approximation
In its current form this would almost be enough for decent approximations. However, closely tied to the accuracy of an approximation is the convergence rate of the series. To improve convergence, we want to make one of the terms small relative to the other term. Is there anything we can do here?
As a matter of fact, yes. We will employ a second manipulation - multiplying and dividing by the same number (inside and outside of the parentheses). Consider the updated equation $$\cos(\theta)+i\sin(\theta)=a^i\cdot\left[\left(\displaystyle\frac{e}{a^{1/\theta}}-1\right)+1\right]^{i\theta}.$$ If you need to, convince yourself that this is still valid. Now, to build a valid approximation, we need $\displaystyle\frac{e}{a^{1/\theta}}-1$ to be small for a range of values. In our previous post, we determined that you need a range of $\pi/4$ for sine and cosine, which could be extrapolated out to encompass all values.
Ideally we would just pick the range $[0,\pi/4]$, with which a value of $a=e^{\pi/8}$ would be a good place to center the approximation. However, If $\theta=0$, the expression $${(e^{\pi/8})^{1/0}}-1$$ is not defined (although the limit appears to exist).
On the other side of the range, at $\theta=\pi/4$, the expression becomes $$\displaystyle\frac{e}{(e^{\pi/8})^{4/\pi}}-1=e^{1/2}-1\approx.64$$ which is enough for convergence, but maybe not optimal.
On the other hand, if we made the range $[2\pi,2\pi+\pi/4]$ with $a=2\pi+\pi/8$, then our bounding values are roughly $\approx\pm.06$ which will give us much better convergence. Due to the nature of this equation, rapidly increasing our range - for example, the range $[100\pi,100\pi+\pi/4]$ - has rapidly diminishing gains, and improved convergence will be offset by roundoff error elsewhere.
## Implementing Approximations
An example of this code is (written for MATLAB) is
iterations=14;
start=2*pi;
range=pi/4;
increment=.001;
theta=start:increment:(start+range);
a=exp(start+range/2);
z=1i.*theta;
w=exp(1)./a.^(1./theta)-1;
approx=zeros(1,length(theta));
for i=iterations:-1:1
approx=1+approx.*(w.*(z-i+1)/i);
end
approx=approx*a.^(1i);
You can plot the error with the following code
error=real(approx)-cos(theta);
figure;
h=semilogy(theta,abs(error));
outfilename=sprintf('cos_error_%d_iter',iterations);
saveas(h, outfilename, 'png');
error=imag(approx)-sin(theta);
figure;
h=semilogy(theta,abs(error));
outfilename=sprintf('sin_error_%d_iter',iterations);
saveas(h, outfilename, 'png');
## Error Plots
See below for the error for approximations for cosine with 4, 8, 12, and 14 terms. With 14 terms, we are more or less bounded by machine precision (~15 significant figures)
Once again we have generated some exciting looking, but nonetheless useless approximations. We were able to use binomial theorem and some trickery to generate approximations with great accuracy, but once again they are nowhere near as speedy to calculate as the Maclaurin series.
Some text some message..
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CC-MAIN-2024-18
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www.prajaanetra.com
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Example: Brand new empirical formula of your material sugar (C
Example: Brand new empirical formula of your material sugar (C
O = $$\frac$$ ? Mass = $$\frac$$ ? Molecule wt
Empirical formula The empirical formula of a compound may be defined as the formula which gives the simplest whole number ratio of atoms of the various elements present in the molecule of the compound. 6HtwelveO6), is CH2O which shows that C, H, and O are present in the simplest ratio of 1 : 2 : 1. Rules for writing the empirical formula The empirical formula is determined by the following steps :
1. Separate the portion of for every aspects by their nuclear mass. This gives the latest cousin level of moles of several points introduce about compound.
2. Separate the newest quotients gotten on over action by littlest of them to get an easy proportion out-of moles of various factors.
3. Multiply the data, thus acquired of the the right integer, if necessary, so you can receive whole amount ratio.
4. In the end jot down the signs of the various issues front because of the side and put the aforementioned number due to the fact subscripts with the straight down right hand spot of each icon. This can represent this new empirical formula of your material.
Example: A substance, on investigation, gave the following constitution : Na = cuatro3.4%, C = eleven.3%, O = forty five.3%. Determine the empirical algorithm [Nuclear people = Na = 23, C = twelve, O = 16] Solution:
O3
Determination molecular formula : Molecular formula = Empirical formula ? n n = $$\frac$$ Example 1: What is the simplest formula of the compound which has the following percentage composition : Carbon 80%, Hydrogen 20%, If the molecular mass is 30, calculate its molecular formula. Solution: Calculation of empirical formula :
? Empirical formula is CH3. Calculation of molecular formula : Empirical formula mass = 12 ? 1 + 1 ? 3 = 15 n = $$\frac =\frac$$ = 2 Molecular formula = Empirical formula ? 2 = CH3 ? 2 = C2H6.
Example 2: On heating a sample of CaC, volume of CO2 evolved at NTP is 112 cc. Calculate (i) Weight of CO2 produced (ii) Weight of CaC taken (iii) Weight of CaO remaining Solution: (i) Mole of CO2 produced $$\frac =\frac$$ mole mass of CO2 = $$\frac \times 44$$ = 0.22 gm (ii) CaC > CaO + CO2(1/200 mole) mole of CaC = $$\frac$$ mole ? mass of CaC = $$\frac \times 100$$ = 0.5 gm (iii) mole of CaO produced = $$\frac$$ mole mass of CaO = $$\frac \times 56$$ = 0.28 gm * Interesting by we can apply Conversation of mass or wt. of CaO = wt. of CaC taken – wt. of CO2 produced = 0.5 – 0.22 = 0.28 gm
Example 3: If all iron present in 1.6 gm Fe2 is converted in form of FeSO4. (NH4)2SO4.6H2O after series of reaction. Calculate mass of product obtained. Solution: If all iron will be converted then no. of mole atoms of Fe in reactant product will be same. ? Mole of Fe2 = $$\frac =\frac$$ mole atoms of Fe = 2 ? $$\frac =\frac$$ mole of FeSO4. (NH4)2SO4.6H2O will be same as mole atoms of Fe because one atom of Fe is present in one molecule. ? Mole of FeSO4.(NH4)2.SO4.6H2 = $$\frac \times 342$$ = 7.84 gm www.datingranking.net/sugar-daddies-usa/pa/philadelphia/.
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https://mtp.tools/converters/data-storage/exbibyte-to-megabit-calculator
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# Exbibyte to Megabit calculator (EiB to Mb)
Convert exbibytes to megabits (EiB to Mb) by typing the amount of exbibytes in the input field below and then clicking in the "Convert" button. If you want to convert from megabits to exbibytes, you can use our megabit to exbibyte converter.
## Formula
Formula used to convert EiB to Mb:
F(x) = x * 9223372036854.775
For example, if you want to convert 1 EiB to Mb, just replace x by 1 [EiB]:
1 EiB = 1 * 9223372036854.775 = 9223372036854.775 Mb
## Steps
1. Multiply the amount of exbibytes by 9223372036854.775.
2. The result will be expressed in megabits.
## Exbibyte to Megabit Conversion Table
The following table will show the most common conversions for Exbibytes (EiB) to Megabits (Mb):
Exbibytes (EiB) Megabits (Mb)
0.001 EiB 9223372036.854776 Mb
0.01 EiB 92233720368.54776 Mb
0.1 EiB 922337203685.4775 Mb
1 EiB 9223372036854.775 Mb
2 EiB 18446744073709.55 Mb
3 EiB 27670116110564.33 Mb
4 EiB 36893488147419.1 Mb
5 EiB 46116860184273.875 Mb
6 EiB 55340232221128.66 Mb
7 EiB 64563604257983.43 Mb
8 EiB 73786976294838.2 Mb
9 EiB 83010348331692.98 Mb
10 EiB 92233720368547.75 Mb
20 EiB 184467440737095.5 Mb
30 EiB 276701161105643.25 Mb
40 EiB 368934881474191 Mb
50 EiB 461168601842738.75 Mb
60 EiB 553402322211286.5 Mb
70 EiB 645636042579834.2 Mb
80 EiB 737869762948382 Mb
90 EiB 830103483316929.8 Mb
100 EiB 922337203685477.5 Mb
A exbibyte is a unit of measurement for digital information and computer storage. The binary prefix exbi (which is expressed with the letters Ei) is defined in the International System of Quantities (ISQ) as a multiplier of 2^60. Therefore, 1 exbibyte is equal to 1,024 pebibytes and equal to 1,152,921,504,606,846,976 bytes (around 1.152 exabytes). The symbol used to represent a exbibyte is EiB.
A megabit is a unit of measurement for digital information and computer storage. The prefix mega (which is expressed with the letter M) is defined in the International System of Units (SI) as a multiplier of 10^6 (1 million). Therefore, 1 megabit is equal to 1,000,000 bits and equal to 1,000 kilobits. The symbol commonly used to represent a megabit is Mb (sometimes as Mbit).
## FAQs for Exbibyte to Megabit converter calculator
### What is Exbibyte to Megabit converter calculator?
Exbibyte to Megabit converter is a free and online calculator that converts Exbibytes to Megabits.
### How do I use Exbibyte to Megabit converter?
You just have to insert the amount of Exbibytes you want to convert and press the "Convert" button. The amount of Megabits will be outputed in the input field below the button.
### Which browsers are supported?
All mayor web browsers are supported, including Internet Explorer, Microsoft Edge, Firefox, Chrome, Safari and Opera.
### Which devices does Exbibyte to Megabit converter work on?
Exbibyte to Megabit converter calculator works in any device that supports any of the browsers mentioned before. It can be a smartphone, desktop computer, notebook, tablet, etc.
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http://www.thebestcasescenario.com/forum/archive/index.php/t-23397.html?s=c45f583f321f02b3763b9a51c3fa63d3
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PDA
View Full Version : Military issued MRE's.
BuzzKillington
06-02-2010, 07:36 PM
Not the nasty civvy versions lol.
Kayin
06-02-2010, 09:38 PM
I don't have any, but MAN I love those.
pheenix
09-29-2010, 04:54 AM
I work on a .mil installation, next time I'm near the comissary I can look to see if they sell them (my last base did for people who went camping or something). Shoot me a PM and remind me if you'd like.
destinydesignlabs
10-06-2010, 01:35 PM
I am also interested in some if you can find a price and hit me up, i might be interested.
d_stilgar
10-06-2010, 03:46 PM
One year at scout camp I ate in a cafeteria and didn't get enough to eat, so I lost 20lbs that week.
The next year we had military MREs for almost every meal. I gained 20lbs that year.
BuzzKillington
10-06-2010, 10:24 PM
hahaha, ya, they're not meant to be meal replacements for breakfast lunch and dinner... There's like 1200 calories in each meal so unless you're doing some hard core training you're bound to gain weight from 3600cal a day.
nevermind1534
10-06-2010, 10:32 PM
hahaha, ya, they're not meant to be meal replacements for breakfast lunch and dinner... There's like 1200 calories in each meal so unless you're doing some hard core training you're bound to gain weight from 3600cal a day.
I usually take in at least that much in a day...
x88x
10-07-2010, 02:25 AM
Wow, that's enough energy to raise all the water in my body by 55.9C! ...wait, calories in food? What's that? :P
...in all seriousness though, that's 15.12MJ! :eek: That is an obscene amount of energy! To borrow an analogy from Wikipedia, 1MJ is equal to "approximately the kinetic energy of a one-ton vehicle moving at 160 km/h (100 mph)"...it's no wonder our bodies can't convert all of that chemical potential energy out in one day. ...actually, that got me curious...so, the average human body apparently consumes 14 Cal per one pound of lean muscle per day..just sitting around doing nothing.. So in a person with 100lbs of lean muscle, that's 1400 Cal/day, or 5.88MJ/day. That's 68J/s! Just sitting around maintaining an equilibrium! That is absurd! ...ok, sure, the human body is an amazing machine, but...damn! That is crazy inefficient! ...suddenly the Matrix power system starts to make more sense...
BuzzKillington
10-07-2010, 03:59 AM
Did you eat any cookies laying around in a plastic bag by any chance? :P
x88x
10-07-2010, 11:24 AM
lol. No, I just get a little...focussed?...late at night. :P
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http://merkury-spb.ru/microeconomic-theory-nicholson-10th-edition-pdf.html
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# Microeconomic Theory Nicholson 10th Edition Pdf
But it could have all the answers. But the converse is not true. Slope of this equation is. How do I view solution manuals on my smartphone?
Insightful graphic presentations help you visually grasp the connections between the calculus and the algebraic and geometric approach to the same material. Total differentiation is written with the help of partial differentiation. Amazon Music Stream millions of songs. Just post a question you need help with, and one of our experts will provide a custom solution. Partial differentiation with respect to x is and with respect to y is.
Learn more about Amazon Prime. It means trade-off between x and y is not linear. Share your thoughts with other customers. Amazon Second Chance Pass it on, trade it in, give it a second life. The optimal solution requires solving three nonlinear simultaneous equations, a task not undertaken here.
First, use the substitution method. Sell on Amazon Start a Selling Account. You can check your reasoning as you tackle a problem using our interactive solutions viewer. Good for advanced individuals who are at the graduate level. Proof of envelope theorem a.
## Microeconomic Theory 10th Edition Textbook Solutions
Asking a study question in a snap - just take a pic. An Empiricist's Companion. You will learn about this increasingly important area of economic study, and then apply what you have learned with behavioral economics problems in the end-of-chapter problem sets.
## Sanjana Gupta
What is the slope of that line? The more time spent in MindTap, the better the results.
Why buy extra books when you can get all the homework help you need in one place? Amazon Inspire Digital Educational Resources. Bookmark it to easily review again before an exam.
The tangent to g x at the point E x will have the form c dx g x for all values of x and c dE x g E x. So we have shown that any concave function must lie on or below the tangent to the function at that point.
So, It means trade-off between x and y is not linear. Behavioral economics is emphasized.
You can also find solutions immediately by searching the millions of fully answered study questions in our archive. The book came in perfect condition and is exactly as described. This would require a solution using the Lagrangian method. The basic approach is to focus on building intuition about economic models while providing students with the mathematical tools needed to go further in their studies.
Thorough summaries, memorable examples, and instructive figures are weaved throughout the text to clarify and expand upon key microeconomic principles. Intermediate Microeconomics by Samiran Banerjee Aboutbooks. In consumer theory, kamasutra fotografico pdf this function can be used to illustrate how diminishing marginal usefulness can be modeled in a very simple setting.
He also has enjoyed showing students some of the stranger things that economists have sought to model. Theories and Policies Richard T. But the right-hand side of this equation is.
Frequently bought together. Learning features strengthen understanding. Intuitively, because concave functions lie below any tangent plane, their level curves must also be convex. Alexa Actionable Analytics for the Web.
One person found this helpful. But the right-hand side of this equation is b. Book was in great condition! Note, however, that it is still optimal to have some of the higher risk asset because asset returns are independent. There's a problem loading this menu right now.
## Frequently bought together
PillPack Pharmacy Simplified. Write the total differential for U. See, for example, Mas Colell et al. English Choose a language for shopping.
Hit a particularly tricky question? If you are a seller for this product, would you like to suggest updates through seller support? Using MindTap throughout your course matters. Get fast, free shipping with Amazon Prime.
Amazon Renewed Refurbished products with a warranty. This shows how much an extra unit of perimeter would raise the enclosed area.
Walter Nicholson, Christopher M. Nicholson received his Ph. Basic Principles and Extensions Walter Nicholson. New developments continue to keep the field exciting, and we hope this edition manages to capture that excitement.
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| 2.90625
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CC-MAIN-2019-30
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https://oeis.org/A134335
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text/html
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crawl-data/CC-MAIN-2022-05/segments/1642320304309.59/warc/CC-MAIN-20220123202547-20220123232547-00331.warc.gz
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The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A134335 Numbers such that the arithmetic mean of their prime factors (counted with multiplicity) is integer, but not prime. 2
15, 35, 39, 42, 50, 51, 55, 65, 77, 78, 87, 91, 92, 95, 110, 111, 114, 115, 119, 123, 140, 141, 143, 155, 159, 161, 164, 170, 183, 185, 186, 187, 189, 201, 203, 204, 209, 215, 219, 221, 222, 225, 230, 235, 236, 242, 247, 258, 259, 264, 267, 284, 285, 287, 290 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 LINKS Hieronymus Fischer, Table of n, a(n) for n = 1..10000 EXAMPLE a(1) = 15, since 15 = 3*5 and (3+5)/2 = 4 is not prime. a(5) = 50, since 50 = 2*5*5 and (2+5+5)/3 = 4 is not prime. CROSSREFS Cf. A000040, A001222, A100118, A046363, A133620, A133621. Cf. A133880, A133890, A133900, A133910, A133911, A134330, A134331, A134332, A134333, A134334. Sequence in context: A253055 A338063 A111170 * A257591 A284406 A329589 Adjacent sequences: A134332 A134333 A134334 * A134336 A134337 A134338 KEYWORD nonn AUTHOR Hieronymus Fischer, Oct 23 2007 EXTENSIONS Definition clarified by the author, May 06 2013 STATUS approved
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Last modified January 23 17:33 EST 2022. Contains 350514 sequences. (Running on oeis4.)
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| 2.875
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CC-MAIN-2022-05
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latest
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https://science.blurtit.com/143619/how-do-you-read-a-mm-ruler
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crawl-data/CC-MAIN-2019-43/segments/1570986660323.32/warc/CC-MAIN-20191015205352-20191015232852-00044.warc.gz
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# How Do You Read A Mm Ruler?
The markings on a millimetre ruler are such that it is not hard at all to take the measurements to one decimal place. Whether it is a 15, 30 or 100-centimetre ruler, it is much the same case with every ruler. The 10 indentations between each centimetre denote each millimetre, of which there are ten in every one centimetre. Every fifth millimetre is seen to be longer than the first, second, third, fourth, sixth, seventh, eighth and ninth. This is to make markings of every half-centimetre easier to read. An object 8.5cm long has 85 millimeters. It is easier to count in centimetres, then work out the decimal point of the last centimetre value by means of looking at the marking on the ruler; if it comes up to the fifth marker of that set of 10 millimetres, the decimal is .5. These rulers are useful in architecture to get exact measurements without rounding up or down to the nearest centimetre, which of course you can do with the millimetre ruler; you round up if it goes beyond or is on the .5 marker, and down if it is below it. It is thus very easy to read a mm ruler.
thanked the writer.
| 282
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| 2.890625
| 3
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CC-MAIN-2019-43
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longest
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en
| 0.946173
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https://jp.mathworks.com/matlabcentral/cody/problems/15-find-the-longest-sequence-of-1-s-in-a-binary-sequence/solutions/725192
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| 15,824
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Cody
# Problem 15. Find the longest sequence of 1's in a binary sequence.
Solution 725192
Submitted on 28 Aug 2015 by Jacob
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% x = '0'; y_correct = 0; assert(isequal(lengthOnes(x),y_correct))
2 Pass
%% x = '1'; y_correct = 1; assert(isequal(lengthOnes(x),y_correct))
3 Pass
%% x = '01'; y_correct = 1; assert(isequal(lengthOnes(x),y_correct))
4 Pass
%% x = '10'; y_correct = 1; assert(isequal(lengthOnes(x),y_correct))
5 Pass
%% x = '00'; y_correct = 0; assert(isequal(lengthOnes(x),y_correct))
6 Pass
%% x = '11'; y_correct = 2; assert(isequal(lengthOnes(x),y_correct))
7 Pass
%% x = '1111111111'; y_correct = 10; assert(isequal(lengthOnes(x),y_correct))
8 Pass
%% x = '100101011111010011111'; y_correct = 5; assert(isequal(lengthOnes(x),y_correct))
9 Pass
%% x = '01010101010101010101010101'; y_correct = 1; assert(isequal(lengthOnes(x),y_correct))
10 Pass
%% x = '0101010111000101110001011100010100001110110100000000110001001000001110001000111010101001101100001111'; y_correct = 4; assert(isequal(lengthOnes(x),y_correct))
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# Some booly functions
(2017-07-27. Index.)
This post is part of a list: Some lambda-notes
Next thing: Pairs
Previous thing: What do the lambdas?
Bit of prelude, first. Some stuff from an earlier numbery post:
(And in the earlier post we said something like: The number n is a function that, if given two arguments, applies the first argument n times to the second. The +-function takes the numbers a and b as arguments and gives back a new λf.λx.-function that applies fa times” to the result of applying fb times” to x.)
We can do like multiuplication as well, before moving on. Multiplication is like addition, just more. We will make a function that, if given the numbers a and b as arguments, will like, start with the number zero, and add ab times” to it:
Soo. We have several numbers and also a couple of ways to make more numbers. So we pretty much have business: We can decide that one of the numbers is the number of monies and another one is the number of products. If we also have booleans we can do business logic.
Booleans are used for if-then-else. We have a stuff we maybe wanna do and another stuff we maybe wanna do instead, and we use a boolean to pick one of the stuffs. So, we would like to have a true-value and a false-value, and we wanna set things up so that something like if true stuff otherstuff will evaluate to stuff, and something like if false stuff otherstuff will evalute to otherstuff.
We will make the two boolean values be functions. (A reasonably easy choice, so long as we can only make functions in our language.) true will, if given two arguments, return the first one. false, if given two arguments, will return the second one.
Okay so it looks like true stuff otherstuff evaluates to stuff just fine on it’s own, and false stuff otherstuff evaluates to otherstuff, and we don’t really need if. But like if we want an if we can have one. It can take a boolean as its first arguments and then two more arguments, and then like, just, hand those two last arguments over to the boolean.
Should get same results as with just the booleans. (And if we want to, we can redefine if to be the identity function, λx.x. Will work fine. The if true part of if true stuff otherstuff wil evaluate to true. So we get true stuff otherstuff, which we already know that works.)
Okay, some boolean logic bits:
• not takes one boolean b. If b is true, not returns false. If b is false, not returns true.
• and takes two booleans a and b. If a is true, and returns b. If a is false, and returns false.
• or takes two booleans a and b. If a is true, or returns true. If a is false, or returns b.
We’ll do some boolean stuff with numbers. We can check if a number is zero by giving it two arguments. The first argument is a function that always returns false: λ_.false. If the number is not zero this function will be applied to something and we’ll get false back. The second argument is true. If the number is zero this true will be returned, without any applying of the λ_.false-function.
Good. Now we can do everyday business logic things. Like you know when boss is like hey we need a program that checks if the number of monies times the number of products is zero and if it is zero we should add five to the number of monies and if it is not zero we should multiply the number of products by 2. And like, no problem.
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# DCF valuation model
Document Sample
``` DCF valuation model
DCF valuation model
Posted :2009-6-17 9:07:41 author: Fei-fei on: Sohu blog
?
Experience: DCF more rigorous analytical methods, but involves a lot of produce,
poor choice of words, will miss is as good, hack away. In contrast, the relatively
stable income (income generation), the cost is relatively stable (mainly depreciation)
of water more suitable for valuation of such companies.
DCF / DiscountCashFlow / discounted cash flow model)
(1) free cash flow (FreeCashFlow) definition
Free cash flow is a financial method used to measure the company&#39;s actual
holdings of cash to reward shareholders. Means without endangering the survival and
development company available for distribution to shareholders under the premise
(and creditors) the maximum amount of cash.
Free cash flow in operating cash flow model based on the consideration of capital
expenditure and dividend payments. Although you might think that payout is not
required, but this expenditure is expected by shareholders, but also a cash payment.
Free cash flow equals operating cash.
Free cash flow that the company&#39;s discretionary cash. If the free cash flow
rich, the company can repay the debt, develop new products, buy back stock, increase
dividend payments. Meanwhile, abundant free cash flow also allows the company to
become acquisition targets.
Free cash flow Free cash flow can be divided into the overall business and corporate
equity free cash flow. Free cash flow refers to the whole enterprise after deducting all
operating expenses, after the investment needs and tax, before debt remaining in cash
flow; equity free cash flow is net of all expenses, tax payments, investment needs and
debt service expenditures After the remaining cash flow. Free cash flow is used to
calculate the overall value of enterprise as a whole, including the equity value and
debt value; equity free cash flow is used to calculate the value of corporate equity.
Equity free cash flow can be simply expressed as &quot;profits + depreciation -
investment.&quot;
Calculation of Free Cash Flow
Professor Copeland (1990) described the calculation of free cash flow:
&quot;Free cash flow equals after-tax net operating profit (excluding interest
payments the company will soon operating profit after deducting taxes paid the
amount of income taxes) plus depreciation and amortization and other non-cash
charges, minus the additional working capital and property, plant equipment and other
assets of the investment. It is the company generated total after-tax cash flow can be
provided to the company&#39;s capital, all suppliers, including Zhaiquan Ren
and shareholders. &quot;
Free cash flow = (after-tax net operating profit + depreciation and amortization) of a
(capital expenditures + increase in working capital)
Net operating profit - taxes
-------------------------------------------------- ----
= NOPAT [net operating profit after tax] - net investment - net working capital
changes
-------------------------------------------------- ----
= Free Cash Flow
The form of free cash flow is
With the definition of free cash flow derived in two ways: equity free cash flow
(FCFE, FreeCashFlowofEquity) and the company free cash flow (FCFF,
FreeCashFlowofFirm), FCFE is the company to pay all operating costs, re-investment
expenditures, income taxes and net debt payments (ie, interest, principal payments
reduce the net issuance of new debt) to shareholders after distribution of surplus cash
flow, which is calculated as follows:
FCFE = Net income 10 A capital expenditure, depreciation, working capital additional
principal amount of a debt repayment + new issue of debt
FCFF is the company to pay all operating costs for the necessary investment in fixed
assets and operating assets can be distributed to all investors in the after-tax cash flow.
FCFF is the company all claims, including common shareholders, preferred
shareholders and creditors, the sum of the cash flow, which is calculated as follows:
FCFF = EBIT x (1 - tax rate) + depreciation of a capital expenditure of an additional
working capital
(2) FCFF model (Freecashflowforthefirmfirm / company free cash flow model)
DDM model
V representative of the intrinsic value of common stock, Dt s t for the common stock
of the dividends paid or the dividend, r is the discount rate
Different assumptions on dividend growth, dividend discount model can be divided
into: a zero-growth model, constant growth model (Gordon growth model), two-stage
dividend growth model (H model), three-stage dividend growth model and multiple
forms of growth models .
The most basic model; dividend discount is the most stringent definition of intrinsic
value; DCF DDM method borrows heavily from some of the logic and calculations
(based on the assumption that the same / the same restrictions).
DCF model
2.DCF/DiscountCashFlow / discounted cash flow model) DCF valuation method is
the most rigorous of the company and stock valuation method, in principle, the model
is applicable to any type of company.
Free cash dividend alternative, more scientific, less vulnerable to human impact.
When all the equity free cash flow for dividend payments, FCFE model and DDM
model is no different; But in general, dividends are not the same as equity free cash
flow, high and sometimes low, for four reasons:
Stability requirements (uncertain future ability to pay high dividends);
Future investment needs (estimated future capital expenditure / financing for the
inconvenience and expensive);
Tax factor (a progressive personal income tax system is high);
Signal characteristics (dividend increase / prospects; dividends down / outlook
bearish)
Advantages: recommendations than other commonly used models include a more
complete assessment of the evaluation model, the framework of the most rigorous and
relatively complex evaluation model. The amount of information required more
comprehensive perspective, take into account the long-term development of the
company. A more detailed forecast a longer time, and to consider more variables, such
as earnings growth, capital costs, can provide the appropriate model of thinking.
Disadvantages: takes a long time to be the case with the company&#39;s
operations have in-depth understanding of industry characteristics. Consider the
company&#39;s future earnings, growth and risk of a complete evaluation model,
but the data are highly subjective estimates and uncertainties. Complex models, may
be difficult to estimate because the data can not be used, even if we manage to
estimate the error in the data set into a perfect model, can not get correct results.
Small changes in input may lead to large changes in the company&#39;s value.
The accuracy of the model by the input of large (can be used for sensitivity analysis of
remedies).
China to apply the formula:
Free cash flow company
= Cash flow from operating activities - Net capital expenditure
= Cash flow from operating activities - Net (acquisition or construction of fixed,
intangible and other long-term assets to pay cash - disposal of fixed, intangible and
other long-term assets, net of cash recovered)
Capital expenditure
Capital expenditure: for the purchase of fixed assets (land, plant and equipment)
investment, investment in intangible assets and long-term equity investments and
other capacity expansion, process improvement with long-term benefit of the cash
expenditures.
The form of capital expenditures:
1. Cash purchase or disposal of long-term assets, cash recovered
2. By issuing bonds or shares in the form of non-cash transactions to obtain long-term
assets,
3. To achieve long-term assets through mergers and acquisitions.
Among them, the subject of &quot;long-term disposal of assets in cash to
purchase or cash back&quot; capital expenditures.
The current cash flow statement in the &quot;cash flow from investing
activities&quot; section, has shown the &quot;acquisition or construction of
fixed, intangible and other long-term assets to pay the cash&quot; and
&quot;disposal of fixed assets and intangible and other long-term recovery Net
cash. &quot;
Therefore: the acquisition and construction of capital expenditures = fixed, intangible
and other long-term assets to pay cash - disposal of fixed, intangible and other
long-term assets, net of cash recovered
Economic significance of free cash flow
All enterprises operating activities cash &quot;net output&quot; to form
&quot;free cash flow,&quot; &quot;free cash flow&quot; to some
extent determine the size of a company&#39;s survival. Long-term output of a
business can not be &quot;free cash flow&quot;, it will eventually run out
of all the original investors to provide capital, and will go bankrupt.
1. &quot;Free cash flow,&quot; abundant, the enterprise can use
&quot;free cash&quot; to pay interest on repayment of principal, dividends
or buy back stock and so on.
2. &quot;Free cash flow&quot; is negative, the company earned with
interest costs are not coming back, but not yet put into operation only use (including
investment) activities, the remaining investors (shareholders, creditors) to provide the
seed capital (assuming also no previous annual &quot;free cash&quot;
surplus) to pay interest, principal payments, distribution of dividends or share
3. When the remainder of the original capital investment to provide enough to pay
interest, principal payments, dividends, the firm can only rely on &quot;拆东墙
business operations. When no &quot;east wall&quot; removable, the
enterprise fund strand breaks, the end result can only seek to be acquired or filed for
bankruptcy reorganization.
```
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https://practicaldev-herokuapp-com.global.ssl.fastly.net/johnkyalo/exploratory-data-analysis-using-data-visualization-techniques-2ok5
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John Kyalo
Posted on
Exploratory Data Analysis using Data Visualization Techniques
Exploratory Data Analysis involves initial investigation and examination of datasets to summarize the main characteristics often with the help of graphical representations.
Basically, EDA helps you get a feel of the data you are working with. This is by identifying the structure and patterns involved. EDA helps generate hypotheses about relationships and trends in data which guide further analysis.
EDA with data visualization involves creating various plots and charts, such as histograms, box plots, scatter plots, bar charts and heatmaps, to visualize the distribution and relationship within the data. With this, you are able to uncover styles, pick out relationships, and gain insights. In case of anomalies, you also get to identify them.
Mostly Matplotlib and Seaborn are Python libraries used for visualization. There exist various sorts of EDA strategies hired depending on the nature of records and desires of evaluation. This includes:
1. Univariate analysis-makes a specialty of analyzing character variables inside the records set. It involves visualizing an unmarried variable at a time to understand its distribution. Examples include: Histogram -displays distribution of a single numerical variable. Useful for understanding data's central tendency.
2. Bivariate analysis- from the name bi which means two, you explore two variables by finding their correlation, association and dependencies. Examples include: Scatter plot - which explores relationship between two variables.
3. Multivariate analysis- extends bivariate evaluation to encompass greater variables. It ambitions to apprehend the complex interactions and dependencies among the many variables in a record set.
More different plots which are also considered as techniques include:
• Box plots (Box and Whisker) which provides a summary of visual distribution of data. Good for identifying outliers.
• Line plots used for time-series data. They show how a variable change over time identifying trends.
• Bar charts mainly used for comparison of categorical data.
• Heatmaps which visualize correlation matrix of numerical variables. They use color intensity to represent the strength of correlations.
• Pie charts which shows the composition of categorical variable.
Many others include: pair plots, violin plots, density plots, word cloud ...
All these visualization techniques can be done in both Python, and other visualization tools such as Excel, PowerBi and Tableau. Choice of a tools depends on the user's and interest and what they would like to achieve.
Therefore, EDA through visualization is a key step in the data analysis process that helps leverage insight into data understanding that results to appropriate business decision making.
It's Data Allday Everyday
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https://www.convertunits.com/from/kilotonne/to/tical
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Convert kilotonne to tical [Asia]
kilotonne tical
How many kilotonne in 1 tical? The answer is 1.64E-8. We assume you are converting between kilotonne and tical [Asia]. You can view more details on each measurement unit: kilotonne or tical The SI base unit for mass is the kilogram. 1 kilogram is equal to 1.0E-6 kilotonne, or 60.975609756098 tical. Note that rounding errors may occur, so always check the results. Use this page to learn how to convert between kilotonnes and tical [Asia]. Type in your own numbers in the form to convert the units!
Quick conversion chart of kilotonne to tical
1 kilotonne to tical = 60975609.7561 tical
2 kilotonne to tical = 121951219.5122 tical
3 kilotonne to tical = 182926829.26829 tical
4 kilotonne to tical = 243902439.02439 tical
5 kilotonne to tical = 304878048.78049 tical
6 kilotonne to tical = 365853658.53659 tical
7 kilotonne to tical = 426829268.29268 tical
8 kilotonne to tical = 487804878.04878 tical
9 kilotonne to tical = 548780487.80488 tical
10 kilotonne to tical = 609756097.56098 tical
Want other units?
You can do the reverse unit conversion from tical to kilotonne, or enter any two units below:
Enter two units to convert
From: To:
Definition: Kilotonne
The SI prefix "kilo" represents a factor of 103, or in exponential notation, 1E3.
So 1 kilotonne = 103 tonnes.
The definition of a tonne is as follows:
A tonne (also called metric ton) is a non-SI unit of mass, accepted for use with SI, defined as: 1 tonne = 1000 kg (= 106 g).
Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more!
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https://oyoads.in/100-freemaster-discrete-math-2020-complete-basic-to-advanced-course/
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Note:
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Approximately 30-40% of questions in premier national middle and high school mathematics competitions, such as the AMC (American Mathematics Competitions), focus on discrete mathematics. More than half of the problems in the high level math contests, such as the AIME (American Invitational Mathematics Examination), are associated with discrete mathematics. Students not having enough knowledge and skills in discrete mathematics can’t do well on these competitions. Our AMC prep course curriculum always includes at least one-third of the studies in discrete mathematics, such as number theory, combinatorics, and graph theory, due to the significance of these topics in the AMC contests
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Discrete mathematics has become popular in recent decades because of its applications to computer science. Discrete mathematics is the mathematical language of computer science. Concepts and notations from discrete mathematics are useful in studying and describing objects and problems in all branches of computer science, such as computer algorithms, programming languages, cryptography, automated theorem proving, and software development. Conversely, computer implementations are tremendously significant in applying ideas from discrete mathematics to real-world applications, such as in operations research.
The set of objects studied in discrete mathematics can be finite or infinite. In real-world applications, the set of objects of interest are mainly finite, the study of which is often called finite mathematics. In some mathematics curricula, the term “finite mathematics” refers to courses that cover discrete mathematical concepts for business, while “discrete mathematics” courses emphasize discrete mathematical concepts for computer science majors.
Discrete math plays the significant role in big data analytics.
The Big Data era poses a critically difficult challenge and striking development opportunities: how to efficiently turn massively large data into valuable information and meaningful knowledge. Discrete mathematics produces a significant collection of powerful methods, including mathematical tools for understanding and managing very high-dimensional data, inference systems for drawing sound conclusions from large and noisy data sets, and algorithms for scaling computations up to very large sizes. Discrete mathematics is the mathematical language of data science, and as such, its importance has increased dramatically in recent decades.
IN SUMMARY, discrete mathematics is an exciting and appropriate vehicle for working toward and achieving the goal of educating informed citizens who are better able to function in our increasingly technological society; have better reasoning power and problem-solving skills; are aware of the importance of mathematics in our society; and are prepared for future careers which will require new and more sophisticated analytical and technical tools. It is an excellent tool for improving reasoning and problem-solving abilities.
Starting from the 6th grade, students should some effort into studying fundamental discrete math, especially combinatorics, graph theory, discrete geometry, number theory, and discrete probability. Students, even possessing very little knowledge and skills in elementary arithmetic and algebra, can join our competitive mathematics classes to begin learning and studying discrete mathematics.
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Yes, I am in 700 Club - 710 (Q46,V41)-Long Debrief : Share GMAT Experience - Page 2
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# Yes, I am in 700 Club - 710 (Q46,V41)-Long Debrief
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Joined: 21 Jun 2007
Posts: 149
Schools: UCLA Anderson School of Mgmt (FEMBA Class of 2013)
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Re: Yes, I am in 700 Club - 710 (Q46,V41)-Long Debrief [#permalink]
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27 Nov 2009, 08:12
Hello Shohadegu,
As LUGO said one way or other one or two of the sections you should do it fast, and that way you have time to concentrate on spend more time on your hardest section, be it RC or CR or SC. SC is probably the easiest one to get it fast. That is why it is important to know the patterns in SC. I can tell you by practice you surely can increase the RC speed. But you need to really work on it. May be do like 3 passages each day and concentrate fully, mostly with the structure and flow of the passage. Once you get this you can always go back and find where to look and get the answer.
Indian Jaguar
LUGO wrote:
shohagedu wrote:
Reading passage for 2/3/4 minutes is impossible for me. Moreover, for SC, I do not get the overall meaning of the sentence if i read it first time. Some times it is very tough for me to understand actual intention of the sentence. Most of the time, I can not concentrate deeply to get the idea as well.
For two/three months' preparation, how can i improve my scoring ability?
Hi Shohadegu
You have around 5-8 minutes to read each passage of the GMAT and more than once as long as you work on CR and SC on about 1-2 minutes per question. There are 4 passages in the GMAT and only one of them is longer than usual. If you want to improve your score I suggest you concentrate on learning how quickly to answer both CR and SC, so that you have enough time to tackle your weakest part, i.e. RC.
Last edited by IndianJaguar on 03 Dec 2009, 06:56, edited 1 time in total.
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Joined: 21 Jun 2007
Posts: 149
Schools: UCLA Anderson School of Mgmt (FEMBA Class of 2013)
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Re: Yes, I am in 700 Club - 710 (Q46,V41)-Long Debrief [#permalink]
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27 Nov 2009, 08:16
Hello aimenheli,
I would say if you are thoroughly done with all OG and prep qns, then go for it. But, important is to keep looking at the repeated mistakes and finding a way to avoid it. This should take some time and when you are done with this, you could try Jeff's. But I don't think this is critical since more than volume quality is the key.
Indian Jaguar
aimenhelmi wrote:
congratulations for the great score. I scored 40 in Q, would you recommend Jeff Q banks for more practice questions.
Re: Yes, I am in 700 Club - 710 (Q46,V41)-Long Debrief [#permalink] 27 Nov 2009, 08:16
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# Yes, I am in 700 Club - 710 (Q46,V41)-Long Debrief
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# CAT 2022 Question Paper | Quant Slot 2
###### CAT Previous Year Paper | CAT Quant Questions | Question 9
CAT 2022 Quant was dominated by Arithmetic followed by Algebra. In Arithmetic, the questions were dominated by topics like Speed-time-distance, Mixture and Alligations. This year, there was a surprise. The questions from Geometry were relatively on the lower side as compared to the previous years. There were 8 TITA Qs this year. Overall this section was at a medium level of difficulty.
Question 9 : Let $r$ and $c$ be real numbers. If $r$ and $-r$ are roots of $5 x^3+c x^2-10 x+9=0$, then $c$ equals
1. $-$frac{9}{2}$ 2. $$frac{9}{2}$ 3. $-4$ 4. $4$ ## Best CAT Online Coaching Try upto 40 hours for free Learn from the best! #### 2IIM : Best Online CAT Coaching. ### Video Explanation ## Best CAT Coaching in Chennai #### CAT Coaching in Chennai - CAT 2022Limited Seats Available - Register Now! ### Explanatory Answer Since r and$-r$ are the roots of the cubic equation $5 x ^ { 3 } + c x ^ { 2 } - 10 x + 9 = 0$
$5 ( r ) ^ { 3 } + c ( r ) ^ { 2 } - 10 ( r ) + 9 = 0$
$5 ( - r ) ^ { 3 } + c ( - r ) ^ { 2 } - 10 ( - r ) + 9 = 0$
$5 r ^ { 3 } + c r ^ { 2 } - 10 r + 9 = 0$
$- 5 r ^ { 3 } + c r ^ { 2 } + 10 r + 9 = 0$
10 r3 - 20 r = 0
r3 - 2r = 0
r (r2 - 2) = 0
r = $$pm $sqrt { 2 }$ Assuming the third root of the cubic equation as k, we can reconstruct it as… $$ x - $sqrt { 2 }$$ x + $sqrt { 2 } )$ x - k ) = 0$
$$left$ x ^ { 2 } - 2 $right)$ x - k ) = 0$
$x ^ { 3 } - k x ^ { 2 } - 2 x + 2 k = 0$
$5 x ^ { 3 } + c x ^ { 2 } - 10 x + 9 = 0$ can be re-written by making the coefficient of x3 as 1.
$x ^ { 3 } + $frac { c } { 5 } x ^ { 2 } - 2 x + $frac { 9 } { 5 } = 0$ $x ^ { 3 } + \frac { c } { 5 } x ^ { 2 } - 2 x + \frac { 9 } { 5 } = 0$ and $x ^ { 3 } - k x ^ { 2 } - 2 x + 2 k = 0$ are the same. $k = \frac { - c } { 5 }$ $2 k = \frac { 9 } { 5 }$ This implies, $c = \frac { - 9 } { 2 }$ The question is " Let $r$ and $c$ be real numbers. If $r$ and $-r$ are roots of $5 x^3+c x^2-10 x+9=0$, then $c$ equals " ##### Hence, the answer is '$-\frac{9}{2}$' Choice A is the correct answer. ###### Best CAT Online Coaching Try upto 40 hours for free Learn from the best! ###### Prepare for CAT 2024 with 2IIM's Daily Preparation Schedule ###### Know all about CAT Exam Syllabus and what to expect in CAT ###### Already have an Account? ###### CAT Coaching in ChennaiCAT 2024 Classroom Batches Starting Now! @Gopalapuram and @Anna nagar ###### Best CAT Coaching in Chennai Attend a Demo Class ###### Best Indore IPM & Rohtak IPM CoachingSignup and sample 9 full classes for free. Register now! ##### Where is 2IIM located? 2IIM Online CAT Coaching A Fermat Education Initiative, 58/16, Indira Gandhi Street, Kaveri Rangan Nagar, Saligramam, Chennai 600 093 ##### How to reach 2IIM? Mobile:$91$ 99626 48484 / 94459 38484
WhatsApp: WhatsApp Now
Email: info@2iim.com
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Godzilla Gets Bigger Each Year
Part of the Godzilla 2014 movie poster. Image courtesy Warner Bros.
I bet you thought Godzilla was godzilla-sized. Well, you are wrong. Godzilla is BIGGER THAN GODZILLA. Check out the new poster from the next version of Godzilla movie.
That’s a big Godzilla. BIG. io9 has an interesting post that includes a graphic of the size of Godzilla as a function of year. Of course, the exact size of Godzilla is difficult to determine, so most of the times it’s just an estimate. In fact, the Wikipedia page on Godzilla has a short discussion of the inconsistent size of Godzilla.
How Big Will Godzilla Be in the 2025 Movie?
Instead of re-estimating the Godzilla sizes, let me just start with the values posted in the io9 graph. From that, get the following plot.
These are mostly guesses for the years. I could have looked up the actual movie dates, but I’m too lazy. So, this is just an approximation of Godzilla’s size not the size of the actual and REAL Godzilla.
Actually, I plotted the natural log of height as a function of time. Why? Well, this is a more generic plot that I can then add a linear function to. Otherwise, I am assuming that Godzilla grows linearly with time. Why would anyone think that? And yes, I cheated. You can’t take the natural log of “50 meters”. Really this is the natural log of the height per meter (so that the stuff inside has no units).
Now, when I fit a linear function to this data I get the following (where a and b are the fit parameters from plotly).
