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### kut86916_ch01
Course: STAT 4315, Spring 2009
School: Columbia
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Linear Part Simple Regression I Chapter 1 Linear Regression with One Predictor Variable Regression analysis is a statistical methodology that utilizes the relation between two or more quantitative variables so that a response or outcome variable can be predicted from the other, or others. This methodology is widely used in business, the social and behavioral sciences, the biological sciences, and many other...
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CHAPTER 14LONG-TERM LIABILITIESMULTIPLE CHOICEConceptualAnswera a b d d b a d d c d d d d c b d d d d c d d d c c. c d b b cNo.1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. *27. *28. *29. *30. *31.De
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Chapter 5(B) Homework Solution and etc.P 5-4: Percentage-of-Completion method Contract price (i.e. total revenue for this project) is \$10,000,000 1. Calculate the amount of gross profit to be recognized in each year: 2006: Total actual costs incurred to
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# The copper pipe has an outer diameter of 40 mm and an
## Problem 5-4 Chapter 5
Mechanics of Materials | 9th Edition
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Problem 5-4
The copper pipe has an outer diameter of 40 mm and an inner diameter of 37 mm. If it is tightly secured to the wall at A and three torques are applied to it as shown, determine the absolute maximum shear stress developed in the pipe.
Step-by-Step Solution:
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##### ISBN: 9780133254426
The full step-by-step solution to problem: 5-4 from chapter: 5 was answered by , our top Engineering and Tech solution expert on 09/29/17, 04:53PM. Since the solution to 5-4 from 5 chapter was answered, more than 294 students have viewed the full step-by-step answer. This full solution covers the following key subjects: diameter, pipe, Outer, determine, developed. This expansive textbook survival guide covers 14 chapters, and 1502 solutions. The answer to “The copper pipe has an outer diameter of 40 mm and an inner diameter of 37 mm. If it is tightly secured to the wall at A and three torques are applied to it as shown, determine the absolute maximum shear stress developed in the pipe.” is broken down into a number of easy to follow steps, and 46 words. Mechanics of Materials was written by and is associated to the ISBN: 9780133254426. This textbook survival guide was created for the textbook: Mechanics of Materials, edition: 9.
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# 98773 (number)
98,773 (ninety-eight thousand seven hundred seventy-three) is an odd five-digits prime number following 98772 and preceding 98774. In scientific notation, it is written as 9.8773 × 104. The sum of its digits is 34. It has a total of 1 prime factor and 2 positive divisors. There are 98,772 positive integers (up to 98773) that are relatively prime to 98773.
## Basic properties
• Is Prime? Yes
• Number parity Odd
• Number length 5
• Sum of Digits 34
• Digital Root 7
## Name
Short name 98 thousand 773 ninety-eight thousand seven hundred seventy-three
## Notation
Scientific notation 9.8773 × 104 98.773 × 103
## Prime Factorization of 98773
Prime Factorization 98773
Prime number
Distinct Factors Total Factors Radical ω(n) 1 Total number of distinct prime factors Ω(n) 1 Total number of prime factors rad(n) 98773 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) -1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 11.5006 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 98,773 is 98773. Since it has a total of 1 prime factor, 98,773 is a prime number.
## Divisors of 98773
2 divisors
Even divisors 0 2 2 0
Total Divisors Sum of Divisors Aliquot Sum τ(n) 2 Total number of the positive divisors of n σ(n) 98774 Sum of all the positive divisors of n s(n) 1 Sum of the proper positive divisors of n A(n) 49387 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 314.282 Returns the nth root of the product of n divisors H(n) 1.99998 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 98,773 can be divided by 2 positive divisors (out of which 0 are even, and 2 are odd). The sum of these divisors (counting 98,773) is 98,774, the average is 49,387.
## Other Arithmetic Functions (n = 98773)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 98772 Total number of positive integers not greater than n that are coprime to n λ(n) 98772 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 9475 Total number of primes less than or equal to n r2(n) 8 The number of ways n can be represented as the sum of 2 squares
There are 98,772 positive integers (less than 98,773) that are coprime with 98,773. And there are approximately 9,475 prime numbers less than or equal to 98,773.
## Divisibility of 98773
m n mod m 2 3 4 5 6 7 8 9 1 1 1 3 1 3 5 7
98,773 is not divisible by any number less than or equal to 9.
• Arithmetic
• Prime
• Deficient
• Polite
• Prime Power
• Square Free
## Base conversion (98773)
Base System Value
2 Binary 11000000111010101
3 Ternary 12000111021
4 Quaternary 120013111
5 Quinary 11130043
6 Senary 2041141
8 Octal 300725
10 Decimal 98773
12 Duodecimal 491b1
20 Vigesimal c6id
36 Base36 247p
## Basic calculations (n = 98773)
### Multiplication
n×y
n×2 197546 296319 395092 493865
### Division
n÷y
n÷2 49386.5 32924.3 24693.2 19754.6
### Exponentiation
ny
n2 9756105529 963639811415917 95181595092984369841 9401371692119345162305093
### Nth Root
y√n
2√n 314.282 46.2253 17.728 9.97534
## 98773 as geometric shapes
### Circle
Diameter 197546 620609 3.06497e+10
### Sphere
Volume 4.03649e+15 1.22599e+11 620609
### Square
Length = n
Perimeter 395092 9.75611e+09 139686
### Cube
Length = n
Surface area 5.85366e+10 9.6364e+14 171080
### Equilateral Triangle
Length = n
Perimeter 296319 4.22452e+09 85539.9
### Triangular Pyramid
Length = n
Surface area 1.68981e+10 1.13566e+14 80647.8
## Cryptographic Hash Functions
md5 5f67cf29b119ca3b97f2c05ca0567f99 61ad0b70565a45b7a17767eef0a5a82fa5606cac b4acdabe0e86bdf085bd0eafb6d2ed84eba31d018740555153aee408f8c93f90 39f88b5e071b22ee5f859dc8bc7ab7da040ca1a5b84abf62da35c76e188a4c377fcff36fb81baf49b564906b05f2f1eb8399b1eca610d641df30572990d3e528 a224be30d72fba4dcd6d780ec52339b4e00ff5a7
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When I first left undergraduate school I took a job teaching 8th and 9th grade math. My 8th graders were introduced to integer multiplication for the first time at that level. They certainly knew natural number multiplication and even knew integer addition which includes subtraction by the time we took up integer multiplication.
When teaching integer addition I presented it as addition of vectors directed along the number line. So +2 + -4 would be an arrow starting at zero of length 2 pointing to the right then append an arrow of length 4 pointing to the left. The tip of the second arrow wound up at the point -2 on the number line.
To extend that idea to integer multiplication I introduced items I called ‘Number Line Frogs’. Each frog had a number associated with it and a direction. So a +2 frog could jump two places at a time and it pointed to the right. The number -3 was represented by a frog that could jump three places and pointed to the left.
I then asked the students to think of integer multiplication problems in two parts. The problem a x b was to be visualized as an ‘a-frog’ being asked to do something and that something was to perform ‘b’ jumps. The problem always started at the origin or ‘0’ on the number line. The sign of the number ‘b’ indicated which direction the frog was to jump. The absolute value of the number ‘b’ indicated how many jumps the frog was to take. The sign of the number ‘a’ indicated which direction the frog was pointing and the absolute value of the number ‘a’ indicated how far the frog could jump.