That’s it. I have a, I have b so I get use this function to get the Godzilla height for any year. Now, let’s just plug in the year of t = 2025 (yes, t is in years). This gives:
That is 121.6 meters but I left off the meters because of my mathematical cheat above. Yes, that’s not so big. The reason is that although the 2014 Godzilla is 140 meters, this is bigger than the other trend. Really, it’s too big. A better value would be around 100 meters (this is based on the fit that includes the 140 m height though).
What about the year 2050? In this year, Godzilla would be 170 meters tall. Also, in this year Godzilla 2050 will be in 3D and include the much anticipated smell-o-vision (we have been promised this since the 1960s).
Of course having a ginormous Godzilla brings in other problems. Big things aren’t the same as small things. If you double the height of an object, but keep the same shape you will increase the mass by a factor of 8.
Rhett Allain is an Associate Professor of Physics at Southeastern Louisiana University. He enjoys teaching and talking about physics. Sometimes he takes things apart and can't put them back together.
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Logical reasoning question types
Each type of Logical Reasoning problem presents a unique challenge, and in order to have success on the Logical Reasoning section, it is essential to develop a strong understanding of the individual question types, as well as specific strategies that align with the different tasks that they present. This infographic and article are here to help.
## Group One: Argument-Based Questions That Require You to Evaluate Reasoning
(Includes: Identify the Flaw, Basic Assumption, Required Assumption, Strengthen the Argument, Weaken the Argument, Supporting Principle & Conform to Principle, Sufficient Assumption, Match the Flaw)
For all of the question types in this group, our task is to identify an argument in the stimulus, figure out why the support given is not enough to justify the point made, and then to address this problem in one way or another, per what is being asked of us in that specific question stem.
Identify the Flaw (6 - 10 per exam)
Identify the Flaw is the most common Logical Reasoning problem type, and these questions simply ask us to select the answer that most accurately represents the reasoning issue in the argument.
Basic Assumption (0 - 1 per exam)
These are very similar to Flaw problems, but instead of expressing the gap in the argument as a reasoning mistake, they represent it as an “assumption” made by the author.
Required Assumption (2 - 8 per exam)
Required Assumption questions, also commonly known as Necessary Assumption questions, ask us to identify the one answer that must be true in order for the argument itself to be valid. Note that the right answer does not have to “fix” the argument in any way—it is simply something that must be true in order for the argument to work at all.
An effective way to test whether an answer is indeed necessary is to try considering the opposite of it. If the opposite of the answer choice would destroy the reasoning in the argument, it’s a great sign that the answer choice is indeed something that is required for the argument to work.
Strengthen & Weaken (5 - 11 per exam)
Strengthen questions ask us to identify an answer that helps bridge the gap between the given point and the given support, and Weaken questions ask us to identify an answer that hurts connection or exposes a problem with it.
Keep in mind that Strengthen and Weaken question stems are written in terms of which answer “most strengthens” or “most weakens,” but for all such problems there will only be one answer that actually strengthens or one answer that actually weakens. Our task is never to see which of two answers strengthens more, or weakens more. Our task is to recognize why four answers simply do not perform the given task (most commonly because the wrong answer is not relevant to the given argument), and to find the one answer that does indeed strengthen or does weaken.
Supporting Principle & Conform to Principle (3 - 5 per exam)
These two types of principle questions are very similar. We can think of a principle as being a more general description of the reasoning structure that exists between the support and the conclusion in any given argument. Supporting Principle questions ask us to identify a general reasoning rule that would best justify the author’s reasoning, and Conform to a Principle questions ask us to identity the reasoning rule with which the given argument matches up best.
Sufficient Assumption (1 - 4 per exam)
Sufficient Assumption questions require us to identify one answer that would completely solve the reasoning issue that exists between the given conclusion and support. We want to look for that one right answer that will, when inserted into the argument, make it so that the support does indeed make the conclusion 100% valid.
Match the Flaw (1 - 2 per exam)
These problems require us to identify the answer choice that has the same reasoning issue as the original argument. Each of the five answers for these problems will themselves have arguments, with conclusion and support that must be evaluated, and so you should expect for these problems to take a little bit longer to solve than others.
## Group Two: Argument-Based Questions That Do Not Require You to Evaluate Reasoning
(Includes: Method of Reasoning, Identify the Conclusion, Identify the Role, Match the Reasoning)
These problems are very similar to those in Group One except for one major difference: for these problems, our job is not to evaluate the validity of the reasoning. We are not being asked to think about why the reasoning is flawed, and doing so can be an unnecessary distraction.
These problems are great indicators of your ability to read for reasoning structure. If you are a great reader, you will find these to be, on average, some of the easiest questions in the section. If you are struggling with the questions in this group, chances are that making improvements here will have a positive effect on how you perform on other types of problems as well.
Method of Reasoning (0-2 per exam)
Method of Reasoning questions ask us to identify the answer that most accurately represents the support-conclusion relationship presented in the stimulus. Some arguments will combine two premises to reach a conclusion (Tom likes rice and Tom likes beans so he should like rice and beans), others link one premise to another to reach a conclusion (All mice are happy animals and all happy animals smile therefore all mice smile), some arguments contain conditional reasoning that deals in absolute truths (such as a conclusion that states “All tenants must sign the new agreement”) while others reach a conclusion about cause and effect (“Since Jill was driving, she must have caused the accident”), and the list of descriptives goes on and on. The right answer will accurately describe the given argument, and the wrong ones will misrepresent it.
Identify the Conclusion (1 -3 per exam)
For these problems, our job is to correctly identify the author’s main point in the given argument. It’s important to be careful not to over-infer—our job is not to come up with our own conclusion, but rather to find the answer that best represents the conclusion given to us in the text.
Identify the Role (0 - 4 per exam)
For these problems, we are asked to identify the role that is played by a particular part of the given stimulus. The role will relate to the structure of the included argument: the point made, support, opposition, or background.
Match the Reasoning (0 - 2 per exam)
Match the Reasoning questions require us to first identify the method of reasoning (as we did for Method of Reasoning questions) then identify an answer choice that most similarly matches that reasoning structure. Just as with Match the Flaw problems, we should expect to spend a bit more time on Match the Reasoning problems, for they require a few extra layers of work from us.
## Group Three: Outliers
(Includes: Inference, Explain this, Identify the Disagreement, Give an Example)
Finally, there are a minority of questions that are not defined by arguments.
Inference (5 - 9 per exam)
This is a very common question type and perhaps deserves a group all for itself. Inference questions ask us to find the one answer that is supported by the information given in the stimulus.
The “level of provability” depends on the exact wording of the question stem. A question that asks you to find an answer that “must be true” will allow for less flexibility than one that asks for an answer that is “most supported.”
The inference made in the right answer can be derived from an individual piece of information given in the stimulus, or it can be derived by combining various parts of the stimulus together. It’s important to note that typically the right answer does not have to be a “main point” of the given stimulus—any answer that is justifiable based on the given information will be correct. On a related note, because there are numerous inferences that can be made off of any given stimulus, the right answers to these questions are not predictable.
Thus Inference questions, even more than others, require us to utilize our elimination skills. The most absolute thing about many of these Inference problems, especially the hardest ones, is that the wrong answers are clearly not justifiable based on the given text. Right answers are often very difficult to prove and will sometimes leave you feeling a bit wishy-washy, but wrong answers are almost always clearly, absolutely, and obviously wrong.
Explain This (2 - 5 per exam)
Explain This problems (a.k.a. “Explain a Discrepancy”) will present you with two pieces of information that seemingly don’t go together (for example, “Fred is very handsome, but no one will ever kiss him.”), and our job is to pick an answer that provides the most plausible explanation (for example, “Fred’s breath is so back people can’t get within a foot of him.”)
A great way to approach these problems is by asking, “How come?” as in “How come no one will kiss Fred even though he is handsome?” Only one of the five choices will provide a plausible explanation.
Identify the Disagreement (0 - 2 per exam)
For these questions, we will be given brief statements by two different individuals, then asked to find the answer that best represents what the two people disagree about.
It’s important to note and remember that these two people are commonly not on the same page—the first person is thinking about one thing and the second person another—there is just a small area of overlap between the two statements, and it is in this overlap that you will find the disagreement.
The most tempting wrong answers for these problems give information about which one person has a very strong opinion, but the other person does not. You want to verify your answer to an Identify the Disagreement problem by making sure that each person does indeed have an opinion about the information in a particular answer choice, and that, of course, these opinions contrast.
Give an Example (0 - 3 per exam)
These problems are essentially the mirror twins of “Conform to a Principle” questions. For “Conform to a Principle” problems, we needed to find a principle with which the situation in the scenario fits. For “Give an Example” questions, we are given a principle in the stimulus, then asked to identify among the answer choices a situation that best fits that principle. The skills you’ve developed in order to master other question types should fairly easily translate to the challenges of this question type.
The Trainer Question Breakdown organizes all Logical Reasoning questions from exams 29 - 75 by type. Please click to open.
Thanks for reading and I hope you've found this information useful. For more free Logical Reasoning instruction, please check out these links:
Mike Kim is the author of The LSAT Trainer, the most popular and acclaimed new LSAT learning product to be released in over a decade. Previously, he co-created ManhattanLSAT. Inspired by self-study students who prepare for the exam on their own, Mike set out to write the ultimate self-study guide, and The LSAT Trainer is the result.
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A343830 a(n) = numerator of (1/e) * Sum_{a_1>=1, a_2>=1, ... , a_n>=1} a_1 * a_2 * ... * a_n / (a_1 + a_2 + ... + a_n)!. 3
%I #30 Jun 11 2021 05:05:56
%S 1,2,31,179,787,6631,2456299,33235913,158433901,17980176031,
%T 2794938616471,8546650588601,5595650767265101,35480190026972501,
%U 15523069639558351459,455264603021602214213,57023540590242398853649,949437664962426221725789,5469912218467062529961407
%N a(n) = numerator of (1/e) * Sum_{a_1>=1, a_2>=1, ... , a_n>=1} a_1 * a_2 * ... * a_n / (a_1 + a_2 + ... + a_n)!.
%D O. Furdui, Limits, Series and Fractional Part Integrals. Problems in Mathematical Analysis, Springer, New York, 2013. See Problem 3.114 and 3.118.
%H Seiichi Manyama, <a href="/A343830/b343830.txt">Table of n, a(n) for n = 1..367</a>
%H Math StackExchange, <a href="https://math.stackexchange.com/questions/2344699/compute-s-n-sum-limits-a-1-a-2-cdots-a-n-1-infty-fraca-1a-2-cdots-a-n?rq=1">Compute S_n = Sum_{a_1 a_2 ... a_n >=1} a_1 a_2 ... a_n/(a_1+a_2+...+a_n)!</a>
%F b(n) = (1/e) * Sum_{a_1>=1, a_2>=1, ... , a_n>=1} a_1 * a_2 * ... * a_n / (a_1 + a_2 + ... + a_n)! = Sum_{j=0..n} (-1)^(n+j-1) * binomial(n,j) * Sum_{k=0..n+j-1} (-1)^k/k! = Sum_{k=0..n-1} binomial(n-1,k)/(k+n)!.
%F a(n) = numerator of b(n).
%e 1, 2/3, 31/120, 179/2520, 787/51840, 6631/2494800, 2456299/6227020800, ...
%t a[n_] := Numerator @ Sum[Binomial[n - 1, k]/(k + n)!, {k, 0, n - 1}]; Array[a, 20] (* _Amiram Eldar_, May 01 2021 *)
%o (PARI) a(n) = numerator(sum(j=0, n, (-1)^(n+j-1)*binomial(n, j)*sum(k=0, n+j-1, (-1)^k/k!)));
%o (PARI) a(n) = numerator(sum(k=0, n-1, binomial(n-1, k)/(k+n)!));
%Y Cf. A343831 (denominator), A343832.
%K nonn,frac
%O 1,2
%A _Seiichi Manyama_, Apr 30 2021
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Last modified May 24 23:50 EDT 2024. Contains 372782 sequences. (Running on oeis4.)
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https://math.stackexchange.com/questions/14383/how-many-right-angled-triangles-can-a-circle-have
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# How many right angled triangles can a circle have?
Here's what I recall of the question from CNML Grade 11, 2010/2011 Contest #3, Question 7:
There are 2010 points on a circle, evenly spaced. Ford Prefect will* randomly choose three points on the circle. He will* connect these points to form a shape. What is the probability that the resulting shape will* form a right angled triangle?
I answered $\frac{1}{4} = 25\%$, but that's probably incorrect. (Right?)
When I got home, I thought it out in my head, and I got this:
$\frac{2010 * (2010/1005)}{2010 \choose 3}$
$\frac{2020050}{1351414120} = \frac{3015}{2017036} = 0.149476756984010201\%$
I'm probably wrong ...again. Can anyone tell how to get the right answer (if I'm not wrong :) )?
*in the past of the future of the perfect present present time double into ripple fluctuater byer doininger of the past future continuum...
EDIT: Realized my mistake in copying the question.
• it seems to me that probability is infinitely close to 0, as there are uncountably many points on the circle, though I am probably wrong, too. :) – InterestedGuest Dec 15 '10 at 6:16
• @TonyK: but does the fact that we can actually construct a right triangle in a circle contradict that? Or is it one of those thing where the probability is 0, but it's possible? – InterestedGuest Dec 15 '10 at 7:08
• @muntoo: I am not sure what you mean. The only way to inscribe a triangle in a circle is by having two vertices diametrically opposite. – Timothy Wagner Dec 15 '10 at 7:17
• @muntoo: See this en.wikipedia.org/wiki/Thales'_theorem – Timothy Wagner Dec 15 '10 at 7:30
• But muntoo...If you're using the past present pluperfect continuous tense, WTF is Ford Prefect doing in the year 2010? WTF is he doing here anyway? – TonyK Dec 15 '10 at 21:40
Let the points be $P_1$, $P_2$, and $P_3$. Letting $P_1$ be arbitrary, we win if $P_2$ is the point diametrically opposite to $P_1$, probability $1/2009$. Otherwise, with probability $2008/2009$, we still win if $P_3$ is diametrically opposite either $P_1$ or $P_2$, which it is with probability $2/2008$. So the probability of a win is $1/2009 + (2008/2009)(2/2008) = 3/2009$.
You can also get this as follows: The number of possible wins is the number of diameters times the number of remaining points, or 1005*2008. The number of possible triples is $2010\choose 3$. Dividing the first by the second gives $3/2009$.
Even easier: The probability that a given pair of points lie on a diameter is $1/2009$. With three points, you have three pairs, hence $3/2009$. Here it's OK to add probabilities because it's not possible to have overlap; if one of the pairs lies on a diameter, no other pair can lie on a diameter.
BTW, I do not agree with the comment thread that says we must consider the possibility that the points are not distinct. Sometimes there can be an ambiguity, but in common language "choose three points from 2010 points" means "without replacement". For example, I am not incorrect in saying that $n\choose r$ is the number of ways of choosing $r$ objects out of a set of $n$. Although I admit that if I were grading the test and this issue were pointed out to me, I might accept the "with replacement" answer as well.
The only other alternative answer I might accept is 1, i.e., that the points are certain to lie on a right triangle. After all, it is Ford Prefect, so I have no way of knowing whether the Infinite Improbability Drive has been activated.
• "I have no way of knowing whether the Infinite Improbability Drive has been activated." - I've already given this answer a +1 before you wrote that, and it's a shame I can't give another... – J. M. is a poor mathematician Dec 15 '10 at 12:25
• To be exact, Ford was trying to come up with the probability that Arthur's cup of not entirely unlike tea will entirely become tea when he would know whether the Improbability Drive would have been activated or not. – Mateen Ulhaq Dec 16 '10 at 0:28
• I'm pretty sure that the problem meant "choose different points", but it wasn't "clearly" said. – Mateen Ulhaq Dec 16 '10 at 0:31
Hint: You can choose $3$ points in ${2010}\choose {3}$ ways. To form a right angled triangle, we need to choose a pair of diametrically opposite points (how many ways to do this?) and then any third point will form a right angled triangle with the chosen pair.
• You have to be a bit careful: firstly, the problem didn't specify that the random points have to be distinct, so there are $2010^3$ ways of choosing those. Secondly, that doesn't quite give you the number of ways of choosing the points. – Alex B. Dec 15 '10 at 8:02
• @Alex: I agree with your first point, I assumed $3$ distinct points were chosen. I am not sure about your second: you can choose a pair of diametrically opposite points in 1050 ways and then choose any other point in 2008 ways. So you can form a right angled triangle with 3 distinct points in $2008\times 1050$ ways. I suppose I interpreted this question to mean how many right angles triangle can be formed using any three distinct of these $2010$ points without regards to the order in which they were chosen. – Timothy Wagner Dec 15 '10 at 8:09
• You calculate the number of triangles correctly, but not the number of ways of choosing a triangle. As you say, the order in which the points were chosen doesn't matter. There are 6 possible orders in which the points of a triangle can be chosen, so you have to multiply by those, if you want an honest probability. – Alex B. Dec 15 '10 at 8:14
• @Alex: Perhaps I am having a brain freeze here, but if you account for the order in which the points were chosen, wouldn't you multiply both the numerator and denominator by $6$. So the probability would be the same under either assumption. – Timothy Wagner Dec 15 '10 at 8:19
• Well, maybe it's me who has a brain freeze. We first count the number of all possible choices. That is $2010^3$, agreed? Now, how many of these choices produce a right-angled triangle? Suppose ABC is a right angled triangle. Then there are several routes to choosing it, by picking the points in different orders. The way we have counted all the choices, the choice (A,B,C) is different from (B,A,C), but they both produce a triangle of the desired shape. Ergo... As I say, it might well be me, who has a brain freeze. – Alex B. Dec 15 '10 at 8:44
Simplify it. There are 2010 points, but if you start with four points instead and pick 3 at random, you see the probability of getting a right triangle is 1 (draw it out, it helps). You can do it with six points too, and if you look at the probability for both figures you can create a formula to give the probability regardless of the number of points. 3/(n-1), where n is the number of points (I think it has to be even for this to work). With four points, 3/(4-1) = 1, six points 3/(6-1) = 3/5. If you have 2010 points then the probability would be 3/(2010-1) = 3/2009.
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## CS302 Lecture Notes - Network Flow Supplemental Lecture
### Some extra help with Network Flow
• April 2, 2008. Latest revision, November, 2014.
• James S. Plank
• Directory: /home/plank/cs302/Notes/Netflow-All
This is an old lecture that I include now as supplemental material. In this lecture, I give an example of calculating network flow with three different path-finding algorithms. At the end, I show how to find the minimum cut.
### Greedy Depth-First Search
The pathelogical example from lecture 1 highlights a problem -- choosing paths with little flow. One way to combat this problem is to perform your DFS so that you prefer edges with a lot of flow to edges with a little flow. You can implement that by having your adjacency lists be sorted by flow, and then running a standard DFS on them. I'll show the augmenting paths that arise with this method on the example graph below. The source is A an the sink is G:
When we perform a greedy DFS on this graph, we start with edge from A to B since its flow is greater than the edge from A to D. There is only one edge leaving B. When we read node C, we traverse the edge to E, since it has the maximum flow of C's three edges. Finally, E goes to G. Thus, the first path we get is A->B->C->D->G with a flow of one.
Below, I show the flow and residual graphs when this path is processed. In the residual, the edges whose flow is reduced are colored red, and the reverse edges are colored green:
Flow Residual
The next greedy path through the residual is A->B->C->D->F->G, with a flow of two:
Flow Residual
The last path is A->D->F->G with a flow of three. Here are the final flow and residual graphs:
Flow Residual
### Finding the Maximum Flow Path through the Graph
Instead of using a depth-first search, you can use a modification of Dijkstra's algorithm to find the path through the residual that has the maximum flow. When processing a node, Dijkstra's algorithm traverses the node's edges, and if shorter paths to any other nodes is discovered, the nodes are updated. At each step, the node with the shortest known path is processed next.
The modification works as follows. When processing a node, again the algorithm traverses the node's edges, and if paths with more flow to any other nodes are discovered, then the nodes are updated. At each step, the node with the maximum flow is processed next.
We'll work an example with the same graph as above:
The maximum flow path here is A->D->F->G with a flow of three:
Flow Residual
The next maximum flow path is A->B->C->D->F->G with a flow of 2:
Flow Residual
The final path is A->B->C->E->G with a flow of one:
Flow Residual
### Edmonds-Karp: Finding the minimum hop path
Finally, the Edmonds-Karp algorithm uses a straightforward breadth-first search to find the minimum hop path through the residual graph. This is equivalent to treating the residual as an unweighted and performing a shortest path search on it.
Again, we use the same graph as an example:
There are two minimum hop paths: A->D->E->G and A->D->F->F. Suppose we process the former of these, with a flow of one:
Flow Residual
Now, there is only one minimum hop path through the residual: A->D->F->G, with a flow of two:
Flow Residual
At this point, there are only two paths through the residual: A->B->C->D->F->G and A->B->C->E->D->F->G. The first of these has fewer hops, so we process it. It has a flow of two:
Flow Residual
The final path through the residual is A->B->C->E->D->F->G with a flow of one. When we process it, we get the same flow and residual graphs as the other two algorithms:
Flow Residual
### Bottom Lines
We've shown three ways to calculate the augmenting paths. Each way yields a different ordering of augmenting paths; however, all three ways lead to the same flow and residual. They differ in what they try to optimize.
• The greedy DFS and modified Dijkstra algorithms attempt to minimize the number of paths that you find by finding paths with a lot of flow. They each have an expensive component -- in greedy DFS, processing the residual graph is O(|V|log|V|) rather than O(|V|). In the modified Dijkstra, finding the augmenting path is O(|E|log|V|) rather than O(|E|).
• Edmonds-Karp attempts to find a small number of paths, but its path-finding algorithm is fast -- O(|E|) -- as is its residual processing algorithm, which is O(|V|).
There's no nice closed form solution for either the greedy DFS or the Modified Dijkstra algorithm. When we evaluate performance, the greedy DFS is significantly slower, and Modified Dijkstra runs on par with Edmonds-Karp. Edmonds Karp does have a closed form running time: O(|V||E|2).
One thing that you should remember about network flow is that it is quite a bit slower than all of the other graph algorithms we've studied so far.
## Finding the Minimum Cut
As detailed in the first network flow lecture, you can use the final residual graph to find the minimum cut. First, you find all nodes reachable from the source in the residual graph. This is one set of nodes, which we color purple below:
Now, in the original graph, we divide our nodes into two sets: the set determined above, and all of the remaining nodes. They are drawn purple and yellow in the original graph below:
The minimum cut is composed of all the edges that go from the source set to the sink set. These are edges AD, CD and EG, which I've drawn in red above. The sum of their capacities equals the maximum flow of six.
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# Fundamentals of Physics Halliday & Resnic
【CH31】Alternating Fields and Current 7, 9, 11, 15, 19, 21, 25, 29, 33, 36, 37, 45, 51, 55, 57, 63, 65 【Problem 31-7】 The energy in an oscillating LC circuit containing a 1.25 H inductor is 5.7 µ J . The maximum charge on the capacitor is 175µ C . For a mechanical system with the same period, find the (a) mass, (b) spring constant, (c) maximum displacement, and (d) maximum speed. <解>：(a) The mass m corresponds to the inductance, so m = 1.25 kg. (b) The spring constant k corresponds to the reciprocal of the capacitance. Since the total energy is given by U = Q2/2C, where Q is the maximum charge on the capacitor and C is the capacitance, 175 × 10−6 C Q2 C= = = 2.69 × 10−3 F −6 2U 2 5.70 × 10 J And k = 1 = 372 N / m. 2.69 × 10−3 m / N
c c
h h
2
(c) The maximum displacement corresponds to the maximum charge, so xmax = 1.75 ×10−4 m. (d) The maximum speed vmax corresponds to the maximum current. The maximum current is I = Qω = Q = LC 125 c . 2 b Hg.69 × 10 Fh
−3
175 × 10−6 C
= 3.02 × 10−3 A.
Consequently, vmax = 3.02 × 10–3 m/s. 【Problem 31-9】 In an oscillating LC circuit with L = 50mH and C = 4 µ F , the current is initially a maximum. How long will it take before the capacitor is fully charged for the first time? <解>：The time required is t = T/4, where the period is given by T = 2 π / ω = 2 π LC. Consequently, T 2 π LC 2 π t= = = 4 4 【Problem 31-11】
0. 4 b050 Hg.0 × 10 Fh 7.0 × 10 c =
−6
−4
4
s.
the capacitor is charged to V = 34. (a) What is the ratio of maximum frequency to minimum frequency that can be obtained with such a capacitor? If this circuit is to obtain frequencies from 0.2µ F .20 × 10−6 F c h c h (b) When the switch is thrown. and L = 54mH . Consequently.0 × 10 −3 H 6.2 ×10−6 F )(34V ) = 2. The switch is kept in position a for a long time and then thrown to position b. Thus. What are the (a) frequency and (b) current amplitude of the resulting oscillations? （圖 31-28） <解>：(a) After the switch is thrown to position b the circuit is an LC circuit. LC f = ω 1 1 = = = 275 Hz. this range can be adjusted. The angular frequency 1 of oscillation is ω = . and the ideal battery has emf ε = 34V . the maximum charge on the capacitor is Q = CV = (6.11×10−4 C .0 V and the current is zero. b g c h 【Problem 31-15】 A variable capacitor with a range from 10 to 365 pF is used with a coil to form a variable-frequency LC circuit to tune the input to a radio.Fundamentals of Physics Halliday & Resnic 東海大學物理系 98 學年度第 2 學期 In Fig. R = 14Ω . 2 π 2 π LC 2 π 54. Consequently. C = 6.60 MHz. The current amplitude is I = ωQ = 2 πfQ = 2 π 275 Hz 2. the smaller value of C gives the larger value of f.11 × 10−4 C = 0. and 第 2 頁，共 13 頁 . (b) what capacitance should be added and (c) what inductance should the coil have? <解>：(a) Since the frequency of oscillation f is related to the inductance L and capacitance C by f = 1 / 2 π LC . f max = 1 / 2 π LCmin .365 A. By adding a capacitor in parallel to the variable capacitor. To obtain the desired frequency range. the ratio computed in (a) is too large. f min = 1 / 2 π LCmax . 31-28.54 MHz to 1.
【Problem 31-19】 In an oscillating LC circuit.98 ×10−6 J. b96g− 1 2 2 (c) We solve f = 1 / 2 π LC for L. f min Cmin 10 pF (b) An additional capacitance C is chosen so the ratio of the frequencies is 160 MHz . (b) the maximum charge on the capacitor. r= = 2.80 ×10 F ) −6 A) 2 ( 25. If C is in picofarads (pF). the charge on the capacitor is 3. then C + 365 pF C + 10 pF = 2.0. What then is φ ? <解>：(a) The total energy U is the sum of the energies in the inductor and capacitor: U =U E +UB = −6 q2 i2 L + 2C 2 2 −3 ( 3. The solution for C is 365 − 2.8µ F .54 × 10 Hzh 2π c 0 c 2 2 2 −12 6 2 1 = 1 = 2. and (c) the maximum current? (d) If the charge on the capacitor is given by q = Q cos(ωt + φ ) .20 ×10 = 2 ( 7. its capacitance adds to that of the tuning capacitor.8µ C . . and the capacitor is charging.54 MHz Since the additional capacitor is in parallel with the tuning capacitor.54 MHz. Thus L= 2 π Cf bg bg401 × 10 Fh. .2 × 10−4 H.80 × 10 −6 F 198 × 10 −6 J = 556 × 10 −6 C. 第 3 頁，共 13 頁 c c h h . except that the capacitor is discharging at t = 0 . For the minimum frequency C = 365 pF + 36 pF = 401 pF and f = 0. b pFgb96g10 pFg 36 pF. What are (a) the total energy in the circuit.Fundamentals of Physics Halliday & Resnic 東海大學物理系 98 學年度第 2 學期 Cmax 365 pF f max = = = 6.80 ×10 C ) + ( 9. 0.96.96. At time t = 0 the current is 9.20 mA. what is the phase angle φ ? (e) Suppose the data are the same. L = 25mH and C = 7. (b) We solve U = Q2/2C for the maximum charge: Q = 2CU = 2 7. b = C= 2.0 ×10 −3 H) 2 = 1.
q 380 GJ G × 10 C J Q 556 . The current (as a function of time) is i= dq = ωQ cos ωt . L = 3mH and C = 2. for φ = –46.00 × 10 Hh. b gc. and (c) what is that greatest rate? <解>：(a) The charge (as a function of time) is given by q = Q sin ωt .9°) is negative. the correct value for increasing charge is φ = –46.00 A 3 2 = . which we wish to be positive.Fundamentals of Physics Halliday & Resnic 東海大學物理系 98 學年度第 2 學期 (c) From U = I2L/2. At t = 0 the charge on the capacitor is zero and the current is 2. it is I = ω Q.9° . Since sin(+46. (b) The energy stored in the capacitor is given by q 2 Q 2 sin 2 ωt = 2C 2C and its rate of change is UE = dU E Q 2ω sin ωt cos ωt = dt C 1 We use the trigonometric identity cosωt sin ωt = 2 sin 2ωt to write this as bg 第 4 頁，共 13 頁 . A sine function was chosen so that q = 0 at time t = 0. 2U I= = = 126 × 10−2 A. 【Problem 31-21】 In an oscillating LC circuit. then q0 = Q cos φ and c h φ = cos−1 F I = cos F. we take φ = +46.00 A.7 µ F . Thus.9°. (e) Now we want the derivative to be negative and sin φ to be positive. .9°. × 10 C I = ±46. We obtain –ω Q sin φ . HK H K −1 −6 −6 For φ = +46. where Q is the maximum charge on the capacitor and ω is the angular frequency of oscillation. (a) What is the maximum charge that will appear on the capacitor? (b) At what earliest time t > 0 is the rate at which energy is stored in the capacitor greatest. Q = I LC = 2.0 × 10 −3 H (d) If q0 is the charge on the capacitor at time t = 0. Since ω = 1/ LC . we find the maximum current: 2 198 × 10−6 J . To check this.9° it is increasing.9° the charge on the capacitor is decreasing. we calculate the derivative of q with respect to time.9°) is positive and sin(– 46.70 × 10 Fh 180 × 10 c −3 −6 −4 C. L 25. evaluated for t = 0. dt and at t = 0.
we obtain ln F I = − Rt q G J 2L Q HK max which leads to 2 L qmax R=− ln t Q 【Problem 31-29】 第 5 頁，共 13 頁 2 ( 220 ×10−3 H ) ln ( 0.70 × 10 Fh 5.00 × 10 Hh.7 W.0% of its initial value in 50.66 × 10−3 Ω .655 × 10 c 180 . dt max 2TC TC t= F I G J H K E Now T = 2 π LC = 2 π 3 2 = c.99 ) = 8.655 × 10 sh. 【Problem 31-25】 What resistance R should be connected in series with an inductance L = 220mH and capacitance C = 12 µ F for the maximum charge on the capacitor to decay to 99. 4ω 4 4 (c) Substituting ω = 2π/T and sin(2ω t) = 1 into dUE/dt = (ω Q2/2C) sin(2ω t). The time required for 50 cycles (with 3 significant figures understood) is 2π t = 50T = 50 = 50 2π LC = 50 2π ω = 0. dU F I = πc × 10 Ch = 66.70 ×10−6 F) = 7.07 ×10−5 s.0 ×10 F) ) −3 −6 The maximum charge on the capacitor decays according to q max = Qe − Rt / 2 L (this is called the exponentially decaying amplitude in §31-5). where Q is the charge at time t = 0 (if we take φ = 0 in Eq. 31-25).5104s. Dividing by Q and taking the natural logarithm of both sides. This means bg π π π = LC = ( 3. we obtain dU E 2 πQ 2 πQ 2 = = . =− 0.0 cycles? (Assume ω ′ ≈ ω .70 × 10 Fh H K 5 dt 2 c −3 −6 −4 2 max −4 −6 −4 s.5104s . ( ) ( ( 220 ×10 H ) ( 12.Fundamentals of Physics Halliday & Resnic 東海大學物理系 98 學年度第 2 學期 dU E ωQ 2 = sin 2ωt . G J c. indicating that the energy in the capacitor is indeed increasing at t = T/8. we may write T = 2π/ω as the period and ω = 1/ LC as the angular frequency.00 ×10−3 H ) ( 2.) <解>：Since ω ≈ ω '. dt 2C The greatest rate of change occurs when sin(2ω t) = 1 or 2ω t = π/2 rad. so We note that this is a positive result.
At what time after t = 0 4 does (a) the generator emf first reach a maximum and (b) the current first reach a maximum? (c) The circuit contains a single element other than the generator.0×10 H)(10 ×10 F) (b) The inductive reactance is XL = ω dL = 2πfdL = 2π(650 Hz)(6. the same as we found in part (a). inductance. Therefore. an inductor. (d) What is the value of the capacitance.73 × 10−3 s . or a resistor? Justify your answer. 4ω d 4(350 rad / s) (b) The current is a maximum when sin(ω dt – 3π/4) = 1.Fundamentals of Physics Halliday & Resnic 東海大學物理系 98 學年度第 2 學期 (a) At what frequency would a 6. The current 4 produced in a connected circuit is i (t ) = I sin(ωd t − 3π ) .0 × 10–3 H) = 24 Ω. where I = 620mA . −3 −6 2π 2π LC 2π (6. . where ε m = 30V and ωd = 350rad / s . or resistance. The first time this occurs after t = 0 is when ω dt – 3π/4 = π/2 (as in part (a). or ω dt – 3π/4 = (π/2) ± 2nπ [n = integer]. or ω d = 1/ LC . and the capacitive reactance is given by XC = 1/ω dC. Therefore. The frequency is fd = ωd 1 1 = = = 6.5 ×102 Hz. n = 0). The two reactances are equal if ω dL = 1/ω dC. The capacitive reactance has the same value at this frequency. 5π 5π t= = = 112 × 10−2 s . as the case may be? <解>：(a) The generator emf is a maximum when sin(ω dt – π/4) = 1 or ω dt – π/4 = (π/2) ± 2nπ [n = integer]. <解>：(a) The inductive reactance for angular frequency ω d is given by X L = ωd L . Is it a capacitor. n = 0). The first time this occurs after t = 0 is when ω dt – π/4 = π/2 (that is. 4ω d 4(350 rad / s) 第 6 頁，共 13 頁 .0 mH inductor and a 10 _F capacitor have the same reactance? (b) What would the reactance be? (c) Show that this frequency would be the natural frequency of an oscillating circuit with the same L and C. 3π 3π t= = = 6. (c) The natural frequency for free LC oscillations is f = ω / 2π = 1/ 2π LC . 【Problem 31-33】 π An ac generator has emf ε = ε m sin(ωd t − ) .
Therefore.0 = 2. but at resonance the impedance Z becomes purely resistive (Z = R) so that we can divide the emf amplitude by the current amplitude at resonance to find R: 8. and ε m = 36V . the amplitude of the potential difference across the element must be the same as the amplitude of the generator emf: VL = εm.7 Ω remain unchanged. εm = Iω dL and ε 30. Thus. (d) The current amplitude I is related to the voltage amplitude VL by VL = IXL. so the circuit element must be an inductor. where the vertical axis scale is set by I s = 4 A . (b) φ . 31-7 and set R = 200Ω . 31-4) C = (ω 2L)−1 = [(25000)2 ×200 × 10−6]−1 = 8. where XL is the inductive reactance. which means (using Eq. What are (a) Z . 【Problem 31-37】 Remove the capacitor from the circuit in Fig.0/4.0 A. the impedance is 第 7 頁，共 13 頁 . and the emf amplitude is 8. given by XL = ω dL. Furthermore. while R = 200 Ω and XL = ω L = 2πfdL = 86.0 µF. 31-29.0 V. What are (a) C and (b) R? （圖 31-29） <解>：(a) The graph shows that the resonance angular frequency is 25000 rad/s. . and (c) I ? (d) Draw a phasor diagram.0V L= m = = 0138 H. since there is only one element in the circuit. (b) The graph also shows that the current amplitude at resonance is 4. −3 Iω d (620 × 10 A)(350 rad / s) 【Problem 31-36】 The current amplitude I versus driving angular frequency ωd for a driven RLC circuit is given in Fig.Fundamentals of Physics Halliday & Resnic 東海大學物理系 98 學年度第 2 學期 (c) The current lags the emf by +π / 2 rad. f d = 60 Hz .0 Ω. （圖 31-7） <解>：(a) Now XC = 0. The inductance is 200 µ H . L = 230mH .
(b) The phase angle is.165 A)(200 Ω) ≈ 33V VL = IX L = (0.165 A . and ε m = 30V . 31-7 has R = 5Ω . Z 218 Ω (d) We first find the voltage amplitudes across the circuit elements: VR = IR = (0.3V This is an inductive circuit. X − XC φ = tan −1 L R −1 86. so εm leads I. 31-16? (b) What is this maximum value? At what (c) lower angular frequency ωd 1 and (d) higher angular frequency ωd 2 will the current amplitude be half this maximum value? (e) What is the fractional half-width ωd 1 − ωd 2 of the resonance curve for this circuit? ω （圖 31-7） <解>：(a) For a given amplitude εm of the generator emf.7Ω) ≈ 14.165A)(86. = tan 200 Ω (c) The current amplitude is now found to be ε 36. C = 20 µ F .7 Ω)2 = 218 Ω .4° . The phasor diagram is drawn to scale below. L = 1H . (a) At what angular frequency ωd will the current amplitude have its maximum value. the current amplitude is given by 第 8 頁，共 13 頁 .0 V I= m = = 0.7 Ω − 0 = 23. as in the resonance curves of Fig. 31-65. from Eq. 【Problem 31-45】 An RLC circuit such as that of Fig.Fundamentals of Physics Halliday & Resnic 東海大學物理系 98 學年度第 2 學期 2 Z = R 2 + X L = (200 Ω)2 + (86.
it does so for ω d = 1/ LC = ω .0 ×10−6 F) 3(20.00 A .0 × 10−6 F) .00 Ω)2 + 4(1. Z R 2 + (ω d L −1/ ωd C )2 We find the maximum by setting the derivative with respect to ω d equal to zero: dI 1 = − ( E ) m [ R 2 + (ω d L − 1 / ω d C ) 2 ]−3/ 2 ω d L − dω d ωdC L M N OL + 1 O L .00 Ω) ω2 = = 2 LC 2(1. we obtain ω 2 ( LC ) ± ω d d d3CRi− 1 = 0 . ωd = = 224 rad / s . For this 1 = LC 1 (100 H)(20. (d) The largest positive solution 第 9 頁，共 13 頁 . Using the quadratic formula.00 Ω (c) We want to find the (positive) values of ω d for which I = ε m / 2 R : εm R 2 + (ω d L − 1/ ω d C )2 = εm . ω G ω CJ H K 2 2 d d Taking the square root of both sides (acknowledging the two ± roots) and multiplying by ω d C .00 H)(20.00 H)(20. we find the smallest positive solution − 3CR + 3C 2 R 2 + 4 LC − 3(20. 2R This may be rearranged to yield F L − 1 I = 3R . and the current amplitude is ε 30. P ω CP M Q Q N 2 d The only factor that can equal zero is ω d L − (1 / ω d C ) . (b) When ω d = ω .0 V I= m = = 6.0 × 10−6 F) = 219 rad/s . the impedance is Z = R.Fundamentals of Physics Halliday & Resnic 東海大學物理系 98 學年度第 2 學期 I= εm εm = .0 ×10−6 F)2 (5.0 ×10−6 F) + 2(1. R 5.00 H)(20.0 × 10−6 F)(5.
(a) What is the resonant frequency of the circuit? (Hint: See Problem 47 in Chapter 30. (c) L1 is increased.30 × 10−3 H 4. inductances L1 = 1. ω0 224 rad/s 【Problem 31-51】 In Fig.00 Ω) ω1 = = 2 LC 2(1.) What happens to the resonant frequency if (b) R is increased. (b) The resonant frequency does not depend on R so it will not change as R increases. (e) The fractional width is ω1 − ω 2 228rad/s − 219rad/s = = 0. C2 = 2. −1/2 (d) Since ω ∝ Ceq and Ceq decreases as C3 is removed.5µ F . 31-33.50 × 10−6 F .5µ F . ω will increase.0 × 10−6 F) + 2(1.50 × 10−6 F + 3.00 H)(20. a generator with an adjustable frequency of oscillation is connected to resistance R = 100Ω .00 Ω)2 + 4(1.00 H)(20. it will decrease as L1 increases.040.0 × 10−6 F)2 (5.00 H)(20.00 × 10−6 F + 2. and (d) C3 is removed from the circuit? （圖 31-33） <解>：(a) Since Leq = L1 + L2 and Ceq = C1 + C2 + C3 for the circuit. c h c 1 h = 796 Hz. the resonant frequency is ω= = 1 = 2 π Leq Ceq 2 π L C b+ L g +C +C g b 1 2 1 2 3 1 2 π 170 × 10−3 H + 2.3Ω inductive reactance in series.7 mH and L2 = 2. and capacitances C1 = 4µ F . 【Problem 31-55】 An air conditioner connected to a 120 V rms ac line is equivalent to a 12Ω resistance and a 1.Fundamentals of Physics Halliday & Resnic 東海大學物理系 98 學年度第 2 學期 + 3CR + 3C 2 R 2 + 4 LC + 3(20. (c) Since ω ∝ (L1 + L2)–1/2. and C3 = 3.0 × 10−6 F) = 228 rad/s .0 × 10−6 F) 3(20.3mH .0 × 10−6 F)(5. 第 10 頁，共 13 頁 . Calculate (a) the impedance of the air conditioner and (b) the average rate at which energy is supplied to the appliance.
1 Ω. (b) Since φ < 0. which leads to XC > XL. The current leads the emf. (f) The inductive reactance may be zero. ω t – φ > ω t.0 Ω ) 2 + ( 1. φ = – 42. This means the box must contain a capacitor and a resistor.30 Ω − 0 ) = 12. (c) The phase constant is related to the reactance difference by tan φ = (XL – XC)/R. tan φ would be zero.19 ×103 W. and φ would be zero. 2 (b) The average rate at which energy has been supplied is ε 2 R ( 120 V ) ( 12. so there need not be an inductor.0° and cos φ = cos(– 42.186 ×103 W ≈ 1. we conclude the circuit is not in resonance.900. The box contains an RLC circuit. Measurements outside the box reveal that ε (t ) = (75V )sin ωd t and i (t ) = (1. Therefore. possibly even a multiloop circuit. (e) Since tan φ is negative and finite. neither the capacitive reactance nor the resistance are zero. the impedance is Z= ( 12.0°) = –0. Since φ is not zero. We have tan φ = tan(– 42. 第 11 頁，共 13 頁 .2 A) sin(ωd t + 420 ) (a) What is the power factor? (b) Does the current lead or lag the emf? (c) Is the circuit in the box largely inductive or largely capacitive? (d) Is the circuit in the box in resonance? (e) Must there be a capacitor in the box? (f) An inductor? (g) A resistor? (h) At what average rate is energy delivered to the box by the generator? (i) Why don’t you need to know ωd to answer all these questions? （圖 31-35） <解>：(a) The power factor is cos φ . (g) Yes. (d) If the circuit were in resonance XL would be the same as XC. there is a resistor. 31-61. whose elements and connections we do not know. XL – XC is negative. 2 Z ( 12. a negative number. The circuit in the box is predominantly capacitive.07 Ω ) 2 【Problem 31-57】 Figure 31-35 shows an ac generator connected to a “black box” through a pair of terminals.743.0 Ω ) Pavg = rms2 = = 1.0°) = 0. where φ is the phase constant defined by the expression i = I sin(ω t – φ ).Fundamentals of Physics Halliday & Resnic 東海大學物理系 98 學年度第 2 學期 <解>：(a) Using Eq. Thus.
what are (a) the voltage decrease ∆V along the transmission line and (b) the rate Pd at which energy is dissipated in the line as thermal energy? If Vt = 8kV . L and C then the value of the frequency would also be needed to compute the power factor.4 W. what is the current in the (b) primary and (c) secondary? <解>：(a) The stepped-down voltage is Vs = V p F I = b Vg 10 I = 2. which is given. .30 Ωg19 V . the current in the secondary is I s = We find the primary current from Eq.3Ω / cable . the current in the secondary is I s = 0.4 V = = 0. The transmission line resistance is 0.16A. Therefore.16 A. Rs 15 Ω F I = b Ag10 I = 3. If values were given for R.0 V 120 A 0. 3125 bb . what are (e) ∆V and (f) Pd ? 3 3 .2 × 10 N G J 016 F J H N 500 HK . what is Vs with an open circuit? If the secondary now has a resistive load of 15Ω . <解>：(a) The rms current in the cable is I rms = P / Vt = 250 × 10 W / 80 × 10 V = 3125 A. GK s p −3 A.Fundamentals of Physics Halliday & Resnic 東海大學物理系 98 學年度第 2 學期 (h) The average power is Pavg = 1 1 ε m I cos φ = 75. (a) If V p is 120 V (rms). (c) As shown above. 【Problem 31-63】 A transformer has 500 primary turns and 10 secondary turns. the rms voltage drop is ∆V = I rms c h R = b A gg.4 V. 2 2 b g g g b b (i) The answers above depend on the frequency only through the phase constant φ . 第 12 頁，共 13 頁 . If Vt = 80kV . and the power of the generator is 250 kW. N GK G J 120 F J H N 500 HK s p (b) By Ohm’s law.743 = 33.8kV . what are (c) ∆V and (d) Pd ? If Vt = 0. 2 0 = . 【Problem 31-65】 An ac generator provides emf to a resistive load in a remote factory over a two-cable transmission line. 31-80: I p = Is Vs 2. At the factory a step-down transformer reduces the voltage from its (rms) transmission value Vt to a much lower value that is safe and convenient for use in the factory.
80 ×10 V ) = 312. (b) The rate of energy dissipation is Pd = I rms R = 3125 A 2 0.60 Ω = 5. (f) Pd = ( 312.0 × 10 V = 3125 A .60 Ω ) = 19V.5 A ) ( 0.25A ) ( 0. so ∆V = ( 312. b g0.9 × 10 W.5 A ) ( 0. so ∆V = ( 31. 第 13 頁，共 13 頁 .9 W. b = 2 2 2 3 3 (e) I rms = 250 ×10 W/ ( 0.9 ×104 W.60 Ωg5. b gg g bb c h (d) Pd = 3125 A .60 Ω ) = 5.5 A .60 Ω ) = 1. (c) Now I rms = 250 × 10 W / 8.Fundamentals of Physics Halliday & Resnic 東海大學物理系 98 學年度第 2 學期 2 . 3 3 .9 × 102 V .
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### Advent Calendar 2006
##### Stage: 2 Challenge Level:
NRICH December 2006 advent calendar - a new tangram for each day in the run-up to Christmas.
### Tubular Path
##### Stage: 2 Challenge Level:
Can you make the green spot travel through the tube by moving the yellow spot? Could you draw a tube that both spots would follow?
### Coordinate Cunning
##### Stage: 2 Challenge Level:
A game for 2 people that can be played on line or with pens and paper. Combine your knowledege of coordinates with your skills of strategic thinking.
### Calculator Bingo
##### Stage: 2 Challenge Level:
A game to be played against the computer, or in groups. Pick a 7-digit number. A random digit is generated. What must you subract to remove the digit from your number? the first to zero wins.
### Round Peg Board
##### Stage: 1 and 2 Challenge Level:
A generic circular pegboard resource.
### Sorting Symmetries
##### Stage: 2 Challenge Level:
Find out how we can describe the "symmetries" of this triangle and investigate some combinations of rotating and flipping it.
### World of Tan 14 - Celebrations
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of Little Ming and Little Fung dancing?
### Got It
##### Stage: 2 and 3 Challenge Level:
A game for two people, or play online. Given a target number, say 23, and a range of numbers to choose from, say 1-4, players take it in turns to add to the running total to hit their target.
### Lost
##### Stage: 3 Challenge Level:
Can you locate the lost giraffe? Input coordinates to help you search and find the giraffe in the fewest guesses.
### Times Tables Shifts
##### Stage: 2 Challenge Level:
In this activity, the computer chooses a times table and shifts it. Can you work out the table and the shift each time?
### First Connect Three for Two
##### Stage: 2 and 3 Challenge Level:
First Connect Three game for an adult and child. Use the dice numbers and either addition or subtraction to get three numbers in a straight line.
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PDA
View Full Version : Maths
HannahEloise
03-03-2015, 11:55 AM
Are you good at Maths?
Reminder: Be modest at all times
Solsagann
03-03-2015, 11:57 AM
That's why I mostly studied philosophy, literature and foreign languages in highschool.
GummiBear64
03-03-2015, 11:59 AM
Terrible.
Dude, it took me ages to memorise Pythagoras Theorem and now I have to do something with my calculator about degrees, hours, minutes, seconds.
a^2 + b^2 = c^2
HannahEloise
03-03-2015, 01:19 PM
Terrible.
Dude, it took me ages to memorise Pythagoras Theorem and now I have to do something with my calculator about degrees, hours, minutes, seconds.
a^2 + b^2 = c^2
Really? XDDDD
- - - Updated - - -
That's why I mostly studied philosophy, literature and foreign languages in highschool.
I suck at foreign languages. I'm so-so with Math.
ElectroDream
03-03-2015, 01:31 PM
I'm good at it.
i'd usually get 90-95 (out of 100).
well, i guess that the materials here, in Indonesia, is just too easy.
I'm 13 (8th grade) and we just learnt Phytagoras Theorem last semester. (is that how to spell it? Pyhtagoras? Pythagoras? Waaah!)
HannahEloise
03-03-2015, 01:31 PM
I'm good at it.
i'd usually get 90-95 (out of 100).
well, i guess that the materials here, in Indonesia, is just too easy.
I'm 13 (8th grade) and we just learnt Phytagoras Theorem last semester. (is that how to spell it? Pyhtagoras? Pythagoras? Waaah!)
Pythagorean.
Sozes Isbored
03-03-2015, 01:35 PM
Im okay at math...not interested in it.
ElectroDream
03-03-2015, 01:51 PM
Pythagorean.
I don't remember spelling it that way, haha.
Perhaps it's different... :prophet:
Well what I know for sure: In indo its Phytagoras.
My teacher's question were pretty easy...
Find the length of right angled triangle ABC if A is 8cm and if B is 6cm.
and he puts a figure according to the question.
._.
DumbOldDoor
03-03-2015, 02:49 PM
Love math!
FincyBoy
03-03-2015, 02:57 PM
Are you good at Maths?
Reminder: Be modest at all times
Im just bad with math xD
Sam Jones
03-03-2015, 04:09 PM
Are you good at Maths?
Reminder: Be modest at all times
I'm really bad :crazy: i dunno anything
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# GMLCreating a direction vector from object direction
#### Dr.Spankenstein
##### Member
In order to get a vector from a direction I could do the following:
Code:
``````var dirX = dsin(direction)
var dirY = dcos(direction)``````
The problem is that this relies on
Code:
``````0 = Up
90 = Right
180 = Down
270 = Left``````
and GameMaker uses
Code:
``````0 = Right
90 = Up
180 = Left
270 = Down``````
How can I translate from gamemaker anticlockwise 90 = Up directions to the clockwise 0 = Up directions?
#### chamaeleon
##### Member
In order to get a vector from a direction I could do the following:
Code:
``````var dirX = dsin(direction)
var dirY = dcos(direction)``````
The problem is that this relies on
Code:
``````0 = Up
90 = Right
180 = Down
270 = Left``````
and GameMaker uses
Code:
``````0 = Right
90 = Up
180 = Left
270 = Down``````
How can I translate from gamemaker anticlockwise 90 = Up directions to the clockwise 0 = Up directions?
You could experiment with using the correct x/y and sin/cos combination.
#### SnotWaffle Studios
##### Member
Make it negative, mod 360, add 90?
-0 mod 360 +90 = 90
-90 mod 360 + 90 = 0
-180 mod 360 + 90 = 270
-270 mod 360 + 90 = 180
You might need to do the reverse.
Edit:
This is the reverse.
So taking gamemaker's direction, just do this:
Code:
``````var newDirection = -(direction-90) mod 360;
var dirX = dsin(direction);
var dirY = dcos(direction);``````
#### Dr.Spankenstein
##### Member
Actually, that doesn't appear to handle direction = 180 correctly?
#### chamaeleon
##### Member
You do know cos is for calculating x, and sin for y, right?
#### flyingsaucerinvasion
##### Member
If I understand you, it should just be
x = dsin(angle)
y = -dcos(angle)
results in:
0: 0,-1
90: 1,0
180: 0,1
270: -1,0
#### Dr.Spankenstein
##### Member
You do know cos is for calculating x, and sin for y, right?
Ah. I'm an idiot for getting them the wrong way around! Works as expected now. Thank you, all.
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What is the best way to parse differential equations and boundary conditions to a custom function?
I have written a method to turn systems of linear differential equations into matrix equations (discretisation). This handles boundary conditions using the row replacement method.
At the moment, I am trying to find the best way to build this method into a custom NDSolve type of callable function with both equations and boundary conditions provided as arguments.
For concreteness, imagine I want to solve $$y''(x)=y(x)$$ on the interval $$[-1,1]$$ with conditions $$y(0)=1$$ and $$y(1)=1$$ (notice that the first condition is not at the boundary). If I were to use NDSolve, I would simply input:
NDSolve[{y''[x] == y[x], y[0] == 1, y[1] == 1}, y[x], {x, -1, 1}]
And NDSolve would immediately interpret the first element of the list as the differential equation, and the two others as constraints ("boundary" conditions).
For a system of two equations:
NDSolve[{y''[x] - z[x] == 0, y[0] == 1, y[1] == 1, z''[x] - y[x] == 0,
z[0] == 1, z[1] == 1}, {y[x], z[x]}, {x, -1, 1}]
would work equally well.
My question is thus : how can I reproduce this (smart) behaviour in my home made function ? How does NDSolve handle the parsing of arguments ?
• how can I reproduce this (smart) behaviour in my home made function by parsing the API. These things are not easy. There are two stages to parsing. First you have to make sure the input is valid. Then you have to pick the pieces you want from the input so they can be used internally. First stage is basically verification stage. It checks the input is all valid (right types, no conflicts, etc...). Once this is done, then parser collects all the pieces andsend them in right order to be processed by NDSolve (may be in different structure) So I am not sure what is it you asking to obtain here. Commented May 14, 2020 at 9:25
• ... there was a question about parsing for differential equation not long ago here pattern-to-match-any-differential-of-a-certain-variable What I am saying, is that if you want to make sure the input is all valid, this is not trivial task. As you have to check for many different combinations and possibilities. Commented May 14, 2020 at 9:28
• @Nasser I am interested in both stages but primarily in the first one, as I expect that the information obtained during the verification stage will make the second stage much easier. Commented May 14, 2020 at 9:33
• I wrote a small ode solver, just for 1st order ode's and I think 30% of the code was just in parsing. I think WRI internally uses pre-build templates and other internal specialized functions and not how I did his by pattern matching each possibility. In the C days, we used Lex and Yacc to do all the parsing of input. But not now. Commented May 14, 2020 at 9:36
• btw, your last example actually does not work. it gives error from NDSolve. NDSolve::ndnco: The number of constraints (3) (initial conditions) is not equal to the total differential order of the system plus the number of discrete variables (4). Commented May 14, 2020 at 9:38
Assuming everything is syntactically correct, the function that does what you want is InternalProcessEquationsSeparateEquations:
InternalProcessEquationsSeparateEquations[{y''[x] == y[x],
y[0] == 1, y[1] == 1}, {x}, {y}]
(*
{{}, {y[0] == 1, y[1] == 1}, {}, {y''[x] == y[x]}}
*)
InternalProcessEquationsSeparateEquations[
{y''[x] - z[x] == 0, y[0] == 1, y[1] == 1, z''[x] - y[x] == 0,
z[0] == 1, z[1] == 1}, {x}, {y, z}]
(*
{{}, {y[0] == 1, y[1] == 1, z[0] == 1, z[1] == 1},
{}, {-z[x] + y''[x] == 0, -y[x] + z''[x] == 0}}
*)
It's undocumented, and this appears to be its syntax and return value:
InternalProcessEquationsSeparateEquations[
{ equations },
{ indendent variables },
{ dependent variables }] (* N.B. No args: y not y[x] *)
(*
{ constraints on independent variables,
constraints on dependent variables, (* e.g BCs *)
algebraic equations,
differential equations }
*)
I've used this to write a parser that returns a data structure like @Nasser's. I don't mind sharing the code, but it is darn long, and I don't want to do too much refactoring to narrow its focus on your requirements.
Appendix: Parser Code Dump
The parser parseDiffEq[] is a somewhat pared-down version of the one alluded to above. It works with standard NDSolve input (omitting options):
myDE = parseDiffEq[{y''[x] == y[x], y[0] == 1, y[1] == 1},
y[x], {x, -1, 1}]
(*
<|"de" -> {y''[x] == y[x]},
"dependentVars" -> {y},
"independentVars" -> {x},
"completeDependentVars" -> {{y,y'}},
"bcs" -> {y[0] == 1, y[1] == 1},
"domain" -> {-1., 1.},
"return" -> y[x],
"firstorder" -> {y[1]'[x] == y[0][x], y[0]'[x] == y[1][x]},
"order" -> {{2}},
"type" -> "ODE"|>
*)
I cut some data structure items but left some that are not needed here but might be of interest. The utility linearQ[], which will check if the DE is a linear system, seemed worth including given the OP's goal.
linearQ@myDE
(* True *)
Second example, a system:
my2DE = parseDiffEq[{y''[x] - z[x] == 0, y[0] == 1, y[1] == 1,
z''[x] - y[x] == 0, z[0] == 1, z[1] == 1}, {y[x], z[x]}, {x, -1, 1}]
(*
<|"de" -> {-z[x] + y''[x] == 0, -y[x] + z''[x] == 0},
"dependentVars" -> {y, z},
"independentVars" -> {x},
"completeDependentVars" -> {{y, y'}, {z, z'}},
"bcs" -> {y[0] == 1, y[1] == 1, z[0] == 1, z[1] == 1},
"domain" -> {-1., 1.},
"return" -> {y[x], z[x]},
"firstorder" -> {
-z[0][x] +y[1]'[x] == 0, -y[0][x] + z[1]'[x] == 0,
y[0]'[x] == y[1][x], z[0]'[x] == z[1][x]},
"order" -> {{2}, {2}},
"type" -> "ODE"|>
*)
linearQ@my2DE
(* True *)
Parser and utility code
There are internal, undocumented helper functions used that might be of interest:
InternalProcessEquationsSeparateEquations
InternalProcessEquationsFindDependentVariables
InternalProcessEquationsFirstOrderize
InternalProcessEquationsDifferentialOrder
Since they are undocumented, my ability to explain them is limited. The input to parseDiffEq[] is validated to some extend, but there are some checks I haven't gotten around to writing. The parser might occasionally fail on bad input without indicating why.
$parseKeys = { (* just a way for me to remember the data structure *) "de", (* the diff. eqns. *) "dependentVars", (* the "X" argument *) "independentVars", (* the "Y" argument *) "completeDependentVars", (* including lower-order derivatives *) "bcs", (* boundary/initial conditions *) "domain", (* interval of integration *) "return", (* return expression *) "firstorder",(* first-order equivalent system -- unnecessary *) "order", (* differential orders of the DEs *) "type" (* ODE, PDE,... -- unnecessary *) }; ClearAll[parseDiffEq]; SetAttributes[parseDiffEq, HoldAll]; Options[parseDiffEq] = Thread[$parseKeys -> Automatic];
parseDiffEq::ndnl = NDSolve::ndnl;
parseDiffEq::dsvar = NDSolve::dsvar;
parseDiffEq::ndlim = NDSolve::ndlim;
(*
* Utilities
*)
ClearAll[
parseInterval, (* check indep var spec *)
validVariableQ, (* check whether an expression is a valid var *)
cullArgs, (* returns arguments of vars: y'[2]==0 -> {2} *)
varsToIndexedVars, (* convert Derivative[n][y] to y[n] *)
linearQ]; (* test whether a DE is linear *)
(* converts derivative y^(n) to y[n] *)
(* Used here for constructing the first order system
* and therefore unnecessary. Useful in other use cases
* for replacing derivatives by independent variables. *)
varsToIndexedVars[vars_][expr_] := varsToIndexedVars[expr, vars];
varsToIndexedVars[expr_, vars_] :=
With[{v = Alternatives @@ Flatten@{vars}},
expr /. {Derivative[n_][y : v] :> y[n], y : v :> y[0]}
];
(* taken from somewhere I've lost track of *)
validVariableQ[var_] := ! NumericQ[var] &&
FreeQ[var,
DirectedInfinity | Indeterminate] &&
(MemberQ[{Symbol, Subscript, K, C},
Context[Evaluate@Head[var]] =!= "System") &&
(* cullArgs - cull args of functions ff: {{args f1}, {args f2},..} *)
(* cullArgs[{y[0]==0,y[1]==0,z[0]==1},{y,z}] --> {{{0},{1}},{{0}}} *)
cullArgs[expr_, ff_] := DeleteDuplicates /@ Flatten[
Last@Reap[
Cases[
expr, (f : Alternatives @@ ff)[
args__] | _Derivative[f : Alternatives @@ ff][args__] :>
Sow[{args}, f], Infinity],
ff
],
1];
cullArgs[ff_][expr_] := cullArgs[expr, ff];
(* Checks if data structure de represents a linear equation or system *)
linearQ::usage = "linearQ[de] returns whether de is linear.";
linearQ[de_] := AllTrue[
Lookup[de, "de"],
InternalLinearQ[
#,
Through[Flatten@{Lookup[de, "completeDependentVars"],
(Derivative @@ #2)@# &,
{Lookup[de, "dependentVars"], Lookup[de, "order"]}]} @@
Lookup[de, "independentVars"]]
] &
];
(* breaks down iterator {x,...} to {x, interval} and
* checks that x is a valid variable *)
parseInterval[xx : {x_, a___}] :=
If[! validVariableQ@x,
Message[parseDiffEq::dsvar, x];
Return[$Failed], {x, {a}} ]; parseInterval[x_] := parseInterval@{x}; (*** end of utilities ***) (* * Main function: parses DE, vars, interval into an association * * Part I parses NDSolve style input into a sequence of option rules * Part II construct the data struction Association[] from rules *) (* part I: parse equation and args into parts *) parseDiffEq[eqns_List, yy_, xx_, deOpts : OptionsPattern[]] := Module[{ x, y, endpoints, interval, conind, condep, alg, diff}, x = parseInterval@xx; If[x =!= $$Failed, {x, interval} = x; (* split indep var and interval *) y = yy /. v_[x] :> v; (* strip arguments of dep var *) {conind, condep, alg, diff} = InternalProcessEquationsSeparateEquations[eqns, Flatten@{x}, Flatten@{y}]; (* TBD check validity {conind,condep,alg,diff} *) endpoints = cullArgs[condep, Flatten@{y}]; interval = Flatten[{interval, endpoints}]; If[Length@interval == 0, Message[parseDiffEq::ndlim, xx]; x =$$Failed, If[! VectorQ[interval, NumericQ], Message[parseDiffEq::ndnl, First@Cases[interval, x0_?(! NumericQ[#] &)], interval]; x = $$Failed, interval = MinMax@N@interval (* N[] optional; use WorkingPrecision? *) ] ] ]; parseDiffEq[ "de" -> diff, "bcs" -> (condep /. Automatic -> {}), "independentVars" -> Flatten@{x}, "dependentVars" -> Flatten@{y}, "return" -> yy, "domain" -> interval, deOpts] /; FreeQ[x,$$Failed] ]; (* part II: check and process parts given as option rules *) parseDiffEq[opts : OptionsPattern[]] := Module[{asc, alldvars, firstordersys, foRules}, (* TBD: validate option values ??? *) (** set up association from options **) asc = <|Thread[$$parseKeys -> OptionValue@$$parseKeys]|>; (** parses indep var from eqns; NDSolve does not do this -- unnecessary **) If[asc@"independentVars" === Automatic, asc@"independentVars" = DeleteDuplicates@ Cases[Flatten@{asc@"de"}, _[x__Symbol] | Derivative[__][_][x__Symbol] :> x, Infinity] ]; (** check type of DE -- unnecessary **) asc@"type" = Switch[Length@asc@"independentVars" , 0, "Algebraic" (* unsupported *) , 1, "ODE" , n_Integer /; n > 1, "PDE" (* unsupported *) , _, $$Failed]; (** parse dependend variables from equations -- unnecesary **) If[asc@"dependentVars" === Automatic , asc@"dependentVars" = InternalProcessEquationsFindDependentVariables[ Flatten@{asc@"de"}, asc@"independentVars"] ]; (** construct first-order equivalent system -- unnecessary **) firstordersys = InternalProcessEquationsFirstOrderize[#1, #2, 1, #3] & @@ Lookup[asc, {"de", "independentVars", "dependentVars"}]; alldvars = firstordersys[[3]] /. firstordersys[[4]]; If[VectorQ[alldvars], alldvars = List /@ alldvars]; asc@"completeDependentVars" = alldvars; foRules = MapAt[ (* replaces NDSolvey$$nnn$1 by y[1] etc *)
varsToIndexedVars[Lookup[asc, "dependentVars"]],
Flatten@{firstordersys[[4]], # -> # & /@
Lookup[asc, "dependentVars"]},
{All, 2}];
asc@"firstorder" =
Join[firstordersys[[1]], firstordersys[[2]]] /. foRules;
(** store differential order -- unnecessary **)
asc@"order" =
InternalProcessEquationsDifferentialOrder @@
Lookup[asc, {"de", "independentVars", "dependentVars"}];
asc
];
• Thanks for showing these. I did not know about these internal functions. Yes if you could share your long parsing code, that will be good. You do not have to do any customizing for it for this question. I am sure we all will learn from it. Currently for my small ODE solver, I parse things using patterns, but it is a pain and I am always not sure if I missed one case or not and it is only now for just 1 ODE and first order only. Commented May 14, 2020 at 17:32
• @Nasser Updated. NDSolveProcessEquations can be used to parse an IVP. However for a BVP, it will invoke the shooting method to convert it to an IVP, which can be an expensive and problematic process. Then you get all the syntax checking done for you. But it does not seem a good solution for the OP. Commented May 14, 2020 at 19:56
• great thanks I will study your code. Very useful. Commented May 14, 2020 at 21:34
• Many thanks. This was exactly the sort of things that I was looking for ! Commented May 15, 2020 at 8:38
• @Nasser I suppose there's no internal function for independent variables because the user is required to specify them. (Other seemingly independent variables are taken to be parameters.) IntegrategetAllVariables[expr, {}] does not quite do what you want, I think. (The second argument is used to exclude user-specified independent variables, I think.) I've used something like DeleteDuplicates@ Cases[eqn, _[x__Symbol] | Derivative[__][_][x__Symbol] :> x, Infinity], but it's not robust. Compare eqn = y[x, t] - q and eqn = y[x, t] - t q and eqn = y[x, t] - 2 t q.z Commented Sep 27, 2021 at 18:45
I'll just give an idea to make it easier to do this. Which is not to use the same API as NDSolve, as that requires much much more work to parse it.
Instead, have the caller pass the input in Association.
Yes, this might be a little bit more work for the user, but not much. On the other hand, this greatly simplifies the parsing and checking inside your ndsolve, because now all the entries can be accessed directly by field names from the association instead of using pattern search.
This is actually how number of other software do it. The user fills in a "record" or a "struct" in C talk, and passes this struct to the function to process.
The function then just reads the values directly from the record by name.
There is a quick prototype. This will work for many number of odes.
You build one association for each ode
ClearAll[y, x, z, ode1, ode2];
ode1 = <|"depVar" -> y,
"indepVar" -> x,
"ode" -> y''[x] - z[x] == 0,
"ic" -> {y[0] == 1, y[1] == 1}|>;
ode2 = <|"depVar" -> z,
"indepVar" -> x,
"ode" -> z''[x] - y[x] == 0,
"ic" -> {z[0] == 1, z[1] == 1}|>;
domain = {{x, -1, 1}};
setOfODES = {ode1, ode2};
ndsolve[setOfODES, domain]
And this is ndsolve
ndsolve[odes_List, domain_List] := Module[{n = Length@odes, m, currentODE},
Print["You entered ", n, " odes"];
Do[
currentODE = odes[[m]];
Print["\nODE ", m, " is ", currentODE["ode"],
"\nthe dependent variable is ", currentODE["depVar"],
"\nthe independent variable is ", currentODE["indepVar"]
]
, {m, 1, n}
];
(*example how to read fields from association*)
If[n > 1,
If[ Length@Union["indepVar" /. odes] > 1,
Return["Error, independent variable must be the same", Module]
]
];
(*many many more additional checks and balances*)
(*check domain is valid*)
(*check initial conditions are valid and using same symbols,etc...*)
Print["Parsed OK"]
(*now you can go on and actually numerically solve them. But the hard work*)
(*has been done above, which is parsing, the rest is easy :) *)
]
And it gives this output
You entered 2 odes
ODE 1 is -z[x]+y''[x]==0
the dependent variable is y
the independent variable is x
ODE 2 is -y[x]+z''[x]==0
the dependent variable is z
the independent variable is x
Parsed OK
The above is just the start. But the main point, it is much easier now to handle, since you do not have to do too much parsing, compared to the way NDSolve takes its input as lists, where you'd have to parse the content of each list, pick which part is which, and so on. This is at the cost, the caller has to set up an association for each ODE. But I think it is not a big deal to do.
• Thank you @Nasser, associations look very useful in this context. Commented May 15, 2020 at 8:38
Here's a simpler way (simpler than my first answer) that I came up with today exploring a problem with DSolve. It calls DSolveValue and intercepts the DSolve parser and returns an association with the equations broken down by type, before the system is solved:
parseODE@NDSolve[{y''[x] == y[x], y[0] == 1, y[1] == 1}, y[x], {x, -1, 1}]
(*
<|"OtherEquations" -> {}, (* nonempty => error (probably) *)
"BoundaryConditions" -> {y[0] == 1, y[1] == 1},
"Algebraic" -> {}, (* algebraic equations in terms of y and x *)
"Differential" -> {y''[x] == y[x]},
"Dependent" -> {y},
"Independent" -> {x},
"Region" -> {x, -1, 1}|> (* see the PDE example below *)
*)
Code for function:
ClearAll[parseODE];
SetAttributes[parseODE, HoldFirst];
$$dsolvers = DSolve | DSolveValue | NDSolve | NDSolveValue | ParametricNDSolve | ParametricNDSolveValue; parseODE[ _?(MatchQ[$$dsolvers])[
eqns_, v_, t : Longest[{_, _?NumericQ, _?NumericQ} ..] | _, ___]
] := parseODE[eqns, v, t];
parseODE[eqns_, v_, t__] :=
Block[{DSolveDSolveParser =
Function[{eqs, dependent, independent, stuff},
Return[
With[{independents =
Flatten@{independent /.
{{x_, _?NumericQ, _?
NumericQ} :> x, vv_ \[Element] _ :> vv}
}},
Join[
{"Other", "Initial", "Algebraic", "Differential"} ->
InternalProcessEquationsSeparateEquations[
Flatten@eqs, independents, dependent]],
<|"Dependent" -> dependent,
"Independent" -> independents,
"Region" -> independent|>
]],
Block]
]},
DSolveValue[eqns, v, t]
]
More examples. Note that the domain {x, 0, 1}, {t, 0, 1} for the PDE in the first example is rewritten by DSolveValue into an ImplicitRegion. The others show variation in input type (x instead of {x, 0, 1}, a system instead of a single ODE).
weqn = D[u[x, t], {t, 2}] == D[u[x, t], {x, 2}];
ic = {u[x, 0] == E^(-x^2), Derivative[0, 1][u][x, 0] == 1};
parseODE@DSolveValue[{weqn, ic}, u[x, t], {x, 0, 1}, {t, 0, 1}]
(*
<|"OtherEquations" -> {},
"BoundaryConditions" -> {{u[x, 0] == E^-x^2, Derivative[0, 1][u][x, 0] == 1}},
"Algebraic" -> {},
"Differential" -> {Derivative[0, 2][u][x, t] == Derivative[2, 0][u][x, t]},
"Dependent" -> {u},
"Independent" -> {x, t},
"Region" -> {{x, t} \[Element]
ImplicitRegion[0 <= x <= 1 && 0 <= t <= 1, {x, t}]}|>
*)
parseODE@DSolve[{y''[x] == y[x], y[0] == 1, y[1] == 1}, y[x], x]
(*
<|"OtherEquations" -> {},
"BoundaryConditions" -> {y[0] == 1, y[1] == 1}, "Algebraic" -> {},
"Differential" -> {y''[x] == y[x]},
"Dependent" -> {y}, "Independent" -> {x}, "Region" -> {x}|>
*)
parseODE@NDSolveValue[{a'[t] == 1, y'[t] == 1, a[0] == 0,
y[0] == 0}, {a[t], y[t]}, {t, 0, 1}]
(*
<|"OtherEquations" -> {},
"BoundaryConditions" -> {a[0] == 0, y[0] == 0}, "Algebraic" -> {},
"Differential" -> {Derivative[1][a][t] == 1,
Derivative[1][y][t] == 1}, "Dependent" -> {a, y},
"Independent" -> {t}, "Region" -> {t, 0, 1}|>
*)
If the differential order(s) of the variables would be useful, one could add a line to the association:
"Order" -> InternalProcessEquationsDifferentialOrder[
Flatten@eqs, independents, dependent]
The best available off-the-shelf solution for this task shall be:
WolframAlpha["y'[x]\[Equal]x^3", IncludePods -> "ODEClassification",
AppearanceElements -> {"Pods"}]
__
This can be driven further:
{WolframAlpha["y'[x]\[Equal]x^3", {{"ODENames", 1}, "Content"}],
WolframAlpha["y'[x]\[Equal]x^3", {{"ODENames", 2}, "Content"}]}
WolframAlpha["y'[x]\[Equal]x^3", IncludePods -> "ODENames",
AppearanceElements -> {"Pods"}]
It is somehow a harsh reduction to life with names because this is a type of ODE and applicable to a lot of equation. Names would have been nice for the typing
G[y',x]==0
with
G=y'-x^3
Or both functions in y' and x^3 are polynomials of positive integer order.
The step-by-step solution in Mathematica is somewhat too short.
x^3dx
integrates in general to (this might need the Integrate built-in)
x^4/4+constant
dy
integrates in general to (this might need the Integrate built-in)
y
So in the equation
y[x_]:=x^4/4+constant
Since exact and separable are the more important typings, attribute of the given ODE they have to be favoured over
first-order linear ordinary differential equation
Exact implies there is a closed solution that is critical case is based on inverse functions. That does not follow from this type classification.
There is a need to discuss the symmetry of the two function. In this case both odd, so the resulting solution functions will be even.
If a comparison is initiated with DETools from Maple there are many question open:
info = WolframAlpha[
"(x^2+x Exp[x])D[y[x],x,x,x,x]+(4 Exp[x]+6 x+ 3 Exp[x] \
x)D[y[x],x,x,x]+(6 + 3 Exp[x] x + 9 Exp[x])D[y[x],x,x]+(Exp[x] x+ 6 \
Exp[x])D[y[x],x]+Exp[x]y[x]\[Equal]0", "PodInformation"];
ids = Rest[DeleteDuplicates[#[[1, 1, 1]] & /@ info]];
titles = Map[{{#, 0}, "Title"} &, ids] /. info;
contents =
Column[Cases[info, _[{{#, _}, "Content"}, val_] :> val]] & /@ ids;
This is following odeadvisor a fully exact ordinary differential equation, linear and higher-order. Mathematica avoids the pod with the ODE names. There are more differences.
Using
WolframAlpha["y''[x] == y[x],y[0] == 1, y[1] == 1"]
is simply the best. The advantages are superior to everthing else on the market.
It name the autonomous equation. It classifies the ODE. It solve the differential equation. You get option to calculate an approximate form and get a step by step solution. Depending on the importance of the ODE there is more information. Even a plot and some variation on solutions parameters can be presented. This is traditional mathematica notation and therefore more common. The solution of the parsing address much more audiance than Mathematica code. Using the pods as I introduced some of them does much more than all the other answerers to the question do. The knowledge is much more rich and versatile.
It depends on the interpretation of parsing. My answer follows the ideation and concepts of Wolfram Alpha complete. Not other solution does this. And the other answer are not parsing in the sense of using knowledge database. They use information otherwise hidden by the built-ins because they are tautologies derive solely from the input by the user.
Make use for example of the definition from Parsing. With my answer the collection of answers to the question start to match what parsing is. This is the step from simple computational language to knowledge databased natural language parsing. The step from 4GL to 4GL knowledge based natural parsing.
The point can be that the other use backtracking of the input, while my solution is real parsing of tokens in natural language mapped on a rich knowledge database.
My solution offers access to the step-by-step-solution of Wolfram Alpha while all the others only envolve the Notebook-In-Out-schema.
WolframAlpha["y''[x] - z[x] == 0, z''[x] - y[x] == 0"]
WolframAlpha give the ODE classification system of ordinary differential equation. Indeed this is second order and linear too. This classification is interpreted the rest is automated knowledge retrieval from the knowledge database about differential equations.
Since none of the is probably new a solution is always a reproduction of some to human mankind known problem of ordinary differential equations. The reproduction of solutions may either be done by algorithms rearranging in the input or graph based search algorithms.
The internal built-ins
InternalProcessEquationsSeparateEquations
InternalProcessEquationsFindDependentVariables
InternalProcessEquationsFirstOrderize
InternalProcessEquationsDifferentialOrder
are examples of such categories of graph keys to organize knowledge in need a priori for a efficient and correct solution. They are introduced earlier but reorganized in the knowledge methodology with the introduction into Mathematica. Both are not only compatible they rely on the very same methodologies internally.
It should be by the degree of maturity to Mathematica that Association are not very fast but most optimal for knowledge representation. Both represenation can be transformed into each without loss of information.
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0
TECHNICAL PAPERS
# Using Series-Series Iwan-Type Models for Understanding Joint Dynamics
[+] Author and Article Information
D. Dane Quinn
Department of Mechanical Engineering, The University of Akron, Akron, OH 44325-3903quinn@uakron.edu
Daniel J. Segalman1
Sandia National Laboratories, P. O. Box 5800, MS 0847, Albuquerque, NM 87185-0847djsegal@sandia.gov
1
Sandia is a multi-program laboratory operated by Sandia Corporation, a Lockheed Martin Company, for the United States Department of Energy under Contract No. DE-AC04-94AL85000.
J. Appl. Mech 72(5), 666-673 (Aug 10, 2004) (8 pages) doi:10.1115/1.1978918 History: Received January 03, 2003; Revised August 10, 2004
View Large
## Figures
Figure 1
Iwan elements
Figure 2
Physical system. The rod slips over the interval 0⩽x̃<ℓ̃(t̃).
Figure 3
Force-displacement curve. The dashed curve represents the force-displacement curve generated from loading into undeformed material [Eq. 28]. In each panel the loading amplitude is 0.30.
Figure 4
Discrete model
Figure 5
Interfacial behavior with n=64(ω=0.25,α=0.25). The displacements have been marked according to the slip velocity—for the lightest points ∣u̇i∣>v0 and for the darker points v0>∣u̇i∣>v02∕n. The velocity of the darkest points is v02∕n>∣u̇i∣, with v0=2ω(Δt)=7.8125×10−6. The slip velocity is in the same direction as that of the end of the interface.
Figure 6
Power dissipated over one steady-state cycle, t=ωτ(ω=0.25,α=0.25)
Figure 7
Frictional dissipation per unit cycle as n varies (ω=0.25,α=0.25)
Figure 8
Frictional work per unit cycle (n=256). The quantity m, represents the slope of this curve, as determined from linear regression
## Errata
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## Trying to find math inside everything else
### Steepest Stairs and Wacky Measurements
After reading Dan Meyer’s post mentioning a Steepest/Shallowest stairs contest, I decided to go for it. But Dan had them do it for homework after they knew what slope was. I decided that I thought steepness of stairs would be a great way to introduce the concept, and then we can have the contest after. So Ms. Barnett (my co-teacher) and I went around the area and took lots of pictures of stairs we can find. Then I put them up as a warm-up and asked them which were steepest and which were shallowest.
In every class, there was near-universal agreement on which stairs were the shallowest (the top-right), but lots of different votes for the steepest. So then I asked them, “How can you know? What does it mean to be steep?” I got a lot of good, intuitive answers from that (My favorite was that something is steeper when it is closer to being vertical). I asked them what they needed to know to find out which was steeper, and they said we should measure it.
But what exactly should we measure? That took a little cajoling and probing, until we eventually decided on the height of the step and how deep it was. So I gave it to them:
Alright, now we have these measurements, what can we do with them? I lead them on a discussion on how best to use these numbers (a ratio), and we looked at another example. This is a pretty clear example (1/2), but not all of them are. So we used our estimating skills.
And my personal favorite…
(They really asked if I had 11 cell phones. I guess my Photoshop skills are better than I thought.)
The best part of these pictures is that they so naturally prompted them to question the units of measurement. “That one is cell phones, but the other one is hands. How can we compare them?” And so it’s natural to talk about slope as a ratio with no units. I didn’t have to artificially insert it. I even had a picture of a curved slide at the end, so we could theorize about the steepness of that.
Finally at the end I mentioned the contest. Unfortunately, I’m afraid Ms. Barnett and I did too well finding stairs. I’ve had students say they’ve been looking, and some say they found some (but don’t have pictures yet, though they have one more week). I hope someone can knock us off our thrones:
## SLIDES:
Steep Stairs (PDF)
Steep Stairs (PPT)
### No Better Feeling
At the beginning of yesterday’s lesson, I threw up this monster of a problem:
I told my students that, by the end of the lesson, they would be able to solve it. They flipped and freaked out. “No way, Mr. Cleveland, not going to happen.”
In all 4 sections, 1 hour later, every student correctly solved the problem. And they were all so proud of themselves for doing so. There’s no better feeling than that.
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Unit: Chemical Quantities
1 / 8
# Unit: Chemical Quantities - PowerPoint PPT Presentation
Day 4 - Notes. Unit: Chemical Quantities. Volume Conversions with the Mole. After today, you will be able to…. Calculate the volume of a gas at STP given the number of moles and vice versa Perform two step volume conversions (Ex: atoms to liters). Mole Volume Conversions.
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Presentation Transcript
Day 4 - Notes
### Unit: Chemical Quantities
Volume Conversions with the Mole
### After today, you will be able to…
Calculate the volume of a gas at STP given the number of moles and vice versa
Perform two step volume conversions (Ex: atoms to liters)
Mole Volume Conversions
For one mole of any gas at STP, the volume occupied will be 22.4 liters.
STP = standard temperature and pressure
= 0 degrees Celsius and 1 atmosphere
1 mole gas = 22.4 L gas
Example: How many moles of Helium gas will occupy a volume of 149.3L at STP?
K: 149.3L He
U: ? mol He
149.3 L He
1 mol He .
x
6.665 mol He
=
22.4 L He
1
2-Step Conversions with Volume, Mass, and Particles
How to convert moles into particles and vice versa by using the conversion factor…
1 mole = 6.02x1023 atoms, molec. or F.U.
How to convert moles into grams and vice versa by using the conversion factor…
1 mole = atomic mass in grams
How to convert moles into liters and vice versa by using the conversion factor…
1 mole = 22.4 L
# of liters
1 mole = 22.4 Liters
1 mole = 6.02 x 1023 particles
MOLE
# of particles
(atoms, molecules or formula units)
1 mole = (Molar Mass) grams
# of grams
K: 4.32x1023 atoms Kr
U: ? L Kr
4.32x1023 atoms Kr
1 mol Kr .
22.4 L Kr .
x
x
=
6.02x1023 atoms Kr
1 mol Kr.
1
16.1 L Kr
HOMEWORK:
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The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A197605 Floor( ( n + 1/n )^6 ). 2
64, 244, 1371, 5892, 19770, 54992, 132810, 287700, 572042, 1061520, 1861242, 3112580, 5000730, 7762992, 11697770, 17174292, 24643050, 34646960, 47833242, 64966020, 86939642, 114792720, 149722890, 193102292, 246493770, 311667792, 390620090, 485590020 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 LINKS Vincenzo Librandi, Table of n, a(n) for n = 1..10000 Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-7,1). FORMULA For n > 3, a(n) = n^6 + 6n^4 + 15n^2 + 20. [Charles R Greathouse IV, Dec 27 2011] G.f.: x*(64-204*x+1007*x^2-821*x^3+1017*x^4-455*x^5+125*x^6-15*x^7+3*x^8-x^9)/(1-x)^7. - Vincenzo Librandi, Dec 18 2014 MATHEMATICA Table[Floor[(n + 1/n)^6], {n, 40}] (* T. D. Noe, Dec 27 2011 *) CoefficientList[Series[(64 - 204 x + 1007 x^2 - 821 x^3 + 1017 x^4 - 455 x^5 + 125 x^6 - 15 x^7 + 3 x^8 - x^9) / (1 - x)^7, {x, 0, 40}], x] (* Vincenzo Librandi, Dec 18 2014 *) PROG (Magma) [Floor((n+1/n)^6): n in [1..40]] (PARI) a(n)=if(n>3, n^6+6*n^4+15*n^2+20, [64, 244, 1371][n]) \\ Charles R Greathouse IV, Dec 27 2011 CROSSREFS Cf. A014052, A197602, A197603, A197604. Sequence in context: A223332 A340696 A333428 * A198071 A218902 A197905 Adjacent sequences: A197602 A197603 A197604 * A197606 A197607 A197608 KEYWORD nonn,easy AUTHOR Vincenzo Librandi, Oct 18 2011 STATUS approved
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Last modified April 14 18:28 EDT 2024. Contains 371667 sequences. (Running on oeis4.)
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Differential equiations
1. Sep 5, 2009
intenzxboi
the general solution for ye^x dy/dx = [e^-y ] + [ e^ -2x-y ]is:
so i tried to see if this was a homogenous equation but it is not.
next i tried to simplify and got:
y e^y dy= (1+e^-2x) dx / e^x
2. Sep 5, 2009
CFDFEAGURU
Could you please re-post this using the homework layout? Also, you are missing the equation that describes the general solution.
Thanks
Matt
3. Sep 6, 2009
HallsofIvy
Staff Emeritus
It is very difficult to understand what you mean. Do you mean that you are asked to find the general solution? And is "e^-2x- y" supposed to be "e^(-2x-y)" or "e^(-2x) - y"? Guessing at what you mean:
Once you have
$$y e^y dy= \frac{1+ e^{-2x}}{e^x} dx$$
just integrate both sides.
Use integration by parts, letting u= y and $dv= e^ydy$ on the left.
You can rewrite the right side as e-x+ e-3x which should be easy to integrate.
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# Determine offset points of polygon
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I'm wondering what the best way of going about this would be: I have a path, a series of x,y coords defining a polygon. I want to turn this path into 'walls' offset from this path so that I can move a sprite within but enclosed by these 'walls'. Here is an Image, the redline being the original polygon and the white line the 'wall' Any ideas on how this might work?
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Do you want to actually construct the walls, or do you just want the sprite to not leave the track more than the distance where virtual walls would be?
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It may be easier to think of as having the sprite constrained to the path rather than having 'walls'. I'd still like to draw the walls though also it may be diffucult to control the sprite at intersections.
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I don't know whether the following is the most efficient method of constructing the walls. The main problem seems me that it is possible that more than 2 lines may emanate from a vertex. Here we go ... err, I handle the problem in 2 dimensions, just for the case you wonder.
Iterate over all vertices of the path. Make the vertex to the local origin and compute all line segments with this origin. Compute the angle that each segment makes with e.g. the positive x axis. Then sort the lines accordingly to this angle, so that you have the line segments in circular order around the vertex.
Now iterate the line segments, e.g. starting with those with the smallest angle and advancing to the one with the next greater angle. Look at the segments so that they start at the local origin and end at the distant point, i.e. as a vector emanating from the origin. Then the unknown edge to the left of the current segment and the unknown edge to the right of the following segment hit at a point that denotes one of the new vertices.
The direction of the vector to that vertex is obviously given by half the angle between the two segments. One way to compute that direction is to use the 2 angles of the belonging segments as follows:
β := ( αn + αn+1 ) / 2
d := [ cos( β ), sin( β ) ]T
where αn denotes the angle of segment n.
(Please notice that n+1 has to wrap around to 0 for the last segment.)
Another way would be to use the perp vectors. If
vn := [ xn, yn ]T
denotes the distant vertex of segment n, then
ln := [ -yn, xn ]T / |vn|
is the normalized perp vector to the left, and
rn := [ yn, -xn ]T / |vn|
is the normalized perp vector to the right. The said direction vector would then be
d := ( ln + rn+1 ) / |ln + rn+1|
Now that the direction is known, the missing part is the distance of the new vertex. What we know is that the wall should be W units away from the path segments. Then the length of the vector would be
|v'| := 1 / sin( ( αn+1 - αn ) / 2 )
You'll have to handle the wrap around at 0° accordingly; I left this to you ;)
Then the vector to the new vertex (still in the local space!) would be
v' := d * |v'|
and undoing the translation by the current vertex gives you the actual new vertex position.