In this way of thinking about it the problem +2 x -3 was the same as saying ‘Allow a +2 frog to jump three jumps backward (-3)’. I stressed that the problem -3 x +2 was the same problem as could be verified by saying ‘Allow a -3 frog to take two jumps forward (+2)’ In both cases the end result was the number -6.
The rule for ‘positive-a times positive-b’ was to say ‘let a postive-a-frog take b-jumps-forward’ which came out to be the same problem as saying ‘let a positive-b-frog take a-jumps-forward’.
The rule for ‘positive-a times negative-b’ was to say ‘let a positive-a-frog take b-jumps-backward, while the rule for ‘negative-b times positive-a’ was to say ‘let a negative-b frog take a-jumps-forward’.
Finally the rule for ‘negative-a times negative-b’ was to say ‘let a ‘negative-a frog take ‘b-jumps-backward’.
When viewed this way and taking into account the ‘frog’ always started at zero the sign table for integer multiplication was justified. Additionally the commutative property of multiplication was illustrated by noting any number that could be decomposed into ‘b-groups of length a’ was equivalent to the observation that the same number could be decomposed into ‘a-groups of length b’.
I will say that the 8th grade group I had were sorted in an inhomogeneous manner with all the top students put into one class and that class was assigned to me. I did worry about this sorting process and felt that it was wrong on multiple levels but given there was nothing a first year math teacher could do to change the situation I made the most of it and presented many mathematical concepts using non standard methods (i.e. presented not-by-the-book). I think that math class did enjoy my presentations.
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# SOLUTION: Kindly Solve this Problem Please. Sorry There was no ISBN in this Book.... 1. A person invested \$ 15,000. The greater part is invested at 8% and smaller part at 6%. The a
Algebra -> Customizable Word Problem Solvers -> Finance -> SOLUTION: Kindly Solve this Problem Please. Sorry There was no ISBN in this Book.... 1. A person invested \$ 15,000. The greater part is invested at 8% and smaller part at 6%. The a Log On
Ad: Over 600 Algebra Word Problems at edhelper.com Ad: Mathway solves algebra homework problems with step-by-step help! Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations!
Word Problems: Money, Business and Interest Solvers Lessons Answers archive Quiz In Depth
Click here to see ALL problems on Money Word Problems Question 85264: Kindly Solve this Problem Please. Sorry There was no ISBN in this Book.... 1. A person invested \$ 15,000. The greater part is invested at 8% and smaller part at 6%. The annual income in the greater part is \$ 200 more than the annual income of the smaller part. What is hte smaller part of the investment?Answer by checkley75(3666) (Show Source): You can put this solution on YOUR website!.08*X=.06(15000-X)+200 .08X=900-.06X+200 .08X+.06X=1100 .14X=1100 X=1100/.14 X=7857.14 INVESTED @ 8%. 15000-7857.14=7142.86 INVESTED @ 6%. PROOF .08*7857.14=.06*7142.86+200 628.57=428.57+200 628.57=628.57
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# DoITPoMS
TLP Library
Teaching and learning packages (TLPs) are self-contained, interactive resources, each focusing on one area of Materials Science.
Introduction To Anisotropy
It is common in basic analysis to treat bulk materials as isotropic - their properties are independent of the direction in which they are measured. However the atomic scale structure can result in properties that vary with direction. This teaching and learning package (TLP) looks into typical examples of such anisotropy and gives a brief mathematical look into modelling the behaviour.
Atomic Force Microscopy
Provides a brief introduction to atomic force microscopy (AFM), some of the ways it is commonly used and some of the problems faced.
Atomic Scale Structure of Materials
This teaching and learning package provides an introduction to crystalline, polycrystalline and amorphous solids, and how the atomic-level structure has radical consequences for some of the properties of the material. It introduces the use of polarised light to examine the optical properties of materials, and shows how a variety of simple models can be used to visualise important features of the microstructure of materials.
Batteries
This TLP investigates the basic principles, design and applications of batteries. It covers both primary and rechargeable batteries, how they work and how they may be used.
Brillouin Zones
This teaching and learning package provides an introduction to Brillouin zones in two and three dimensions and is aimed at developing familiarity with Brillouin Zones. It will not cover any specific applications. Brillouin Zones are particularly useful in understanding the electronic and thermal properties of crystalline solids.
Casting
This TLP introduces a number of important processes through which metallic items can be fabricated from molten metal. As well as detailing the practical aspects of these manufacturing processes, attention is given to the important parameters which determine the microstructure of the finished items.
Creep Deformation of Metals
Creep is a major concern in engineering, since it can cause materials to fail well below their yield stress. This package outlines the mechanisms of creep and the associated equations. It is largely based around a first year Materials Science practical at the University of Cambridge, which is concerned with the creep of solder at different temperatures. It also includes a case study of a creep-resistant material to illustrate how materials can be designed to prevent creep.
Crystallinity in Polymers
An understanding of polymer crystallinity is important because the mechanical properties of crystalline polymers are different from those of amorphous polymers. Polymer crystals are much stiffer and stronger than amorphous regions of polymer.
Introduction To Deformation Processes
This teaching and learning package covers the fundamentals of metal forming processes.
Dielectric Materials
This teaching and learning package will introduce you to the properties and uses of dielectric materials.
Diffraction and Imaging
A brief summary of diffraction and imaging using an optical system.
Diffusion
An introduction to the mechanisms and driving forces of diffusion, and some of the processes in which it is observed.
Electromigration
Electromigration is an ever-increasing problem as integrated circuits are pushed towards further miniaturization. The theory of the phenomenon is explained, including electromigration-induced failure and how it has been and can be minimized.
Ellingham Diagrams
The Ellingham diagram is a tool most often used in extraction metallurgy to find the conditions necessary for the reduction of the ores of important metals. This Teaching and Learning Package incorporates an interactive Ellingham diagram. This diagram can be used to quickly and simply find a range of thermodynamic data relating to many metallurgical reactions.
Epitaxial Growth
This TLP enables you to explore the way in which perfect thin crystalline layers are deposited epitaxially (i.e. in the same crystal orientation) on semiconductor substrates. This is the way many electronic and opto-electronic devices are now fabricated using techniques such as molecular beam epitaxy (MBE).
Examination of a Manufactured Article
This TLP provides an introduction to the deconstruction and investigation of the materials and processes used in an everyday item or article.
Ferroelectric Materials
Ferroelectrics are an important device in today's world. They are useful both as capacitors, for example in camera flashes, or as non-volatile memory storage. The memory use of which you are most likely to be aware is in the Playstation 2.
Fuel Cells
This teaching and learning package provides a short summary of four of the most promising fuel cell technologies. It gives a general overview of the field with focus on materials used (electrolytes and electrodes) and the mechanism of function (electrochemistry and thermodynamics).
The Glass Transition in Polymers
This teaching and learning package is based on a lecture demonstrations used within the Department of Materials Science and Metallurgy at the University of Cambridge. The package is aimed at first year undergraduate Materials Science students and focuses on the glass transition in polymers.
Indexing Electron Diffraction Patterns
An introduction to the indexing of diffraction patterns.
The Jominy End Quench Test
Discusses the aims, method and use of results of a test for the hardenability of steel.
Kinetics of Aqueous Corrosion
This teaching and learning package (TLP) introduces the mechanism of aqueous corrosion and the associated kinetics.