Another way, totally ignoring trigonometry but using vector math, would be to demand that the projection of the vector to the new vertex onto the both perp vectors should be W:
EDIT: Hmmm, no longer sure whether this works as wanted, so it is removed for now! /EDIT
What is left is to concatenate the new vertices in the correct way. You can do so because the topology of the path can be followed by using the ordering computed at the beginning of the algorithm. So the sequence of new vertices at the inner side of each path segment loop can be found and used to build edges for the walls.
[Edited by - haegarr on March 8, 2010 2:03:05 PM]
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Thanks for the help. I'm a bit new at this - could you tell me what the superscript T stands for?
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1. Maybe consider computing the Minkowski sum of your path and some small polygon which more-or-less approximates a disk? There are efficient algorithms for computing Minkowski sums of polygons...
2. Alternatively, it may be easier to constrain your sprite to be within a certain distance of the path, as haegarr said, and then solve the drawing problem separately. For instance, you might be able to just draw overlapping capsules around your line segments and let the stencil or z-buffer sort them out to make outlines...
E.g., off the top of my head (you might be able to do this more efficiently):
1. Clear stencil buffer to zero; enable stencil buffer writes but disable stencil tests.
2a. Set stencil draw mode to "one, always."
2b. Draw large capsules around line segments, of radius R.
3a. Set stencil draw mode to "zero,always."
3b. Draw small capsules around line segments, of radius r.
4a. Disable stencil writes and enable stencil tests; set the test to only draw where the buffer is 1.
4b. Draw large capsules (or just quads large enough to bound the capsules) of radius R+epsilon around your segments, in the color that you want your walls to be drawn.
Now you will have outlines around your segments at a distance (R+r)/2 and with thickness R-r.
[EDIT: Actually, depth sprites may be the better way to do this: Render distance fields into your depth buffer; then draw a quad with depth-testing, only drawing those pixels at a depth equal to the desired radius plus-or-minus half the desired outline width... (My inspiration here is the old method for drawing Voronoi diagrams using graphics hardware by rendering intersecting cones... here it is... but I'm noting that you don't actually need to use polygonal representations of cones; you can instead use depth sprites or even pixel shaders (to analytically compute distance for each pixel).)]
[Edited by - Emergent on March 27, 2010 10:43:33 AM]
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This can be a complicated problem to solve, if you want to explicitly calculate the inflated/offset geometry, and you want it to work in general (i.e not just for "well behaved" nice input).
For instance, your example image doesn't have any really acute corners. At acute corners, you'll need to bevel or round the inflated geometry, because otherwise the inflated surface shoots off to infinity. Or for example imagine if your input path was made of segments which were much shorter/smaller than the inflation radius.. this can also cause problems.
http://fcacciola.50webs.com/Offseting%20Methods.htm
http://www.gamedev.net/community/forums/topic.asp?topic_id=552378&PageSize=25&WhichPage=2
http://www.gamedev.net/community/forums/topic.asp?topic_id=562159
As mthicke pointed out, if you can avoid requiring the actual offset geometry explicitly, that would make life easier. But less interesting :)
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# Math in Art: The Golden Ratio
Golden Ratio
Composition is an important aspect of any art piece–how the subjects within a painting or photograph are arranged is just as important as the colors used to represent them. Apart from being a useful concept to keep in mind to help balance a painting, composition can often inform the viewer of a picture about the importance of the subjects within (Think of Leonardo da Vinci’s iconic Last Supper–where is Jesus Christ in the painting?) One important tool to help with the composition of a painting is the golden ratio.
Given two quantities a and b, the golden ratio is when the ratio of the sum of the quantities to the larger quantity (a) is equal to the ratio of the larger quantity to the smaller one (b). So, a: a+b= a: b. When this ratio is achieved, in painting or in architecture, it is more aesthetically pleasing as well as more balanced. Artists and architects use this ratio in the form of the golden rectangle, where the length of the sides are in a golden ratio, as well as the golden triangle, where the length of the legs of the triangles exist in a golden ratio as well.
The Golden Triangle in “The Mona Lisa”
The Golden Rectangle in “The Mona Lisa"
The golden ratio is also seen photography. The infamous rule-of-thirds (where it is desirable to line up subjects at the 1/3 and 2/3 intersections) is a product of the intersection of several golden triangles, as shown in this graphic.
Rule of Thirds
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decorators as a special case of an @ operator?
Dan Christensen jdc at uwo.ca
Mon Aug 9 16:32:21 CEST 2004
```I wonder what people think of the following crazy idea, which has two
parts.
1) Allow anonymous multi-line functions. For definiteness, I'll use
the following syntax, but lambda or something else could be used
too:
f = def (a,b,c):
d = a*b
return d + c
2) Define a *binary* operator @ which is just a shorthand for
function application: f @ x = f(x)
Note that f(x) is sometimes pronounced "f at x", so @ is a
reasonable symbol to use. But it could also be something else.
This @ is not associative; make the rule that it associates
from right to left, so g @ f @ x = g(f(x))
Presto, you have decorators:
f = decorator1 @
decorator2 @
def (a,b,c):
d = a*b
return d + c
And the function name is at the top, like some people prefer.
Notes:
- Multi-line anonymous functions are something that have been
- With 1) alone, you already can do:
f = decorator1(
decorator2(
def (a,b,c):
d = a*b
return d + c
))
The sole purpose of @ is to avoid all the closing parens.
- To avoid trailing backslashes, the parser would have to
automatically continue to the next line when a line ends in @.
- Since g @ f @ x = g(f(x)), @ is a bit like function composition,
usually written as a small circle, again suggesting that @ is a
reasonable symbol. (But note that g @ f = g(f) which is not the
same as g composed with f...)
- This @ could be useful in other contexts to avoid deeply
nested parentheses.
- If x is a tuple, f @ *x could mean f(*x), which allows
@ to be used with functions of several arguments. (Not
very relevant here.)
Dan
```
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x86 Assembly Lesson 1 Chapter 3
# Arithmetic Instructions
## Using Them for Simple Calculations
### Welcome
Hi! Welcome to the third chapter of this series. Now, we'll try to learn a bit more about doing math in assembly. I hope that you really grasped the idea discussed in the previous chapter. The arithmatic operations discussed here is done in integers. No reals yet.
Doing math in assembly will likely change the flag. To refresh our mind: Flags are just the way to tag something after the processor executes an operation: whether the last operation holds zero value, negatives, has a carry or borrow (for addition and subtractions), and so on. If you'd like to review, please click here.
Additions and subtractions are straightforward. It takes the following formats:
``` add x, y ; --> means: x = x + y
sub x, y ; --> means: x = x - y
```
Additions and substractions can be done on any registers except the segment registers. Just like the mov command, you can have one of them as memory locations. I encourage you to experiment on how to use this command. What is allowable and what is not. The following is legal in assembly (assuming the variables are already defined):
``` add ax, 5 ; --> means: ax = ax + 5
add bx, cx ; --> means: bx = bx + cx
add [n], ax ; --> means: [n] = [n] + ax
add cx, [n] ; --> means: cx = cx + [n]
sub di, di ; --> means: di = di - di (in other words: di = 0)
sub ax, si ; --> means: ...
sub ax, 4 ; --> means: ...
```
As I pointed out in the previous chapter, for those of you that has 80286 processor or faster may actually add or subtract variables with constants. But don't forget to add the word ptr or dword ptr as appropriate. For example:
``` add [word ptr i], 10
```
Ah... that's simple! However, remember that the operation depends on the registers you are using. For example, if you say add al, cl, you are doing an 8-bit addition. So, the processor actually "expects" (well... sort of) the result to be bounded within 8-bit range (i.e. 0 to 255). Similarly, for 16-bit addition, it should be within 0 to 65535. For substraction, the processor also "expects" it to be non-negative (i.e. no borrows). "Oh? So we have such limitation?" you asked. Well, we actually have a "work around" to do that.
If the result of an addition overflows, the carry flag is set to 1, otherwise it is 0. By detecting the carry flag after doing the addition, we'll know whether the last addition overflows or not. Similarly, if the result of subtraction requires a borrow, then ... (guess what) the carry flag is also set to 1, otherwise it is 0. Wait.... is this a typo? No... The internal circuitry to store carry or borrow is the same. How can this be? Well, this will involve a deeper understanding about computer logic, which is out of scope of this chapter. Don't bother... :-) Just accept it for now.
Now the next question would be: "So, if the last addition overflows, does the next add automatically count the carry flag too?" Good question. The answer is no. However, Intel processor has a special instruction called adc. This command behaves similarly as the add command. The only extra thing is that it also add the value carry flag along. So, this may be very handy to add large integers. Suppose you'd like to add a 32-bit integers with 16-bit registers. How can we do that? Well, let's say that the first integer is held on the register pair DX:AX, and the second one is on BX:CX. This is how:
``` add ax, cx
```
Ah, so first, the lower 16-bit is added by add ax, cx. Then the higher 16-bit is added using adc instead of add. It is because: if there are overflows, the carry bit is automatically added in the higher 16-bit. So, no cumbersome checking. This method can be extended to 64 bits and so on... Note that: If the 32-bit integer addition overflows too at the higher 16-bit, the result will not be correct and the carry flag is set, e.g. Adding 5 billion to 5 billion.
For the subtraction, we have similar instruction called sbb. It is pretty much the counterpart for sub. I won't discuss it much here. You should try it out yourself.
How about adding and subtracting negative values? add, adc, sub, and sbb handles that "automatically". Well, again, the internal logic for addition and subtraction both for positive and negative values are the same. So, don't worry if you have negative operands. These instruction can go smoothly.
### Multiplication, Division, and Remainder
While addition and subtraction can be done on any register, multiplication and division do not. Multiplication and division always assume AX as the place holder. The format is as follows:
``` mul x ; --> If x is 8-bit, then AX = AL * x;
; if x is 16-bit, then DX:AX = AX * x;
div x ; --> If x is 8-bit, then AL = AX / x, AH stores the remainder;
; if x is 16-bit, then AX = DX:AX / x, DX stores the remainder;
```
See, that's different depending on the source. As far as I recall, you cannot have variables for x in 8086. In 80286 or above you could (but again you must mention the xxxx ptr modifier). If you'd like to multiply or divide by constants (i.e. x is a constant), you'll need to load the constants into one of the registers (especially 8086). In 80286, you probably could. Anyone may correct this. I'm not quite remember.
Doing division is the same. The result of division is always rounded down. The nice extra is that we can obtain the remainder for free.
If there is an overflow in multiplication, the overflow flag will be set. There is no extra instruction like adc or sbb. So, you'll want to have an extra caution on this. Similarly, if you divide a number by 0, you'll likely to trigger an error. In windows machines this may cause a blue screen of death. So, watch out.
Note: mul and div will treat every numbers as positive. If you have negative values, you'll need to replace them imul and idiv respectively.
### Increment and Decrement
Often times, we'd like to incrementing something by 1 or decrement thing by 1. You can use add x, 1 or sub x, 1 if you'd like to, but Intel x86 assembly has a special instruction for them. Instead of add x, 1 we use inc x. These are equivalent. Likewise in subtraction, you can use dec x for subtitution. Beware that neither inc nor dec instruction sets the carry flag as add and sub do.
### A Nice Tip
The arithmatic operations can have special properties. For example: add x, x is actually equal to multiplying x by 2. Similarly, sub x, x is actually setting x to 0. In 8086 processor, these arithmatic is faster than doing mul or doing mov x, 0. Ha! Even more, its code size is smaller. No wonder why the assembly wizards often fond of this subtitution.
### Closing
OK, I think that's all for now. See you next time. The next chapter would be about bitwise operation.
### Where to go
Chapter 4
News Page
x86 Assembly Lesson 1 index
Contacting Me
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Symbol
# Calculator search results
Formula
Factorize the expression
$16x ^{ 4 } -625y ^{ 4 }$
$\left ( 2 x - 5 y \right ) \left ( 2 x + 5 y \right ) \left ( 4 x ^ { 2 } + 25 y ^ { 2 } \right )$
Arrange the expression in the form of factorization..
$\color{#FF6800}{ 16 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 4 } } \color{#FF6800}{ - } \color{#FF6800}{ 625 } \color{#FF6800}{ y } ^ { \color{#FF6800}{ 4 } }$
Factorize to use the polynomial formula of sum and difference
$\left ( \color{#FF6800}{ 4 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 25 } \color{#FF6800}{ y } ^ { \color{#FF6800}{ 2 } } \right ) \left ( \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 5 } \color{#FF6800}{ y } \right ) \left ( \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ y } \right )$
$\left ( \color{#FF6800}{ 4 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 25 } \color{#FF6800}{ y } ^ { \color{#FF6800}{ 2 } } \right ) \left ( \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 5 } \color{#FF6800}{ y } \right ) \left ( \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ y } \right )$
Sort the factors
$\left ( \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ y } \right ) \left ( \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 5 } \color{#FF6800}{ y } \right ) \left ( \color{#FF6800}{ 4 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 25 } \color{#FF6800}{ y } ^ { \color{#FF6800}{ 2 } } \right )$
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# Laser Beam Question
## Homework Statement
If a 20-W laser beam, which has an initial diameter of 2.0 mm, spreads out to a diameter of 2.0 m after traveling 10,000 m, what is the final intensity of the beam?
A. 20 x 10-4 W/m2
B. 20 x 10-6 W/m2
C. 20 x 10-8 W/m2
D. 6.4 W/m2
E. 6.4 x 10-6 W/m2
## Homework Equations
I found an equation that states if light hits a surface area A at normal incidence, the intensity I is equal to:
I = <P> / A
## The Attempt at a Solution
If I draw this beam out, I have a frustrum of a right circular cone.
The lateral surface area of a frustrum = PI * S * (R + r) where R and r are the radia and s is the diagonal side.
After a little math, I found s to be about 10000. It is a little over but its neglible.
So the surface area would be = PI * 10000 * (.001 m + 1.0 m) = 31447.3 m^2
Wouldnt that mean the intensity I = 20 W / 31447.3 m^2 = .000636 W / m^2
Intensity = 6.4 * 10^-4 W / m^2
Obviously, that is not one of the answers. What am I doing wrong here?
Any help will be greatly appreciated. Thanks in advance.
Sincerely,
Travis Walters
dlgoff
Gold Member
Think of it this way. How much energy (watts) do you have through a 2m diam. circle?
Hey there,
For reference, the answer key says 6.4 W/m2 is the correct answer.
The answer I got was 6.4 * 10^-4 W / m^2, so I am just wondering the answer is a typo.
I am not sure where you are getting at dlgoff? Please explain :)
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# New to QlikView
Discussion board where members can get started with QlikView.
Not applicable
## Calculating days from date
Hi Experts,
i have a table with following fields.
Table1:
Date,
TransactionID
from Table1.qvd(qvd);
in a pivot table, i need to calculate the no. of days worked by an Employee in a month.
condition for calculation is,
1. if there is a transaction against an employee on a particular date, then it mean that employee has worked on that day.
2. if the weekday(Date) is Saturday, then count is 0.5 else 1
How can i do this in expression.
Tags (1)
1 Solution
Accepted Solutions
MVP
## Re: Calculating days from date
If there are multiple transactions per date, you need to consider & remove the duplicates per employee and date.
You can do it like this:
=Sum(Aggr(WorkDayCount, emp.code, PRINTED_DATE))
Or you would need to create a table in the data model with one record per employee and date and the WorkdayCount value.
19 Replies
MVP
## Re: Calculating days from date
Set DateFormat = 'MM/DD/YYYY'; // Or whatever your date format in Date is
Table1:
LOAD *, If( WeekDay(Date)=5, 0.5,1 ) as WorkDayCount;
Date,
TransactionID
from Table1.qvd(qvd);
Then create chart with dimension EmployeeName and as expression
=Sum(WorkDayCount)
Not applicable
## Re: Calculating days from date
Table1:
IF(weekday(Date) = 'Sat',0.5,1) AS COUNT_WORKING;
Date,
TransactionID
from Table1.qvd(qvd);
Then, in Pivot Table you show SUM(COUNT_WORKING) by Employe
Hope it helps
Regards,
Not applicable
## Re: Calculating days from date
Something Like this ..
Table1:
LOAD *,if(not ISNULL(TransactionID) or num(TransactionID)>0, If( WeekDay(Date)=5, 0.5,1 ),0) as WorkDayCount;
Date,
TransactionID
from Table1.qvd(qvd);
Then create chart with dimension EmployeeName and as expression
=Sum(WorkDayCount)
Not applicable
## Re: Calculating days from date
Try like below UI Expression:
Sum(IF(Workday(Date_Field)=5, 0.5,1))
I would suggest add the flag in the script instead of UI if cond expression.
Not applicable
## Re: Calculating days from date
Hi Rajendran,
Date,
TransactionID,
if(isnull(value)=0 and WeekDay(value)<>5, 1, if(isnull(value) and WeekDay(value)=5,0.5,0)) as weektotal
from Table1.qvd(qvd);
and in the chart use
=sum(weektotal)
Thanks,
Sreeman.
Not applicable
## Re: Calculating days from date
Hi all
i tried all the three, but its not giving required value.
MVP
## Re: Calculating days from date
It's hard to create a correct solution with the information provided.
Not applicable
## Re: Calculating days from date
Please provide sample data and expected result.
MVP
## Re: Calculating days from date
It depends on which days of the week are working days.
If the working week is Monday-Friday then look at the NETWORKDAYS() function as this calculates the number of working days between two dates and you can add a list of holiday dates to be excluded.
If the working week includes Saturday or Sunday then you will need to create your own expression.
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https://pokemondb.net/pokebase/381575/how-fast-would-regieleki-be-in-this-scenario?show=381578
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PokéBase - Pokémon Q&A
# How fast would Regieleki be in this Scenario?
2,106 views
So, let's say I have a Regieleki with a maxed out speed stat, holding a Choice Scarf, +6 in speed, having Tailwind active, and skill swap Swift Swim onto it in the rain. How fast would Regieleki be exactly?
I love this question XD
Turn-order wise, this Regieleki would be treated as having a speed stat of 1808. If its teammate had 1810 speed, it would be slower than that, but faster than an opponent with 1807.
The trick is that the game forces a couple speed limits. 10,000 is the value of the hard cap, where no matter what the normal formula for stats would dictate, anything that would be greater than 10000 gets stopped at that point and refuses to go any higher. (This is because of the current implementation of Trick Room, which replaces all speeds with (10000 - speed) if it's present, and needs to make sure it's never subtracting anything big enough to produce a negative number as a result.) That then gives way to a second cleaning system, this one more of a pit trap: any speed stat that's 8192 or higher gets dropped backward by subtracting 8192, so speeds that were fast enough to slam into that first wall will also be pushed back by this, and 1808 is the point it lands on.
selected by
Yay I love math
Let's see here... First you add the Choice Scarf boost to 548.
548 x 1.5 = 822
Every +1 stat boost in Pokemon is equivalent to 50%, so a +6 Speed would be a 300% boost, or x4.
822 x 4 = 3288
3288 x 2 = 6576
And then, finally, 6576 x 2 = 13152
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http://stackoverflow.com/questions/22578987/interface-mismatch-higher-order-functions
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# Interface mismatch - higher order functions
I'm trying to 'reproduce' higher order functions in fortran.
``````module rk4
contains
pure function f(t,x) result (fx)
real, dimension(1), intent(in) :: x
real, intent(in) :: t
real, dimension(1) :: fx
fx = x
end function f
function step(x,f,dt) result(xn)
real, intent(in) :: dt
real, intent(in), dimension(:) :: x
real, dimension(:), allocatable :: k1,k2,k3,k4,xn
real, external :: f
integer :: N
N = size(x)
allocate(k1(N))
allocate(k2(N))
allocate(k3(N))
allocate(k4(N))
k1 = f(t,x)
k2 = f(t+0.5*dt,x+0.5*k1*dt)
k3 = f(t+0.5*dt,x+0.5*k2*dt)
k4 = f(t+dt,x+dt*k3)
allocate(xn(N))
xn = x + (dt/6.)*(k1 + 2*k2 + 2*k3 + k4)
deallocate(k1)
deallocate(k2)
deallocate(k3)
deallocate(k4)
end function step
end module rk4
``````
When the module is called in the following way
`````` real, dimension(1) :: x0 = 2
x0 = step(x0,f,0.01)
``````
I get the following error
``````\$gfortran -c test_rk4.f95
test_rk4.f95:7.15:
x0 = step(x0,f,0.01)
1
Error: Interface mismatch in dummy procedure 'f' at (1): Type/rank mismatch in function result
``````
What might be causing this?
-
The error message is complaining about the function `f` being incompatible with the dummy argument `f`.
You declare it as
`````` real, external :: f
``````
which means it should return a scalar whereas in reality the function `f` returns an array.
The same names do not really help the understanding here. I changed the name of the dummy argument in the following code to `g`.
The easiest way to solve this is
`````` pure function f(t,x) result (fx)
real, dimension(1), intent(in) :: x
real, intent(in) :: t
real, dimension(1) :: fx
fx = x
end function f
function step(x,g,dt) result(xn)
real, intent(in) :: dt
real, intent(in), dimension(:) :: x
real, dimension(:), allocatable :: xn
procedure(f) :: g
!here call g, not f!!!
``````
The procedure statement comes from Fortran 2003 and causes the dummy argument procedure `g` to have the same interface as the procedure `f`.
Otherwise you can use an interface block:
`````` function step(x,g,dt) result(xn)
real, intent(in) :: dt
real, intent(in), dimension(:) :: x
real, dimension(:), allocatable :: xn
interface
pure function g(t,x) result (fx)
real, dimension(1), intent(in) :: x
real, intent(in) :: t
real, dimension(1) :: fx
end function g
end interface
``````
The external statement should be used in modern code only in some exceptional cases.
-
Thanks. But how can your code be extended to a more generic case. x may be a vector of length 2 (and hence f also has to return a vector of length 2) – tgoossens Mar 22 at 15:41
I figured it out. Thanks.I will post my solution later – tgoossens Mar 22 at 15:55
You chose the explicit length. Normally assumed shape arrays are better. Then just use `fx(size(x))`. But it is not that much connected to the original error. – Vladimir F Mar 22 at 16:13
You are right. I accepted your answer now :) – tgoossens Mar 22 at 16:38
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https://saxafund.org/net-interest-rate-spread-definition-and-use-in/
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# Net Interest Rate Spread: Definition and Use in Profit Analysis
## Net Interest Rate Spread: Definition and Use in Profit Analysis
The net interest rate spread is an indicator of a financial institution’s ability to generate income from its interest-earning assets, such as loans and investments. It is calculated by subtracting the interest expense from the interest income and dividing the result by the average interest-earning assets. The net interest rate spread is typically expressed as a percentage.
A positive net interest rate spread indicates that a financial institution is earning more interest income than it is paying out in interest expense. This is generally considered favorable, as it means that the institution is able to generate a profit from its interest-earning activities. On the other hand, a negative net interest rate spread indicates that the institution is paying out more in interest expense than it is earning in interest income, which can be a sign of financial distress.
### Calculating Net Interest Rate Spread
To calculate the net interest rate spread, you need to gather the necessary financial data from a financial institution’s income statement and balance sheet. The interest income can be found under the “interest income” or “interest earned” line item, while the interest expense can be found under the “interest expense” or “interest paid” line item. The average interest-earning assets can be calculated by taking the average of the beginning and ending balances of the institution’s interest-earning assets.
Once you have gathered the necessary data, you can calculate the net interest rate spread using the following formula:
### Importance of Net Interest Rate Spread in Profit Analysis
The net interest rate spread is a key metric in profit analysis for financial institutions. It provides insights into the institution’s ability to generate income from its interest-earning activities and its overall profitability. A positive net interest rate spread indicates that the institution is able to generate a profit from its interest income, while a negative net interest rate spread may indicate financial difficulties.
By analyzing the net interest rate spread over time, financial institutions can identify trends and make informed decisions about their interest rate management, loan pricing, and overall business strategy. It can also help investors and stakeholders evaluate the financial health and performance of a financial institution.
The net interest rate spread is a financial metric that measures the difference between the interest income generated by a financial institution and the interest expenses it incurs. It is a crucial indicator of a bank’s profitability and efficiency in managing its interest-bearing assets and liabilities.
Net interest rate spread is calculated by subtracting the average interest rate paid on liabilities from the average interest rate earned on assets. The resulting figure represents the net interest income, which is a key component of a bank’s overall profitability.
### Importance of Net Interest Rate Spread
The net interest rate spread is an important metric for both banks and investors. For banks, it provides insight into their ability to generate income from their lending activities and manage their interest rate risk. A wider net interest rate spread indicates that a bank is earning more interest income than it is paying in interest expenses, which is a positive sign of profitability.
Investors also pay close attention to the net interest rate spread when evaluating a bank’s financial health and performance. A wider spread suggests that a bank is effectively utilizing its assets to generate income and may be more capable of withstanding economic downturns or fluctuations in interest rates.
Furthermore, the net interest rate spread can be used as a benchmark for comparing the performance of different banks or financial institutions. It allows investors and analysts to assess the relative profitability and efficiency of these institutions and make informed investment decisions.
### Factors Affecting Net Interest Rate Spread
Several factors can influence the net interest rate spread of a financial institution. One of the primary factors is the interest rate environment. When interest rates are low, banks may struggle to generate sufficient interest income to cover their interest expenses, resulting in a narrower spread. Conversely, in a high-interest-rate environment, banks may benefit from a wider spread.
The composition of a bank’s assets and liabilities can also impact its net interest rate spread. For example, if a bank has a higher proportion of low-yielding assets or high-cost liabilities, it may experience a narrower spread. On the other hand, a bank with a higher proportion of high-yielding assets or low-cost liabilities may enjoy a wider spread.
In addition, the competitive landscape and market conditions can affect a bank’s net interest rate spread. Increased competition among banks can lead to lower interest rates and narrower spreads, while a less competitive market may allow banks to charge higher interest rates and widen their spreads.
## Calculating Net Interest Rate Spread
The net interest rate spread is a key financial metric used in profit analysis. It measures the difference between the interest earned on loans and investments and the interest paid on deposits and borrowings. Calculating the net interest rate spread allows financial institutions to assess their profitability and manage their interest rate risk.
To calculate the net interest rate spread, follow these steps:
Step 1: Determine Interest Income
Start by identifying all sources of interest income, such as loans, investments, and other interest-earning assets. Sum up the interest earned from each source to obtain the total interest income.
Step 2: Calculate Interest Expense
Next, determine all sources of interest expense, including deposits, borrowings, and other interest-bearing liabilities. Add up the interest paid on each source to find the total interest expense.
Step 3: Subtract Interest Expense from Interest Income
Subtract the total interest expense from the total interest income. The result is the net interest income.
Step 4: Divide Net Interest Income by Average Interest-Earning Assets
To calculate the net interest rate spread, divide the net interest income by the average interest-earning assets. This ratio provides a measure of the profitability of a financial institution’s interest-earning activities.
A positive net interest rate spread indicates that the institution is earning more interest income than it is paying in interest expenses, resulting in a profitable operation. Conversely, a negative net interest rate spread suggests that the institution is paying more in interest expenses than it is earning in interest income, indicating potential financial challenges.
Financial institutions closely monitor their net interest rate spread to assess their profitability and make informed decisions regarding their interest rate risk. By analyzing this metric, they can identify opportunities to increase interest income, manage interest expenses, and optimize their overall profitability.
## Importance of Net Interest Rate Spread in Profit Analysis
The net interest rate spread is a crucial metric in profit analysis for financial institutions. It measures the difference between the interest earned on assets and the interest paid on liabilities, such as deposits or borrowings. This spread indicates the profitability of a financial institution’s core lending and borrowing activities.
A positive net interest rate spread signifies that the institution is earning more interest income from its assets than it is paying out in interest on its liabilities. This indicates a healthy profit margin and a strong financial position. It allows the institution to cover operating expenses, loan losses, and still generate a profit.
On the other hand, a negative net interest rate spread indicates that the institution is paying out more in interest on its liabilities than it is earning on its assets. This can be a cause for concern as it suggests that the institution’s interest expenses are higher than its interest income, potentially leading to losses or reduced profitability.
Furthermore, the net interest rate spread provides insights into the institution’s risk profile. A wider spread indicates a higher potential for profit, but it also suggests a higher level of risk. This is because a wider spread often results from charging higher interest rates on loans and investments, which may attract riskier borrowers. Therefore, financial institutions must strike a balance between maximizing profit and managing risk.
Overall, the net interest rate spread is a key indicator of a financial institution’s profitability and risk management. By analyzing this metric, institutions can identify areas of improvement, optimize their interest rate strategies, and make informed decisions to enhance their financial performance.
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https://turningtooneanother.net/2020/09/21/what-is-surface-tension-in-fluid-mechanics/
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# What is surface tension in fluid mechanics?
## What is surface tension in fluid mechanics?
Surface Tension: “The property of the surface of a liquid that allows it to resist an external force, due to the cohesive nature of its molecules.” The cohesive forces between liquid molecules are responsible for the phenomenon known as surface tension.
## How do you calculate surface tension in fluid mechanics?
Here is an example of calculating surface tension by using the formula.
1. Compute the surface tension of a given liquid whose dragging force is 7 N and length in which the force acts is 2 m?
2. Solution:
3. T = F/L.
4. ⇒ T = 7/2.
5. ⇒ T = 3.5 N/m.
What is the surface tension of glass?
The surface tensions of a number of different glasses have been determined by a new method. The magnitude of the surface tension for this range varies from 230 to 360 dynes per cm depending upon the kind of glass. A survey of previous results on surface tension discloses values of widely different magnitudes.
### How do you derive the surface tension?
Or, T = [M1 L1 T-2] × [L-1] = M1 T-2. Therefore, surface tension is dimensionally represented as M1 T-2.
### On what factors surface tension depends?
Surface tension depends mainly upon the forces of attraction between the particles within the given liquid and also upon the gas, solid, or liquid in contact with it. The molecules in a drop of water, for example, attract each other weakly.
How do you find pressure when given surface tension?
The force from surface tension is F = 2γL = 2γ2πr = 4γπr. For the bubble to be stable and not collapse, the pressure inside the bubble must be higher than the pressure on the outside. The force do to the pressure difference must balance the force from the surface tension.
#### What is G in surface tension?
γ is the surface tension of the liquid in dynes per centimeter or newtons per meter. g is the acceleration due to gravity and is equal to 980 cm/s2 or 9.8 m/s2.
#### How do you break the surface tension of water?
In order for water to flow more easily into these small spaces, you need to decrease its surface tension. You can do this by adding soap, which is a surfactant (a material that decreases the surface tension of a liquid).
What is critical surface tension?
The critical surface tension is the highest liquid surface tension that can completely wet a specific solid surface. For adhesive bonding complete wetting is used to maximize the adhesive joint strength.
## What is dimensional formula for surface tension?
Surface Tension (T) = Force × Length-1. Or, T = [M1 L1 T-2] × [L-1] = M1 T-2. Therefore, the surface tension dimension formula is represented as M1 T-2.
## How is surface tension displayed in fluid mechanics?
Surface tension will be displayed by the symbol σ. Let us consider three molecules of water A, B and C as displayed in following figure. Molecule A is well within water and therefore molecule A will be attracted equally in all direction by its surrounding molecules of liquid i.e. water.
How does surface tension affect a liquid droplet?
Liquid droplet entire surface will be under the influence of tensile force due to surface tension. We must note it here that pressure intensity inside the droplet is basically the difference in pressure between inside and outside of the liquid droplet.
### Why does water have the greatest surface tension?
In fact, other than mercury, water has the greatest surface tension of any liquid. (Source: Lakes of Missouri) Within a body of a liquid, a molecule will not experience a net force because the forces by the neighboring molecules all cancel out (diagram).
### How to calculate tensile force due to surface tension?
Tensile force due to surface tension = Surface tension x Circumference Tensile force due to surface tension = σ x П d Force due to pressure inside the droplet = Pressure intensity inside the droplet x Area Force due to pressure inside the droplet = P x (П/4) d2
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https://www.easyerra.com/harmonic-mean-calculator/
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# Harmonic Mean Calculator
Calculate harmonic means efficiently with our Harmonic Mean Calculator. Simplify data analysis and averaging tasks.
Enter Inputs 5,20,40,80,100Enter the Numbers with Comma separated(,)
Result: No.of Inputs Harmonic Mean
## Similar Calculators:
The Harmonic Mean Calculator is a useful tool for calculating the harmonic mean of a set of numbers. The harmonic mean is a type of average that is particularly helpful when dealing with rates, ratios, or values that exhibit a reciprocal relationship. It is defined as the reciprocal of the arithmetic mean of the reciprocals of a set of numbers.
The formula for the harmonic mean ($$H$$) is:
$H = \frac{n}{\left(\frac{1}{x_1} + \frac{1}{x_2} + \ldots + \frac{1}{x_n}\right)}$
Where:
• $$n$$ is the total number of values in the set.
• $$x_1, x_2, \ldots, x_n$$ are the individual values in the set.
## Using the Calculator
This calculator allows you to perform the following tasks:
• Input a set of numbers for which you want to calculate the harmonic mean.
• Receive the calculated harmonic mean as the output.
• Apply the harmonic mean in various contexts, such as physics, engineering, economics, and more.
## Applications
The harmonic mean has specific applications in several fields:
• Physics: It is used in calculations related to wave frequencies, resistances in parallel circuits, and harmonic motion.
• Economics: Applied to compute average rates of return and average growth rates.
• Environmental Science: Utilized for ecological studies, particularly in the calculation of environmental indices.
• Finance: Relevant when dealing with financial ratios and investment performance metrics.
The Harmonic Mean Calculator simplifies the process of finding the harmonic mean of a set of numbers, making it a valuable tool for scientists, engineers, economists, and professionals in various fields.
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Signal Processing and Linear Systems-B.P.Lathi copy
# Signal Processing and Linear Systems-B.P.Lathi copy
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Unformatted text preview: (0.8k) is n ot periodic. 8 Discrete-time Signals a nd Systems 550 6, E xercise E 8.3 S tate w ith reasons if t he following sinusoids are periodic. I f p eriodic, find t he p eriod. ( i) cos k) ( ii) cos (~k) (iii) cos (y'1Tk) Ans: ( i) P eriodic: period No = 14. ( ii) a nd (iii) Aperiodic: fl.j27T i rrational. 'V (¥- <::) C omputer E xample C S.3 S ketch a nd verify if cos (¥-k) is periodic. According t o E q. ( 8.9b), t he s mallest value of m t hat will m ake No = m (~) m ( ¥) a n i nteger is 3. T herefore, No = 14. T his r esult means cos (¥-k) is periodic a nd i ts p eriod is 14 s amples in t hree cycles of i ts envelop. T his a ssertion c an b e verified by t he following M ATLAB c ommands: t =-5*pi:pi/lOO:5*pi; t =t'; f t=cos(3"'pi*t/7); p lot(t,ft,':'), h old o n k =-15:15; k =k'; f k=cos(k * 3*pi/7); s tem(k,fk), h old o ff <::) 2 8.2 Some Useful Discrete-time Signal models 551 T his result shows t hat a sinusoid cos ( nk + 8) can always be expressed a s cos ( nfk + 8), where -11' ::s: nf < 11' ( the fundamental frequency range). T he readel;. s hould get used t o t he fact t hat t he r ange o f discrete-time frequencies is only 211'. We m ay select this range t o b e from -11' t o 11' o r from 0 t o 211', o r a ny other interval of width 211'. I t is most convenient t o use t he r ange from -11' t o 11'. A t times, however, we shall find it convenient t o use t he r ange from 0 t o 211'. T hus, in t he discrete-time world, frequencies can be considered t o lie only in t he f undamental frequency range (from -11' t o 11', for instance). Sinusoids of frequencies outside t he f undamental frequencies do exist technically. B ut physically, t hey c annot be distinguished from t he sinusoids of frequencies within t he f undamental range. Thus, a discrete-time sinusoid of any frequency, no m atter how high, is identical t o a sinusoid of some frequency within t he f undamental range (-11' t o 11'). T he above results, derived for discrete-time sinusoids, are also applicable t o d iscrete-time exponentials of t he form e jrlk . For example m, integer N onuniqueness o f D iscrete-Time Sinusoid Waveforms A c ontinuous-time sinusoid cos w t has a unique waveform for every value of w in t he r ange 0 to 0 0. Increasing w results in a sinusoid of ever increasing frequency. Such is n ot t he case for t he discrete-time sinusoid cos Ok because (8.12) Here we have used t he fact t hat e±j27Tn = 1 for all integral values of n . T his result means t hat discrete-time exponentials of frequencies separated by integral multiples of 211' a re identical. F urther R eduction in t he F requency Range o f D istinguishable D iscrete-Time S inusoids cos ( 0 ± 211'm)k = cos ( nk ± 211'mk) Now, if m is a n integer, m k is also a n integer, a nd t he above equation reduces to cos ( 0 ± 211'm)k = cos n k m integer (8.10) T his r esult shows t hat a discrete-time sinusoid of frequency 0 is indistinguishable from a sinusoid of frequency n plus or minus a n integral multiple of 211'. This s ta...
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# Trigonometry - How to figure out translation of a graph
I have been wracking my brain and can not figure this out so I've broken down and am asking this question on here in hopes that someone can show me how to do this.
I need to figure out if this sine graph has been translated left or right and by how many units.
I can't figure it out from the information provided by the graph. I'm stuck on this idea that I need to figure out the period but I can't do that because I can't determine what the x-values of the min. and max. are. Well, it's looks like the x-value of the min. is 0 but I can't figure out the max.
At any rate, anyone know how to do this?
• It could've been shifted left or right, but it doesn't matter. – Simply Beautiful Art Feb 20 '17 at 1:15
• Given that you know the graph was merely translated, not changed in any other way, the minimum is enough information. If you really want to know the maximum you can figure it out from the minimum. – David K Feb 20 '17 at 1:32
• Note that the positive maximum occurs a little past $3$ and the next minimum, though off the edge appears to be a bit more than $6$. This supports the idea that the period of this function has not been modified, which your question also seems to suggest, since it only talks about translation. The amplitude clearly is still $1$, so that hasn't changed either. As you've noted, there is a minimum at $0$. Where is the nearest minimum of $\sin$? That will tell you how far it has been translated and in which direction. – Paul Sinclair Feb 20 '17 at 3:49
Period = distance between two peaks or maxima. It appears the peaks or maxima are at a little more than $\pm 3$ so $x =\pm \pi \approx \pm 3.14$ would be a reasonable guess. Then period = $\pi - (- \pi) = 2\pi$, unchanged from the basix function sin$(x)$.
Sin$(x)$ = 0 at x = 0, and sin$(x)$ is increasing at $x = 0.$
The two points visible on this graph which fit both criteria are at x close to or a little above 1.5, and $x$ close to -4.6 or -4.7. Since it appears we are dealing with radian measure, a period of $2 \pi$ and maxima and minima at multiples of pi, then a reasonable assumption would be that $x = 0$ and $f(x)$ is increasing ($f'(x)$ positive) at $x = \pi/2 \approx 1.57$ and at $x = - 3 \pi/2 \approx -4.71$
Thus the graph could be sin$(x)$ shifted $\pi/2$ to the right OR $3 \pi/2$ to the left. (Since the period is $2\pi$ adding any multiple of $2\pi$ to the phase shift will give a correct result.)
The function would be $f(x) = \sin (x - \pi/2)$ OR $f(x) = \sin (x + 3\pi/2)$
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## Rules of Thumb: Estimating Range
JennyB 1 kW
Posts: 448
Joined: Jan 25 2008 12:51pm
Location: Northern Ireland
### Rules of Thumb: Estimating Range
Rules of Thumb are not meant to be totally accurate, but they do help to ensure that your ideas are not wildly astray. These are some that I have come up with for estimating range, based on a 1992 article by Bill Bushnell and Chris Hull. Please let me know if they are helpful or misleading. I would also be grateful if others could post similar rules for topics of which I am even more ignorant: motor, batteries, controller, charging etc.
Estimating the Energy Requirement of a route
The climbing index (CI) is the watt-hour equivalent height in metres of climbing to a kilometre on the flat for the riders cruising speed.
For typical ebikes this will lie somewhere between 20 and 50 metres, lower for slower or heavier bikes, higher for lighter bikes or those with a faster cruising speed. An estimate to the nearest 10 metres is close enough.
This index may be estimated by comparing energy drain over various routes, or by using a simulator such as that at http://www.noping.Net/english/ . Note that these show the power required at the wheels, which needs to be checked against actual watt hours consumed, which may be double.
To estimate the energy (and therefore the battery capacity) required for a given trip:
C = cruising speed in kilometres per hour on the flat with no wind
W = the watt hour/kilometre cost of C
D = total distance in kilometres
N = net height gained in metres - the difference between start and finish altitude.
G = total height climbed
ED = the equivalent distance that could be travelled on the flat for the same energy expended.
If the route was uphill all the way ED would be D + ( N / CI ) kilometres and the energy used would be W*ED watt hours. The time taken would be ED / C hours. More normally, there would be G - N metres of downhills. As a general rule, coasting down hills saves 50% of the extra energy needed to to climb them. Maintaining power saves perhaps 30%.
Some examples:
CI = 24 C=27 W=20
Using power lightly on the downhills, so saving 40%
Riding a 100k loop with 2000 metres of climbing.
ED = 100 + ( 2000 / CI * 60% ) = 150
Watt hours needed = 150 * 20 = 300
If the ride ended 360 metres above the start
ED = 100 + ( 360 / CI ) + ( (2000-360)/CI * 60%)
= 100 + 15 + 41 = 156
Estimated time taken = 156/C = about 5 hours 50 minutes.
Limitations
These are, of course, only rough estimates. A strong wind can affect range by as much as 40% either way. Air resistance increases as the cube of speed through the air, so at typical ebike speeds an increase of 10 kph almost halves the range. A route with short climbs and long descents is much easier to ride than the same route in the opposite direction. It follows that the most economical way to ride is to get as close to cruising speed as possible through the slower sections, so that you do not need to go as fast elsewhere. On hilly routes a "good big'n" capable of holding speed on most climbs should have a greater range than a good little'n." Is that borne out by experience?
Samba 100 W
Posts: 139
Joined: Sep 26 2010 8:11pm
Location: Pennsylvania
### Re: Rules of Thumb: Estimating Range
JennyB wrote: The time taken would be ED / C hours.
Estimated time taken = 156/C = about 5 hours 50 minutes.
You lost me there.
C is cruising speed on flat 27kph. Its not the same speed used for climbing? I'd expect time to be 100KM/C irregardless of heights. Theres nothing in there about differing speed for uphills just additional power.
Its for calculating energy use -- 156*W -- same WH used for 156KM on flat is needed for this 100KM route with climbs.
JennyB 1 kW
Posts: 448
Joined: Jan 25 2008 12:51pm
Location: Northern Ireland
### Re: Rules of Thumb: Estimating Range
Samba wrote:
JennyB wrote: The time taken would be ED / C hours.
Estimated time taken = 156/C = about 5 hours 50 minutes.
You lost me there.
C is cruising speed on flat 27kph. Its not the same speed used for climbing? I'd expect time to be 100KM/C irregardless of heights. Theres nothing in there about differing speed for uphills just additional power.
Its for calculating energy use -- 156*W -- same WH used for 156KM on flat is needed for this 100KM route with climbs.
I wondered about that too. It does work out surprisingly well for touring cyclists, which is the original application, because that is a constant power scenario. a tourist cannot produce much more than their mean power for very long, so they gear down and go slower on the hills. Whether it would still work for a 1000W hub ridden at a constant 27kph is another question.
John in CR 100 GW
Posts: 13690
Joined: May 20 2008 12:58am
### Re: Rules of Thumb: Estimating Range
That's the most complicated "rule of thumb" I've ever seen. I'd do something more like rolling hills on your route are likely to double your consumption over flat terrain consumption, so it's typically better to look for a flatter way around. You can't apply a cyclist rule of thumb with a hub motor, because current will increase significantly, so the power used up hills is likely to max out your controller limits. If you know the hills on your route, then there are some bike calculators that will give you a pretty accurate picture of the energy requirements to climb them.
John
dogman dan 100 GW
Posts: 34889
Joined: May 17 2008 12:53pm
Location: Las Cruces New Mexico USA
### Re: Rules of Thumb: Estimating Range
My rule of thumb for the typical hubmotor kit has been 1 ah for 1 mile on 36v bikes. This is NOT how far you go, but rather how big a battery you should get for a commute. There is up to 30% reserve there in this rule of thumb to allow for that day when the wind blows in your face 20 mph. It's good for battery lifespan anyway.