Liquid Crystals
This Teaching and Learning Package provides an introduction to liquid crystals, their physical properties and their modern-day applications.
Microstructural Examination
This teaching and learning package (TLP) looks at how what we see in micrographs relates to equilibrium phase diagrams and cooling routes for alloy systems.
The Nernst Equation and Pourbaix Diagrams
This teaching and learning package (TLP) investigates the Nernst equation and Pourbaix diagrams, which are both important parts of electrochemistry and corrosion science.
Optical Microscopy
An introduction to the use of optical microscopes. It introduces the different types of microscope used to examine specimens and how to set them up correctly. There is also an introduction to specimen preparation.
Tags: microscopy
Phase Diagrams and Solidification
Phase diagrams are a useful tool in metallurgy and other branches of materials science. They show the mixture of phases present in thermodynamic equilibrium. This teaching and learning package looks at the theory behind phase diagrams, and ways of constructing them, before running through an experimental procedure, and presenting the results which can be obtained.
Introduction To Photoelasticity
This tutorial is based on lab work within the Department of Materials Science and Metallurgy at the University of Cambridge. The tutorial provides an introduction to the topic of photoelasticity and preparation for lab work. Photographs illustrate many features of birefringence in polymers under polarised light.
Piezoelectric Materials
This teaching and learning package (TLP) provides an introduction to piezoelectric materials.
Pyroelectric Materials
Pyroelectric materials are found in almost every home, in the form of intrusion detectors and other devices, and this TLP will consider how they work, and what the most common ones are made of.
Raman Spectroscopy
An introduction to the analysis of materials and chemicals by the Raman scattering of light.
Recycling of Metals
The next time you drain a canned beverage or take a journey in a car, you might like to think about what will happen to it when it reaches the end of its useful life. This teaching and learning package will look at metals recycling from a materials science viewpoint – not simply outlining the need for recycling, but explaining the complex scientific principles behind some aspects of the recycling process itself.
Introduction To Semiconductors
This teaching and learning package provides a very basic introduction to semiconductors. These materials are essential to the operation of solid state electronic devices.
Solid Solutions
This teaching and learning package is based on a practical used within the Department of Materials Science and Metallurgy at the University of Cambridge. The package is aimed at first year undergraduate Materials Science students and focuses on the different types of solid solution and the thermodynamic principles involved in understanding them.
Solidification of Alloys
This teaching and learning package (TLP) is an introduction to how solute affects the solidification of metallic alloys.
Superconductivity
Electrons in pairs? Levitating trains? Superconductivity - the combination of lossless electrical conduction and the ability of a material to expel a magnetic field - is a property that excites interest in fundamental science whilst offering tantalising prospects for a range of applications. In this teaching and learning package (TLP), we trace the history of superconductivity, outline some fundamental properties of superconductors, and describe current and potential applications of materials with this unusual property.
Transmission Electron Microscopy
Transmission electron microscopy is a very important tool in materials science for investigating the fine-scale structure of materials. This TLP serves as an introduction to the basic concepts and structure of the transmission electron microscope.
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Lecture 4
# The second x x line is the assumption that x
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Unformatted text preview: ions, oftentimes constraints are linear. In that case GCQ is automatically satisfied, so you don’t need to check it (exercise). It is known that the GCQ is the weakest possible condition [1]. 5 Sufficient condition The Karush-Kuhn-Tucker theorem provides necessary conditions for optimalSecond Order ity: if the constraint qualification holds, then a local solution must satisfy Sufficient Condition the Karush-Kuhn-Tucker conditions (first-order conditions and complementary Not needed slackness conditions). Note that the KKT conditions are equivalent to ∇x L(¯, λ, µ) = 0, x (4) where L(x, λ, µ) is the Lagrangian. (4) is the first-order necessary condition of the unconstrained minimization problem min L(x, λ, µ). (5) x ∈ RN Below I give a sufficient condition for optimality. Proposition 6. Suppose that x is a solution to the unconstrained minimization ¯ problem (5) for some λ ∈ RI and µ ∈ RJ . If gi (¯) ≤ 0 and λi gi (¯) = 0 for all x x + i and hj (¯) = 0 for all j , then x is a solution to the constrained minimization x ¯ problem (1). Proof. Take any x such that gi (x) ≤ 0 for all i and hj (x) = 0 for all j . Then J I µj hj (¯) x λi gi (¯) + x f (¯) = f (¯) + x x j =1 i=1 = L(¯, λ, µ) ≤ L(x, λ, µ) x J I µj hj (x) ≤ f (x). λi gi (x) + = f (x) + i=1 j =1 The first line is due to λi gi (¯) = 0 for all i and hj (¯) = 0 for all j . The second x x line is the assumption that x minimizes L(·, λ, µ). The third line is due to λi ≥ 0 ¯ and gi (x) ≤ 0 for all i and hj (x) = 0 for all j . 6 Constrained maximization Finally, we briefly discuss maximization. Although maximization is equivalent to minimization by flipping the sign of the objective function, doing so every time is awkward. So consider the maximization problem maximize f (x) subject to gi (x) ≥ 0 hj (x) = 0 (i = 1, . . . , I ) (j = 1, . . . , J ). (6) 2014W Econ 172B Operations Research (B) Alexis Akira Toda (6) is equivalent to the minimization problem minimize − f (x) subject to − gi (x) ≤ 0 − hj (x) = 0 (i = 1, . . . , I ) (j = 1, . . . , J ). (7) Assuming that x is a local solution and the constraint qualification holds, then ¯ the KKT conditions are I − ∇f (¯) − x J λi ∇gi (¯) − x i=1 µj ∇hj (¯) = 0, x (8a) j =1 (∀i) λi (−gi (¯)) = 0. x (8b) But (8) is equivalent to (3). For this reason, it is customary to formulate a maximization problem as in (6) so that the inequality constraints are always “greater than or equal to zero”. As an example, consider a consumer with utility function u(x) = α log x1 + (1 − α) log x2 , where 0...
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## This document was uploaded on 02/18/2014 for the course ECON 172b at UCSD.
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# Related formulas
Go!
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## Basic Integrals
· Power Rule of Integration
Apply the power rule for integration, $\displaystyle\int x^n dx=\frac{x^{n+1}}{n+1}$, where $n$ represents a constant function
$\int x^ndx=\frac{x^{\left(n+1\right)}}{n+1}+C$
### Problem Analysis
$\int_{ }^{ }\left(10-9x\right)^3dx$
Calculus
~ 0.04 seconds
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# If f1=f2 and f3 are the frequencies of corresponding Kα,Kβ and Lα X-rays of an element, then :-
A
f1=f2=f3
B
f1f2=f3
C
f2=f1+f3
D
f22=f1f3
Video Solution
Text Solution
Verified by Experts
## For kα emmision ⇒ transition ⇒ L shell to k - shell For kβ emission ⇒ tranisition ⇒ M shell to k-shell For Lα emmision =⇒ transition ⇒ M shell to L-shell EM−EK=(EM−EL)+(EL−EK) ⇒hf2=hf3+hf1⇒f2=f1+f3
|
Updated on:21/07/2023
### Knowledge Check
• Question 1 - Select One
## In X-ray experiment Kα,Kβ denotes
ACharacteristic
BContinuous wavelength
Cα,β emissions respectively
DNone of these
• Question 2 - Select One
## If λKα,λKβ and λLα are the wavelengths of Kα,Kβ and Lα, lines respectively , then
AλKβ=λKαλLαλKα+λLα
BλLα=λKαλKβλKα+λKβ
CλLα=λKαλKβλKβ+λKα
Dnone of these
• Question 3 - Select One
## If the frequency of Kα,Kβ and Lα , X-ray lines of a substance are vKα,vKβ, and vLβ
AvKα+vKβ=vLα
BvKαvKβ=vLα
CvKα+vLα=vKβ
Dnone of these
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# Modular exponentiation
Modular exponentiation
You are encouraged to solve this task according to the task description, using any language you may know.
Find the last 40 decimal digits of ${\displaystyle a^{b}}$, where
• ${\displaystyle a=2988348162058574136915891421498819466320163312926952423791023078876139}$
• ${\displaystyle b=2351399303373464486466122544523690094744975233415544072992656881240319}$
A computer is too slow to find the entire value of ${\displaystyle a^{b}}$. Instead, the program must use a fast algorithm for modular exponentiation: ${\displaystyle a^{b}\mod m}$.
The algorithm must work for any integers ${\displaystyle a,b,m}$ where ${\displaystyle b\geq 0}$ and ${\displaystyle m>0}$.
Using the big integer implementation from a cryptographic library [1].
procedure Mod_Exp is
``` A: String :=
"2988348162058574136915891421498819466320163312926952423791023078876139";
B: String :=
"2351399303373464486466122544523690094744975233415544072992656881240319";
```
``` D: constant Positive := Positive'Max(Positive'Max(A'Length, B'Length), 40);
-- the number of decimals to store A, B, and result
Bits: constant Positive := (34*D)/10;
-- (slightly more than) the number of bits to store A, B, and result
package LN is new Crypto.Types.Big_Numbers (Bits + (32 - Bits mod 32));
-- the actual number of bits has to be a multiple of 32
use type LN.Big_Unsigned;
```
``` function "+"(S: String) return LN.Big_Unsigned
renames LN.Utils.To_Big_Unsigned;
```
``` M: LN.Big_Unsigned := (+"10") ** (+"40");
```
begin
``` Ada.Text_IO.Put("A**B (mod 10**40) = ");
```
end Mod_Exp;</lang>
Output:
`A**B (mod 10**40) = 1527229998585248450016808958343740453059`
## Bracmat
Translation of: Icon_and_Unicon
<lang bracmat> ( ( mod-power
``` = base exponent modulus result
. !arg:(?base,?exponent,?modulus)
& !exponent:~<0
& 1:?result
& whl
' ( !exponent:>0
& ( ( mod\$(!exponent.2):1
& mod\$(!result*!base.!modulus):?result
& -1
| 0
)
+ !exponent
)
* 1/2
: ?exponent
& mod\$(!base^2.!modulus):?base
)
& !result
)
& ( a
= 2988348162058574136915891421498819466320163312926952423791023078876139
)
& ( b
= 2351399303373464486466122544523690094744975233415544072992656881240319
)
& out\$("last 40 digits = " mod-power\$(!a,!b,10^40))
)</lang>
```
Output:
`last 40 digits = 1527229998585248450016808958343740453059`
## BBC BASIC
Uses the Huge Integer Math & Encryption library. <lang bbcbasic> INSTALL @lib\$+"HIMELIB"
``` PROC_himeinit("")
PROC_hiputdec(1, "2988348162058574136915891421498819466320163312926952423791023078876139")
PROC_hiputdec(2, "2351399303373464486466122544523690094744975233415544072992656881240319")
PROC_hiputdec(3, "10000000000000000000000000000000000000000")
h1% = 1 : h2% = 2 : h3% = 3 : h4% = 4
SYS `hi_PowMod`, ^h1%, ^h2%, ^h3%, ^h4%
PRINT FN_higetdec(4)</lang>
```
Output:
```1527229998585248450016808958343740453059
```
## C
Given numbers are too big for even 64 bit integers, so might as well take the lazy route and use GMP:
Library: GMP
<lang c>#include <gmp.h>
int main() { mpz_t a, b, m, r;
mpz_init_set_str(a, "2988348162058574136915891421498819466320" "163312926952423791023078876139", 0); mpz_init_set_str(b, "2351399303373464486466122544523690094744" "975233415544072992656881240319", 0); mpz_init(m); mpz_ui_pow_ui(m, 10, 40);
mpz_init(r); mpz_powm(r, a, b, m);
gmp_printf("%Zd\n", r); /* ...16808958343740453059 */
mpz_clear(a); mpz_clear(b); mpz_clear(m); mpz_clear(r);
return 0; }</lang>
## Common Lisp
<lang lisp>(defun rosetta-mod-expt (base power divisor)
``` "Return BASE raised to the POWER, modulo DIVISOR.
This function is faster than (MOD (EXPT BASE POWER) DIVISOR), but
only works when POWER is a non-negative integer."
(setq base (mod base divisor))
;; Multiply product with base until power is zero.
(do ((product 1))
((zerop power) product)
;; Square base, and divide power by 2, until power becomes odd.
(do () ((oddp power))
(setq base (mod (* base base) divisor)
```
power (ash power -1)))
``` (setq product (mod (* product base) divisor)
```
power (1- power))))
(let ((a 2988348162058574136915891421498819466320163312926952423791023078876139)
``` (b 2351399303373464486466122544523690094744975233415544072992656881240319))
(format t "~A~%" (rosetta-mod-expt a b (expt 10 40))))</lang>
```
Works with: CLISP
<lang lisp>;; CLISP provides EXT:MOD-EXPT (let ((a 2988348162058574136915891421498819466320163312926952423791023078876139)
``` (b 2351399303373464486466122544523690094744975233415544072992656881240319))
(format t "~A~%" (mod-expt a b (expt 10 40))))</lang>
```
### Implementation with LOOP
<lang lisp>(defun mod-expt (a n m)
``` (loop with c = 1 while (plusp n) do
(if (oddp n) (setf c (mod (* a c) m)))
(setf n (ash n -1))
(setf a (mod (* a a) m))
finally (return c)))</lang>
```
## D
Translation of: Icon_and_Unicon
<lang d>import std.stdio, std.bigint;
BigInt powMod(BigInt base, BigInt exponent, BigInt modulus) in {
``` assert(exponent >= 0);
```
} body {
``` BigInt result = 1;
while (exponent > 0) {
if (exponent % 2 == 1)
result = (result * base) % modulus;
exponent /= 2;
base = base ^^ 2 % modulus;
}
return result;
```
}
void main() {
``` powMod(BigInt("29883481620585741369158914214988194" ~
"66320163312926952423791023078876139"),
BigInt("235139930337346448646612254452369009" ~
"4744975233415544072992656881240319"),
BigInt(10) ^^ 40).writeln();
```
}</lang>
Output:
`1527229998585248450016808958343740453059`
## Dc
<lang Dc>2988348162058574136915891421498819466320163312926952423791023078876139 2351399303373464486466122544523690094744975233415544072992656881240319 10 40^|p</lang>
## Emacs Lisp
Library: Calc
<lang lisp>(let ((a "2988348162058574136915891421498819466320163312926952423791023078876139")
``` (b "2351399303373464486466122544523690094744975233415544072992656881240319"))
;; "\$ ^ \$\$ mod (10 ^ 40)" performs modular exponentiation.