JennyB 1 kW
Posts: 448
Joined: Jan 25 2008 12:51pm
Location: Northern Ireland
### Re: Rules of Thumb: Estimating Range
John in CR wrote:That's the most complicated "rule of thumb" I've ever seen. I'd do something more like rolling hills on your route are likely to double your consumption over flat terrain consumption, so it's typically better to look for a flatter way around. You can't apply a cyclist rule of thumb with a hub motor, because current will increase significantly, so the power used up hills is likely to max out your controller limits. If you know the hills on your route, then there are some bike calculators that will give you a pretty accurate picture of the energy requirements to climb them.
John
If you think that's complicated, you should read the original article !
That's how we did things pre-Web, and that's the maths the calculators use. Web mapping apps don't work too well for elevation here, so I just count the number of 10-metre contour lines crossed on an OS map. Compare two trips, one hilly and one flat. If you know the watt-hours expended on each then simple algebra give you your Climbing Index. Since I use power mainly on the ascents, it turns out to be a lot lower than it would be if I was only pedalling or using the motor continuously - about 10 metres. It's probably much the same for most Euro-legal bikes ridden in this fashion.
So the rule for me simplifies even further: to estimate the difficulty of a new route, add a kilometre for every 20 metres climbed, and add or subtract another kilometre for each 20 metres the finish is above or below the start.
That's not so difficult, is it?
busted_bike 100 mW
Posts: 42
Joined: Jun 24 2010 11:18am
### Re: Rules of Thumb: Estimating Range
JennyB wrote: So the rule for me simplifies even further: to estimate the difficulty of a new route, add a kilometre for every 20 metres climbed, and add or subtract another kilometre for each 20 metres the finish is above or below the start.
That's not so difficult, is it?
Now that's a rule of thumb!
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An Approximate Probabilistic Model for Structured Gaussian Elimination
Edward A. Bender
Department of Mathematics, University of California, San Diego
Algorithms Seminar
November 2, 1998
[summary by François Morain]
A properly typeset version of this document is available in postscript and in pdf.
If some fonts do not look right on your screen, this might be fixed by configuring your browser (see the documentation here).
## 1 Introduction
Modern algorithms [5, 7] for factoring integers or for solving discrete logarithms problems work in two phases. In the first one, one collects a huge amount of data that help create a large matrix that is triangularized in the second phase. For instance, the largest non-trivial number ever factored as of today is (10211 - 1)/9, which involved finding dependencies in a 4, 820, 249× 4, 895, 741 boolean matrix (see [2]). The easiest way to solve the problem is to find a computer with enough memory so that the matrix fits in core and Gaussian elimination can be used. If such a behemoth is not available, alternative methods have to be used. A method that is widely used relies on the fact that the matrix we are interested in is sparse. For instance the matrix referred to above has only 48.1 non-zero coefficients per row on average. Moreover the structure of the matrix is very peculiar: the leftmost columns are very dense, while the rightmost ones are very sparse. This has led several authors to work out what is called the Structured Gaussian Elimination (SGE) method. Apart from this approach, one can use two probabilistic iterative methods to solve the problem, namely Wiedemann's method [9] or Lanczos's [6]. In practice, these approaches are commonly combined.
We will concentrate here on linear algebra for integer factorization. For the discrete logarithm problem, the same ideas can be used with some (sometimes not so trivial) modifications. The aim of the talk is to describe one variant of SGE and try to analyse it.
## 2 Integer Factorization and Linear Algebra
Suppose we want to factor a large integer N. The basic idea is to look for two integers X and Y such that X2 º Y2 mod N, but X ¬ º± Ymod N. If this is the case, then gcd(X-Y, N) is a non-trivial factor of N. How do we proceed to find X and Y? This is usually done using a combination of congruences of the type:
xi2 º
k Õ j=1
p
a(i, j) j
mod N (1)
where the pj's are prime numbers forming the factor basis B = {p1, p2, ..., pk}. Now we look for a subset I of these such that Õi Õj pja(i, j) is the square of an integer, say Y2, leading to (Õi Î I xi)2 º Y2 mod N and we have solved our problem. The method of generation of the auxiliary congruences (1) vary from an algorithm to the other, see the references given above for more details.
Finding a subset I boils down to a linear algebra problem. Indeed, Õi Õj pja(i, j) is the square of an integer if and only if Õj pjåi a(i, j) is, or equivalently, åi a(i, j) is an even integer for all j, which in turn is equivalent to the fact that we have found a relation between some rows of the matrix M = (mi, j) where mi, j = a(i, j) mod 2, that is a vector x such that x M = 0.
What is the shape of the matrix? It is rather easy to guess that a prime number pj occurs in a factorization with probability O(1/pj). If we number our primes w.r.t. their magnitude, the left columns of M are seen to be much more dense than the right ones.
## 3 Structured Gaussian Elimination
The idea of the method [4, 8] is to try to perform Gaussian elimination on the sparse part of the matrix, as long as the fill-in is not too important. We will present here the version given in [1]1. The weight of a row (resp. column) is the number of its non-zero coefficients.
• Step 0. [Initial deactivation] deactivate some (small) fraction of the columns and call all remaining columns light.
• Step 1. [Initial clean up] repeat the following steps (a) and (b) until all columns have weight greater than 1:
• (a) Eliminate columns of weight 0 (it is of no use).
• (b) If a column has weight 1, eliminate it and the row intersecting it (since it cannot be part of a dependency).
• Step 2. [Deactivation] repeat steps (a) and (b) until all columns have been either deactivated or eliminated:
• (a) If any row has weight 1, eliminate it and the light column intersecting it. Repeat as often as possible.
• (b) Deactivate a column of high weight and repeat (a).
• Step 3. [Final step] Find the dependencies of the small matrix and build back the solutions of the initial system.
This algorithm is an iterative process that decreases the size of the matrix. It can happen that more and more rows are eliminated in Step 2a, leading to what is called a ``catastrophe'' in [8]. Among the questions we ask are: what is the size of the smallest reduced matrix that we can obtain before the catastrophe begins?
## 4 Branching Processes and the Critical Product
Let us review the basic properties of a Galton-Watson branching process, in the view of applying it to SGE. Theoretical properties on this topic can be found in [3, pp. 3--7].
In a Galton-Watson branching process, time is divided into generations. We start with one object alive in the 0th generation and at generation i, the number of objects depends on the number of objects in generation i-1. Let f(x) denote the probability generating function of this number. The probability generating function for the number of objects in the nth generation is f(n)(x) = f(f(··· f(f(x))···)). Therefore, the expected number of objects in the nth generation is
(fn(x))' |
x=1
= (f'(1))n = jn.
This quantity j acts as a threshold. If j £ 1, the process terminates with probability 1, whereas if j > 1, the process has a nonzero probability of surviving forever.
How do we apply this theory to our problem? In Step 2a of the algorithm, an object is a row of weight 1 that disappears and may create new objects of weight 1 for the next generation. The associated value j (called the critical product) of this branching process is given by:
Theorem 1 Assuming that the rows and columns are independent, we have
j » æ
ç
ç
ç
ç
ç
è
å k
k (k-1) ck
å k
k ck
ö
÷
÷
÷
÷
÷
ø
æ
ç
ç
ç
ç
ç
è
2 r2
å k
k rk
ö
÷
÷
÷
÷
÷
ø
,
where ck (resp. rk) is the probability that a randomly chosen column (resp. row) contains exactly k nonzero entries and the approximation is due to the fact that the matrix is finite.
For our purpose, we see that if j > 1, then our algorithm loops forever and the catastrophe occurs.
## 5 Critical Parameters and their Analysis
### 5.1 Probability Model
Let M = (mi, j) denote our m× n matrix. We make the assumption that Prob(mi, j ¹ 0) = D/j for some fixed parameter D. This sounds realistic since the column j of M is related to the prime pj » j log j dividing some number, thus with probability 1/pj » D/j since log is a slowly increasing function. On the other hand, a reasonable model for the weight of rows is that of a Poisson model. We will let C = D m / n.
### 5.2 Effect of the Initial Clean Up
This step causes n a1 columns (and corresponding rows) of weight 1 and n a0 columns (and no rows) of weight 0 to be eliminated. At the end of this step, one gets a new matrix with m-n a1 rows and n (1-a0-a1) columns.
Theorem 2 One has:
a1 »
a
1 =
C E1(C)
1-D(E0(C)-E1(C))
, a0 » C E2(C) + D E1(C) ×
a
1,
where Er(C) = òC¥ e-t t-r dt is the exponential integral.
Numerically, for m/n = 1.0 and D = 3.0, this yields a0 = 0.012 and a1 = 0.044. More generally, the values of a0 and a1 are rather small. Moreover, the new matrix will have the same shape as the original one, meaning that the Poisson model will still apply.
### 5.3 Matrix After Reduction
What happens in our case is that the value of j increases from one iteration of the algorithm to the other, reaching a value >1 and thus causing a catastrophe. We are interested in the parameters associated with the matrix when this occurs. We suppose we enter Step 2 of the algorithm with an m× n matrix M = (mi, j). At Step 2b, we suppose that some fraction t of the columns have been deactivated and eliminated. For simplicity, assume these are the leftmost t n columns of M. We want to relate t and j.
We suppose that our model is still valid for M, and in particular the model used for the row weight is again a (truncated, since there are no rows of weight 0 or 1) Poisson model of parameter l. We first have:
Theorem 3 With the notations above:
-D log t » l æ
ç
ç
ç
è
e
l
-1
e
l
-l-1
ö
÷
÷
÷
ø
.
From this, one gets:
Theorem 4 Letting C = D m / n, one has
j »
C (1-t)
-tlogt
l
e
l
-1
.
This formula enables one to compute the value t0 corresponding to j =1, i.e., when a catastrophe occurs. For instance, when m/n=1.0 and D=3.0, t0 = 0.318. In any case, t0 is always far from 0, which indicates that the SGE algorithm ends up with a matrix of size proportional to that of the original matrix.
### 5.4 Final Matrix
By final, we understand the matrix which is rebuilt after SGE and to which we need apply another linear algebra algorithm. Among the t n columns discarded in Step2 of the algorithm, a fraction d of them were deactivated (a remaining t-d having been eliminated), therefore being eligible to the final matrix.
Theorem 5 With the notations above:
d(t) » ó
õ
t 0
æ
ç
ç
ç
è
1+
C l
t(1-j)(e
l
-1)
ö
÷
÷
÷
ø
-1
dt.
For the same numerical values, m/n = 1.0 and D=3.0, one finds d0 = 0.186. This again shows that the final matrix is of size proportional to that of the original matrix.
## 6 Conclusions
The authors have attempted an analysis of the structured Gaussian elimination. With rather crude assumptions, they were able to derive an approximate model that is close, at least qualitatively, to that observed by their own simulations as well as by people who really factor numbers. Going further, that is have a model close to reality in quantity would require more work.
## References
[1]
Bender (Edward A.) and Canfield (E. Rodney). -- An approximate probabilistic model for structured Gaussian elimination. Journal of Algorithms, vol. 31, n°2, 1999, pp. 271--290.
[2]
CABAL. -- 211-digit SNFS factorization. -- ftp://ftp.cwi.nl/pub/herman/NFSrecords/SNFS-211, April 1999.
[3]
Harris (T. E.). -- The theory of branching processes. -- Dover Publications, 1989.
[4]
LaMacchia (B. A.) and Odlyzko (A. M.). -- Solving large sparse linear systems over finite fields. In Menezes (A. J.) and Vanstone (S. A.) (editors), Advances in Cryptology. Lecture Notes in Computer Science, vol. 537, pp. 109--133. -- Springer-Verlag, 1990. Proceedings Crypto '90, Santa Barbara, August 11--15, 1988.
[5]
Lenstra (A. K.) and Lenstra, Jr. (H. W.) (editors). -- The development of the number field sieve. -- Springer, Lecture Notes in Mathematics, vol. 1554, 1993.
[6]
Montgomery (P. L.). -- A block Lanczos algorithm for finding dependencies over GF(2). In Guillou (L. C.) and Quisquater (J.-J.) (editors), Advances in Cryptology -- EUROCRYPT '95, Lecture Notes in Computer Science, vol. 921, pp. 106--120. -- 1995. International Conference on the Theory and Application of Cryptographic Techniques, Saint-Malo, France, May 1995, Proceedings.
[7]
Pomerance (Carl) (editor). -- Cryptology and computational number theory. -- American Mathematical Society, Providence, RI, 1990, xii+171p. Lecture notes prepared for the American Mathematical Society Short Course held in Boulder, Colorado, August 6--7, 1989, AMS Short Course Lecture Notes.
[8]
Pomerance (Carl) and Smith (J. W.). -- Reduction of huge, sparse matrices over finite fields via created catastrophes. Experimental Mathematics, vol. 1, n°2, 1992, pp. 89--94.
[9]
Wiedemann (Douglas H.). -- Solving sparse linear equations over finite fields. IEEE Transactions on Information Theory, vol. 32, n°1, 1986, pp. 54--62.
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There is no canonical algorithm for SGE: from the same idea, details can differ in the implementation and the choice of some parameters.
This document was translated from LATEX by HEVEA.
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# ATLAS events
## Curriculum Learning Objective: Understand decay of particles in terms of the quark model
[e.g. OCR A-level 6.4.2(I)]
Beams of protons are accelerated around the ring of the Large Hadron Collider (LHC). These beams collide at the centre of the ATLAS detector. These collisions produce debris in the form of new particles which fly out in all directions and are recorded by the detector. Over a billion particle interactions take place in the ATLAS detector every second.
## Curriculum Learning Objective: Understand energy transfer (in context of particle acceleration)
Accelerating particles links to the high-school physics concept that when particles are accelerated the work done on them is converted to and therefore equal to their kinetic energy (this is covered e.g. in A-levels). Work is done on the protons by accelerating them through an electric field.
The protons within the two beams are grouped in ‘bunches’ which are squeezed down in size to increase the chances of a collision. Squeezing the protons into a smaller space increases the amount of them passing through a given volume per unit time, giving them a better chance of colliding. In the released data, the bunches crossed every 50 ns and there were about 30 collisions on average per bunch-crossing. This means, on average, there were about 20 million bunch crossings and 600 million collisions per second! Lots of very clever algorithms are used to pick out the interesting collisions to be analysed. This is because it would take far too long to process and gather all the data so it is better to pick the best ones and store them only.
An event is the data resulting from a particular bunch-crossing.
Pile-up is defined as the average number of particle interactions per bunch-crossing. It is directly correlated with something known as ‘instantaneous luminosity’.
Luminosity is one of the most important parameters of the LHC. Its details are hard to explain briefly, but it essentially gives a measure of how many collisions are happening in a particle accelerator. The higher the luminosity, the more data the experiments can gather increasing the likelihood of observing rare processes. However, as mentioned, increasing luminosity of course increases pile-up. This presents a challenge for physics analyses as it makes successfully identifying collisions of interest harder because so many particle interactions occur.
We use the term vertex (plural vertices) to refer to a point in the detector where particles collide or decay to produce new particles. In a typical collision event, several vertices are produced along the beam from proton-proton collisions. In order to have a manageable amount of data and get rid of a lot of uninteresting events, we choose one primary vertex to analyse. This primary vertex is defined as the inelastic collision of two protons with the highest overall transverse momentum - the collisions with lower transverse momentum are less likely to include interesting rare events.
It is important to correctly identify these vertices, and which of the observed particles come from which vertex, in order to suppress the effects from pile-up by removing particles from vertices we aren’t interested in. The event display at the top of the page shows a candidate Z boson decaying into two muons with 11 reconstructed vertices. This event was recorded on April 24th and is typical for the 2011 environment with high pile-up (lots of particle interactions per bunch crossing), the high pile-up can be seen by the large number of different lines in the picture. The reconstruction of vertices is important for many physics studies. This includes searches for new particles, identifying jets containing b-quarks or taus, and reconstruction of exclusive b-quark decays. (See e.g. OCR A Kerboodle textbook for the basics on Standard Model Particles: Pages 478-479)
## Recap: What is an event?
Data from when a bunch of (around 30) protons collide
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What happens to the supply curve when there is a decrease in production? a. Price will increase. c. Producers will produce more at each and every price. b. Price will decrease.
d. Consumers will increase purchases at each and every price.
What happens to the supply curve when there is a decrease in production: a. Price will increase
Question
Asked 336 days ago|8/21/2013 12:33:56 PM
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Weegy: (9x + 2)(9x - 2) = 81x^2 - 4 User: Select the product of (8x + 7)(x2 - 6x + 4).
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Switch to:
GuruFocus has detected 4 Warning Signs with Molson Coors Brewing Co \$TAP.
More than 500,000 people have already joined GuruFocus to track the stocks they follow and exchange investment ideas.
Molson Coors Brewing Co (NYSE:TAP)
E10
\$3.79 (As of Dec. 2016)
E10 is a concept invented by Prof. Robert Shiller, who uses E10 for his Shiller P/E calculation. E10 is the average of the inflation adjusted earnings of a company over the past 10 years.
Molson Coors Brewing Co's adjusted earnings per share data for the three months ended in Dec. 2016 was \$6.730. Add all the adjusted EPS for the past 10 years together and divide 10 will get our e10, which is \$3.79 for the trailing ten years ended in Dec. 2016.
As of today, Molson Coors Brewing Co's current stock price is \$96.30. Molson Coors Brewing Co's E10 for the quarter that ended in Dec. 2016 was \$3.79. Molson Coors Brewing Co's Shiller P/E Ratio of today is 25.41.
During the past 13 years, the highest Shiller P/E Ratio of Molson Coors Brewing Co was 34.34. The lowest was 13.26. And the median was 19.67.
Definition
E10 is a concept invented by Prof. Robert Shiller, who uses E10 for his Shiller P/E calculation. When we calculate the today's Shiller P/E ratio of a stock, we use todays price divided by E10.
What is E10? How do we calculate E10?
E10 is the average of the inflation adjusted earnings of a company over the past 10 years. Lets use an example to explain.
If we want to calculate the E10 of Wal-Mart (WMT) for Dec. 31, 2010, we need to have the inflation data and the earnings from 2001 through 2010.
We adjusted the earnings of 2001 earnings data with the total inflation from 2001 through 2010 to the equivalent earnings in 2010. If the total inflation from 2001 to 2010 is 40%, and Wal-Mart earned \$1 a share in 2001, then the 2001s equivalent earnings in 2010 is \$1.4 a share. If Wal-Mart earns \$1 again in 2002, and the total inflation from 2002 through 2010 is 35%, then the equivalent 2002 earnings in 2010 is \$1.35. So on and so forth, you get the equivalent earnings of past 10 years. Then you add them together and divided the sum by 10 to get E10.
For example, Molson Coors Brewing Co's adjusted earnings per share data for the three months ended in Dec. 2016 was:
Adj_EPS = Earnigns per Share / CPI of Dec. 2016 (Change) * Current CPI (Dec. 2016) = 6.73 / 241.432 * 241.432 = 6.730
Current CPI (Dec. 2016) = 241.432.
Molson Coors Brewing Co Quarterly Data
201409 201412 201503 201506 201509 201512 201603 201606 201609 201612 per share eps -0.19 0.5 0.43 1.23 0.09 0.18 0.78 0.8 0.94 6.73 CPI 238.031 234.812 236.119 238.638 237.945 236.525 238.132 241.018 241.428 241.432 Adj_EPS -0.193 0.514 0.44 1.244 0.091 0.184 0.791 0.801 0.94 6.73 201203 201206 201209 201212 201303 201306 201309 201312 201403 201406 per share eps 0.44 0.57 1.09 0.33 0.16 1.45 0.73 0.74 0.88 1.56 CPI 229.392 229.478 231.407 229.601 232.773 233.504 234.149 233.049 236.293 238.343 Adj_EPS 0.463 0.6 1.137 0.347 0.166 1.499 0.753 0.767 0.899 1.58 200909 200912 201003 201006 201009 201012 201103 201106 201109 201112 per share eps 1.26 1.19 0.56 1.27 1.37 0.58 0.44 1.18 1.06 0.95 CPI 215.969 215.949 217.631 217.965 218.439 219.179 223.467 225.722 226.889 225.672 Adj_EPS 1.409 1.33 0.621 1.407 1.514 0.639 0.475 1.262 1.128 1.016 200703 200706 200709 200712 200803 200806 200809 200812 200903 200906 per share eps 0.03 1.02 0.74 0.92 0.19 0.42 0.92 0.51 0.41 1.01 CPI 205.352 208.352 208.49 210.036 213.528 218.815 218.783 210.228 212.709 215.693 Adj_EPS 0.035 1.182 0.857 1.058 0.215 0.463 1.015 0.586 0.465 1.131
Add all the adjusted EPS together and divide 10 will get our e10.
Explanation
If a company grows much fast than inflation, E10 may underestimate the company's earnings power. Shiller P/E Ratio can seem to be too high even the actual P/E is low.
For the Shiller P/E, the earnings of the past 10 years are inflation-adjusted and averaged. The result is used for P/E calculation. Since it looks at the average over the last 10 years, the Shiller P/E is also called PE10.
The Shiller P/E was first used by professor Robert Shiller to measure the valuation of the overall market. The same calculation is applied here to individual companies.
Molson Coors Brewing Co's Shiller P/E Ratio of today is calculated as
Shiller P/E Ratio = Share Price / E10 = 96.30 / 3.79 = 25.41
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
During the past 13 years, the highest Shiller P/E Ratio of Molson Coors Brewing Co was 34.34. The lowest was 13.26. And the median was 19.67.
Be Aware
Shiller P/E Ratio works better for cyclical companies. It gives you a better idea on the company's real earnings power.
Related Terms
Shiller P/E Ratio
Historical Data
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
Molson Coors Brewing Co Annual Data
Dec07 Dec08 Dec09 Dec10 Dec11 Dec12 Dec13 Dec14 Dec15 Dec16 e10 2.06 2.15 2.43 2.67 2.90 2.91 2.94 2.92 3.06 3.79
Molson Coors Brewing Co Quarterly Data
Sep14 Dec14 Mar15 Jun15 Sep15 Dec15 Mar16 Jun16 Sep16 Dec16 e10 2.99 2.92 3.03 3.15 3.06 3.06 3.19 3.19 3.24 3.79
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https://math.stackexchange.com/questions/3302411/dimension-of-domain-is-greater-than-less-than-equal-to-dimension-of-range-for-a?noredirect=1
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# Dimension of domain is greater than/less than/equal to dimension of range for a smooth surjection/injection/submersion/immersion?
My book is An Introduction to Manifolds by Loring W. Tu.
Let $$N$$ and $$M$$ be smooth manifolds with dimensions. Let $$p \in N$$. Let $$F: N \to M$$ be a smooth map.
Question 1. Are these correct?
A. If $$F$$ is injective, then $$\dim N \le \dim M$$, by this (from Momentum Maps and Hamiltonian Reduction by Juan-Pablo Ortega, Tudor Ratiu)
B. If $$F$$ is open, then $$\dim N \ge \dim M$$, by this (from Momentum Maps and Hamiltonian Reduction by Juan-Pablo Ortega, Tudor Ratiu).
C. If $$F$$ is a submersion, then $$F$$ is open and so $$\dim N \ge \dim M$$. Alternatively, we may use this.
D. If $$F$$ is an immersion, then $$\dim N \le \dim M$$, by this.
Question 2. Given that injections, immersions and submersions imply either $$\dim N \le \dim M$$ or $$\ge$$, I guess surjections imply one of those too. Which if any does surjection imply, and why?
• I think it's the same as submersion (and open): $$\dim N \ge \dim M$$.
• An example would be retraction $$r(x) = \frac{x}{||x||}, r: \mathbb R^2 \setminus 0 \to S^1$$ (I recall this is smooth. Not sure). At the very least, I think the example (if correct) proves that surjections definitely do not imply $$\dim N \le \dim M$$.
• 1C and 1D follow just by definition of submersion/immersion, considering one single tangent space. 1A and 1B are right (I’d prefer a complete argument based on constant-rank theorem, but okay). For 2, if $F:N \rightarrow M$ is smooth surjective, then $\dim\,N \geq \dim\,M$ (there is an issue with Sard’s theorem otherwise). – Mindlack Jul 24 '19 at 9:30
• @Mindlack Thanks! – user636532 Jul 24 '19 at 9:32
• @Mindlack Sard's Theorem is related to 1A or 1B? 1A looks like an injective version for Sard's Theorem. – user636532 Jul 24 '19 at 9:39
• Sard is for your question 2. I suppose it also works for question 1B (if the dimension of the source is smaller, the only regular values are the values not reached by the function, which is a full-measure closed subset, thus the entire manifold, a contradiction). – Mindlack Jul 24 '19 at 10:01
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Here is the list of online maths exercises on numbers. Each corrected exercise is accompanied by hints, course reminders and methodological advice to help you practise independently.
72 exercises
• N°1110 (Numbers) : Countdown is an arithmetic exercise that allows you to practice quick mental calculation in an efficient way. The goal is to reconstruct an integer using other integers and basic arithmetic operations (+,-,*,/).
Exercise example :
Use yellow numbers and blue arithmetic operators to calculate the target number.
## Countdown
1110 arithmetic_solver
• N°1111 (Numbers) : The purpose of this exercise is to compare two integers, using the appropriate comparison operator.
Exercise example :
Compare the following numbers: 751 ... 956.
1111
• N°1112 (Numbers) : The goal of this corrected exercise is to find the missing sign of an arithmetic expression, course reminders accompany this exercise.
Exercise example :
What is the missing sign in the expression: 15 ... 5=3.
1112
• N°1113 (Numbers) : The purpose of this math activity is to calculate the percentage of an integer.
Exercise example :
What is the result of 10% of 80.
1113
• N°1114 (Numbers) : The objective of this numerical activity is to multiply a decimal number by 10,100,1000 or 0.1,0.01, or 0.001.
Exercise example :
What is the result of the following product: 836.912 * 0.001 ?
• N°1115 (Numbers) : The purpose of this exercise is to find the result of arithmetic operations (+, -, *, /) with integers.
Exercise example :
What is the result of the following operation 473+295 ?
• N°1116 (Numbers) : The purpose of this exercise on Euclidean division of two integers is to find the quotient and remainder.
Exercise example :
Compute the quotient and remainder of the Euclidean division of 621 by 45.
1116 6th Grade euclidean_division
• N°1117 (Numbers) : The purpose of this exercise is to round a decimal number to a given precision.
Exercise example :
Give the rounding to 2 decimal places of : 11.445631.
1117 6th Grade round
• N°1118 (Numbers) : This graphical exercise allows to learn how to locate in the plane, to succeed, you just have to place at the right place a point of given abscissa.
Exercise example :
Point A is on the x-axis, its x-coordinate is 7. Place it correctly.
• N°1119 (Numbers) : The objective of this activity is to complete a "hole" operation, otherwise known as solving a simple equation.
Exercise example :
What is the missing number in the following expression : ...-5 = 4?
1119
• N°1124 (Numbers) : This exercise corrected of calculation makes it possible to practise on line to transform a decimal number into percentage.
Exercise example :
Express 0.97 as a percentage.
1124 percentage
• N°1125 (Numbers) : This corrected math exercise allows you to practice transforming a fraction into a percentage online.
Exercise example :
Express 24/25 as a percentage.
1125 percentage
• N°1129 (Numbers) : This activity on fractions allows to practice simplifying an integer fraction using the GCD.
Exercise example :
Simplify the following fraction 7/14.
1129 6th Grade fraction
• N°1130 (Numbers) : This corrected exercise on fractions aims to complete a fractional expression so that an equality is verified.
Exercise example :
By what value must the following expression 10/4=.../36 be completed for the equality to be verified?
• N°1131 (Numbers) : The purpose of this corrected fraction exercise is to simplify a fractional expression.
Exercise example :
Write in the form of a fraction or a whole number the following expression: 1/4*14.
1131 6th Grade fraction
• N°1132 (Numbers) : The purpose of this exercise is to give an approximate value to one decimal place of an algebraic.
Exercise example :
Give an approximate value to one decimal place of the following expression: 10/17*19
• N°1211 (Numbers) : The goal of this whole number math exercise is to practice multiplying and adding them.
Exercise example :
What is the result of 87*34+7027.
1211 7th Grade calculator
• N°1212 (Numbers) : The objective of this corrected calculus exercise is to add a fraction with a whole number.
Exercise example :
What is the result of 14+1040/80.
1212 7th Grade fraction
• N°1213 (Numbers) : The objective of this corrected math exercise is to simplify a fraction after calculating its denominator.
Exercise example :
What is the result of 3768/(91+66).
1213 7th Grade fraction
• N°1214 (Numbers) : The objective of this math exercise is to simplify a fraction that has a fraction as the denominator.
Exercise example :
What is the result of 96/(32/4).
1214 7th Grade fraction
• N°1215 (Numbers) : The objective of this corrected math exercise is to calculate a written expression in natural language.
Exercise example :
What is the result of 5 plus 4 plus 2.
• N°1217 (Numbers) : This exercise allows you to apply the techniques of simplification of fractions to calculate the product of two fractions.
Exercise example :
What is the result of the following fraction product: (24/2)*(14/3).
1217 7th Grade fraction
• N°1219 (Numbers) : The purpose of this exercise is to compare two fractions, choosing the appropriate comparison operator from the list provided.
Exercise example :
Compare the following fractions 6/4...5/4.
1219 7th Grade compare_fractions
• N°1220 (Numbers) : This corrected exercise aims to practice addition, subtraction or multiplication of 2 fractions.
Exercise example :
Perform the following calculation 3/5*4/6.
1220 7th Grade fraction
• N°1223 (Numbers) : The purpose of this exercise is to calculate an arithmetic expression composed of addition and subtraction, taking into account the priorities of the operations.
Exercise example :
What is the result of (6-4)+(9+6)+(3-10).
1223 7th Grade calculator
• N°1225 (Numbers) : The purpose of this corrected math exercise is to find the denominator of a fraction from an equality.
Exercise example :
What is the missing number in the following expression 27/?=3.
• N°1226 (Numbers) : The purpose of this exercise is to verify an equality, the goal being to learn about equation solving.
Exercise example :
Is the following equality 4+x-5 = 2 true for x = 2?
• N°1236 (Numbers) : The purpose of this number math exercise is to practice adding relative numbers.
Exercise example :
What is the result of 8+(-10)?
1236 7th Grade evaluate
• N°1237 (Numbers) : The purpose of this corrected math exercise is to complete an equality that involves relative numbers.
Exercise example :
What is the missing number in the following expression: 2+...=-4.
• N°1238 (Numbers) : The purpose of this math exercise is to string together a series of additions and subtractions of relative numbers.
Exercise example :
What is the result of 10+6+4-6-7-5.
1238 7th Grade evaluate
• N°1241 (Numbers) : The purpose of this exercise is to learn about solving equations using magic squares.
Exercise example :
In a magic square, the sum of the numbers in each row, in each column and on each diagonal is the same.
Complete the following magic square:
1241
• N°1311 (Numbers) : The objective of this algebraic calculus exercise is to expand an algebraic expression.
Exercise example :
Expand the following expression -10*(-12).
1311 8th Grade evaluate
• N°1312 (Numbers) : The objective of this exercise is to put the product of a number by the inverse of another number in fraction form.
Exercise example :
Put in the form a*1/b the following fraction 19/6
• N°1313 (Numbers) : This exercise allows to practice calculations with decimals, the goal is to give the approximate value of a fraction to one decimal place.
Exercise example :
Give a value approximating to 0.1 of the following fraction 19/6.
• N°1314 (Numbers) : The objective of this activity is to simplify a fraction whose numerator and denominator are composed of products of whole numbers.
Exercise example :
Simplify the following fraction (3*3)/(8*3).
1314 8th Grade fraction
• N°1315 (Numbers) : This exercise allows to learn how to simplify a fraction whose numerator and denominator are composed of product of relative numbers.
Exercise example :
Put in the form of an irreducible fraction: -5*8/(9*6).
1315 8th Grade fraction
• N°1316 (Numbers) : The objective of this exercise is to simplify a fraction whose numerator and denominator are fractions, in other words, a fraction of fractions.
Exercise example :
Put in the form of an irreducible fraction: ((2)/(-4))/((6)/(9)).
1316 8th Grade fraction
• N°1317 (Numbers) : The objective of this numerical computation exercise is to write in the form of a power of 10 a product of numbers.
Exercise example :
Write the following product as a power of 10: 10^(-10)*10^(-3).
• N°1318 (Numbers) : The purpose of this exercise is to practice finding the writing in the form of a power of 10, of a number raised to a power.
Exercise example :
Write the following product as a power of 10: (10^(-6))^(-9).
• N°1319 (Numbers) : The purpose of this corrected math exercise is to write a number with scientific notation.
Exercise example :
Write using scientific notation the following number: -0.7539046.
• N°1321 (Numbers) : This exercise allows to practice putting a product of numbers in the form of a power.
Exercise example :
Write in the form of a single power the following expression: 20^(-3)*20^(6).
• N°1322 (Numbers) : The purpose of this math exercise is to complete an equality that involves powers.
Exercise example :
With what value must the question mark be replaced for the following equality to be true? 94^(-3)*94^(?)=94^(-8)
• N°1324 (Numbers) : The objective of this exercise is to give an approximate value of a square root to one decimal place.
Exercise example :
Give a value approximated to 10^-1 of sqrt(91).
1324 8th Grade sqrt
• N°1328 (Numbers) : The objective of this solved exercise is to compare two fractions using the correct operator.
Exercise example :
Compare the following two expressions 90/10 ... 27/10.
1328 8th Grade compare_fractions
• N°1329 (Numbers) : The objective of this math exercise is to frame a fraction using two decimal numbers.
Exercise example :
Give a frame to the nearest 10^-1 of the following fraction 25/4.
• N°1330 (Numbers) : The objective of this math exercise is to equate a simple problem to solve it.
Exercise example :
I have a number, I add 27 to it, I subtract 22 from the result and I get 43, what is this number?
1330 8th Grade equation_solver
• N°1331 (Numbers) : The purpose of this exercise on proportionality is to check that a table is indeed a proportionality table.
Exercise example :
Does the following table correspond to a proportionality table?
643
604132
1331
• N°1332 (Numbers) : The purpose of this corrected calculus exercise is to find the coefficient of a proportionality table.
Exercise example :
What is the coefficient of proportionality that allows us to go from the first to the second line of the following table?
971
54426
1332
• N°1333 (Numbers) : The purpose of this exercise is to find the missing value of a proportionality table.
Exercise example :
By what value must we replace the ? so that the following table is a proportionality table?
7348
28?1632
1333
• N°1334 (Numbers) : The purpose of this exercise is to complete using the cross product, a simple table so that it is a proportionality table.
Exercise example :
What is the missing value in the following proportionality table?
x80
294560
1334
• N°1335 (Numbers) : The goal of this math exercise is to solve a first degree inequation with one unknown.
Exercise example :
We suppose that 3*x-2<5. Choose the expression that is true from the list below.
1335 8th Grade inequality_solver
• N°1427 (Numbers) : The objective of this corrected arithmetic exercise is to calculate the gcd of two integers.
Exercise example :
Calculate the gcd of 225 and 84.
1427 9th Grade gcd
• N°1428 (Numbers) : The objective of this exercise is to use the GCD to check if two integers are prime to each other.
Exercise example :
Are the following numbers 205 and 318 prime to each other ?
1428 9th Grade gcd
• N°1429 (Numbers) : The purpose of this exercise is to put a fraction into its irreducible form.
Exercise example :
Put the following fraction in its irreducible form: 144/210.
1429 9th Grade fraction
• N°1434 (Numbers) : The objective of this corrected calculus exercise is to simplify an expression containing fractions.
Exercise example :
Write D as an irreducible fraction: D = ((9/6+5/7)*2)/8
1434 9th Grade fraction
• N°1435 (Numbers) : The purpose of this corrected math exercise is to calculate the value of a function for a given number.
Exercise example :
Let f be the application of R in R defined by f(x)=2*(3*x+1)^2-6*(2*x+3)^2 :
Compute f(2).
1435 9th Grade array_values
• N°1436 (Numbers) : The objective of this exercise is to reduce a fraction, simplify a square root, and calculate a fraction that contains powers.
Exercise example :
Let A=3/8+10/6-8/4 , B=-8*sqrt(24)-4*sqrt(216)-5*sqrt(54)-1*sqrt(54) and C=(18*10^5*3*10^6)/(6*10^11)
1. Compute A and give the result as an irreducible fraction.
2. Write B in the form a*(sqrt(6)) where a is a relative integer.
3. Write C as an integer.
1436 9th Grade fraction
• N°1437 (Numbers) : The objective of this exercise is to calculate the gcd of two numbers and simplify a fraction.
Exercise example :
1. Compute the GCD of the numbers 5104 and 968.
2. Write the fraction 5104/968 in irreducible form.
1437 9th Grade fraction
• N°1511 (Numbers) : This corrected exercise consists simply in calculating the absolute value of a numerical expression.
Exercise example :
Calculate the absolute value of C=8+9.
1511 10th Grade abs
• N°1512 (Numbers) : This corrected exercise consists simply in calculating the absolute value of an algebraic expression composed of fractions.
Exercise example :
Calculate the absolute value of F=2/3-3/7.
1512 10th Grade abs
• N°1515 (Numbers) : The purpose of this corrected exercise is to complete the decomposition of a number into prime numbers.
Exercise example :
Indicate by which number the "question mark" must be replaced in the prime decomposition of 60 so that the following equality is verified.
60 = 3*5*?*?
1515 10th Grade prime_factorization
• N°1516 (Numbers) : The purpose of this exercise is to find the ordered decomposition of a number into primes.
Exercise example :
Give the decomposition of 854 into a product of prime numbers by ordering the factors and using the power operator ^ if necessary.
1516 10th Grade prime_factorization
• N°1517 (Numbers) : The purpose of this corrected arithmetic exercise is to determine if a number is a prime number.
Exercise example :
51 is an integer, is it prime ?
1517 10th Grade prime_factorization
• N°1518 (Numbers) : The goal of this exercise is to find the ordered decomposition of a product of numbers into primes.
Exercise example :
Give the product decomposition of the following expression 30*16 by ordering the factors and using the power operator ^ if necessary.
1518 10th Grade prime_factorization
• N°1520 (Numbers) : The purpose of this exercise is to simplify a fraction using the decomposition of a number into a product of prime factors.
Exercise example :
Write in the form of an irreducible fraction the following fraction (40*28)/(35*24) using the decomposition into prime factors.
1520 10th Grade fraction
• N°1539 (Numbers) : The purpose of this exercise is to use algebraic computation techniques to determine the irreducible form of a division of fractions.
Exercise example :
Put into irreducible fraction form: ((-9)/(20))/((-36)/(-15)).
1539 10th Grade fraction
• N°1541 (Numbers) : The purpose of this corrected calculus exercise is to use algebraic calculus techniques to simplify a product of fractions.
Exercise example :
Put into irreducible fraction form: ((-9)/(20))/((-36)/(-15)).
1541 10th Grade fraction
• N°11171 (Numbers) : This exercise allows you to practice rounding a decimal number to the nearest whole number.
Exercise example :
Rounding to the nearest whole number 71.710931
11171 6th Grade round
• N°11172 (Numbers) : To pass this problem on rounding, you just have to round correctly a decimal number to one digit after the decimal point.
Exercise example :
Rounding to one digit after the decimal point 72.597345.
11172 6th Grade round
• N°11211 (Numbers) : This graphical reading activity consists in reading correctly the coordinates of a point placed in a reference frame.
Exercise example :
What are the coordinates of the following point ?
11211
• N°11212 (Numbers) : To do this exercise of location in the plane, you must place a point in a reference frame from its abscissa and its ordinate.
Exercise example :
The point A has coordinates (7;0). Place it correctly.
11212
• N°14110 (Numbers) : The purpose of this corrected math exercise is to simplify a square root.
Exercise example :
Write in the form sqrt(a)/b, the following expression 1/sqrt(360), where a and b represent two integers.
14110 9th Grade simplify_surd
The Numbers topic is available for : 6th Grade, 7th Grade, 8th Grade, 9th Grade, 10th Grade
List of exercises by class : middle and high schools, 6th Grade, 7th Grade, 8th Grade, 9th Grade, 10th Grade, 11th Grade, 12th Grade.
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A178785 a(n) is the smallest n-perfect number of the form 2^(n+1)*l, where l is an odd number with exponents<=n in its prime power factorization, and a(n)=0 if such n-perfect number does not exist 0
60, 6552, 222768, 288288, 87360, 49585536, 25486965504, 203558400, 683289600, 556121548800 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS Let k>=1. In the multiplicative basis Q^(k)={p^(k+1)^j, p runs A000040, j=0,1,...} every positive integer m has unique factorization of the form m=Prod{q is in Q^(k)}q^(m_q), where m_q is in {0,1,...,k}. In particular, in the case of k=1, we have the unique factorization over distinct terms of A050376. Notice that the standard prime basis is the limiting for k tending to infinity, and, by the definition, Q^(infinity)=A000040. Number d is called a k-divisor of m if the exponents d_q in its factorization in basis Q^(k) do not exceed m_q. A number m is called k-perfect if it equals to the sum of its proper positive k-divisors. Conjecture. a(11)=0. Note that we also know n-perfect numbers for n=12,14,15,16 and 18. LINKS S. Litsyn and V. S. Shevelev, On factorization of integers with restrictions on the exponent, INTEGERS: Electronic Journal of Combinatorial Number Theory, 7 (2007), #A33, 1-36. FORMULA m = Prod{q is in Q^(k)}q^(m_q) is k-perfect number iff Prod{q is in Q^(k)}(q^((m_q)+1)-1)/(q-1)=2*m. EXAMPLE In case of n=2, we have the basis ("2-primes"): 2,3,5,7,8,11,13,...By the formula, we construct from the left m and from the right 2*m. By the condition, m begins from "2-prime" 8. From the right we have 8+1=3^2, therefore from the left we have 8*3^2 and from the right 3^2*(3^3-1)/(3-1)=3^2*13. Thus from the left it should be 8*3^2*13 and from the right 3^2*13*14. Finally, from the left we obtain m=8*3^2*13*7=6552 and from the right we have 2*m=3^2*13*14*8. By the construction, it is the smallest 2-perfect number of the required form. Thus a(2)=6552. CROSSREFS Cf. A000396, A050376, A007357, A092356. Sequence in context: A289307 A091032 A328951 * A091753 A303790 A327678 Adjacent sequences: A178782 A178783 A178784 * A178786 A178787 A178788 KEYWORD nonn,more AUTHOR Vladimir Shevelev, Jun 14 2010, Jun 18 2010 STATUS approved
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Depdendent Variable
Number of equations to solve: 23456789
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Equ. #2:
Equ. #3:
Equ. #4:
Equ. #5:
Equ. #6:
Equ. #7:
Equ. #8:
Equ. #9:
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1464. Maximum Product of Two Elements in an Array
Easy
1.6K
188
Given the array of integers `nums`, you will choose two different indices `i` and `j` of that array. Return the maximum value of `(nums[i]-1)*(nums[j]-1)`.
Example 1:
```Input: nums = [3,4,5,2]
Output: 12
Explanation: If you choose the indices i=1 and j=2 (indexed from 0), you will get the maximum value, that is, (nums[1]-1)*(nums[2]-1) = (4-1)*(5-1) = 3*4 = 12.
```
Example 2:
```Input: nums = [1,5,4,5]
Output: 16
Explanation: Choosing the indices i=1 and j=3 (indexed from 0), you will get the maximum value of (5-1)*(5-1) = 16.
```
Example 3:
```Input: nums = [3,7]
Output: 12
```
Constraints:
• `2 <= nums.length <= 500`
• `1 <= nums[i] <= 10^3`
Accepted
193.9K
Submissions
241.9K
Acceptance Rate
80.1%
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# Confidence Intervals and Standard Deviation
In a manufacturing process, the diameter of the steel rods is the product characteristic. A random sample of 25 steel rods is taken and sample mean is 8.88 cm. The diameter is assumed to be normally distributed with standard deviation 1.25.
(i) Find a 95% confidence interval for the true diameter. If the standard deviation is
unknown and the sample standard deviation is 1.25, will the 95% confidence interval be narrower or wider?
(ii) If we would like the difference between the true mean and the sample mean to be
less than 0.15, what is the minimum number of steel rods that we need to sample. Use alpha = 0.05.
#### Solution Preview
Please see the attached file for the complete solution.
Thanks for using BrainMass.
Inference on mean - known variance
In a manufacturing process, the diameter of the steel rods is the product characteristic. A random sample of 25 steel rods is taken and sample mean is 8.88 cm. The diameter is assumed to be normally distributed with standard deviation 1.25.
(i) Find a 95% confidence interval for the true diameter. If the standard deviation is
unknown and the ...
#### Solution Summary
Confidence intervals and standard deviation are investigated.
\$2.19
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This 8 part flip book covers the following Properties of Exponents concepts and will help your students with Laws of Exponents Activities: • Zero Exponents - 4 examples • Negative Exponents - 5 examples • Parenthesis & Exponents - 3 examples • Adding & Subtracting Exponents - 3 examples • Mu
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This set of foldable notes is designed to use in an interactive notebook to be a reference for the rules or properties of exponents. There is an example in variable terms on the flap and students write a brief explanantion under the flap. ***This resources is also included in the Properties of Exp
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This foldable for the multiplication properties of exponents has three flaps. Under each flap the property is listed along with four examples. A student practice sheet is included. It could be used as an additional practice page in student's notebooks or it could be used as homework. An answer k
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This foldable for the division properties of exponents has two flaps. Under each flap the property is listed along with eight examples. Under each flap, six of the examples only use the new property, while the other two use previous properties to make a more complicated example. An answer key is
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This package includes a template you can print to create a foldable on the properties of exponents as well as a smartboard file that matches the foldable and can be used as a walk-through in creating the foldable. Great for students with IEPs who need to have some of their notes already written for
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This foldable is designed for interactive math notebooks. This foldable walks students through the following concepts: 1. Quotient of Powers (example followed by 8 practice problems) 2. Power of a Quotient (example followed by 4 practice problems) 3. Zero Exponent (example followed by 6 pract
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Also included in: Algebra 1 - Second Half of the School Year Foldable GROWING BUNDLE
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This is a foldable designed for math interactive notebooks. This foldable guides students through the process of: 1. a Product of Powers - 1 example and 5 practice problems 2. a Power of a Power - 1 example and 4 practice problems 3. a Power of a Product - 1 example and 4 practice problems An
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These pages create two foldable pages on the multiplication properties of exponents. One foldable covers the product of powers property. The other foldable is on the power of a power property & the power of a product property. Interactive Notebooks are a great way to share information with stud
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These pages create two foldable pages on the division properties of exponents. One is on the quotient of powers property and the power of a quotient property. Instructions on how to put the pages together are also included. Interactive Notebooks are a great way to share information with students an
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This foldable reinforces the properties of exponents with a fun foldable! Use in your Interactive Notebook or just for funsies! Directions included!
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This package includes a template you can print to create a foldable on the properties of exponents as well as a smartboard file that matches the foldable and can be used as a walk-through in creating the foldable. Great for students with IEPs who need to have some of their notes already written for
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This foldable shows all exponent properties, including product of like bases, quotient of like bases power to a power, product to a power, quotient to a power, zero exponent, and negative exponents, with an example and the expanded form. Copy front and back and have students follow the directions t
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Exponent rules foldable style notes. About this resource : This foldable style activity allows students to have a compact and engaging way to learn or review the give basic rules of exponents. Great for Interactive Notebooks! Rules covered: Product Rule Quotient Rule Power Rules (Power of a pow
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Students love getting practice with the properties of exponents with this variety of games and activities. These 7 resources include a presentation, notes, foldable, mazes, tic tac toe game, a whole class review game, puzzles, and more! Supports CCSS 8.EE.A.1 The following resources are included
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A fun foldable to teach writing rational exponents in radical form, writing radicals in rational exponent form, and properties of rational exponents.
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Now with TWO options! This foldable allows students to discover, take notes, and practice skills associated with zero and negative exponents. Option 1: It begins with a chart where students extend a pattern of positive exponents to discover the way zero and negative exponents work. The followin
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Properties of Exponents Rules Foldable Interactive Notebook Graphic Organizer This is a double sided PDF foldable that explains the Properties of Exponents. Each “rule” is explained in written form and includes an example. When printed the foldable is slightly larger than a composition notebook p
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This 12- question scavenger hunt provides students with practice applying the following properties of exponents: • Zero Exponents • Negative Exponents • Product Property • Power to a Power • Quotient Property Works great to use as a review before a quiz or test! This product is also included in t
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Use this foldable to teach students the properties of integer exponents. Print two-sided, fold in half on the line and cut on the dotted lines. Now includes 22 practice problems and answer key. Don't forget to rate me to earn your TPT credits and follow me to be the first to see my new products an
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This foldable provides an introduction to exponents. With organized notes, and practice problems, students will be exposed to the following vocabulary and types of problems: • Vocabulary (power, base, exponent) • Zero Product Property • Rewriting repeated multiplication using exponents • Filling
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In this activity, students will follow the solution path to discover how to apply properties of integer exponents. This is done by verbalizing the solution process — or sharing thoughts with students so that they can effectively crawl inside a teacher's brain — it is what´s known as a "think-alou
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Not everything has to be a foldable! This is a great notebook "tape-in" graphic organizer to help your students during instruction involving properties of exponents. Each page contains 4 tape-in organizers.
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***** SEE PREVIEW FOR SAMPLES & INSTRUCTIONS ***** Student-Centered, Guided/Scaffolded Notes, Bell Ringer, Warmup This printable can be used by all levels of math students from grades 6-12. Included with this foldable are the following resources/pages: 1. Completed Exponents Rules w/Flap Ti
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This is a foldable graphic organizer that has the steps for simplifying numbers with negative exponents to their fraction form. It works great in your interactive notebook. This will help your students with one of the properties of exponents. Also, there are example problems for students to comple
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https://instalearn.ai/interview-playground/data-structure-algorithm/maximum-unit-of-boxes-894
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894. Maximum Unit of Boxes
1
Easy
You are assigned to put some amount of boxes onto one truck. You are given a 2D array boxTypes, where
boxTypes[i] = [numberOfBoxes(i) , numberOfUnitsPerBox(i)]: * NumberOfBoxes(i) is the number of boxes of type i. * numberOfUnitsPerBox(i) is the number of units in each box of the type i.
You are also given an integer truckSize, which is the maximum number of boxes that can be put on the truck. You can choose any boxes to put on the truck as long as the number of boxes does not exceed truckSize. Print the maximum total number of units that can be put on the truck.
Input Format
First Line contains two space separated integers N and K(truck size) .
Next N Lines contains boxtype[i] where each boxtype is of size 2 .
Output Format
An integer denoting the maximum unit of boxes .
Example
Input
4 10 5 10 2 5 4 7 3 9
Output
91
Constraints
1 <= boxTypes.length <= 1000
1 <= numberOfBoxes(i), numberOfUnitsPerBox(i) <= 1000
1 <= truckSize <= 106
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CC-MAIN-2023-50
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https://homeworkhelplocal.com/joliet-junior-college-mathematics-homework-help/
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# Joliet junior college | Mathematics homework help
The campus at Joliet Junior College has a lake. A student used a Secchi disk to measure the clarity of the lake’s water by lowering the disk into the water and measuring the distance below the water surface at which the disk is no longer visible. The following measurements (in inches) were taken on the lake at various points in time over the course of a year.
(a) Use the data to compute a point estimate for the population mean and population standard deviation.
(b) Because the sample size is small, we must verify that the data are normally distributed and do not contain any outliers. The figures show the normal probability plot and boxplot. Are the conditions for constructing a confidence interval about m satisfied?
(c) Construct a 95% confidence interval for the mean Secchi disk measurement. Interpret this interval.
(d) Construct a 99% confidence interval for the mean Secchi disk measurement. Interpret this interval.
(e) Construct a 95% confidence interval for the population standard deviation Secchi disk measurement. Interpret this interval.
(550 words)
Approximate price: \$22
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# Kerodon
$\Newextarrow{\xRightarrow}{5,5}{0x21D2}$ $\newcommand\empty{}$
### 4.8.8 Relative Higher Homotopy Categories
Let $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ be a functor between categories. Recall that the essential image of $F$ is the full subcategory $\operatorname{\mathcal{D}}' \subseteq \operatorname{\mathcal{D}}$ spanned by those objects which are isomorphic to $F(X)$, for some object $X \in \operatorname{\mathcal{C}}$. The functor $F$ then factors as a composition
$\operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}_0 \hookrightarrow \operatorname{\mathcal{D}},$
where the functor on the left is essentially surjective and the functor on the right is fully faithful. It is sometimes useful to consider a different factorization.
Proposition 4.8.8.1. Let $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ be a functor between categories. Then $F$ factors as a composition
$\operatorname{\mathcal{C}}\xrightarrow {F'} \operatorname{\mathcal{D}}' \xrightarrow {G} \operatorname{\mathcal{D}}$
where $G$ is faithful and $F'$ is both full and essentially surjective.
Proof. We construct the category $\operatorname{\mathcal{D}}'$ as follows:
• The objects of $\operatorname{\mathcal{D}}'$ are the objects of $\operatorname{\mathcal{C}}$. To avoid confusion, for each object $X \in \operatorname{\mathcal{C}}$, we write $\overline{X}$ for the corresponding object of $\operatorname{\mathcal{D}}'$.
• For every pair of objects $X,Y \in \operatorname{\mathcal{C}}$, we take $\operatorname{Hom}_{ \operatorname{\mathcal{D}}' }( \overline{X}, \overline{Y} )$ to be image of the map $F_{X,Y}: \operatorname{Hom}_{\operatorname{\mathcal{C}}}(X,Y) \rightarrow \operatorname{Hom}_{\operatorname{\mathcal{D}}}( F(X), F(Y) )$. To avoid confusion, if $u: F(X) \rightarrow F(Y)$ is a morphism of $\operatorname{\mathcal{D}}$ which belongs to the image of $F_{X,Y}$, we write $\overline{u}: \overline{X} \rightarrow \overline{Y}$ for the corresponding morphism of $\operatorname{\mathcal{D}}'$.
• For every pair of objects $X,Y, Z \in \operatorname{\mathcal{C}}$, the composition law
$\circ : \operatorname{Hom}_{\operatorname{\mathcal{D}}'}( \overline{Y}, \overline{Z} ) \times \operatorname{Hom}_{\operatorname{\mathcal{D}}'}( \overline{X}, \overline{Y} ) \rightarrow \operatorname{Hom}_{ \operatorname{\mathcal{D}}'}( \overline{X}, \overline{Z} )$
is the restriction of the composition law $\operatorname{Hom}_{\operatorname{\mathcal{D}}}( F(Y), F(Z) ) \times \operatorname{Hom}_{\operatorname{\mathcal{D}}}( F(X), F(Y) ) \rightarrow \operatorname{Hom}_{\operatorname{\mathcal{D}}}( F(X), F(Z) )$ for the category $\operatorname{\mathcal{D}}$: that is, it satisfies the formula $\overline{v} \circ \overline{u} = \overline{v \circ u}$.
Let $F': \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}'$ be the functor which carries each object $X \in \operatorname{\mathcal{C}}$ to the object $\overline{X} \in \operatorname{\mathcal{D}}'$, and each morphism $u: X \rightarrow Y$ of $\operatorname{\mathcal{C}}$ to the morphism $\overline{ F(u) }: \overline{X} \rightarrow \overline{Y}$ of $\operatorname{\mathcal{D}}'$. Let $G: \operatorname{\mathcal{D}}' \rightarrow \operatorname{\mathcal{D}}$ be the functor which carries each object $\overline{X} \in \operatorname{\mathcal{D}}'$ to the object $F(X) \in \operatorname{\mathcal{D}}$, and each morphism $\overline{u}: \overline{X} \rightarrow \overline{Y}$ of $\operatorname{\mathcal{D}}'$ to the morphism $u: F(X) \rightarrow F(Y)$ of $\operatorname{\mathcal{D}}$. Then the functor $G$ is faithful, the functor $F'$ is full and essentially surjective, and the composition $G \circ F'$ is equal to $F$. $\square$
Exercise 4.8.8.2 (Uniqueness). Let $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ be a functor between categories. The proof of Proposition 4.8.8.1 constructs a factorization
$\operatorname{\mathcal{C}}\xrightarrow {F'} \operatorname{\mathcal{D}}' \xrightarrow {G} \operatorname{\mathcal{D}}$
where $G$ is faithful and $F'$ is both full and bijective on objects. Show that these properties characterize the category $\operatorname{\mathcal{D}}'$ up to (unique) isomorphism.
Our goal in this section is to prove the following $\infty$-categorical generalization of Proposition 4.8.8.1:
Theorem 4.8.8.3. Let $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ be a functor of $\infty$-categories and let $n$ be an integer. Then $F$ admits a factorization $\operatorname{\mathcal{C}}\xrightarrow {F'} \operatorname{\mathcal{D}}' \xrightarrow {G} \operatorname{\mathcal{D}}$ with the following properties:
• The functor $G$ is essentially $n$-categorical: that is, it is $m$-full for $m \geq n+2$.
• The functor $F'$ is categorically $(n+1)$-connective: that is, it is $m$-full for $m \leq n+1$.
Example 4.8.8.4. For $n \leq -2$, Theorem 4.8.8.3 asserts that every functor of $\infty$-categories $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ admits a factorization $\operatorname{\mathcal{C}}\xrightarrow {F'} \operatorname{\mathcal{D}}' \xrightarrow {G} \operatorname{\mathcal{D}}$, where the functor $G$ is an equivalence of $\infty$-categories. This is trivial: we can take $\operatorname{\mathcal{D}}' = \operatorname{\mathcal{D}}$ and $G$ to be the identity functor.
Example 4.8.8.5. When $n = -1$, Theorem 4.8.8.3 asserts that every functor of $\infty$-categories $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ admits a factorization $\operatorname{\mathcal{C}}\xrightarrow {F'} \operatorname{\mathcal{D}}' \xrightarrow {G} \operatorname{\mathcal{D}}$, where the functor $G$ is fully faithful and the functor $F'$ is essentially surjective. For example, we can take $\operatorname{\mathcal{D}}' \subseteq \operatorname{\mathcal{D}}$ to be the essential image of the functor $F$, and $G: \operatorname{\mathcal{D}}' \hookrightarrow \operatorname{\mathcal{D}}$ to be the inclusion map. See Remark 4.6.2.12.
Example 4.8.8.6. When $n = 0$, Theorem 4.8.8.3 asserts that every functor of $\infty$-categories $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ admits a factorization $\operatorname{\mathcal{C}}\xrightarrow {F'} \operatorname{\mathcal{D}}' \xrightarrow {G} \operatorname{\mathcal{D}}$, where the functor $G$ is faithful and the functor $F'$ is both full and essentially surjective. When $\operatorname{\mathcal{C}}$ and $\operatorname{\mathcal{D}}$ are (nerves of) ordinary categories, this follows from Proposition 4.8.8.1. To handle the general case, we can use (the proof of) Proposition 4.8.8.1 to factor the functor $\mathrm{h} \mathit{F}$ as a composition $\mathrm{h} \mathit{\operatorname{\mathcal{C}}} \xrightarrow {F'_0} \operatorname{\mathcal{D}}'_0 \xrightarrow {G_0} \mathrm{h} \mathit{\operatorname{\mathcal{D}}}$ where $G_0$ is a faithful functor and $F'_0$ is a full functor which is essentially surjective (or even bijective on objects). To prove Theorem 4.8.8.3, we can take $\operatorname{\mathcal{D}}'$ to be the fiber product $\operatorname{N}_{\bullet }( \operatorname{\mathcal{D}}'_0 ) \times _{ \operatorname{N}_{\bullet }(\mathrm{h} \mathit{\operatorname{\mathcal{D}}}) } \operatorname{\mathcal{D}}$, and $G: \operatorname{\mathcal{D}}' \rightarrow \operatorname{\mathcal{D}}$ to be the functor given by projection onto the second factor (which is faithful by virtue of Proposition 4.8.5.8).
Example 4.8.8.7. Let $\operatorname{\mathcal{C}}$ be an $\infty$-category and let $n$ be an integer. Then the projection map $\operatorname{\mathcal{C}}\rightarrow \Delta ^0$ factors as a composition
$\operatorname{\mathcal{C}}\xrightarrow {F'} \mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}})} \xrightarrow {G} \Delta ^0,$
where $\mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}})}$ is the homotopy $n$-category constructed in §4.8.4. This factorization satisfies the requirements of Theorem 4.8.8.3: the functor $G$ is essentially $n$-categorical because $\mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}})}$ is an $(n,1)$-category (Example 4.8.6.4), and the functor $F'$ is categorically $(n+1)$-connective by Example 4.8.5.12.
Remark 4.8.8.8 (Uniqueness). Let $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ be a functor of $\infty$-categories. Then, for every integer $n$, the factorization of Theorem 4.8.8.3 is well-defined up to equivalence. More precisely, if the functor $F$ admits two factorizations
$\operatorname{\mathcal{C}}\xrightarrow {F'_0} \operatorname{\mathcal{D}}'_0 \xrightarrow {G_0} \operatorname{\mathcal{D}}\quad \quad \operatorname{\mathcal{C}}\xrightarrow {F'_1} \operatorname{\mathcal{D}}'_1 \xrightarrow {G_1} \operatorname{\mathcal{D}}$
where the functors $F'_0$ and $F'_1$ are essentially $n$-categorical, and the functors $G_0$ are $G_1$ are categorically $(n+1)$-connective, then we can find a commutative diagram
$\xymatrix@R =50pt@C=50pt{ \operatorname{\mathcal{C}}\ar [r]^-{F'_0} & \operatorname{\mathcal{D}}'_0 \ar [r]^-{ G_0 } & \operatorname{\mathcal{D}}\\ \operatorname{\mathcal{C}}\ar@ {=}[u] \ar@ {=}[d] \ar [r] & \operatorname{\mathcal{D}}'_{01} \ar [u]_{\sim } \ar [d]^{\sim } \ar [r] & \operatorname{\mathcal{D}}\ar@ {=}[u] \ar@ {=}[d] \\ \operatorname{\mathcal{C}}\ar [r]^-{ F'_1 } & \operatorname{\mathcal{D}}'_{1} \ar [r]^-{ G_1 } & \operatorname{\mathcal{D}}}$
where the vertical maps are equivalences of $\infty$-categories. To prove this, we can use Corollary 4.5.2.23 to reduce to the case where $F'_0$ is a monomorphism of simplicial sets and $G_1$ is an isofibration. In this case, Corollary 4.8.7.18 (and Remark 4.8.7.19) guarantee that the functors $F'_0$ and $G_1$ induce a trivial Kan fibration
$\operatorname{Fun}( \operatorname{\mathcal{D}}'_0, \operatorname{\mathcal{D}}'_1 ) \rightarrow \operatorname{Fun}( \operatorname{\mathcal{C}}, \operatorname{\mathcal{D}}'_1 ) \times _{ \operatorname{Fun}( \operatorname{\mathcal{C}}, \operatorname{\mathcal{D}}) } \operatorname{Fun}( \operatorname{\mathcal{D}}'_0, \operatorname{\mathcal{D}}).$
In particular, this map is surjective on vertices, so the lifting problem
$\xymatrix@R =50pt@C=50pt{ \operatorname{\mathcal{C}}\ar [r]^-{F'_1} \ar [d]^{F'_0} & \operatorname{\mathcal{D}}'_1 \ar [d]^{G_1} \\ \operatorname{\mathcal{D}}'_0 \ar [r]^-{G_0} \ar@ {-->}[ur] & \operatorname{\mathcal{D}}}$
has a solution. A choice of solution determines a commutative diagram
$\xymatrix@R =50pt@C=50pt{ \operatorname{\mathcal{C}}\ar [r]^-{F'_0} \ar@ {=}[d] & \operatorname{\mathcal{D}}'_0 \ar [r]^-{ G_0 } \ar [d]^{H} & \operatorname{\mathcal{D}}\ar@ {=}[d] \\ \operatorname{\mathcal{C}}\ar [r]^-{ F'_1 } & \operatorname{\mathcal{D}}'_{1} \ar [r]^-{ G_1 } & \operatorname{\mathcal{D}}. }$
It follows from Proposition 4.8.7.12 that the functor $H$ is categorically $(n+1)$-connective, and from Remark 4.8.6.7 that $H$ is essentially $n$-categorical. Applying Remark 4.8.5.11, we conclude that $H$ is an equivalence of $\infty$-categories.
Corollary 4.8.8.9. Let $f: X \rightarrow Z$ be a morphism of Kan complexes and let $n$ be an integer. Then $f$ factors as a composition $X \xrightarrow {f'} Y \xrightarrow {f''} Z$, where $f''$ is $n$-truncated and $f'$ is $(n+1)$-connective.
Proof. Using Theorem 4.8.8.3, we can factor $f$ as a composition $f'' \circ f'$, where $f'': \operatorname{\mathcal{C}}\rightarrow Z$ is an essentially $n$-categorical functor of $\infty$-categories and $f': X \rightarrow \operatorname{\mathcal{C}}$ is categorically $(n+1)$-connective. If $n \leq -1$, then $f''$ induces an equivalence from $\operatorname{\mathcal{C}}$ to a summand of $Z$, so that $\operatorname{\mathcal{C}}$ is a Kan complex. If $n \geq 0$, Remark 4.8.7.10 guarantees that $\operatorname{\mathcal{C}}$ is a Kan complex. Setting $Y = \operatorname{\mathcal{C}}$, we observe that $f''$ is $n$-truncated (Example 4.8.6.3) and $f'$ is $(n+1)$-connective (Example 4.8.7.3). $\square$
We will prove Theorem 4.8.8.3 in general by reducing to the special case studied in Example 4.8.8.7. For this, we will need a relative version of the construction $\operatorname{\mathcal{C}}\mapsto \mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}})}$ introduced in §4.8.4.
Construction 4.8.8.10 (Relative Homotopy $n$-Categories). Let $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ be an inner fibration of simplicial sets and let $n \geq 0$ be an integer. For every $m$-simplex $\sigma$ of $\operatorname{\mathcal{D}}$, let $\operatorname{\mathcal{C}}_{\sigma }$ denote the fiber product $\Delta ^ m \times _{\operatorname{\mathcal{D}}} \operatorname{\mathcal{C}}$. We let $\mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}}/\operatorname{\mathcal{D}})}_{m}$ denote the collection of pairs $(\sigma , \tau )$, where $\sigma$ is an $m$-simplex of $\operatorname{\mathcal{D}}$ and $\tau$ is a section of the projection map
$\mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}}_{\sigma })} \rightarrow \mathrm{h}_{\mathit{\leq n}}\mathit{(\Delta ^ m)} \simeq \Delta ^{m}.$
If $f: [m'] \rightarrow [m]$ is a nondecreasing function, we let $f^{\ast }: \mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}}/\operatorname{\mathcal{D}})}_{m} \rightarrow \mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}}/\operatorname{\mathcal{D}})}_{m'}$ denote the map given by $f^{\ast }( \sigma , \tau ) = (\sigma ', \tau ' )$, where $\sigma '$ is the composite map $\Delta ^{m'} \xrightarrow {f} \Delta ^{m} \xrightarrow {\sigma } \operatorname{\mathcal{D}}$ and $\tau '$ is given by the composition
\begin{eqnarray*} \Delta ^{m'} & \simeq & \Delta ^{m'} \times _{ \Delta ^{m} } \Delta ^{m} \\ & \xrightarrow {(\operatorname{id}, \tau )} & \Delta ^{m'} \times _{ \Delta ^ m } \mathrm{h}_{\mathit{\leq n}}\mathit{( \operatorname{\mathcal{C}}_{\sigma } )} \\ & \simeq & \mathrm{h}_{\mathit{\leq n}}\mathit{( \Delta ^{m'} \times _{\Delta ^{m}} \operatorname{\mathcal{C}}_{\sigma } )} \\ & \simeq & \mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}}_{\sigma '} )}, \end{eqnarray*}
where the second isomorphism is provided by Proposition 4.8.4.20. By means of this construction, we can view the assignment $[m] \mapsto \mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}}/\operatorname{\mathcal{D}})}_{m}$ as a simplicial set, which we will denote by $\mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}}/\operatorname{\mathcal{D}})}$. Note that the construction $(\sigma , \tau ) \mapsto \sigma$ determines a comparison map of simplicial sets $G: \mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}}/\operatorname{\mathcal{D}})} \rightarrow \operatorname{\mathcal{D}}$.
It will be useful to extend this construction to the case where $n < 0$. If $n = -1$, we define $\mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}}/\operatorname{\mathcal{D}})}$ to be the full simplicial subset of $\operatorname{\mathcal{D}}$ whose vertices belong to the image of $F$, and we take $G: \mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}}/\operatorname{\mathcal{D}})} \hookrightarrow \operatorname{\mathcal{D}}$ to be the inclusion map. If $n \leq -2$, we define $\mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}}/\operatorname{\mathcal{D}})}$ to be the simplicial set $\operatorname{\mathcal{D}}$, and $G$ to be the identity morphism $\operatorname{id}_{\operatorname{\mathcal{D}}}$.
Example 4.8.8.11. Let $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ be an inner fibration of $\infty$-categories and let $n$ be an integer. Then there is a comparison map from the simplicial set $\mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}}/\operatorname{\mathcal{D}})}$ to the homotopy $n$-category $\mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}})}$. For $n \geq 0$, this map carries an $m$-simplex $(\sigma ,\tau )$ of $\mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}}/\operatorname{\mathcal{D}})}$ to the $m$-simplex of $\mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}})}$ given by the composite map
$\Delta ^ m \xrightarrow {\tau } \mathrm{h}_{\mathit{\leq n}}\mathit{( \operatorname{\mathcal{C}}_{\sigma } )} \rightarrow \mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}})}.$
If $\operatorname{\mathcal{D}}$ is an $(n,1)$-category, then this comparison map is an isomorphism (Proposition 4.8.4.20).
Example 4.8.8.12. Let $\operatorname{\mathcal{C}}$ be an $\infty$-category, so that the projection map $F: \operatorname{\mathcal{C}}\rightarrow \Delta ^0$ is an inner fibration. Since $\Delta ^0$ is an $(n,1)$-category, Example 4.8.8.11 supplies an isomorphism of simplicial sets $\mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}}/\Delta ^0)} \simeq \mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}})}$.
Remark 4.8.8.13 (Base Change). Suppose we are given a pullback diagram of simplicial sets
$\xymatrix@C =50pt@R=50pt{ \operatorname{\mathcal{C}}' \ar [d] \ar [r] & \operatorname{\mathcal{C}}\ar [d] \\ \operatorname{\mathcal{D}}' \ar [r] & \operatorname{\mathcal{D}}, }$
where the vertical maps are inner fibrations. Then, for every integer $n$, the simplicial set $\mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}}'/\operatorname{\mathcal{D}}')}$ can be identified with the fiber product $\operatorname{\mathcal{D}}' \times _{\operatorname{\mathcal{D}}} {\, }\mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}}/\operatorname{\mathcal{D}})}$. In particular, for every vertex $D \in \operatorname{\mathcal{D}}$, we have a canonical isomorphism
$\{ D\} \times _{\operatorname{\mathcal{D}}} \mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}}/\operatorname{\mathcal{D}})} \simeq \mathrm{h}_{\mathit{\leq n}}\mathit{( \{ D\} \times _{\operatorname{\mathcal{D}}} \operatorname{\mathcal{C}})}.$
Proposition 4.8.8.14. Let $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ be an inner fibration of simplicial sets and let $n$ be an integer. Then the comparison map $G: \mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}}/\operatorname{\mathcal{D}})} \rightarrow \operatorname{\mathcal{D}}$ of Construction 4.8.8.10 is an $n$-categorical inner fibration (see Definition 4.8.6.24).
Proof. For $n < 0$, this is immediate from the construction. We may therefore assume without loss of generality that $n \geq 0$. Using Remarks 4.8.6.33 and 4.8.8.13, we can reduce to the case where $\operatorname{\mathcal{D}}= \Delta ^ m$ is a standard simplex. In particular, $\operatorname{\mathcal{D}}$ is an $(n,1)$-category. In this case, Example 4.8.8.11 guarantees that the simplicial set $\mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}}/\operatorname{\mathcal{D}})} \simeq \mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{D}})}$ is an $(n,1)$-category. The desired result now follows from Proposition 4.8.6.32. $\square$
Remark 4.8.8.15. Let $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ be an inner fibration of simplicial sets and let $n$ be an integer. Then the comparison map $G: \mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}}/\operatorname{\mathcal{D}})} \rightarrow \operatorname{\mathcal{D}}$ of Construction 4.8.8.10 fits into a commutative diagram
$\xymatrix@C =50pt@R=50pt{ & \mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}}/\operatorname{\mathcal{D}})} \ar [dr]^{G} & \\ \operatorname{\mathcal{C}}\ar [ur]^{F'} \ar [rr]^{F} & & \operatorname{\mathcal{D}}. }$
For $n \geq 0$, the morphism $F'$ carries each $m$-simplex of $\operatorname{\mathcal{C}}$ to the $m$-simplex $( F(\sigma ), \tau )$ of $\mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}}/\operatorname{\mathcal{D}})}$, where $\tau$ is the composite map
$\Delta ^{m} \xrightarrow {(\operatorname{id},\sigma )} \Delta ^{m} \times _{\operatorname{\mathcal{D}}} \operatorname{\mathcal{C}}= \operatorname{\mathcal{C}}_{\sigma } \rightarrow \mathrm{h}_{\mathit{\leq n}}\mathit{( \operatorname{\mathcal{C}}_{\sigma } )}.$
The simplicial set $\mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}}/\operatorname{\mathcal{D}})}$ of Construction 4.8.8.10 can be characterized by a universal mapping property:
Proposition 4.8.8.16. Let $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ be an inner fibration of simplicial sets and let $n$ be an integer. Then, for every $n$-categorical inner fibration $\operatorname{\mathcal{D}}' \rightarrow \operatorname{\mathcal{D}}$, the comparison map of Remark 4.8.8.15 induces an isomorphism of simplicial sets
$\theta : \operatorname{Fun}_{ / \operatorname{\mathcal{D}}}( \mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}}/\operatorname{\mathcal{D}})}, \operatorname{\mathcal{D}}' ) \rightarrow \operatorname{Fun}_{/\operatorname{\mathcal{D}}}( \operatorname{\mathcal{C}}, \operatorname{\mathcal{D}}' ).$
Proof. We may assume without loss of generality that $n \geq 0$ (otherwise, the result follows immediately from the construction). For every morphism of simplicial sets $K \rightarrow \operatorname{\mathcal{D}}$, Remark 4.8.8.15 determines a comparison map
$\theta _{K}: \operatorname{Fun}_{/\operatorname{\mathcal{D}}}( K \times _{\operatorname{\mathcal{D}}} \mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}}/\operatorname{\mathcal{D}})}, \operatorname{\mathcal{D}}' ) \rightarrow \operatorname{Fun}_{/\operatorname{\mathcal{D}}}( K \times _{\operatorname{\mathcal{D}}} \operatorname{\mathcal{C}}, \operatorname{\mathcal{D}}' ).$
We will prove that each $\theta _{K}$ is an isomorphism of simplicial sets; Proposition 4.8.8.16 then follows by taking $K = \operatorname{\mathcal{D}}$. Note that the construction $K \mapsto \theta _{K}$ carries colimits (in the category of simplicial sets with a morphism to $\operatorname{\mathcal{D}}$) to limits (in the arrow category $\operatorname{Fun}( [1], \operatorname{Set_{\Delta }})$). By virtue of Remark 1.1.3.13, we can assume without loss of generality that $K$ is a standard simplex. Replacing $F$ by the projection map $K \times _{\operatorname{\mathcal{D}}} \operatorname{\mathcal{C}}\rightarrow K$ and $\operatorname{\mathcal{D}}'$ by the fiber product $K \times _{\operatorname{\mathcal{C}}} \operatorname{\mathcal{D}}'$, we are reduced to proving Proposition 4.8.8.16 in the special case where $\operatorname{\mathcal{D}}$ is a standard simplex: in particular, it is an $(n,1)$-category. In this case, $\operatorname{\mathcal{D}}'$ is also an $(n,1)$-category (Proposition 4.8.6.32), and we can identify $\mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}}/\operatorname{\mathcal{D}})}$ with the homotopy $n$-category of $\operatorname{\mathcal{C}}$ (Example 4.8.8.11). Applying Proposition 4.8.4.7, we see that the horizontal maps in the commutative diagram
$\xymatrix@C =50pt@R=50pt{ \operatorname{Fun}( \mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}}/\operatorname{\mathcal{D}})}, \operatorname{\mathcal{D}}' ) \ar [r] \ar [d] & \operatorname{Fun}( \operatorname{\mathcal{C}}, \operatorname{\mathcal{D}}' ) \ar [d] \\ \operatorname{Fun}( \mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}}/\operatorname{\mathcal{D}})}, \operatorname{\mathcal{D}}) \ar [r] & \operatorname{Fun}( \operatorname{\mathcal{C}}, \operatorname{\mathcal{D}}) }$
are isomorphisms. The desired result now follows by passing to fibers of the vertical maps. $\square$
Remark 4.8.8.17. Let $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ be an inner fibration of simplicial sets, let $n$ be an integer, and let $A \subseteq B$ be simplicial sets. If $B$ has dimension $\leq n+1$, then every lifting problem
$\xymatrix@C =50pt@R=50pt{ A \ar [d] \ar [r] & \operatorname{\mathcal{C}}\ar [d]^{F'} \\ B \ar [r] \ar@ {-->}[ur] & \mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}}/\operatorname{\mathcal{D}})} }$
has a solution. Moreover, if $B$ has dimension $\leq n-1$, then the solution is unique. To prove this, we can assume without loss of generality that $B = \Delta ^ m$ is a standard simplex for some $m \leq n+1$, and that $A = \operatorname{\partial \Delta }^ m$ is its boundary (see Proposition 1.1.4.12). The case $n \leq -2$ is vacuous, and the case $n = -1$ is immediate from the definition. We may therefore assume that $n \geq 0$. Replacing $F$ by the projection map $\Delta ^ m \times _{\operatorname{\mathcal{D}}} \operatorname{\mathcal{C}}\rightarrow \Delta ^ m$, we can reduce to the case where $\operatorname{\mathcal{D}}$ is a standard simplex, so that $U'$ identifies $\mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}}/\operatorname{\mathcal{D}})}$ with the homotopy $n$-category $\mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}})}$ (Example 4.8.8.11). In this case, the desired result follows from Corollary 4.8.4.17.
Proposition 4.8.8.18. Let $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ be an inner fibration of simplicial sets and let $n$ be an integer. Then the comparison map $F': \operatorname{\mathcal{C}}\rightarrow \mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}}/\operatorname{\mathcal{D}})}$ of Remark 4.8.8.15 is an inner fibration.
Proof. Without loss of generality, we may assume that $n \geq 0$. Using Remarks 4.1.1.13 and 4.8.8.13, we can reduce to the case where $\operatorname{\mathcal{D}}= \Delta ^ m$ is a standard simplex. In this case, $F'$ identifies with the tautological map $\operatorname{\mathcal{C}}\rightarrow \mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}})}$ (Example 4.8.8.11), so the desired result follows from Corollary 4.8.4.16. $\square$
Corollary 4.8.8.19. Let $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ be an inner fibration of $\infty$-categories and let $n$ be an integer. Then the simplicial set $\mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}}/\operatorname{\mathcal{D}})}$ is an $\infty$-category. Moreover, the functor $F': \operatorname{\mathcal{C}}\rightarrow \mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}}/\operatorname{\mathcal{D}})}$ of Remark 4.8.8.15 is categorically $(n+1)$-connective.
Proof. Since $\operatorname{\mathcal{D}}$ is an $\infty$-category and the comparison map $G: \mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}}/\operatorname{\mathcal{D}})} \rightarrow \operatorname{\mathcal{D}}$ is an inner fibration (Proposition 4.8.8.14), the simplicial set $\mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}}/\operatorname{\mathcal{D}})}$ is also an $\infty$-category (Remark 4.1.1.9). Fix an integer $m \leq n+1$; we wish to show that the functor $F'$ is $m$-full. For $n = -2$, there is nothing to prove. If $n = -1$, then $U'$ is surjective on objects (by construction) and therefore essentially surjective. We may therefore assume without loss of generality that $n \geq 0$. Since $U'$ is an inner fibration (Proposition 4.8.8.18), it will suffice to show that for every morphism $\Delta ^1 \rightarrow \mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{E}}/\operatorname{\mathcal{C}})}$, the projection map $\Delta ^{1} \times _{ \mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{E}}/\operatorname{\mathcal{C}})} } \operatorname{\mathcal{E}}\rightarrow \Delta ^1$ is $m$-full (Proposition 4.8.5.27. Using Remark 4.8.8.13, we can replace $F$ by the projection map $\Delta ^1 \times _{\operatorname{\mathcal{D}}} \operatorname{\mathcal{C}}\rightarrow \Delta ^1$, and thereby reduce to the situation where $\operatorname{\mathcal{D}}= \Delta ^1$ is an $(n,1)$-category. In this case, the functor $F'$ exhibits $\mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}}/\operatorname{\mathcal{D}})}$ as a homotopy $n$-category of $\operatorname{\mathcal{C}}$ (Example 4.8.8.11), so the desired result follows from Example 4.8.5.12. $\square$
Proof of Theorem 4.8.8.3. Let $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ be a functor of $\infty$-categories and let $n$ be an integer. We wish to show that $F$ factors as a composition $G \circ F'$, where $G$ is essentially $n$-categorical and $F'$ is categorically $(n+1)$-connective. Using Proposition 4.1.3.2, we can reduce to the case where $F$ is an inner fibration. In this case, the factorization
$\operatorname{\mathcal{C}}\xrightarrow {F'} \mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}}/\operatorname{\mathcal{D}})} \xrightarrow {G} \operatorname{\mathcal{C}}$
of Remark 4.8.8.15 has the desired properties: Proposition 4.8.8.14 guarantees that $G$ is an $n$-categorical inner fibration (and is therefore essentially $n$-categorical, by virtue of Proposition 4.8.6.35), and Corollary 4.8.8.19 guarantees that $F'$ is categorically $(n+1)$-connective. $\square$
Warning 4.8.8.20. Let $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ be an inner fibration of $\infty$-categories. In the case $n = 0$, our proof of Theorem 4.8.8.3 shows that $F$ factors as a composition
$\operatorname{\mathcal{E}}\xrightarrow {F'} \mathrm{h}_{\mathit{\leq 0}}\mathit{(\operatorname{\mathcal{C}}/\operatorname{\mathcal{D}})} \xrightarrow {G} \operatorname{\mathcal{D}},$
where $F'$ is fully faithful and essentially surjective, and $G$ is a $0$-categorical inner fibration (in particular, $G$ is faithful). Beware that generally does not coincide with the factorization constructed in Example 4.8.8.6. If $u: X \rightarrow Y$ is an isomorphism in the $\infty$-category $\operatorname{\mathcal{C}}$ having the property that $F(u)$ is an identity morphism in $\operatorname{\mathcal{D}}$, then the functor $F'$ carries $X$ and $Y$ to the same object of $\mathrm{h}_{\mathit{\leq 0}}\mathit{(\operatorname{\mathcal{C}}/\operatorname{\mathcal{D}})}$. Consequently, the functor $F'$ is generally not bijective on objects.
A related phenomenon occurs in the case $n = -1$. By construction, $\mathrm{h}_{\mathit{\leq -1}}\mathit{(\operatorname{\mathcal{C}}/\operatorname{\mathcal{D}})}$ is the full subcategory of $\operatorname{\mathcal{D}}$ spanned by objects of the form $F(X)$, where $X$ is an object of $\operatorname{\mathcal{C}}$. If the inner fibration $U$ is not an isofibration, this subcategory might be smaller than the essential image of $F$.
We close this section with a few additional observations about Construction 4.8.8.10.
Proposition 4.8.8.21. Let $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ be an inner fibration of simplicial sets, let $n$ be an integer, and let $G: \mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}}/\operatorname{\mathcal{D}})} \rightarrow \operatorname{\mathcal{D}}$ be the comparison map of Construction 4.8.8.10. Then:
$(1)$
If $F$ is a left fibration, then $G$ is a left fibration.
$(2)$
If $F$ is a right fibration, then $G$ is a right fibration.
$(3)$
If $F$ is a Kan fibration, then $G$ is a Kan fibration.
$(4)$
If $F$ is an isofibration of $\infty$-categories, then $G$ is an isofibration of $\infty$-categories.
Proof. We first prove $(1)$. Assume that $F$ is a left fibration, and suppose we are given integers $0 \leq i < n$; we wish to show that every lifting problem
4.88
$$\begin{gathered}\label{equation:relative-homotopy-other-fibrations} \xymatrix@C =50pt@R=50pt{ \Lambda ^{m}_{i} \ar [r]^-{\sigma _0} \ar [d] & \mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}}/\operatorname{\mathcal{D}})} \ar [d]^{G} \\ \Delta ^{m} \ar@ {-->}[ur]^{\sigma } \ar [r]^-{ \overline{\sigma } } & \operatorname{\mathcal{D}}} \end{gathered}$$
admits a solution. If $m \leq n+2$, then $\sigma _0$ can be lifted to a morphism $\Lambda ^{m}_{i} \rightarrow \operatorname{\mathcal{C}}$ (Remark 4.8.8.17), so the desired result follows from our assumption that $F$ is a left fibration. We may therefore assume that $m \geq n+3$. If $n = -2$, then $G$ is an isomorphism and there is nothing to prove. If $n = -1$, then $G$ identifies $\mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}}/\operatorname{\mathcal{D}})}$ with a full simplicial subset of $\operatorname{\mathcal{D}}$, and the desired result follows from the observation that $\Lambda ^{m}_{i}$ contains every vertex of $\Delta ^ m$. We may therefore assume that $n \geq 0$. Replacing $F$ by the projection map $\Delta ^ m \times _{\operatorname{\mathcal{D}}} \operatorname{\mathcal{C}}\rightarrow \Delta ^ m$, we can reduce to the case where $\operatorname{\mathcal{D}}= \Delta ^ m$ is a standard simplex. In this case, $\mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}}/\operatorname{\mathcal{D}})}$ is an $(n,1)$-category (Example 4.8.8.11). In particular, it is an $(n+1)$-coskeletal simplicial set, so the lifting problem (4.88) has a unique solution (since $\Lambda ^{m}_{i}$ contains the $(n+1)$-skeleton of $\Delta ^ m$).
Assertion $(2)$ follows by applying $(1)$ to the opposite inner fibration $U^{\operatorname{op}}: \operatorname{\mathcal{C}}^{\operatorname{op}} \rightarrow \operatorname{\mathcal{D}}^{\operatorname{op}}$. Assertion $(3)$ follows by combining $(1)$ and $(2)$ with Example 4.2.1.5. It remains to prove $(4)$. Fix an object $Y \in \mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}}/\operatorname{\mathcal{D}})}$ and an isomorphism $\overline{e}: \overline{X} \rightarrow V(Y)$ in the $\infty$-category $\operatorname{\mathcal{D}}$; we wish to show that $\overline{e}$ can be lifted to an isomorphism $e: X \rightarrow Y$ of $\mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}}/\operatorname{\mathcal{D}})}$. If $n \leq -2$, then $G$ is an isomorphism and the result is obvious. Otherwise, the comparison map $F': \operatorname{\mathcal{C}}\rightarrow \mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}}/\operatorname{\mathcal{D}})}$ is surjective on vertices, so we can choose an object $\widetilde{Y} \in \operatorname{\mathcal{C}}$ satisfying $F'( \widetilde{Y} ) = Y$. If $F$ is an isofibration, then there exists an isomorphism $\widetilde{e}: \widetilde{X} \rightarrow \widetilde{Y}$ of $\operatorname{\mathcal{C}}$ satisfying $F( \widetilde{e} ) = \overline{e}$. It follows that $e = F'( \widetilde{e} )$ is an isomorphism in $\mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}}/\operatorname{\mathcal{D}})}$ satisfying $G(e) = \overline{e}$. $\square$
Proposition 4.8.8.22. Let $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ be an inner fibration of $\infty$-categories and let $n$ be an integer. The following conditions are equivalent:
$(1)$
The functor $F$ is essentially $n$-categorical.
$(2)$
The comparison map $F': \operatorname{\mathcal{C}}\rightarrow \mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}}/\operatorname{\mathcal{D}})}$ of Remark 4.8.8.15 is an equivalence of $\infty$-categories.
Proof. It follows from Proposition 4.8.8.14 (and Proposition 4.8.6.35) that the comparison map $G: \mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}}/\operatorname{\mathcal{D}})} \rightarrow \operatorname{\mathcal{D}}$ is essentially $n$-categorical. By virtue of Remark 4.8.6.8, we can replace $(1)$ by the following condition:
$(1')$
The functor $F'$ is essentially $n$-categorical: that is, it is $m$-full for $m \geq n+2$.