;; "unpack(-5, x)_1" unpacks the integer from the modulo form.
(message "%s" (calc-eval "unpack(-5, \$ ^ \$\$ mod (10 ^ 40))_1" nil a b)))</lang>
```
## Go
<lang go>package main
import (
``` "fmt"
"math/big"
```
)
func main() {
``` a, _ := new(big.Int).SetString(
"2988348162058574136915891421498819466320163312926952423791023078876139", 10)
b, _ := new(big.Int).SetString(
"2351399303373464486466122544523690094744975233415544072992656881240319", 10)
m := big.NewInt(10)
r := big.NewInt(40)
m.Exp(m, r, nil)
```
``` r.Exp(a, b, m)
fmt.Println(r)
```
}</lang> Output:
```1527229998585248450016808958343740453059
```
Kind of a hack. We partially implement a "modular arithmetic" instance of `Num`, so that we can take advantage of the efficient built-in exponentiation-by-squaring operation without implementing it ourselves. Since there are no "local" instances, we must keep the modulo base around with us in the type, which makes the code inelegant. <lang haskell>-- Private type. Do not use outside of the modPow function newtype ModN = ModN (Integer, Integer) deriving (Eq, Show) instance Num ModN where
``` -- actually only multiplication needs to be implemented
-- but we do some of the other ones too for good measure
ModN (x, m) + ModN (y, m') | m == m' = ModN ((x + y) `mod` m, m)
| otherwise = undefined
ModN (x, m) * ModN (y, m') | m == m' = ModN ((x * y) `mod` m, m)
| otherwise = undefined
negate (ModN (x, m)) = ModN ((- x) `mod` m, m)
abs _ = undefined
signum _ = undefined
fromInteger _ = undefined
```
modPow :: Integer -> Integer -> Integer -> Integer modPow _ 0 m = 1 `mod` m modPow a b m = c
``` where a' = ModN (a, m)
ModN (c, _) = a' ^ b
```
main :: IO () main = print \$ modPow a b m
``` where a = 2988348162058574136915891421498819466320163312926952423791023078876139
b = 2351399303373464486466122544523690094744975233415544072992656881240319
m = 10 ^ 40</lang>
```
Output:
`1527229998585248450016808958343740453059`
## Icon and Unicon
This uses the exponentiation procedure from RSA Code an example of the right to left binary method. <lang Icon>procedure main()
``` a := 2988348162058574136915891421498819466320163312926952423791023078876139
b := 2351399303373464486466122544523690094744975233415544072992656881240319
write("last 40 digits = ",mod_power(a,b,(10^40))
```
end
procedure mod_power(base, exponent, modulus) # fast modular exponentation
``` if exponent < 0 then runerr(205,m) # added for this task
result := 1
while exponent > 0 do {
if exponent % 2 = 1 then
result := (result * base) % modulus
exponent /:= 2
base := base ^ 2 % modulus
}
return result
```
end</lang>
Output:
`last 40 digits = 1527229998585248450016808958343740453059`
## J
Solution:<lang j> m&|@^</lang> Example:<lang j> a =: 2988348162058574136915891421498819466320163312926952423791023078876139x
``` b =: 2351399303373464486466122544523690094744975233415544072992656881240319x
m =: 10^40x
```
``` a m&|@^ b
```
1527229998585248450016808958343740453059</lang> Discussion: The phrase m&|@^ is the natural expression of a^b mod m in J, and is recognized by the interpreter as an opportunity for optimization, by avoiding the exponentiation.
## Java
`java.math.BigInteger.modPow` solves this task. Inside OpenJDK, BigInteger.java implements `BigInteger.modPow` with a fast algorithm from Colin Plumb's bnlib. This "window algorithm" caches odd powers of the base, to decrease the number of squares and multiplications. It also exploits both the Chinese remainder theorem and the Montgomery reduction.
<lang java>import java.math.BigInteger;
public class PowMod {
``` public static void main(String[] args){
BigInteger a = new BigInteger(
"2988348162058574136915891421498819466320163312926952423791023078876139");
BigInteger b = new BigInteger(
"2351399303373464486466122544523690094744975233415544072992656881240319");
BigInteger m = new BigInteger("10000000000000000000000000000000000000000");
System.out.println(a.modPow(b, m));
}
```
}</lang> Output:
`1527229998585248450016808958343740453059`
## Maple
<lang Maple>a := 2988348162058574136915891421498819466320163312926952423791023078876139: b := 2351399303373464486466122544523690094744975233415544072992656881240319: a &^ b mod 10^40;</lang> Output:
`1527229998585248450016808958343740453059`
## Mathematica
<lang Mathematica>a = 2988348162058574136915891421498819466320163312926952423791023078876139; b = 2351399303373464486466122544523690094744975233415544072992656881240319; m = 10^40; PowerMod[a, b, m] -> 1527229998585248450016808958343740453059</lang>
## Maxima
<lang maxima>a: 2988348162058574136915891421498819466320163312926952423791023078876139\$ b: 2351399303373464486466122544523690094744975233415544072992656881240319\$ power_mod(a, b, 10^40); /* 1527229998585248450016808958343740453059 */</lang>
## PARI/GP
<lang parigp>a=2988348162058574136915891421498819466320163312926952423791023078876139; b=2351399303373464486466122544523690094744975233415544072992656881240319; lift(Mod(a,10^40)^b)</lang>
## Pascal
Works with: Free_Pascal
Library: GMP
A port of the C example using gmp. <lang pascal>Program ModularExponentiation(output);
uses
``` gmp;
```
var
``` a, b, m, r: mpz_t;
fmt: pchar;
```
begin
``` mpz_init_set_str(a, '2988348162058574136915891421498819466320163312926952423791023078876139', 10);
mpz_init_set_str(b, '2351399303373464486466122544523690094744975233415544072992656881240319', 10);
mpz_init(m);
mpz_ui_pow_ui(m, 10, 40);
```
``` mpz_init(r);
mpz_powm(r, a, b, m);
```
``` fmt := '%Zd' + chr(13) + chr(10);
mp_printf(fmt, @r); (* ...16808958343740453059 *)
mpz_clear(a);
mpz_clear(b);
mpz_clear(m);
mpz_clear(r);
```
end.</lang> Output:
```% ./ModularExponentiation
1527229998585248450016808958343740453059
```
## Perl
<lang perl>use bigint;
my \$a = 2988348162058574136915891421498819466320163312926952423791023078876139; my \$b = 2351399303373464486466122544523690094744975233415544072992656881240319; my \$m = 10 ** 40; print \$a->bmodpow(\$b, \$m), "\n";</lang> Output:
`1527229998585248450016808958343740453059`
## Perl 6
This is specced as a built-in, but here's an explicit version: <lang perl6>sub expmod(Int \$a is copy, Int \$b is copy, \$n) {
``` my \$c = 1;
repeat while \$b div= 2 {
(\$c *= \$a) %= \$n if \$b % 2;
(\$a *= \$a) %= \$n;
}
\$c;
```
}
say expmod
``` 2988348162058574136915891421498819466320163312926952423791023078876139,
2351399303373464486466122544523690094744975233415544072992656881240319,
10**40;</lang>
```
Output:
`1527229998585248450016808958343740453059`
## PHP
<lang php><?php \$a = '2988348162058574136915891421498819466320163312926952423791023078876139'; \$b = '2351399303373464486466122544523690094744975233415544072992656881240319'; \$m = '1' . str_repeat('0', 40); echo bcpowmod(\$a, \$b, \$m), "\n"; ?></lang> Output:
`1527229998585248450016808958343740453059`
## PicoLisp
The following function is taken from "lib/rsa.l": <lang PicoLisp>(de **Mod (X Y N)
``` (let M 1
(loop
(when (bit? 1 Y)
(setq M (% (* M X) N)) )
(T (=0 (setq Y (>> 1 Y)))
M )
(setq X (% (* X X) N)) ) ) )</lang>
```
Test: <lang PicoLisp>: (**Mod
``` 2988348162058574136915891421498819466320163312926952423791023078876139
2351399303373464486466122544523690094744975233415544072992656881240319
10000000000000000000000000000000000000000 )
```
-> 1527229998585248450016808958343740453059</lang>
## Python
<lang python>a = 2988348162058574136915891421498819466320163312926952423791023078876139 b = 2351399303373464486466122544523690094744975233415544072992656881240319 m = 10 ** 40 print(pow(a, b, m))</lang> Output:
`1527229998585248450016808958343740453059`
## REXX
This REXX program attempts to handle any a,b, or m, but there are limits for any computer language.
For REXX, it's around eight million digits, unless ${\displaystyle a^{2}}$ or ${\displaystyle 10^{m}}$ exceeds that. <lang rexx>/*REXX program to show modular exponentation: a**b mod M */ parse arg a b mm /*get the arguments (maybe).*/ if a== | a==',' then a=,
``` 2988348162058574136915891421498819466320163312926952423791023078876139
```
if b== | b==',' then b=,
``` 2351399303373464486466122544523690094744975233415544072992656881240319
```
if mm== then mm=40 say 'a=' a; say ' ('length(a) "digits)" say 'b=' b; say ' ('length(b) "digits)"
``` do j=1 for words(mm); m=word(mm,j); say copies('─',linesize()-1)
say 'a**b (mod 10**'m")=" powerModulated(a,b,10**m)
end /*j*/
```
exit /*stick a fork in it, we're done.*/ /*──────────────────────────────────────POWERMODULATED subroutine───────*/ powerModulated: procedure; parse arg x,p,n /*fast modular exponentation*/ if p==0 then return 1 /*special case. */ if p==1 then return x /*special case. */ if p<0 then do; say '***error!*** power is negative:' p; exit 13; end parse value max(x**2,p,n)'E0' with "E" e /*pick biggest of the three.*/ numeric digits max(20,e*2) /*big enough to handle A² */ _=1
``` do while p\==0; if p//2==1 then _=_*x//n
p=p%2; x=x*x // n
end /*while*/
```
return _</lang> output when using the input of: 40 80 180 888
Note the REXX program was executing within a window of 600 bytes wide.
```a= 2988348162058574136915891421498819466320163312926952423791023078876139
(70 digits)
b= 2351399303373464486466122544523690094744975233415544072992656881240319
(70 digits)
───────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────
a**b (mod 10**40)= 1527229998585248450016808958343740453059
───────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────
a**b (mod 10**80)= 53259517041910225328867076245698908287781527229998585248450016808958343740453059
───────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────
a**b (mod 10**180)= 31857295076204937005344367438778481743660325586328069392203762862423884839076695547212682454523811053259517041910225328867076245698908287781527229998585248450016808958343740453059
───────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────
a**b (mod 10**888)= 2612849643808365153970307063634422265713972370574889513136845452410856423299436762487557161242604471887885300171829510516527484255607339748359444160694661767131561827274483018385170003434853270016569482853811730383390737793312301323406698998964489388587853627711904603124125798753498716559994462054260496622614506334484689315735068762556447491553489235236807309998697854727791160093566968169527719659307289405305177993299425901141782840092602984267350865792542825912897568403588118221513074793528568569833937153488707152390200379629380198479929609788498528506130631774711751914442
62586321233906926671000476591123695550566585083205841790404069511972417770392822283604206143472509425391114072344402850867571806031857295076204937005344367438778481743660325586328069392203762862423884839076695547212682454523811053259517041910225328867076245698908287781527229998585248450016808958343740453059
```
## Ruby
Ruby's core library has no modular exponentiation. OpenSSL, in Ruby's standard library, provides OpenSSL::BN#mod_exp. To reach this method, we call Integer#to_bn to convert a from Integer to OpenSSL::BN. The method implicitly converts b and m.
Library: OpenSSL
<lang ruby>require 'openssl'
a = 2988348162058574136915891421498819466320163312926952423791023078876139 b = 2351399303373464486466122544523690094744975233415544072992656881240319 m = 10 ** 40 puts a.to_bn.mod_exp(b, m)</lang>
Or we can implement a custom method, Integer#rosetta_mod_exp, to calculate the same result. This method does exponentiation by successive squaring, but replaces each intermediate product with a congruent value. (Program needs Ruby 1.8.7 for Integer#odd?.)
Works with: Ruby version 1.8.7
<lang ruby>class Integer
``` def rosetta_mod_exp(exp, mod)
exp < 0 and raise ArgumentError, "negative exponent"
prod = 1
base = self % mod
until exp.zero?
exp.odd? and prod = (prod * base) % mod
exp >>= 1
base = (base * base) % mod
end
prod
end
```
end
a = 2988348162058574136915891421498819466320163312926952423791023078876139 b = 2351399303373464486466122544523690094744975233415544072992656881240319 m = 10 ** 40 puts a.rosetta_mod_exp(b, m)</lang>
## Scala
<lang scala>import scala.math.BigInt
val a = BigInt(
``` "2988348162058574136915891421498819466320163312926952423791023078876139")
```
val b = BigInt(
``` "2351399303373464486466122544523690094744975233415544072992656881240319")
```
println(a.modPow(b, BigInt(10).pow(40)))</lang>
## Tcl
While Tcl does have arbitrary-precision arithmetic (from 8.5 onwards), it doesn't expose a modular exponentiation function. Thus we implement one ourselves.
### Recursive
<lang tcl>package require Tcl 8.5
1. Algorithm from http://introcs.cs.princeton.edu/java/78crypto/ModExp.java.html
2. but Tcl has arbitrary-width integers and an exponentiation operator, which
3. helps simplify the code.
proc tcl::mathfunc::modexp {a b n} {
``` if {\$b == 0} {return 1}
set c [expr {modexp(\$a, \$b / 2, \$n)**2 % \$n}]
if {\$b & 1} {
```
set c [expr {(\$c * \$a) % \$n}]
``` }
return \$c
```
}</lang> Demonstrating: <lang tcl>set a 2988348162058574136915891421498819466320163312926952423791023078876139 set b 2351399303373464486466122544523690094744975233415544072992656881240319 set n [expr {10**40}] puts [expr {modexp(\$a,\$b,\$n)}]</lang> Output:
```1527229998585248450016808958343740453059
```
### Iterative
<lang tcl>package require Tcl 8.5 proc modexp {a b n} {
``` for {set c 1} {\$b} {set a [expr {\$a*\$a % \$n}]} {
```
if {\$b & 1} { set c [expr {\$c*\$a % \$n}] } set b [expr {\$b >> 1}]
``` }
return \$c
```
}</lang> Demonstrating: <lang tcl>set a 2988348162058574136915891421498819466320163312926952423791023078876139 set b 2351399303373464486466122544523690094744975233415544072992656881240319 set n [expr {10**40}] puts [modexp \$a \$b \$n]</lang> Output:
```1527229998585248450016808958343740453059
```
## TXR
<lang txr>@(bind result @(exptmod 2988348162058574136915891421498819466320163312926952423791023078876139
``` 2351399303373464486466122544523690094744975233415544072992656881240319
(expt 10 40)))</lang>
```
```\$ ./txr rosetta/modexp.txr
result="1527229998585248450016808958343740453059"```
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# Complemented Banach spaces.