Since $F'$ is also $m$-full for $m \leq n+1$ (Corollary 4.8.8.19), the equivalence $(1') \Leftrightarrow (2)$ follows from Remark 4.8.5.11. $\square$
Corollary 4.8.8.23. Let $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ be a functor of $\infty$-categories. For every integer $n$, the following conditions are equivalent:
$(1)$
The functor $F$ is essentially $n$-categorical.
$(2)$
The functor $F$ factors as a composition $\operatorname{\mathcal{C}}\xrightarrow {F'} \operatorname{\mathcal{D}}' \xrightarrow {G} \operatorname{\mathcal{D}}$, where $F'$ is an equivalence of $\infty$-categories and $G$ is an $n$-categorical isofibration.
Proof. The implication $(2) \Rightarrow (1)$ follows from Proposition 4.8.6.35 (together with Remark 4.8.5.18). To prove the converse, we may assume without loss of generality that $F$ is an isofibration (Corollary 4.5.2.23). In this case, the factorization $\operatorname{\mathcal{C}}\xrightarrow {F'} \mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}}/\operatorname{\mathcal{D}})} \xrightarrow {G} \operatorname{\mathcal{D}}$ of Remark 4.8.8.15 has the desired properties: Proposition 4.8.8.22 guarantees that $F'$ is an equivalence of $\infty$-categories, Proposition 4.8.8.21 guarantees that $G$ is an isofibration, and Proposition 4.8.8.14 guarantees that $G$ is $n$-categorical. $\square$
Proposition 4.8.8.24. Let $n$ be an integer, and suppose we are given a commutative diagram of $\infty$-categories
$\xymatrix@R =50pt@C=50pt{ \operatorname{\mathcal{C}}\ar [rr]^{F} \ar [dr] & & \operatorname{\mathcal{D}}\ar [dl] \\ & \operatorname{\mathcal{E}}& }$
where the vertical maps are inner fibrations. If $F$ is categorically $(n+1)$-connective, then it induces an equivalence of $\infty$-categories $F': \mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}}/\operatorname{\mathcal{E}})} \rightarrow \mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{D}}/\operatorname{\mathcal{E}})}$.
Proof. We have a commutative diagram of $\infty$-categories
$\xymatrix@R =50pt@C=50pt{ \operatorname{\mathcal{C}}\ar [r]^-{F} \ar [d] & \operatorname{\mathcal{D}}\ar [d] \\ \mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}}/\operatorname{\mathcal{E}})} \ar [r]^-{F'} & \mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{D}}/\operatorname{\mathcal{E}})}. }$
Here $F$ is categorically $(n+1)$-connective by assumption, and the vertical maps are categorically $(n+1)$-connective by virtue of Corollary 4.8.8.19. Applying Proposition 4.8.7.12, we see that the functor $F'$ is also categorically $(n+1)$-connective. We also have a commutative diagram
$\xymatrix@R =50pt@C=50pt{ \mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}}/\operatorname{\mathcal{E}})} \ar [rr]^{F'} \ar [dr] & & \mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{D}}/\operatorname{\mathcal{E}})} \ar [dl] \\ & \operatorname{\mathcal{E}}, & }$
where the vertical maps are essentially $n$-categorical (Proposition 4.8.8.14). Using Remark 4.8.6.8, we see that $F'$ is also essentially $n$-categorical. Using Remark 4.8.5.11, we see that $F'$ is an equivalence of $\infty$-categories. $\square$
Corollary 4.8.8.25. Let $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ be an inner fibration of $\infty$-categories and let $n$ be an integer. The following conditions are equivalent:
$(1)$
The comparison map $G: \mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}}/\operatorname{\mathcal{D}})} \rightarrow \operatorname{\mathcal{D}}$ is an equivalence of $\infty$-categories.
$(2)$
The functor $F$ is categorically $(n+1)$-connective.
Proof. The implication $(1) \Rightarrow (2)$ follows from Proposition 4.8.8.14 and Remark 4.8.5.16. The reverse implication follows by applying Proposition 4.8.8.24 in the special case $\operatorname{\mathcal{E}}= \operatorname{\mathcal{D}}$. $\square$
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# Comportement asymptotique de diffusions renforcées sur R^d
Abstract : The first chapter is concerned with some self-interacting diffusions $(X_t,t\geq 0)$
living on $\mathbb{R}^d$. These diffusions are solutions to stochastic differential equations:
$\mathrm{d}X_t = \mathrm{d}B_t - g(t)\nabla V(X_t -\overline{\mu}_t) \mathrm{d}t,$ where $\overline{\mu}_t$ is the empirical mean of the process $X$, $V$ is an asymptotically strictly convex potential and $g$ is a given function. We study the ergodic behavior of $X$ and prove that it is strongly related to $g$. Actually, $X$ and $\overline{\mu}_t$ have the same asymptotic behavior and we will give necessary and sufficient conditions (on $g$ and $V$) for the almost sure convergence of $X$. In chapter 2, we finish the previous study. We have still studied the ergodic behavior of $X$ and proved that it is strongly
related to $g$. We go further and give necessary and sufficient conditions (for small $g$'s) in order that $X$ converges in law to $X_\infty$ (which is related to the global minima of $V$).
In the second part, we begin to situate our study in Chapter 3. Self-interacting diffusions are solutions to SDEs with a drift term depending on the process and its normalized occupation measure $\mu_t$ (via an interaction potential $V$ and a confinement potential $W$):
$\mathrm{d}X_t = \mathrm{d}B_t -\left( \nabla V(X_t)+\frac{1}{t} \int_0^t \nabla_x W(X_t,X_s) \mathrm{d}s \right) \mathrm{d}t \\ \mathrm{d}\mu_t = (\delta_{X_t} - \mu_t)\frac{\mathrm{d}t}{r+t}\\ X_0 = x, \mu_0=\mu.$
We establish a relation between the asymptotic behavior of $\mu_t$ and the asymptotic behavior of a deterministic dynamical flow (defined on the space of the Borel probability measures). We extend previous results on $\mathbb{R}^d$ or more generally a smooth complete connected Riemannian manifold without boundary. We will also give some sufficient conditions for the convergence of $\mu_t$. We then illustrate, in Chapter 5, the previous study of self-interacting diffusions living in $\mathbb{R}^d$ with some examples in the two-dimensional case. The
preceding chapter contains abstract results, and therefore we describe here a simple example and illustrate some of our previous results. We will show in particular that, depending on $W$, either the empirical measure behaves like the Brownian motion" (constructed with respect to the measure $e^{V(x)} \mathrm{d}x$); or the empirical occupation measure converges almost surely to a probability measure, which is approximatively a Gaussian distribution ; or there is enough attraction, and then the term induced by $W$ forces $\mu_t$ to circle around and the limit set of $(\mu_t)$ is a circle of measures $\{\nu(\delta), 0\leq\delta<2\pi\}$.
Keywords :
Document type :
Theses
Domain :
https://tel.archives-ouvertes.fr/tel-00155876
Contributor : Aline Kurtzmann <>
Submitted on : Tuesday, June 19, 2007 - 2:07:51 PM
Last modification on : Wednesday, November 28, 2018 - 2:48:22 PM
Long-term archiving on: : Thursday, April 8, 2010 - 5:42:28 PM
### Identifiers
• HAL Id : tel-00155876, version 1
### Citation
Aline Kurtzmann. Comportement asymptotique de diffusions renforcées sur R^d. Mathématiques [math]. Université de Neuchâtel, 2007. Français. ⟨tel-00155876⟩
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# Tag Info
4
That's not what the formula gives you. As the caption says, the capacity of the augmenting path in the residual network in (b) is $4$. Therefore we send 4 units of flow along the augmenting path from $s$ to $t$, namely, the path $s \to v_2 \to v_3 \to t$. In particular, $f(s,v_2)=8$, $f'(s,v_2)=4$, and $f'(v_2,s)=0$, so the updated flow is $8+4-0=12$.
4
Those answers assume that all edge capacities are integers. Assuming they are, this works. Suppose the min-cut in the original graph has total capacity $x$; then it will have total capacity $x(|E|+1)+k$ in the transformed graph, where $k$ counts the number of edges crossing that cut. Note that if you consider any cut in the original graph with larger ...
3
Let us consider $K_{2,2}$, the complete bipartite graph with two vertices on either side. A valid max flow sends $1/2$ units of flow across each edge of the bipartite graph. This gives a negative answer to your first question. On the other hand, the integral flow theorem guarantees that there exists an integral max flow, and such a max flow can be found ...
3
Edmonds-Karp is a specialisation/elaboration of Ford-Fulkerson, so any bound for the latter also applies to the former. In other words, EK is $O(|E|\min(f_{max}, |V||E|))$ time (and writing it this way does add information, since $f_{max}$ can be much smaller than $|V||E|$ -- and this is the only time when you might otherwise consider using some other ...
3
It is explained in part (b) of the caption of Figure 26.4. The residual network $G_f$ with augmenting path $p$ shaded; its residual capacity is $c_f(p)=c_f(v_2,v_3)=4$. Since the capacity of path $p$ is 4 (not 5), we find a flow $f'$ in the residual network $G_f$ that is defined by $f'(s,v_2)=f'(v_2,v_3)=f'(v_3,t)=4$. So for the network flow $f\uparrow f'... 3 Your problem is NP-hard. There is a reduction from Independent Set to its decision version. Consider an instance$G=(V,E)$of Independent Set, you construct a network with vertices$\{s,t\}\cup V\cup V'$where each vertex in$V'$corresponds to a pair of vertices in$V$. For example, if$V=\{1,2,3\}$, then$V'=\{v_{12},v_{23},v_{13}\}$. Then we construct ... 2 The given algorithm is correct. The flow network constructed need to be directed, and the value of a$S$-$T$cut only considers edges going out of the vertex set$S$. 2 To test correctness: the set of edges belonging to a min-cut is not unique in general, so a dataset like you ask would not really be helpful, apart from checking that the value of the cut is the right one. However, it is easy to check correctness yourself by using the output of your algorithm if you compute both the max-flow and the min-cut; just check that ... 2 Consider a graph of two node$s$and$t$and one edge$(s,t)$with a flow$f$,$f(s,t)=1$. Let$S=\{s\}$and$T=\{t\}$. Then the flow across the cut$(S, T)$is, apparently 1. Or, $$f(S, T) = \sum_{u\in S} \sum_{v\in T} f(u,v) - \sum_{u\in S} \sum_{v\in T} f(v,u)= f(u,v)=1.$$$\sum_{u\in S} \sum_{v\in T} f(u,v)$is the flow from$S$to$T$.$\sum_{u\in S}...
2
No, your gut feeling is not correct. Consider the following flow network with source $s$ and sink $t$, where the capacity of every edge is 1. The max-flow from source to sink is 0. The s-t cut $(\{s,A\}, \{B,t\})$ is a minimum cut since the only connecting edge $(B, A)$ goes from sink side to source side. $$s \longrightarrow A \longleftarrow B \... 2 A full edge, e.g. a \rightarrow c, has a residual capacity of 0 in the residual network. So you can't make an augmenting path over that directed edge. However the reversed edge, c \rightarrow a has a residual capacity of 5 (since c_{c \rightarrow a} = 0 and f_{c \rightarrow a} = -5). Therefore you can create an augmenting path using the reversed ... 2 This is sometimes called the minimum edge-cost flow problem or fixed-cost flow problem. As you suspected, it is indeed NP-hard, even when the network is bipartite. It is listed as problem ND32 in the list of NP-hard problems by Garey and Johnson: M.R. Garey, D.S. Johnson. Computers and Intractability: A Guide to the Theory of NP-Completeness. Freeman, New ... 1 This problem is NP-hard if 0 weight is allowed. We can reduce Not-All-Equal 3SAT to the decision version of this problem. Given an instance of Not-All-Equal 3SAT with n variables and m clauses, for each variable x_i, we create two vertices v_i and v_i' with an edge between them for each variable. In addition, for each clause, for example, x_1\... 1 Yes, you can. If it has, say, no outcoming edge, there can be no flow routed over this node. Otherwise the flow conservation constraint for this node (v)$$\sum_{(u,v) \in E} f_{uv} - \underbrace{\sum_{(v,w) \in E} f_{vw}}_{= 0} = 0 is violated if you have incoming flow.
1
I would do it kind of the other way around as you suggested. First compute a flow that saturates $(u,v)$. This can be done with the Ford--Fulkerson algorithm. Look only for augmenting paths which contain $(u,v)$ and augment the flow until the edge is saturated. In the second step you augment further, but avoid the edge $(u,v)$ when searching for augmenting ...
1
Timetabling is known to be NP-complete. Your more complex variant is too. Don't expect "nice" or "efficient" solutions. Either settle for an approximate solution (good luck in deriving one) or some sort of randomized heuristic. I'd try some variant of genetic algorithms (look around for it's application to time tabling, they use special mutation and ...
1
Apply the transformation $w \mapsto (m+1)w + 1$ to all weights (where $m$ is the number of edges in the graph), and find the minimum weight cut in the new graph. This will give you the minimum weight cut with the minimum number of edges. By computing the value modulo $m+1$, you can determine the number of edges.
1
Remove $u$ and $v$ (as well as all edges connected to them), and for any removed edge $(u,x)$, add an edge from $s$ to $x$ with the same capacity; for any removed edge $(y,v)$, add an edge from $y$ to $t$ with the same capacity. Now find a min cut in this new graph. The partition of nodes in this cut suggests a min cut among those including $e$ in the ...
1
By the max flow min cut theorem, the maximum value of a flow equals the minimum capacity of a cut. The capacity of a cut is of the form 12x+18y, which can’t equal 56 because 56 is not a multiple of 6.
1
Just to respond to the above comment by the OP "why does linear programming for $K_{2,2}$ fail". Perhaps your confusion is because we need to distinguish between "solving an LP" and "solving an LP using a particular algorithm". The LP formulation of maxflow (with real variables) has optimal solution (1,0,1,0) (for the edges of the bipartite graph $K_{2,2}... 1 Flow is an abstraction of how much "stuff" you want to move through the network. Exactly what the stuff is depends on what you're modelling with the network - water pipes, transport networks, computer networks, etc. The problem (or something close to it) can also be used to model other problems, in which case flow could be all kinds of things. 1 There is a significant typo on that slide. "$c_f (u, v) = f (v, u)$if$f (v, u)$not in$E$" should have been "$c_f (u, v) = f (v, u)$if$(u, v)$not in$E$" or, what is equivalent, "$c_f (u, v) = f (v, u)$if$(v, u)$is in$E$". Why do we care about edge that is not in$E$? In fact, we care only edge$(u,v)$if either$(u,v)$is in$E$or$(v,u)$... 1 First a quick note, the flow entering a vertex is equal to the flow leaving. I'll just refer to it as the amount of flow through a vertex. Second, note that we can say$a \leq b \wedge b \leq a$, thus we can add the constraint$a = b$for two vertices. Then we can re-use the NP-hard result from the paper that you quoted: "A negative disjunctive ... 1 I would say that 3. is a special case of 2. instead but it is a point of view. On 3., you can create a new node "source" which has edges to every supply nodes (b(v) > 0). These edges have cost 0 and capacity b(v). Then you create a new node "sink" which has edges from every demand node. These edges have cost 0 and capacity -b(v). All this replaces the b(v)... 1 Here is your definition of reversed edges in the case of flow network given in your comment. Between 2 vertices there is the normal forward edge (u,v) , and another edge (v,u) that goes backward(this is the reversed edge). regardless their capacities. If your definition is used, flow networks may have reversed edges indeed. For example, the flow network (... 1 The problem is NP-complete, because in the special case that all cars have the same capacity, it is just the bin-packing problem. If car A has a higher capacity than car B, and you get an optimal solution (smallest number of cars) containing B but not A, then you can swap cars A and B, you won't need more cars, and because A has more empty capacity, you ... 1 Let$G=(V,E)$be your input graph. Now consider a maximum flow$f$on$G$. Let$f$be a flow in$G$such that the residual network$G_R$has no s-t path, then$f$is a maximum flow. Let's define$G'=(V,E')$to be your graph with$E'=E \cup \{(u_i,s)\}$for$i=1,...,N$. Since$f$is a maximum flow on$G$, then$G_R\$ has no s-t paths. Now you can ...
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Algebra II Honors: Introducing Function Composition
In Algebra II, our second semester begins with function operations and inverses. Since function composition is a very commonly used concept outside of the math classroom, I wanted to introduce the idea within a context.
So I opened the lesson with this picture:
The goal for my students was to determine how much I paid for gas the day before. I gave them some information:
My students wanted to know how much gas was and then proceeded to spend 45 seconds commenting on how inexpensive gas is at the moment (I figured this would happen given that my a lot of my students are starting to drive).
Then I told them how many miles I had driven:
Now that they knew the price and the fact I had driven 117.2 miles, I wanted them to make a prediction.
I got anything from a \$15 prediction to a \$40 prediction. I was surprised to see that when I probed them about why they predicted the price that they did, quite a few of my students used the fact that I was filling up about a half tank of gas. Questions and comments started flying about how many gallons of gas my car’s tank would hold. I didn’t have an answer for them, and I tried to question how that would be a useful piece of information. They said that if they knew how many gallons the tank was in total, they could figure out how many gallons half a tank is and use that to determine how much I paid.
A conversation about how they didn’t know exactly what proportion of my gas tank was being filled. They weren’t happy…they said it was about half of a tank…I said that using the word “about” makes that information unusable…one or two students still weren’t happy.
However, we did come to the conclusion that it would be nice to know how many gallons of gas my car needed…and I could help them with that:
Then, I sent them off in their groups to determine how much I paid. Five minutes later, I saw a lot of work that looked like this:
This was typical of the work I saw from my students. There were a lot of good conversations about keeping track of units and how the process worked. After discussing with students the answer:
We had a conversation about what the process entailed. They decided that they took a mileage and turned it into a number of gallons, then took those gallons and turned it into a price. I then defined this as a composition of functions. Then we decontextualized and practiced.
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https://snucongo.org/which-of-the-following-have-the-same-number-of-valence-electrons/
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Explanation:
The team number is the same as the number of the valence electrons each aspect within that team has. Therefore, two aspects in the same group will have actually the same variety of valence electrons. Because all bonding and/or ionizations show off the valence shell of electrons, atoms with the same variety of valence electrons behave similarly. Both tellurium and sulfur space in team VI, and have 6 valence electrons.
You are watching: Which of the following have the same number of valence electrons
Explanation:
The period number/row number is the power level for each element. Therefore, two atoms found in the same row on the periodic table have actually the same power level. Potassium and selenium are both in row 4 of the routine table. Their highest energy level because that electrons is 4.
Explanation:
Ionization energy is the energy required to remove an electron. Noble gases are special due to the fact that the have a complete valence covering of electrons, which provides them the many stable elements, and also to eliminate an electron needs a many energy. The tendency for ionization energy is together follows: ionization energy increases indigenous left to right within a row and also from bottom to optimal within a team on the routine table. Every one of the answer options are in the exact same row the the periodic table, but fluorine is the furthest to the right. For this reason fluorine has actually the best ionization energy.
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Example concern #14 : The periodic Table
How many valence electrons does the element oxygen have?
Explanation:
For neutral atoms, the variety of valence electron is equal to the atom"s main team number. According to the periodic table, oxygen is in team 6; therefore, it has actually 6 valence electron in its external shell.
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Example inquiry #15 : The routine Table
Which the the adhering to is the most electronegative element?
Fluorine
Oxygen
Chlorine
None that the answer selections are correct
Nitrogen
Fluorine
Explanation:
Generally speaking, as you go across a period and up a group on the routine table, electronegativity increases. Fluorine is the most electronegative element, v a Pauling range electronegativity ranking of about 4.0.
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Example question #16 : The routine Table
Which of these elements is the very least likely to hold on come its valence electrons as soon as in the presence of a extremely electronegative atom?
Iron
Sodium
Cobalt
Cesium
Rubidium
Cesium
Explanation:
Atomic radii boosts from right to left of the periodic table and that decreases bottom to top. For this reason francium, in the bottom left that the regular table, has actually the biggest atomic radius; helium, in the optimal right of the chart, has actually the smallest atomic radius. Based on these trends, cesium would certainly be least likely to organize on come its valence electrons since it has actually a larger atomic radius compared to cobalt or iron. To compare this come helium which has actually a tiny atomic radius and a complete valence shell of electrons, which renders it very stable.
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Example inquiry #17 : The regular Table
Which that the complying with groups of elements would need the greatest influx of power to dislodge an electron from its valence shell?
Transition metals
Alkaline earth metals
Alkali metals
Nonmetals
Halogens
Halogens
Explanation:
Halogens would need the greatest first ionization power to dislodge among their valence shell electrons because they have both the greatest electron affinity and also the smallest atomic radii. Since their electrons space both closer to their nuclei and halogens are more "electron greedy" (electronegative), castle require more energy to remove an electron. However, because the noble gasses have complete valence shells, they have the greatest first ionization energies.
See more: The Major Factor Keeping Forests From Growing In Grassland Areas Is
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Example inquiry #18 : The periodic Table
You space tasked with packing a very tiny jar v as many individual atom of the same aspect as possible. Which facet would you pick to ensure the largest quantity that atoms would certainly fit in to the jar?
Francium
Boron
Helium
Hydrogen
Helium
Explanation:
Atomic radii decrease indigenous left to right across the regular table and increase from optimal to bottom of the periodic table. Based on these trends, helium has the smallest atomic radii and more atoms would fit inside our imaginary container.
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Example inquiry #19 : The regular Table
Which the these facets is most likely to kind a polar bond?
Oxygen
Chlorine
Bromine
Fluorine
Nitrogen
Fluorine
Explanation:
Electronegativity is the propensity of an atom to entice electrons come it. Electronegative atoms room electron "greedy". When they kind covalent bonds, very electronegative atom often form polar covalent bond in i m sorry the electrons spend a greater amount that time near the electronegative atom resulting in a dipole moment. Water is the quintessential example of a polar molecule.
Electronegativity rises from left to right throughout the regular table and also increases native bottom to top as well. Fluorine is the many electronegative atom and also it would be most likely to result in a polar molecule. Save in mind that the development of a polar bond relies on the different electronegativities that the atoms in question. Because that example, two oxygen atoms execute not make a polar bond even though both atom are extremely electronegative.
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Example concern #20 : The routine Table
Which that the complying with lists atomic radius increasing from smallest to biggest?
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## Degree of Operating Leverage Calculator
The degree of operating leverage shows the effect on operating income of the cost structure of a business. The higher the fixed costs the higher the leverage and the more volatile and risky the operating income of a business is viewed.
The degree of operating leverage calculator allows for details relating to two businesses or accounting periods to be entered so that comparisons can be made.
Degree of Operating Leverage Calculator February 1st, 2018
## Operating Leverage
The operating leverage shows the level of fixed cost leverage within a business, and the degree of operating leverage shows the impact of the cost structure on the operating income of the business.
The operating income for a business with high leverage can change dramatically for a given change in the number of units sold, and its earnings are said to be more volatile and therefore more risky
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## Debt to Equity Ratio Calculator
The debt to equity ratio is the ratio of how much a business owes (debt) compared to how much the owners have invested (equity). It is calculated by dividing debt by owners equity.
The Excel debt equity ratio calculator, available for download below, is used to compute this important leverage ratio by entering details relating to the debt and owners equity of the business.
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## Times Interest Earned Calculator
The times interest earned ratio is an important leverage ratio indicating whether a business has sufficient earnings to make interest payments on its borrowings.
This Excel times interest earned calculator, available for download below, calculates the TIE ratio using the times interest earned formula, by entering details of the earnings before interest and tax (EBIT), and the interest expense from the income statement of the business.
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## Interest Coverage Ratio
The interest coverage ratio measures the amount of earnings a business has to make interest payments. It is calculated by dividing the profit before interest and tax by the interest expense. It is sometimes called the interest cover ratio or simply interest coverage or interest cover.
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## What is a Leverage Ratio?
A Leverage Ratio is used to show the capital structure of a business and in particular the level of debt in relation to owners equity. A business with a high level of debt is considered to be more risky but will give greater returns to the owners provided cash and profit are managed correctly.
What is a Leverage Ratio? November 6th, 2016
## Debt Equity Ratio
The debt equity ratio is the ratio of how much a business owes (debt) compared to how much the owners have invested (equity). It is calculated by dividing debt by equity.
Debt Equity Ratio November 6th, 2016
## Gearing Ratio
The Gearing ratio is the ratio of how much a business owes (debt) compared to how much the owners have invested (equity). It is calculated by dividing debt by equity.
Gearing Ratio November 6th, 2016
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# Eventually tossed a coin many times. Part-II
A coin is tossed $$10$$ times. If the probability of getting at least six heads is $$\left( \dfrac{a}{b} \right)$$ where gcd(a,b)=1. Then find the value of $$(a+b)$$.
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# DebbiesAlgebraActivities's Shop
6k+Views
#### Write Linear Equations Activities Bundle 1
This bundle includes five activities for practice writing linear equations at a discounted price. Please click the links to see the description of each product for a good look at the whole package! Sold separately, these items total \$12. Save money and have fun with these classroom activities. This purchase is for one teacher only. This resource is not to be shared with colleagues or used by an entire grade level, school, or district without purchasing the proper number of licenses. If you are a coach, principal, or district interested in a site license, please contact me for a quote at debbiesalgebraactivities@gmail.com. This resource may not be uploaded to the internet in any form, including classroom/personal websites or network drives.
#### Solve Systems of Linear Equations Mixed Review Problem Pass Activity
Twelve rounds include practice or review solving systems of linear equations using the graphing method, the substitution method and the linear combinations / elimination method. This activity includes a problem with “no solution” and another with “infinitely many solutions”. Students work with a partner while seated at their desks. They should write in their own notebook or on the blank Answer Sheet (included). At the same time the first pair of students is working on Problem 1, the second pair works on Problem 2, the third pair works on Problem 3, etc. At the teacher’s signal, all students pass their problem in a specific direction. The students who started with Problem 2 should now pass it to the students who started with Problem 1, the students who started with Problem 3 pass it to the students who started with Problem 2, etc. The first pair of students starting with Problem 1 should deliver their finished problem to the last pair of students, or the teacher may prefer to deliver these each time. Students now flip over their new page to find the ANSWER to the problem they just finished. Students continue to work problems in order, pass problems and check their answers on the back of the next problem page until they have completed all problems included in the activity. This activity works well in the middle of a lesson while students are actively practicing a new skill or can be used as a review. The answer key is built into the activity so students check for accuracy themselves. Teacher Setup: Print single-sided copies and slide pages into plastic page protectors to keep problems and answers together. Put Problem 1 and ANSWER Problem 12 back-to-back in the same plastic page protector; put Problem 2 and ANSWER Problem 1 back-to-back in another plastic page protector, Problem 3 and ANSWER Problem 2 together, etc. Prepare 2 or more complete sets of the activity to have enough pages for each pair of students in the class. Keep complete sets in order. It is very important to hand out problems in numerical order so the page with the answer on the back follows its problem number. As you hand out problems in order, problem side up, tell each pair of students the direction they should pass their problem when finished. This direction may vary by row if you zig-zag or “snake” up and down the rows of desks. The first pair of students starting with Problem 1 should deliver their finished problems to the last pair of students, or you may prefer to deliver these each time. CCSS: Analyze and solve pairs of simultaneous linear equations: 8.EE.C.8, 8.EE.C.8a, 8.EE.C.8b CCSS: Solve systems of equations: HSA.REI.C5, HSA.REI.C.6 This purchase is for one teacher only.
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# Forces involved analysis …
### We try to do a little analysis of what are the forces acting on the section of a blade of our generator, faithful to the principle explained in this post.
#### Static analysis of the distribution of lift and drag forces.
We start from an analysis of the stationary machine: It assumed null the angular velocity of the blade around the main axis ω = 0.
The lift and drag values are dependent by incidence of the blade section relative to the wind direction, for simplicity in the analysis of the breakdown we will write only P and R to indicate the functions P (i) and R (i) with i = angle of incidence.
P(i)=Cp(i) ρ S V^2
R(i)=Cr(i) ρ S V^2
#### Resistance component
We define: b the distance of the blade section from the main axis, α the angle that describes the position of the blade and V the speed of the true wind.
We calculate the length of the arm in function of α:
b = 2 r abs ((sin (α / 2))).
We calculate the resistance of the blade along the tangent of the circle that represents the direction in which the blade is free to move: R ‘= R cos (α), the other orthogonal component is nulled by the constraint .
We calculate the moment (M) respect to the main axis:
M = b*f = b*R’ = – 2 r R cos(α) ∣sen(α/2)∣
From this formula we obtain:
• α = 0° we have b = 0 , then M = 0;
• α = 90° we have R’ = 0, R is perpendicular to the direction of the blade , then M = 0;
• α = 180° => b = 2r , R =R’ and R has max value.
• α = 270° => R’ = 0 , R is perpendicular to the direction of the blade , then M = 0;
• -90°<α<90° the resistance opposed to the motion of the blades and vanishes for α = 0° .
We can observe that in the starting phase, when angular velocity is null, the wind resistance contributes negatively to the displacement of the blade for a half cycle (-90 °> α <90 °) in which the arm is, however, smaller and the resistance the same is lower due to the transverse position to the wind of the blade.
Contrary with 90 °> α <225 ° there is a positive contribution of the resistive component on the half cycle which has longer arm.
Assuming a linear relationship between the coefficient of resistance and angle of incidence with a maximum value of R = 1, setting b (max) = 1 and S = 1 we have the trend as in the figure:
#### Lift component
We try to do a similar analysis to the lift force generated by the blade:
We define
β = α – 90 °
as the angle between the lift vector and the direction of the blade ( tangent line on the α position).
we can write the P component of the lift as P ‘= P cos (β).
then:
• on α = 0° => b = 0 and M = 0;
• α = 45° => M is approximately = 0.268 * 2r* P(i) ==>so it contributes to the motion
• α = 90° => M is approximately = 0.707 * 2r* P(i) ==> contributes to the motion
• α = 180° => M = 0 , P is perpendicular to the direction of the blade .
• α = 225° => M is approximately = -0,65 * 2r *P(i) , P reverses the direction and still has that lift contributes to the motion of the blade in the desired direction.
We can therefore say that: the contribution of the lift to the motion of the blade is always positive except for α = 0 ° and α = 180 °, respectively, in which they cancel the arm length and lift force.
Here, respectively, the graphs relating to the lift contribution (| cos (β) | * | sin (α / 2) |) depending on the angle α and, approximately, the value of the coefficient of lift force (Cp (i)) based on the angle of incidence (α / 2) for a flat surface:
By product of the two graphs you get the trend of the force transmitted to the machine from the lift force.
The peak of lift occurs with an angle of incidence of about 15 ° (flat surface) which corresponds to α = 30 ° where the contribution of the lift blade movement is about 70 ~ 80%.
#### Dynamic operation
In the above image we try to establish how the apparent wind vector (v ‘) is positioned respect to the (section of) blade when the tangential velocity is close ( or equal) to the true wind speed (v).
We can see as, approaching the tangential velocity to the real wind velocity, the vector v’ tends to form a smaller angle respect to the profile of the blade (i => 0 °), and in consequence of this:
The resistance tends to decrease considerably and its vector tends to rotate counterclockwise.
The lift force occupies a major role in generating force to be transferred to the machine and improves the efficiency with the approach to an angle of incidence of 15 ° (maximum lift of the yield, see image on the left) that is obtained with a tangential velocity equal to ~% 96 true wind speed.
The machine behaves as resistive generators (eg Savonius) with vertical axis with resistance operation at startup, this also leads to a lower start-up wind (speed lower cut).
As the speed increases, the behavior is like to wind turbines with horizontal axis, where the lift force becomes the main component in the generation of the force that actively contributes to the motion of the blade.
In next post an detailed analysis on blade profile…stay tuned.
p.s. : Sorry for my bad english.
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Bjorn Lisper lisper at it.kth.se
Thu Jun 22 09:37:42 EDT 2006
```Udo Stenzel:
>Bjorn Lisper wrote:
>> Here is one way to do it. First, you have to interpret operations on
>> matrices as being elementwise applied. E.g, (*) is interpreted as zipWith
>> (zipWith (*)) rather than matrix multiply
>
>What's this, the principle of greatest surprise at work? Nonono, (*)
>should be matrix multiplication, fromInteger x should be (x * I) and I
>should be the identity matrix. Now all we need is an infinitely large
>I, and that gives:
>
>instance Num a => Num [[a]] where
> (+) = zipWith (zipWith (+))
> (-) = zipWith (zipWith (-))
> negate = map (map negate)
> fromInteger x = fix (((x : repeat 0) :) . map (0:))
> m * n = [ [ sum \$ zipWith (*) v w | w <- transpose n ] | v <- m ]
There are pros and cons, of course. Using (*) for matrix multiply is
well-established in linear algebra. But:
- it breaks the symmetry. This particular operator is then overloaded in a
different way than all the others, and
- your definition of fromInteger will behave strangely with the elementwise
extended operations, like (+). 1 + [[1,2],[3,4]] will become
[[2,2],[3,5]] rather than [[2,3],[4,5]]. Array languages supporting this
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The Watering Hole
A zebra is walking to the watering hole when on the way he notices 5 giraffes. Swinging from each of the giraffes’ necks are 12 monkeys. Hanging from the tails of each monkey are 7 birds. On each of the birds’ backs are 4 little frogs. In each of the frogs’ mouths are 2 flies. How many animals were going to the watering hole?
Only 1. The zebra only SAW the animals on the way to the watering hole. But none of them except for the zebra were going to the watering hole.
Reveal Answer
View More Fun Brain Teasers In The Logic Riddles Section!
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Search Result for "crc":
```
V.E.R.A. -- Virtual Entity of Relevant Acronyms (February 2016):CRC
Cyclic Redundancy Check[sum]
The Free On-line Dictionary of Computing (30 December 2018):cyclic redundancy check
CRC
cyclic redundancy code
(CRC or "cyclic redundancy code") A number derived
from, and stored or transmitted with, a block of data in order
to detect corruption. By recalculating the CRC and comparing
it to the value originally transmitted, the receiver can
detect some types of transmission errors.
A CRC is more complicated than a checksum. It is calculated
using division either using shifts and exclusive ORs or
table lookup (modulo 256 or 65536).
The CRC is "redundant" in that it adds no information. A
single corrupted bit in the data will result in a one bit
change in the calculated CRC but multiple corrupted bits may
cancel each other out.
CRCs treat blocks of input bits as coefficient-sets for
polynomials. E.g., binary 10100000 implies the polynomial:
1*x^7 + 0*x^6 + 1*x^5 + 0*x^4 + 0*x^3 + 0*x^2 + 0*x^1 + 0*x^0.
This is the "message polynomial". A second polynomial, with
constant coefficients, is called the "generator polynomial".
This is divided into the message polynomial, giving a quotient
and remainder. The coefficients of the remainder form the
bits of the final CRC. So, an order-33 generator polynomial
is necessary to generate a 32-bit CRC. The exact bit-set used
for the generator polynomial will naturally affect the CRC
that is computed.
Most CRC implementations seem to operate 8 bits at a time by
building a table of 256 entries, representing all 256 possible
8-bit byte combinations, and determining the effect that each
byte will have. CRCs are then computed using an input byte to
select a 16- or 32-bit value from the table. This value is
then used to update the CRC.
Ethernet packets have a 32-bit CRC. Many disk formats
include a CRC at some level.
(1997-08-02)
```
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Cody
# Problem 1053. Create a Multiplication table matrix...
Solution 2185008
Submitted on 31 Mar 2020 by Yaroslav Br
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
m = 5; n = 4; y_correct = [ 1 2 3 4 5; 2 4 6 8 10; 3 6 9 12 15; 4 8 12 16 20]; assert(isequal(Table(m,n),y_correct))
2 Pass
m = 1; n = 1; y_correct = 1; assert(isequal(Table(m,n),y_correct))
3 Pass
m = 3; n = 1; y_correct = [1 2 3]; assert(isequal(Table(m,n),y_correct))
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0
# What is the greatest common factor for 24 and 15?
Updated: 8/30/2023
Wiki User
β 9y ago
The prime factorization of 15 is 3*5. The prime factorization of 24 is 2*2*2*3. The greatest common factor is 3. (This is the product of all common prime factors)
The least common multiple is 2*2*2*3*5 = 120. (This is the product of the two numbers, divided by the greatest common factor)
Wiki User
β 15y ago
Wiki User
β 14y ago
The LCM of 15 and 24 is 120.
The GCF/HCF of 15 and 24 is 3.
120 * 3 = 360.
Wiki User
β 8y ago
The LCM of 15, 20, and 24 is 120
15 = 3 x 5
20 = 22 x 5
24 = 23 x 3
lcm = 23 x 3 x 5 = 120
Wiki User
β 11y ago
Factors of 15: 1, 3, 5, 15
Factors of 24: 1, 2, 3, 4, 6, 8, 12, 24
GCF (15, 24) = 3
Wiki User
β 10y ago
The least common multiple of 15 , 24 , 30 = 120
Wiki User
β 8y ago
15 and 24's highest common factor is three.
Wiki User
β 14y ago
1 would be the greatest common factor.
Wiki User
β 9y ago
The GCF is 3.
Earn +20 pts
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Continue Learning about Movies & Television
### What is the greatest common factor of 24 10 and 15?
The GCF is 1.The Greatest Common Factor is 1
### What is the greatest common factor of 15 twenty-fourths?
We want the GCF or highest common factor of 15 and 24. The factors of 15 and 1,3, and 5.For 24 we have 1,3, 8, 12 and 24.Since 5 is not a factor of 24, the GCF or highest factor is 3
The GCF is 1.
The GCF is 3.
### What is the greatest common factor of 12 and 30 and 15 and 24?
The GCF is 3.
Related questions
### What is the greatest common factor of 15 and 24 and 30?
The greatest common factor of 15, 24, and 30 is 3
### What is the greatest common factor of 15 24?
Greatest Common Factor (GCF) of (15,24) is 3
3
### What is the greatest common factor of 15 18 and 24?
The greatest common factor of 15 18 and 24 is 3.The GCF is 3.
### What is the greatest common factor of 24 and 360?
360 / 24 = 15.Since 24 is a factor of 360, and also obviously a factor of itself,it must be the greatest common factor of 24 and 360.
### What is the greatest common factor of 24 10 and 15?
The GCF is 1.The Greatest Common Factor is 1
The GCF is 3.
The GCF is 3.
It is 3
### What is the greatest common factor for 15 and 24 and 30?
The GCF of 15, 24, and 30 is 3.
### What is the greatest common factor of 15 over 24?
The GCF of 15 and 24 is 3. 15/24 = 5/8
### What is the greatest common factor of 15 twenty-fourths?
We want the GCF or highest common factor of 15 and 24. The factors of 15 and 1,3, and 5.For 24 we have 1,3, 8, 12 and 24.Since 5 is not a factor of 24, the GCF or highest factor is 3
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# Ideal property is centralizer-closed
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
This article gives the statement, and possibly proof, of a Lie subring property (i.e., ideal of a Lie ring) satisfying a Lie subring metaproperty (i.e., centralizer-closed Lie subring property)
View all Lie subring metaproperty satisfactions | View all Lie subring metaproperty dissatisfactions |Get help on looking up metaproperty (dis)satisfactions for Lie subring properties
Get more facts about ideal of a Lie ring |Get facts that use property satisfaction of ideal of a Lie ring | Get facts that use property satisfaction of ideal of a Lie ring|Get more facts about centralizer-closed Lie subring property
## Statement
Suppose $L$ is a Lie ring, $S$ is an ideal of $S$, and $C = C_L(S)$ is the centralizer of $S$ in $L$. Then, $C$ is also an ideal of $S$.
## Related facts
### Generalizations
The general version of this result is: invariance under any set of derivations is centralizer-closed, which reduces to this result when we take the inner derivations (see Lie ring acts as derivations by adjoint action). Another special case of this general result is:
## Proof
### Hands-on proof using Jacobi identity
Given: A Lie ring $L$, an ideal $S$ of $L$. $C = C_L(S)$ is the centralizer of $S$ in $L$.
To prove: $C$ is an ideal of $L$, i.e., for any $x \in L$ and $c \in C$, $[x,c] \in C$.
Proof: To show $[x,c] \in C$ it suffices to show that $[[x,c],s] = 0$ for all $s \in S$.
By the Jacobi identity:
$[[x,c],s] + [[c,s],x] + [[s,x],c] = 0$.
Since $C$ centralizes $S$, $[c,s] = 0$, so the second term is zero. Further, since $S$ is an ideal, $[s,x] \in S$, and since $C$ centralizes $S$, $[[s,x],c] = 0$. Thus, both the second and third term on the left side are zero, forcing $[[x,c],s] = 0$.
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# Lift Gain Calculation
#1
Can anyone help me understand how cells at Lift@Decile column are calculated. I have understood till Cum%Pop
#2
Hello @Akash_Haldankar,
The calculation is:
Example for the 7th decile.
`(number of 1's in each decile(440)/total number of 1s (3850))*100*10.`
However you can also take a look at the below snapshot:
It is using the same calculations but instead of multiplying by 10 it is dividing by 10(or the proportion of data in each decile.)
Hope this helps!!
#3
Thanks @shuvayan
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# Do We Put Too Much Emphasis on Memorizing Multiplication Facts?
Fact fluency might matter more.
A non-teacher friend recently mentioned how, as a kid, he had really loved math class until the fourth grade. Before he even told me, I knew exactly where he was tripped up all those years before: memorizing multiplication facts.
For so long, third and fourth graders have been asked to run the gauntlet of multiplication fact memorization. Students are usually expected to practice their facts at home and are tested with sheets of problems they must solve in a set amount of time.
Some lucky students memorize their math facts with ease. Many students memorize a good chunk of the prescribed facts, but struggle with a handful of the more difficult ones.
Finally, you have students like my friend who never manage to move beyond their 3s or 4s facts. An unfortunate consequence is that these students can see their struggle with facts as an indication that they are not good at math.
Instead of putting in so much time into rote memorization, teachers can take math facts further by shifting their focus to fact fluency.
## Fact Fluency vs. Memorization
When explaining the difference between memorization and fact fluency to students, I often describe memory’s limitations. Memory is all or nothing. If you have something memorized, you simply know the math fact or poem or state capital you set out to memorize.
On the other hand, students often feel stuck when they don’t have a fact memorized. They ask others for the answer instead of having a toolbox of strategies to get at the fact they can’t remember.
Fact fluency, like language fluency, means students can flexibly apply what they already know. They find facts they don’t recall by using the facts they already have memorized. They double their 2s facts to get their 4s facts. And they easily add one group more to get the fact they need. Not only do they see how multiplication facts are related to one another, but they also efficiently use the relationships between multiplication, addition, subtraction, and division.
## 3 Ways to Increase Fact Fluency Today
### 1. Put less emphasis on time and speed
I’ve heard arguments made that timed tests are the only way to build automaticity with facts because they keep students using from less-efficient strategies such as skip-counting. While we want to encourage efficient strategies, timed tests aren’t as effective as direct instruction in developing these skills. Timed tests offer a good snapshot of how well students have memorized their facts, but are not a powerful teaching tool.
### 2. Prompt students to make connections between facts
One deceptively simple strategy to improve fact fluency is to start asking students how they can use one fact they know to find another they may not.
Too often, students reflexively look to teachers and parents to give them the answer to 7 times 8 without first trying to get it on their own. The focus on rote memorization has left our students thinking that math facts are discrete pieces of information that aren’t related to one another. Foster problem-solving skills by prompting them to first search for the product for themselves.
### 3. Explicitly teach strategies
We can’t assume students know how multiplication facts are related. Take time to explain strategies for using an easier fact to find a trickier one. For example, students can use the product of 8 X10 to find the product of 8 X 9 by subtracting one group of 8. Other strategies include doubling doubles to find 4s facts or doubling 4s facts to find 8s facts.
Finally, teach students to apply the concept of partial products to their multiplication facts. Repeatedly show them how to combine two or more products to find an unknown answer. For example, have students use the products of 3 X 5 and 6 X 5 to find the product of 8 X 5.
What’s your take on teaching multiplication? Do you explicitly teach fact fluency? Come in share in our WeAreTeachers HELPLINE group on Facebook. WeAreTeachers HELPLINE is a place for teachers to ask and respond to questions on classroom challenges, collaboration and advice.
Plus…free multiplication spinner games and multiplication anchor charts.
## You Might Also Like
### Free Classroom Poster: Let’s Get Kids Talking About Math
This free poster can help prompt your students with questions to ask and ways to answer them.