Let $X$ be Banach space and $Y$ a closed subspace of $X$. Assume that there exist a closed "subset" $Z$ of $X$ with the properties:
$Z\cap Y=\{0\}$ and every $x\in X$ can be written in a unique form as $x=y+z$ with $y\in Y$ and $z\in Z$
Can we conclude that $Y$ is complemented in $X$?
Edit: I'm not asking if $Z$ is a complement of $Y$ in $X$. Indeed, $Z$ does not need to be linear. What I am asking is if we can find a set closed linear $W\subset X$ such that $W$ is a complement of $Y$ in $X$.
• You'd want $Z\cap Y = \{0\}$, not $\emptyset$. Otherwise you'd have problems with $x \in Y$. My gut feeling says you can't conclude that $Y$ is complemented, but it has been wrong before. – Daniel Fischer Sep 12 '13 at 12:14
• Yes @DanielFischer, let me fix it, thank you. – Tomás Sep 12 '13 at 12:16
• I'm not sure about your question: are you asking if the definition of topologically complementary sets was weakened, then it continues to be true that $Z$ is a complement of $Y$? Otherwise the answer is "no" in general, because by definition $Z$ must be a (closed) linear subspace. – Federico Sep 12 '13 at 14:13
• @Federico, Im using here the usual definition of a space being complemented. I will write it there – Tomás Sep 12 '13 at 14:26
Take $X = \mathbb R^2$ with your favorite norm (say $l^2$ norm). Let $Y$ be the $y$-axis, a closed linear subspace. Let $Z$ be the graph of a continuous function $h$ with $h(0)=0$. Say $h(x) = x^3$. Every point of $X$ is uniquely the sum of a point on the graph plus a point of $Y$. SO ... the set $Z$ need not be the complement.
Can we imitate this using $X$ a Banach space and $Y$ a non-complemented subspace? Make a function $h$ choosing one element from each equivalence class?
phrase this variant in a different way ... If $T : X \to U$ is a surjective continuous linear map of Banach spaces, and if there is a continuous section (that is, a continuous $V : U \to X$ with $V(T(u))=u$ for all $u$), then must there exist also a continuous linear section?
• Sorry, I think there is a misundertanding here. This is my hypothesis, now I want to know if it is possible to find a complement (in the usual definition) of $Y$ in $\mathbb{R}$. In this case this is obvious. – Tomás Sep 12 '13 at 14:19
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# Collection
• ### Penn State STAT 506: Sampling Theory and Methods
The aim of this course is to cover sampling design and analysis methods that would be useful for research and management in many field. A well designed sampling procedure ensures that we can summarize and analyze data with a minimum of assumptions and complications. Perfect for both students and teachers wanting to learn/acquire materials for this topic.
• ### Penn State STAT 505: Applied Multivariate Statistical Analysis
Those who complete this course will be able to select appropriate methods of multivariate data analysis, given multivariate data and study objectives; write SAS and/or Minitab programs to carry out multivariate data analyses; and interpret results of multivariate data analyses. Perfect for students and teachers wanting to learn/acquire materials for this topic.
• ### Penn State STAT 504: Analysis of Discrete Data
The focus of this class is a multivariate analysis of discrete data. We will learn basic statistical methods and discuss issues relevant for the analysis of some discrete distribution, cross-classified tables of counts, (i.e., contingency tables), success/failure records, questionnaire items, judge's ratings, etc. Being familiar with matrix algebra is helpful in completing this course. Perfect for students and teachers wanting to learn/acquire materials for this topic.
• ### Penn State STAT 503: Design of Experiments
Statistics is often taught as though the design of the data collection and the data cleaning have already been done in advance. However, as most practicing statisticians quickly learn, typically problems that arise at the analysis stage, could have been avoided if the experimenter had consulted a statistician before the experiment was done and the data were conducted. This course is created to provide an understanding of how experiments should be designed so that when the data are collected, these shortcomings are avoided. Perfect for students and teachers wanting to learn/acquire materials for this topic.
• ### Penn State STAT 502: Analysis of Variance and Design of Experiments
This is a graduate level course/collection of lessons in analysis of variance (ANOVA), including randomization and blocking, single and multiple factor designs, crossed and nested factors, quantitative and qualitative factors, random and fixed effects, split plot and repeated measures designs, crossover designs and analysis of covariance (ANCOVA). Perfect for students and teachers alike looking to learn/acquire materials on ANOVA.
• ### Penn State STAT 501: Regression Methods
This graduate level course offers an introduction into regression analysis. A researcher is often interested in using sample data to investigate relationships, with an ultimate goal of creating a model to predict a future value for some dependent variable. The process of finding this mathematical model that best fits the data involves regression analysis. STAT 501 is an applied linear regression course that emphasizes data analysis and interpretation and is perfect for both students and teachers of statistics courses.
• ### Data 8: The Foundations of Data Science
This UC Berkeley Foundations of Data Science course combines three perspectives: inferential thinking, computational thinking, and real-world relevance. Given data arising from some real-world phenomenon, how does one analyze that data so as to understand that phenomenon? This course teaches critical concepts and skills in computer programming and statistical inference, in conjunction with hands-on analysis of real-world datasets, including economic data, document collections, geographical data, and social networks. It delves into social issues surrounding data analysis such as privacy and design.
• ### A Compendium of Clean Graphs in R
This compendium facilitates the creation of good graphs by presenting a set of concrete examples, ranging from the trivial to the advanced. The graphs can all be reproduced and adjusted by copy-pasting code into the R console. Almost every example in this compendium is driven by the same philosophy: A good graph is a simple graph, in the Einsteinian sense that a graph should be made as simple as possible, but not simpler. A note for R fans: the majority of our plots have been created in base R, but you will encounter some examples in ggplot.
• ### StatLib -- Datasets Archive
One of the original (and still best) sources for archived data.
• ### The Probability Web
The Probability Web is a collection of probability resources designed to be especially helpful to researchers, teachers, and people in the probability community. Web page links on this site include probabilty/statistics books and journals, information on mathematics and statistics-based careers, statistical software, teaching resources on probabilty topics, and more.
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10217
# Compressive Stress-Strain Relationships of Layered Solid Foams
This Demonstration calculates and plots the sigmoid compressive stress-strain curve of hypothetical slabs composed of three layers of solid foams having different mechanical properties and thicknesses. It also calculates and displays, separately, the individual components’ stress-strain curves. Each layer is characterized by the foam’s compressibility model equation, whose coefficients can be adjusted, and by its initial thickness, which also can be adjusted.