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• # question_answer A spring-mass suspension has a natural frequency of 40 rad/s. What is the damping ratio required if it is desired to reduce this frequency to 20 rad/s by adding a damper to it? A) $\frac{\sqrt{3}}{2}$ B) $\frac{1}{2}$C) $\frac{1}{\sqrt{2}}$ D) $\frac{1}{4}$
${{\omega }_{d}}={{\omega }_{n}}\sqrt{1-{{\zeta }^{2}}}$ $20=40\sqrt{1-{{\zeta }^{2}}}$ $1-{{\zeta }^{2}}=\frac{1}{4}$ ${{\zeta }^{2}}=\frac{3}{4},$ $\zeta =\frac{\sqrt{3}}{2}$
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# 20.83 kg to lbs - 20.83 kilograms to pounds
Before we get to the more practical part - that is 20.83 kg how much lbs calculation - we are going to tell you a little bit of theoretical information about these two units - kilograms and pounds. So let’s move on.
How to convert 20.83 kg to lbs? 20.83 kilograms it is equal 45.9222891746 pounds, so 20.83 kg is equal 45.9222891746 lbs.
## 20.83 kgs in pounds
We will begin with the kilogram. The kilogram is a unit of mass. It is a base unit in a metric system, known also as International System of Units (in short form SI).
At times the kilogram could be written as kilogramme. The symbol of the kilogram is kg.
Firstly the kilogram was defined in 1795. The kilogram was defined as the mass of one liter of water. This definition was not complicated but totally impractical to use.
Then, in 1889 the kilogram was defined using the International Prototype of the Kilogram (in abbreviated form IPK). The IPK was prepared of 90% platinum and 10 % iridium. The International Prototype of the Kilogram was used until 2019, when it was switched by a new definition.
Today the definition of the kilogram is based on physical constants, especially Planck constant. Here is the official definition: “The kilogram, symbol kg, is the SI unit of mass. It is defined by taking the fixed numerical value of the Planck constant h to be 6.62607015×10−34 when expressed in the unit J⋅s, which is equal to kg⋅m2⋅s−1, where the metre and the second are defined in terms of c and ΔνCs.”
One kilogram is 0.001 tonne. It could be also divided into 100 decagrams and 1000 grams.
## 20.83 kilogram to pounds
You know something about kilogram, so now we can move on to the pound. The pound is also a unit of mass. It is needed to point out that there are more than one kind of pound. What are we talking about? For example, there are also pound-force. In this article we want to focus only on pound-mass.
The pound is used in the British and United States customary systems of measurements. Naturally, this unit is in use also in other systems. The symbol of this unit is lb or “.
The international avoirdupois pound has no descriptive definition. It is equal 0.45359237 kilograms. One avoirdupois pound could be divided into 16 avoirdupois ounces and 7000 grains.
The avoirdupois pound was implemented in the Weights and Measures Act 1963. The definition of the pound was written in first section of this act: “The yard or the metre shall be the unit of measurement of length and the pound or the kilogram shall be the unit of measurement of mass by reference to which any measurement involving a measurement of length or mass shall be made in the United Kingdom; and- (a) the yard shall be 0.9144 metre exactly; (b) the pound shall be 0.45359237 kilogram exactly.”
### How many lbs is 20.83 kg?
20.83 kilogram is equal to 45.9222891746 pounds. If You want convert kilograms to pounds, multiply the kilogram value by 2.2046226218.
### 20.83 kg in lbs
Theoretical section is already behind us. In next section we are going to tell you how much is 20.83 kg to lbs. Now you know that 20.83 kg = x lbs. So it is high time to get the answer. Let’s see:
20.83 kilogram = 45.9222891746 pounds.
It is an exact outcome of how much 20.83 kg to pound. It is possible to also round off the result. After rounding off your result will be exactly: 20.83 kg = 45.826 lbs.
You learned 20.83 kg is how many lbs, so see how many kg 20.83 lbs: 20.83 pound = 0.45359237 kilograms.
Obviously, in this case it is possible to also round off this result. After it your result is exactly: 20.83 lb = 0.45 kgs.
We are also going to show you 20.83 kg to how many pounds and 20.83 pound how many kg outcomes in charts. Have a look:
We will start with a table for how much is 20.83 kg equal to pound.
### 20.83 Kilograms to Pounds conversion table
Kilograms (kg) Pounds (lb) Pounds (lbs) (rounded off to two decimal places)
20.83 45.9222891746 45.8260
Now look at a chart for how many kilograms 20.83 pounds.
Pounds Kilograms Kilograms (rounded off to two decimal places
20.83 0.45359237 0.45
Now you know how many 20.83 kg to lbs and how many kilograms 20.83 pound, so we can move on to the 20.83 kg to lbs formula.
### 20.83 kg to pounds
To convert 20.83 kg to us lbs you need a formula. We are going to show you a formula in two different versions. Let’s begin with the first one:
Number of kilograms * 2.20462262 = the 45.9222891746 result in pounds
The first formula give you the most accurate result. In some situations even the smallest difference could be considerable. So if you need an accurate outcome - first formula will be the best solution to calculate how many pounds are equivalent to 20.83 kilogram.
So let’s move on to the shorer version of a formula, which also enables conversions to know how much 20.83 kilogram in pounds.
The shorter formula is down below, have a look:
Number of kilograms * 2.2 = the outcome in pounds
As you see, the second formula is simpler. It could be the best choice if you want to make a conversion of 20.83 kilogram to pounds in fast way, for example, during shopping. Just remember that final outcome will be not so correct.
Now we are going to show you these two formulas in practice. But before we are going to make a conversion of 20.83 kg to lbs we want to show you easier way to know 20.83 kg to how many lbs totally effortless.
### 20.83 kg to lbs converter
Another way to learn what is 20.83 kilogram equal to in pounds is to use 20.83 kg lbs calculator. What is a kg to lb converter?
Converter is an application. Converter is based on longer version of a formula which we gave you above. Thanks to 20.83 kg pound calculator you can easily convert 20.83 kg to lbs. You only have to enter number of kilograms which you need to convert and click ‘calculate’ button. The result will be shown in a flash.
So try to convert 20.83 kg into lbs using 20.83 kg vs pound converter. We entered 20.83 as a number of kilograms. It is the outcome: 20.83 kilogram = 45.9222891746 pounds.
As you see, our 20.83 kg vs lbs calculator is user friendly.
Now we can move on to our primary topic - how to convert 20.83 kilograms to pounds on your own.
#### 20.83 kg to lbs conversion
We are going to start 20.83 kilogram equals to how many pounds calculation with the first formula to get the most correct result. A quick reminder of a formula:
Number of kilograms * 2.20462262 = 45.9222891746 the result in pounds
So what need you do to check how many pounds equal to 20.83 kilogram? Just multiply number of kilograms, this time 20.83, by 2.20462262. It is 45.9222891746. So 20.83 kilogram is equal 45.9222891746.
It is also possible to round off this result, for example, to two decimal places. It is equal 2.20. So 20.83 kilogram = 45.8260 pounds.
It is time for an example from everyday life. Let’s convert 20.83 kg gold in pounds. So 20.83 kg equal to how many lbs? And again - multiply 20.83 by 2.20462262. It is exactly 45.9222891746. So equivalent of 20.83 kilograms to pounds, when it comes to gold, is 45.9222891746.
In this case it is also possible to round off the result. Here is the outcome after rounding off, in this case to one decimal place - 20.83 kilogram 45.826 pounds.
Now we can go to examples converted using a short version of a formula.
#### How many 20.83 kg to lbs
Before we show you an example - a quick reminder of shorter formula:
Amount of kilograms * 2.2 = 45.826 the outcome in pounds
So 20.83 kg equal to how much lbs? As in the previous example you have to multiply amount of kilogram, in this case 20.83, by 2.2. Look: 20.83 * 2.2 = 45.826. So 20.83 kilogram is exactly 2.2 pounds.
Do another calculation with use of shorer version of a formula. Now calculate something from everyday life, for example, 20.83 kg to lbs weight of strawberries.
So let’s convert - 20.83 kilogram of strawberries * 2.2 = 45.826 pounds of strawberries. So 20.83 kg to pound mass is 45.826.
If you learned how much is 20.83 kilogram weight in pounds and are able to calculate it with use of two different versions of a formula, we can move on. Now we are going to show you all results in charts.
#### Convert 20.83 kilogram to pounds
We know that results shown in tables are so much clearer for most of you. It is totally understandable, so we gathered all these results in tables for your convenience. Due to this you can quickly compare 20.83 kg equivalent to lbs outcomes.
Let’s start with a 20.83 kg equals lbs table for the first formula:
Kilograms Pounds Pounds (after rounding off to two decimal places)
20.83 45.9222891746 45.8260
And now see 20.83 kg equal pound table for the second formula:
Kilograms Pounds
20.83 45.826
As you can see, after rounding off, when it comes to how much 20.83 kilogram equals pounds, the results are not different. The bigger amount the more considerable difference. Remember it when you need to make bigger number than 20.83 kilograms pounds conversion.
#### How many kilograms 20.83 pound
Now you learned how to calculate 20.83 kilograms how much pounds but we are going to show you something more. Are you interested what it is? What about 20.83 kilogram to pounds and ounces conversion?
We will show you how you can calculate it little by little. Let’s begin. How much is 20.83 kg in lbs and oz?
First thing you need to do is multiply number of kilograms, in this case 20.83, by 2.20462262. So 20.83 * 2.20462262 = 45.9222891746. One kilogram is 2.20462262 pounds.
The integer part is number of pounds. So in this example there are 2 pounds.
To know how much 20.83 kilogram is equal to pounds and ounces you need to multiply fraction part by 16. So multiply 20462262 by 16. It is exactly 327396192 ounces.
So final result is equal 2 pounds and 327396192 ounces. It is also possible to round off ounces, for instance, to two places. Then final result will be exactly 2 pounds and 33 ounces.
As you can see, conversion 20.83 kilogram in pounds and ounces simply.
The last calculation which we are going to show you is calculation of 20.83 foot pounds to kilograms meters. Both foot pounds and kilograms meters are units of work.
To convert foot pounds to kilogram meters it is needed another formula. Before we show you this formula, let’s see:
• 20.83 kilograms meters = 7.23301385 foot pounds,
• 20.83 foot pounds = 0.13825495 kilograms meters.
Now let’s see a formula:
Amount.RandomElement()) of foot pounds * 0.13825495 = the result in kilograms meters
So to convert 20.83 foot pounds to kilograms meters you have to multiply 20.83 by 0.13825495. It gives 0.13825495. So 20.83 foot pounds is equal 0.13825495 kilogram meters.
You can also round off this result, for instance, to two decimal places. Then 20.83 foot pounds will be equal 0.14 kilogram meters.
We hope that this calculation was as easy as 20.83 kilogram into pounds conversions.
We showed you not only how to do a calculation 20.83 kilogram to metric pounds but also two another calculations - to know how many 20.83 kg in pounds and ounces and how many 20.83 foot pounds to kilograms meters.
We showed you also another way to do 20.83 kilogram how many pounds calculations, this is with use of 20.83 kg en pound converter. This is the best option for those of you who do not like calculating on your own at all or this time do not want to make @baseAmountStr kg how lbs conversions on your own.
We hope that now all of you are able to make 20.83 kilogram equal to how many pounds conversion - on your own or with use of our 20.83 kgs to pounds calculator.
So what are you waiting for? Calculate 20.83 kilogram mass to pounds in the way you like.
Do you want to do other than 20.83 kilogram as pounds calculation? For example, for 10 kilograms? Check our other articles! We guarantee that conversions for other amounts of kilograms are so simply as for 20.83 kilogram equal many pounds.
### How much is 20.83 kg in pounds
To quickly sum up this topic, that is how much is 20.83 kg in pounds , we prepared one more section. Here you can see the most important information about how much is 20.83 kg equal to lbs and how to convert 20.83 kg to lbs . Let’s see.
What is the kilogram to pound conversion? To make the kg to lb conversion it is needed to multiply 2 numbers. How does 20.83 kg to pound conversion formula look? . See it down below:
The number of kilograms * 2.20462262 = the result in pounds
How does the result of the conversion of 20.83 kilogram to pounds? The accurate result is 45.9222891746 lbs.
It is also possible to calculate how much 20.83 kilogram is equal to pounds with second, shortened version of the equation. Have a look.
The number of kilograms * 2.2 = the result in pounds
So in this case, 20.83 kg equal to how much lbs ? The answer is 45.9222891746 lb.
How to convert 20.83 kg to lbs in an easier way? It is possible to use the 20.83 kg to lbs converter , which will make all calculations for you and give you a correct answer .
#### Kilograms [kg]
The kilogram, or kilogramme, is the base unit of weight in the Metric system. It is the approximate weight of a cube of water 10 centimeters on a side.
#### Pounds [lbs]
A pound is a unit of weight commonly used in the United States and the British commonwealths. A pound is defined as exactly 0.45359237 kilograms.
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# Number Sense Blog
By Mathnasium | Added Jun 13, 2018
If you want a plant to grow, you don’t “leaf” it alone, you take care of it.
Likewise, if you want your math skills to grow, you have to take care of them by practicing. In this week’s word problem challenge you’ll need your foundational elementary school math skills such as addition, subtraction, and percents to solve the question of how fast a rogue vine will grow.
Go ahead and give it a try. Take your time, and check below when you're ready for the solution!
Question: A tree is 200 feet tall. A vine growing up the trunk of the tree overtakes 10% of the tree’s remaining height each week. How high does the vine reach after 3 weeks?
Solution: On the first week, the vine grows 10% of 200 = 20 feet, leaving 200 – 20 = 180 feet untouched. On the second week, it grows 10% of 180 = 18 feet, leaving 162 feet untouched. On the third week, it grows 10% of 162 = 16.2 feet. So, in total, the vine has grown 20 + 18 + 16.2 = 54.2 feet up the tree.
(Image above via https://www.goodfreephotos.com)
### Favorites
Bedtime Math: http://bedtimemath.org/
National PTA: http://www.pta.org/
NCTM: http://www.nctm.org/
Numberphile: http://www.numberphile.com/
PTO Today: http://www.ptotoday.com/
STEM Connector: http://blog.stemconnector.org/
Before I came to Mathnasium, I could sum up everything I felt about math in one word: 'EVIL!' I hated math. ... Mathnasium has been my safe haven. They truly have shown me the light when it came to addressing my fear and provided me with the tools that I need to rebuild my prior knowledge so that I won't forget it. Math is no longer a subject I shy away from but it is a subject I can boldly accept and understand.
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stymied by excelMarch 17, 2011 2:32 PM Subscribe
There has to be an answer for this, but I just can't figure it out. In Excel, I've got 5 columns (A-E). I need to return a list of values from column A if either column B or C equals 'foo', *and* if either column D or E equals 'bar'. What's the magical formula (or whathaveyou) that will make this happen? For bonus points, results need to appear on a separate tab than the source data. asking on behalf of mrs spinturtle, I personally have no idea what I'm talking about.
posted by spinturtle to Computers & Internet (6 answers total) 3 users marked this as a favorite
start with
``` =IF(AND(OR(B8="foo",C8="foo"),OR(D8="bar",E8="bar")),"WIN","[null]")```
posted by Prince_of_Cups at 2:36 PM on March 17, 2011
What's the deal with list of values in column A? Do you need to find some sub-set of the contents and put them on another tab? Or is the 'list of values' one cell in column A?
The easy case, where you just want whatever's in cell A given your criteria is go to your new tab and put this in cell a1:
=IF(AND(OR(Sheet1!B1="foo",Sheet1!C1="foo"),OR(Sheet1!D1="bar",Sheet1!E1="bar")), A1, "whatever you should do if the criteria arent true")
posted by jeb at 2:36 PM on March 17, 2011 [1 favorite]
I just saw results need to appear on a separate tab than the source data
and, yeah, jeb has the info on pointing at a cell on a different worksheet.
posted by Prince_of_Cups at 2:37 PM on March 17, 2011
Response by poster: mrs says: yes, I'm looking for a subset of contents. It's a 20k row spreadsheet, I need the value from A to appear on a new list (maybe 9k records) if the criteria for B/C and D/E above are met.
posted by spinturtle at 2:45 PM on March 17, 2011
Best answer: Ok, but only one value from A for each row?
In that case...here's the easiest way to do this:
- put a formula like the one I described above on the second sheet, changing things like the Sheet1 to be the actual name of your sheet
- in the last pair of quotation marks (what gets returned when there's no match), take out all that stuff I wrote ("whatever you should...") and change it to "exclude".
- copy the formula
- paste it all the way down column A in the new sheet
- it will populate cells with either the value from column A or the word "exclude"
- select cell A1
- (This part depends on what version of Excel you are using, its either Tools|Data|Filter or the data ribbon, then clicking filter). Set up autofilter to include only rows where the value is NOT equal to "Exclude"
- select all the resulting data
- select a cell in a new sheet
- Edit|Paste Special...|Values to get just the data, nicely packed together, no more formulas
posted by jeb at 2:52 PM on March 17, 2011
Response by poster: Thanks jeb! She says that sounds like what she's looking for, will try it out when she gets to work tomorrow and report back.
posted by spinturtle at 3:10 PM on March 17, 2011
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5
# Remaining Time: hour; 53 minutes, 09 secondsQuestion Completion Status:QUESTIONCalculate Vo in this circuit R1=20,R2= 20,ZC =-j30,ZL=j1Q,V=2<600 VzV 22t0 fle und...
## Question
###### Remaining Time: hour; 53 minutes, 09 secondsQuestion Completion Status:QUESTIONCalculate Vo in this circuit R1=20,R2= 20,ZC =-j30,ZL=j1Q,V=2<600 VzV 22t0 fle und -ubmwt. ( Yt? n Ui A#er5 () Sa"e all nnst'eru. Clc? Sc: 6t ' ubintAi DType here t0 search
Remaining Time: hour; 53 minutes, 09 seconds Question Completion Status: QUESTION Calculate Vo in this circuit R1=20,R2= 20,ZC =-j30,ZL=j1Q,V=2<600 V z V 22 t0 fle und -ubmwt. ( Yt? n Ui A#er5 () Sa"e all nnst'eru. Clc? Sc: 6t ' ubint Ai D Type here t0 search
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# Video: KS2-M19 • Paper 1 • Question 18
20% of 3000 = ＿.
03:29
### Video Transcript
20 percent of 3000 equals what.
The word percent means out of 100. And it’s a way of showing parts of a whole, with 100 percent being the same as a whole. In our problem, the whole amount is 3000. And in this problem, we’re being asked to find what 20 percent of 3000 is worth. The interesting thing about problems like this is that there are often several ways we can find the answer. So how can we find 20 percent of a number?
Let’s go through two different methods that we could use. The first is to find 10 percent of 3000 and then double it to give us 20 percent. Finding 10 percent of a number is usually quite quick to do, and so is doubling. So this is quite an efficient strategy we could use.
There are 10 lots of 10 in 100. So to find 10 percent of a number, all we need to do is to divide that number by 10. And what is 3000 divided by 10? The digit shift one place to the right. And we get an answer of 300. So if 10 percent of 3000 equals 300, then 20 percent of 3000 must be worth twice this amount. 300 doubled equals 600. And so we know 20 percent of 3000 equals 600.
We did say there was more than one method we could use. So let’s go through a second strategy that we could use to find the answer. This strategy involves thinking about 20 percent. How many lots of 20 percent are there in 100? One, two, three, four, five. There are five lots of 20 in 100. And so another way of thinking about this is that 20 percent is one-fifth of 100.
So to find 20 percent of 3000, we just need to find one-fifth of 3000 or divide 3000 by five. And if we look carefully at our first method, we know that dividing by 10 and then multiplying by two is the same as dividing by five. And that’s why both methods are going to give us the same answer.
So how many fives are in 3000? Well here, we can use a multiplication fact to help us. We know that 30 divided by five equals six. 3000 is 100 times larger than 30. So there must be 100 times as many fives in it. So the answer must be 600.
As we discussed, percentage questions are always interesting because there’s usually more than one way we could find the answer. Here, we found the answer by first finding 10 percent and then doubling it. And the second strategy was to realise that there are five lots of 20 in 100. So all we had to do was to divide by five and find one-fifth. 20 percent is the same as one-fifth.
So 20 percent of 3000 equals 600.
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# Four Step Problem Solving Process
Appreciative Inquiry is designed to help you understand complex problems so that you can start the process of solving them.It uses four stages to help you uncover more details about what's creating the problem, and then define actions that will improve the situation.
Tags: Solving Ratio And Proportion Word ProblemsBusiness Plan For Business DevelopmentAnimal Rights Essays5 Paragraph Essay AliensDissertation Medical PhysicsAsus Transformer Book T100 Customer ReviewsSmall Business Research PaperEssay VocabularyEssay Analyzing A Text
However, if you look a bit deeper, the real issue might be a lack of training, or an unreasonable workload.
help you ask the right questions, and work through the layers of a problem to uncover what's really going on.
There are four basic steps in solving a problem: Steps 2 to 4 of this process are covered in depth in other areas of Mind Tools.
For these, see our sections on Creativity for step 2 (generating alternatives); Decision Making for step 3 (evaluating and selecting alternatives); and Project Management for step 4 (implementing solutions).
Many of the tools in this section help you do just that.
We look at these, and then review some useful, well-established problem-solving frameworks.
Whether you're solving a problem for a client (internal or external), supporting those who are solving problems, or discovering new problems to solve, the problems you face can be large or small, simple or complex, and easy or difficult.
A fundamental part of every manager's role is finding ways to solve them.
For example, consider this problem statement: "We have to find a way of disciplining of people who do substandard work." This doesn't allow you the opportunity of discovering the real reasons for under-performance.
The CATWOE checklist provides a powerful reminder to look at many elements that may contribute to the problem, and to expand your thinking around it.
## Comments Four Step Problem Solving Process
• ###### The Impact of Applying the First Two Steps of Polya's Four.
Polya's Four ProblemSolving Steps in an Advanced. Three themes emerged related to the first two steps Polya's problem-solving strategies; the lack of.…
• ###### What Is Problem Solving? - Problem Solving Skills from.
There are four basic steps in solving a problem Defining the. Steps 2 to 4 of this process are covered in depth in other areas of Mind Tools. For these, see our.…
• ###### Problem Solving Technique 4 Steps to Improve Your Processes
Problem Solving Technique 4 steps to improve your processes. With a free tool, for anybody serious about improving their business or management processes.…
• ###### THE EFFECTS OF THE FOUR-STEP PROBLEM-SOLVING.
On students‟ ability to solve multi-step word problems and students‟ beliefs about. Pólya 1945 broke down problem solving into four steps to consider and.…
• ###### Step Problem Solving Fact Sheet
What do we need to know about the 4-step problem solving process. based problem solving is used to determine the intensity and focus of instruction/.…
• ###### A Four-Step Problem-Solving Model for Conflict Resolution
Identify the problem What are you arguing about? / Have the other person state his or her wants and feelings. / Describe how you feel. / Read the other.…
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# Algebra Motion Problems
In these lessons, we will learn how to solve algebra word problems that involve motion.
What are Motion or Distance Word Problems?
Motion problems are based on the formula
d = rt
where d = distance, r = rate and t = time.
How to solve Motion or Distance Word Problems?
Step 1: Draw a diagram to represent the relationship between the distances involved in the problem.
Step 2: Set up a chart based on the formula: rate × time = distance.
Step 3: Use the chart to set up one or more equations.
Step 4: Solve the equations.
We will look at three types of Motion Word Problems:
1. Two objects going in opposite directions.
2. Both objects going in the same direction, but one goes further.
3. One object going and returning at different rates.
Solve Motion Word Problems: Two objects going in opposite directions
Example:
John and Philip who live 14 miles apart start at noon to walk toward each other at rates of 3 mph and 4 mph respectively. After how many hours will they meet?
Solution:
Let x = time walked.
r t d John 3 x 3x Philip 4 x 4x
3x + 4x = 14
7x = 14
x = 2
They will meet in 2 hours.
How to solve motion word problems with objects traveling in opposite directions?
Example:
Two cars leave from the same place at the same time and travel in opposite direction. One car travels at 55 mph and the other at 75 mph. After how many hours will there be 520 miles apart?
Example:
Two planes leave the same point at 8 AM. Plane 1 heads East at 600 mph and Plane 2 heads West at 450 mph. How long will they be 1400 miles apart? At what time will they be 1400 miles apart? How far has each plane traveled?
Solve Motion Word Problems: Two objects going in the same direction
Example:
Aaron left L.A. to drive at 55 mph towards Las Vegas. Mike left L.A. an hour after Aaron (also towards Las Vegas), driving at 70 mph. How long will it take Mike to overtake Aaron?
How to solve motion word problems with objects traveling in the same direction?
Example:
John left his house at 3.00 pm to drive 60 mph to drive towards Michigan. Phoebe left the same house at 5.00 pm, driving 80 mph in the same direction as John. How long will it take Phoebe to overtake John?
Solve Motion Word Problems: One object going and returning at different rates
Example:
In still water, Peter’s boat goes 4 times as fast as the current in the river. He takes a 15-mile trip up the river and returns in 4 hours. Find the rate of the current.
Solution:
Let x = rate of the current.
r t d down river 4x + x 15 / 5x 15 up river 4x - x 15 / 3x 15
The rate of the current is 2 mph.
Example:
Gordon rode his bike at 15 mph to get his car. He then drove back at 45 mph. If the entire trip took him 8 hours, how far away was his car?
Motion Word Problems
This is how to set up motion problems for Algebra.
Three Types of Problems
1. Both going the same direction but one going further
2. Two going in opposite directions
3. Going in one direction and then returning at a different rate.
Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.
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http://www.opengl.org/discussion_boards/showthread.php/169865-Zoom-to-fit-screen?p=1197359&viewfull=1
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# Thread: Zoom to fit screen
1. ## Re: Zoom to fit screen
Is it possible to see the code that finally worked? I have the same problem to solve.
Thanks much!
2. ## Re: Zoom to fit screen
'sin' does not just give better results, it is the right trigonometric function to use.
For the code you need to know:
* width:height aspect ratio (width / height as floating point number as used by gluPerspective)
* center and radius of the bounding sphere for the scene
* direction of the camera (unit length)
* fov in y (as used by gluPerspective)
The code would be as follows:
Code :
```half_min_fov_in_radians = 0.5 * (fov * PI / 180);
if (aspect < 1.0)
{
// fov in x is smaller
half_min_fov_in_radians = atan(aspect * tan(half_min_fov_in_radians));
}
distance_to_center = radius / sin(half_min_fov_in_radians);
eye = center - dir * distance_to_center;```
You may also take advantage of that information to fit your zfar clipping plane (if you do not want to keep it too far).
Code :
```zfar = distance_to_center + radius;
if (zfar < 1.5 * znear)
{
// Keep zfar always bigger than znear
zfar = 1.5 * znear;
}```
#### Posting Permissions
• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•
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https://finance.yahoo.com/news/ternium-nyse-tx-high-return-143041721.html
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U.S. Markets open in 1 hr 16 mins
# Should You Like Ternium S.A.’s (NYSE:TX) High Return On Capital Employed?
Want to participate in a short research study? Help shape the future of investing tools and you could win a \$250 gift card!
Today we’ll look at Ternium S.A. (NYSE:TX) and reflect on its potential as an investment. Specifically, we’re going to calculate its Return On Capital Employed (ROCE), in the hopes of getting some insight into the business.
First, we’ll go over how we calculate ROCE. Then we’ll compare its ROCE to similar companies. Last but not least, we’ll look at what impact its current liabilities have on its ROCE.
### Return On Capital Employed (ROCE): What is it?
ROCE measures the amount of pre-tax profits a company can generate from the capital employed in its business. All else being equal, a better business will have a higher ROCE. In brief, it is a useful tool, but it is not without drawbacks. Renowned investment researcher Michael Mauboussin has suggested that a high ROCE can indicate that ‘one dollar invested in the company generates value of more than one dollar’.
### How Do You Calculate Return On Capital Employed?
Analysts use this formula to calculate return on capital employed:
Return on Capital Employed = Earnings Before Interest and Tax (EBIT) ÷ (Total Assets – Current Liabilities)
Or for Ternium:
0.20 = US\$1.5b ÷ (US\$12b – US\$2.3b) (Based on the trailing twelve months to September 2018.)
So, Ternium has an ROCE of 20%.
### Does Ternium Have A Good ROCE?
ROCE can be useful when making comparisons, such as between similar companies. Ternium’s ROCE appears to be substantially greater than the 11% average in the Metals and Mining industry. We consider this a positive sign, because it suggests it uses capital more efficiently than similar companies. Independently of how Ternium compares to its industry, its ROCE in absolute terms appears decent, and the company may be worthy of closer investigation.
Our data shows that Ternium currently has an ROCE of 20%, compared to its ROCE of 8.4% 3 years ago. This makes us think about whether the company has been reinvesting shrewdly.
When considering ROCE, bear in mind that it reflects the past and does not necessarily predict the future. Companies in cyclical industries can be difficult to understand using ROCE, as returns typically look high during boom times, and low during busts. ROCE is only a point-in-time measure. Remember that most companies like Ternium are cyclical businesses. Since the future is so important for investors, you should check out our free report on analyst forecasts for Ternium.
### Ternium’s Current Liabilities And Their Impact On Its ROCE
Current liabilities include invoices, such as supplier payments, short-term debt, or a tax bill, that need to be paid within 12 months. Due to the way ROCE is calculated, a high level of current liabilities makes a company look as though it has less capital employed, and thus can (sometimes unfairly) boost the ROCE. To check the impact of this, we calculate if a company has high current liabilities relative to its total assets.
Ternium has total assets of US\$12b and current liabilities of US\$2.3b. As a result, its current liabilities are equal to approximately 19% of its total assets. Low current liabilities are not boosting the ROCE too much.
### Our Take On Ternium’s ROCE
This is good to see, and with a sound ROCE, Ternium could be worth a closer look. Of course, you might find a fantastic investment by looking at a few good candidates. So take a peek at this free list of companies with modest (or no) debt, trading on a P/E below 20.
I will like Ternium better if I see some big insider buys. While we wait, check out this free list of growing companies with considerable, recent, insider buying.
To help readers see past the short term volatility of the financial market, we aim to bring you a long-term focused research analysis purely driven by fundamental data. Note that our analysis does not factor in the latest price-sensitive company announcements.
The author is an independent contributor and at the time of publication had no position in the stocks mentioned. For errors that warrant correction please contact the editor at editorial-team@simplywallst.com.
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# Lots of substitution tilings
Posted 5 years ago
I recently updated Substitution Tilings, one of my many items at the Wolfram Demonstrations Project. Some of these were introduced in my blog Shattering the Plane with Twelve New Substitution Tilings Using 2, φ, ψ, χ, ρ. Here are 26 of the 40 tiling currently in Substitution Tilings. Some but not all of these are at the Tilings Encyclopedia.Wolfram Language code is attached below at the end of this post. Any corrections, suggestions or additions are welcome. Attachments:
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Posted 2 years ago
I'm not sure what you mean by "algebraic tilings." It doesn't seem to match the "Algebraic Tiling" article by S. K. Stein. I just saw your AlgebraicSubstitutionTiling function. ResourceFunction["AlgebraicSubstitutionTiling"]["KitesandDarts", 0] will give the substitutions used to define the "Kites and Darts" tiling. I wish I could define my own substitutions and see what the resulting tilings would look like. For example: This uses relatively simple shapes. It's got an inflation factor of sqrt(3). For a more complex example here is the beginnings of substitution tiling with regular 14-gons, 7-gons and other polygons that fit with them. As of now, I haven't computed it's inflation factor. How easy is it to program a new substitution tiling such as one of these? Thanks. John Berglund
Posted 2 years ago
For the algebraic tilings, with just a few weird shapes with weird angles that somehow work together, they are more discovered than made. They always start with an algebraic root that has strange properties. For example, in 3D space, there is a set of 19 points at power distances, based on the plastic constant. There is a classic adage in math and physics: "If it doesn't work, try 1, 0 or the square root." For algebraic geometry, once you've identified an algebraic root, the square root always seems to simplify things geometrically. That's why I wrote the SqrtSpace function. After discovering that made all of my tilings easy, I went through the literature and looked at all other known substitution tilings to see if there were exceptions. So far, there aren't any exceptions.For non-algebraic with "many" pieces, things are simpler. For example, with multiples of the 5 tetrominoes (Tetris pieces), make larger copies of each of the tetrominoes. This is then a substitution tiling. But there are many ways to do this.I still haven't found a 3D substitution tiling with the plastic constant, but I'm fairly sure one exists.
Posted 2 years ago
I'm new to the community. The examples are great. How would you go about making up a substitution tiling of your own? It would be nice if you could specify what the substitutions would be for each shape, and have the program show what some patches would look like.
Posted 3 years ago
-- you have earned Featured Contributor Badge Your exceptional post has been selected for our editorial column Staff Picks http://wolfr.am/StaffPicks and Your Profile is now distinguished by a Featured Contributor Badge and is displayed on the Featured Contributor Board. Thank you!
Posted 4 years ago
I've fixed a gap error in AlgebraicSubstitutionTiling that didn't handle complex substitutions well. Now it handles them fine. For example, the new Harvest tiling. Here's a few steps of it with my new code. Here's the Dale Walton quintic from above. But it seems with the changes I've added some minor errors, so I need to work on it more. Then I need to add some complicated substitution tilings that went wonky before due to the gap error.EDIT: Fixed all errors, I think. Let me know. Attachments:
Posted 5 years ago
Hi Ed, Yes, I like this one-to-five cross tiling, and years ago I spent some time calculating a set of channels that could probably be encoded to force the substitution hierarchy. Here's a printout from one of my notebooks: However, I can't recall my certainty about that proof, and cross hierarchy is relatively difficult compared to the that of Gosper's island: I seem to recall a few other people agreeing about the matching rules, but probably from a subsequent version. It's already been four or five years ago, so I don't know. Anyways, don't forget the Gosper Island! Later it would probably be worthwhile to try and at least give a second layer to the tiles, which (at least) carries a set of channels sufficient for encoding matching rules.--Brad
Posted 5 years ago
Hi Ed,Good that you got integer inflation factor tilings in the second post, but there are trivial examples Missingone square to four, or one equilateral triangle to four.You might also want to include 3D ABCK tiling by Danzer & Co. Ive already done some exploration of integer coordinatization, see for example:http://demonstrations.wolfram.com/TransformationOfIcosahedralSolidsInZ15/which I think you probably published some time earlier. I still can appreciate the result of this demonstration, and think that it suggests more to be done on your program here. If you decompose tiles to edges, how many unique vectors do you get? How are those vectors written out in a canonical basis?Cheers Brad
Posted 5 years ago
Missingone square to four, or one equilateral triangle to four. How about one square to five? AlgebraicSubstitutionTiling[{1,{{-3,-1},{-3,1},{-1,-3},{-1,-1},{-1,1},{-1,3},{1,-3},{1,-1},{1,1},{1,3},{3,-1},{3,1}}, {{1,6,12,7}-> {{1,2,5,4},{5,6,10,9},{8,9,12,11},{3,4,8,7},{4,5,9,8}}} , {{1,1,1,1,1}}},5,{"N", "ImageSize"->{600,Automatic}}] Here's another one I was just looking at AlgebraicSubstitutionTiling[{Root[-1-#1^2+#1^3&,1],{{{4,0,0},{0,0,0}},{{8,-4,-1},{0,-8,7}},{{0,0,-4},{0,0,0}},{{0,0,0},{0,0,0}},{{-2,-3,3},{-6,3,1}},{{2,1,-1},{-6,3,1}},{{4,5,-6},{4,1,-2}},{{0,-3,2},{4,1,-2}}}/4, {{1,2,3}-> {{4,7,3},{1,6,4},{6,5,2},{7,8,4},{4,5,6},{2,8,7},{8,5,4},{5,8,2}}, {1,2,3}-> {{1,2,3}},{1,2,3}-> {{1,2,3}},{1,2,3}-> {{1,2,3}},{1,2,3}-> {{1,2,3}},{1,2,3}-> {{1,2,3}},{1,2,3}-> {{1,2,3}}} , {{0,2,3,3,4,4,5,5}+1,{1},{2},{3},{4},{5},{6}}},1,{"N", "ImageSize"->{600,Automatic}}] It's in the supergolden ratio, psi. But if you scale the area of the big triangle to the component triangles, you get areas psi^{7.7217, 5, 3, 2, 2, 1, 1, 0, 0} ... the big triangle is out of phase for a smooth substitution tiling system with a fixed number of sizes at each step.
Posted 5 years ago
I should mention the Demonstrations of Dieter Steemann and the tiling demos of Karl Scherer, particularly Rep-tiles and Irreptiles. I'm still gleaning tilings from these, the Tilings Encyclopedia and IFStile. Hopefully I'll be able to improve my Demonstration Substitution Tilings to be stronger with a lot more tiling systems so that all of them are easily investigated.
Posted 5 years ago
So how to make these? Dale Walton sent me a picture of a new tiling. "All edges are powers of x=1.2365057033915... (5,6,7) triangle divides into (0,5,6); (3,4,5); (2,4,5)" Where $x^5-x^3-1=0$. This particular root has discriminant 3017. It's an algebraic number field seen giving extremal solutions in Wheels of Powered Triangles and Degenerate Power Simplices.By the end of the notebook, I get to this image. Not quite there. For a full substitution tiling system there should eventually be a fixed number of colors where every color represents congruent triangles. I haven't solved that yet for this tiling. Attachments:
Posted 5 years ago
May as well put in the rest of what I have so far.
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Math and Arithmetic
Meteorology and Weather
Temperature
# What is 35 degrees Celsius in Fahrenheit?
###### Wiki User
95F
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Anonymous
2020-09-28 22:34:01
OML I thought it was like 55 I’m bad at this :/
###### Wiki User
Converting Celsius to FahrenheitTake the number in Celsius and multiply it by 9. Then divide that number by 5, and then add 32. Fahrenheit = 9 x °C / 5 + 32
In this case, the answer is: 35° Celsius = 95° Fahrenheit.
Conversion: 35°C → 35 × 9/5 + 32 °F = 95°F
F = (9/5 x C) +32
F = (9/5 x 35 ) + 32
F = 63 + 32 = 95
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###### Lillian Beeler
mostly 95F nd i just realized how someone else had the same answer_wiki User
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Lillian Beeler
2020-11-13 16:53:04
(￣﹃￣) hi
###### Bill West
C = 5/9 (F-32)
35 - 32 -= 3
3 x 5 = 15
15 / 9 = 1.6666667
..
Sounds about right. Water freezes at 32 F or 0 C.
..
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Bill West
2020-11-18 06:14:48
Oops. I got the question backward!
F 9/5 C + 32
35 / 5 = 7
7 x 9 =63
32 + 63 = 99
35 C = 99 F
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###### Wiki User
35 deg C = 95 deg F
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###### Anonymous
95 F. The conversion is: 35 * 9 / 5 + 32 = 95.
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###### Anonymous
96 degrees Fahrenheit
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###### Anonymous
What is 36.1 Celsius in Fahrenheit?
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# How to solve equations step-by step
If you're ready to learn How to solve equations step-by step, keep reading! Math can be difficult for some students, but with the right tools, it can be conquered.
## How can we solve equations step-by step
One of the most important skills that students need to learn is How to solve equations step-by step. To solve an equation using square roots, we must first isolate the square root term on one side of the equation. We can then square both sides of the equation to cancel out the square root terms, leaving us with a much simpler equation to solve. This method can be very effective for equations that are otherwise difficult to solve.
To solve for the x intercept, you need to set y = 0 and solve for x. This can be done by using algebraic methods or by graphing the equation and finding the point where it crosses the x-axis.
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Solving complex numbers can be difficult, but there are some methods that can make the process easier. One method is to use the " conjugate method. This involves multiplying the complex number by its conjugate. This will simplify the number and make it easier to solve. Another method is to use the " FOIL method. This involves first, multiplying the complex number by its complex conjugate. This will simplify the number and make it easier to solve.
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this app is awesome! it’s very accurate in reading the numbers and it shows the steps and explanations really well. I’d recommend this to anyone who has a hard time with math (although not when they are studying for a test, you won’t have this during the test so don’t have it while you study). 10/10 lads
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Elimination solver calculator Second order homogeneous differential equation solver Math problem solver picture Solve an algebra equation Math photo answer Solving systems of equations by elimination solver
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physics
posted by jessica
What is the maximum speed (in meters/second) with which a 1300 kg car can round a turn of radius 700 m on a flat road, if the coefficent of friction between tires and road is 0.530
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mondrian.rolap
Interface Modulos
• All Known Implementing Classes:
Modulos.Base, Modulos.Many, Modulos.One, Modulos.Three, Modulos.Two, Modulos.Zero
public interface Modulos
Modulos implementations encapsulate algorithms to map between integral ordinals and position arrays. There are particular implementations for the most likely cases where the number of axes is 1, 2 and 3 as well as a general implementation.
Suppose the result is 4 x 3 x 2, then modulo = {1, 4, 12, 24}.
Then the ordinal of cell (3, 2, 1)
= (modulo[0] * 3) + (modulo[1] * 2) + (modulo[2] * 1)
= (1 * 3) + (4 * 2) + (12 * 1)
= 23
Reverse calculation:
p[0] = (23 % modulo[1]) / modulo[0] = (23 % 4) / 1 = 3
p[1] = (23 % modulo[2]) / modulo[1] = (23 % 12) / 4 = 2
p[2] = (23 % modulo[3]) / modulo[2] = (23 % 24) / 12 = 1
Author:
jhyde
• Method Detail
• getCellPos
int[] getCellPos(int cellOrdinal)
Converts a cell ordinal to a set of cell coordinates. Converse of getCellOrdinal(int[]). For example, if this result is 10 x 10 x 10, then cell ordinal 537 has coordinates (5, 3, 7).
Parameters:
cellOrdinal - Cell ordinal
Returns:
cell coordinates
• getCellOrdinal
int getCellOrdinal(int[] pos)
Converts a set of cell coordinates to a cell ordinal. Converse of getCellPos(int).
Parameters:
pos - Cell coordinates
Returns:
cell ordinal
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https://www.imooc.com/wenda/detail/591050
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/ 猿问
# 在两点之间缩放系列
2019-09-24 16:36:21
# generate series from exponential distr
s = sort(rexp(100))
# scale/interpolate 's' such that it starts at 0 and ends at 1?
# approx(s)
# scale(s)
## 3 回答
s = sort(rexp(100))
range01 <- function(x){(x-min(x))/(max(x)-min(x))}
range01(s)
[1] 0.000000000 0.003338782 0.007572326 0.012192201 0.016055006 0.017161145
[7] 0.019949532 0.023839810 0.024421602 0.027197168 0.029889484 0.033039408
[13] 0.033783376 0.038051265 0.045183382 0.049560233 0.056941611 0.057552543
[19] 0.062674982 0.066001242 0.066420884 0.067689067 0.069247825 0.069432174
[25] 0.070136067 0.076340460 0.078709590 0.080393512 0.085591881 0.087540132
[31] 0.090517295 0.091026499 0.091251213 0.099218526 0.103236344 0.105724733
[37] 0.107495340 0.113332392 0.116103438 0.124050331 0.125596034 0.126599323
[43] 0.127154661 0.133392300 0.134258532 0.138253452 0.141933433 0.146748798
[49] 0.147490227 0.149960293 0.153126478 0.154275371 0.167701855 0.170160948
[55] 0.180313542 0.181834891 0.182554291 0.189188137 0.193807559 0.195903010
[61] 0.208902645 0.211308713 0.232942314 0.236135220 0.251950116 0.260816843
[67] 0.284090255 0.284150541 0.288498370 0.295515143 0.299408623 0.301264703
[73] 0.306817872 0.307853369 0.324882091 0.353241217 0.366800517 0.389474449
[79] 0.398838576 0.404266315 0.408936260 0.409198619 0.415165553 0.433960390
[85] 0.440690262 0.458692639 0.464027428 0.474214070 0.517224262 0.538532221
[91] 0.544911543 0.559945121 0.585390414 0.647030109 0.694095422 0.708385079
[97] 0.736486707 0.787250428 0.870874773 1.000000000
library("scales")
rescale(s)
rescale(s, to=c(0,10))
rescale(s, from=c(0, max(s)))
reshape::rescaler.default(s, type = "range")
> system.time(replicate(100, range01(s)))
user system elapsed
0.56 0.12 0.69
> system.time(replicate(100, reshape::rescaler.default(s, type = "range")))
user system elapsed
0.53 0.18 0.70
range02 <- function(x) {
(x - min(x, na.rm=TRUE)) / diff(range(x, na.rm=TRUE))
}
> system.time(replicate(100, range02(s)))
user system elapsed
0.56 0.12 0.68
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{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.921875
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CC-MAIN-2020-34
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latest
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en
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