### DETAILS
Snapshot 1: layered spongy baked product
Snapshot 2: two soft and one hard layer combination
Snapshot 3: two-layered sponge
This Demonstration calculates the compressive engineering stress-strain relationships of layered arrays of one to three solid foam slabs having the same cross sectional area stacked together. The sigmoid compressive stress-strain relationship of the individual foams is described by the three-parameter model , or , where is the engineering stress in kPa, the engineering strain (dimensionless), and , , and are constants; is primarily a scale factor and has kPa units, , which is dimensionless, primarily controls the shoulder’s height, and , also dimensionless, serves as a marker of the densification stage [1]. For alternative models, see [1] and [2].
The absolute deformation of each layer is given by , being the layer’s initial thickness.
In a layered array of foams , , and under uniaxial compression, the absolute deformation is . Therefore, the corresponding engineering strain of the array is [2]: , where the deformations and strains are calculated by the layers’ compressibility equations.
The compressive engineering stress-strain curves of the individual foams are calculated and displayed together on the top graph and that of the array on the bottom graph. Vary the stress slider to calculate the corresponding strain of the individual layers and of the array as a whole. The chosen conditions will be seen as a moving dot on each of the stress-strain curves and the corresponding values displayed in a table above the plots.
To eliminate a layer from the array, enter zero as its initial height.
To go back to the initial settings, click the plus sign in the upper right corner of the panel, then "Initial Settings".
References
[1] S. Swyngedau, A. Nussinovitch, I. Roy, M. Peleg, and V. Huang, "Comparison of Four Models for the Compressibility of Breads and Plastic Foams," Journal of Food Science, 56, 1991 pp. 756–759.
[2] S. Swyngedau, A. Nussinovitch, and M. Peleg, "Models for the Compressibility of Layered Polymeric Sponges," Polymer Engineering & Science, 31, 1991 pp. 140–144.
### PERMANENT CITATION
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`Program Exercise No. 1: Collatz SequenceThe Collatz sequence of a positive integer n is defined as follows:a) Start with the integer n (the starting number).b) If the integer is even, divide it by 2 (integer division) (that is, n←n/2)c) If the integer is odd, multiply it by 3 and add 1. (that is, n←3n + 1)d) Repeat the process until n becomes 1.For example, let n = 5. Then its Collatz sequence is:5→16→8→4→2→1Since it takes six iterations, the length of the Collatz sequence for starting number 5 is 6.As another example, let n = 13. Then its Collatz sequence is:13→40→20→10→5→16→8→4→2→1The length of the Collatz sequence for starting number 13 is 10.Although it is not proven yet, it is thought that all Collatz sequences end in 1. Program Exercise No. 2: Cousin of Collatz SequenceFor this exercise, you will be writing a program in steps very similar to Exercise No. 1, except that you will be testing a sequence called the “Cousin of Collatz”. It is also called the 7x ± 1 problem and is defined as follows:Start at a positive integer n.If n is even, divide it by 2 (integer division) (that is n <- n/2)If n is odd and n % 4 is 1, then multiply it by 7 and add 1 (that is, n <- 7n+1)If n is odd and n % 4 is 3, then multiply it by 7 and subtract 1 (that is, n <- 7n -1)Repeat until n becomes 1.(Note: As you know from your first year, % represents the remainder of an integer division. 15%4 is 3 whereas 17%4 is 1).For example, if n = 5, then the 7x ± 1 sequence is:5 ->36->18 ->9->64->32->16 ->8 ->4 ->2 ->1The sequence can grow pretty large. For example, if n = 235, the sequence is the following (the odd integers are shown in bold): `
# CSCI 2110 Data Structures and Algorithms Fall 2022 Assignment No. 2 Java Solved
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• CSCI 2110 Data Structures and Algorithms Fall 2022 Assignment No. 2 Java Solution
Assignment No. 2 Solution.
Solution File : PDF
Part 1 and Part 2 complete Java Solution.
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# Rational Expressions and Equations
This content was STOLEN from BrainMass.com - View the original, and get the already-completed solution here!
See attached file for full problem description.
1. 5 -4
___ + ________
c2d3 7cd2
36. 3x + 2 x - 2
_______ + ______
3x + 6 x2 - 4
50. 3(x - 2) 5(2x + 1) 3(x + 1)
_______ + ________ + ________
2x - 3 2x - 3 3 - 2x
18. 3 2
_________ _ _____________
2 2
12 + x - x x - 9
34. 4x - 6 7 - 2x
________ _ ________
x - 5 5 - x
40. a 5
_____________ - _________________________
2 2
a + 11a + 30 a + 9a + 20
Solve for x:
26. x + 1 x - 1
_______ _ _______ = 1
3 2
38. 4 2x 1
_____ + ________ = ____________
X - 3 2 - 9 x + 3
X
40. 5 30
______ - _________ = 1
y - 3 2
y - 9
51. Why is it especially important to check the possible solutions to a rational equation?
52. How can a graph be used to determine how many solutions an equation has?
8. Bobbi can pick a quart of raspberries in 20 min. Blanche can pick a quart in 25min.
How long would it take if Bobbi and Blanche worked together?
14. The speed of a freight train is 15 mph slower than the speed of a passenger train.
The freight train travels 390 mi in the same time that it takes the passenger train to
travel 480 mi. Find the speed of each train.
44. It takes 60 oz. of grass seed to seed 3000 ft of lawn. At this rate, how much would
be needed to seed 2 of lawn?
5000 ft
See attached file for full problem description.
https://brainmass.com/math/basic-algebra/99180
#### Solution Preview
5 -4
___ + ________
c2d3 7cd2
Put both over a common denominator:
5(7) + -4 (cd)
c2d3(7) 7cd2(cd)
35 + -4cd
7c2d3 7c2d3
35 - 4cd
7c2d3
36. 3x + 2 x - 2
_______ + ______
3x + 6 x2 - 4
Put both over a common denominator:
(3x + 2)(x2 - 4) + (x - 2)(3x + 6)
(3x + 6)(x2 - 4) (x2 - 4)(3x +6)
(3x + 2)(x2 - 4) + (x - 2)(3x + 6)
(x2 - 4)(3x +6)
Notice that x2 - 4 = (x - 2)(x + 2) and 3x + 6 = 3(x + 2).
Factor:
(3x + 2)(x - 2)(x + 2) + 3(x - 2)(x + 2)
3(x + 2)(x - 2)(x + 2)
(x - 2)(x + 2) [(3x + 2) + 3]
3(x + 2)(x - 2)(x + 2)
Cancel like terms:
[(3x + 2) + 3]
3(x + 2)
Simplify:
3x + 5
3(x + 2)
50. 3(x - 2) 5(2x + 1) 3(x + 1)
_______ + ________ + ________
2x - 3 2x - 3 3 - 2x
This is the same as above. Put over a common denominator, add, and simplify:
3(x - 2) 5(2x + 1) 3(x + 1)
_______ + ________ + ________
2x - 3 2x - 3 - (2x - 3)
3(x - 2) + 5(2x + 1) - 3(x + 1)
2x - 3
3x - 6 + 10x + 5 - 3x -3
2x - 3
10x - 4
2x - 3
2(5x - 2)
2x - 3
Subtract. Simplify if possible.
18. 3 2
_________ _ _____________
2 2
12 + x - x x - 9
Look at the denominator of the first fraction: -x2 + x + 12. That's the same as -(x2 -x -12), which can be factored as -(x - 4)(x + 3). The denominator of the ...
#### Solution Summary
This problem set has six problems involving simplifying rational expressions (that involve adding and subtracting fractions), three similar problems that involve solving rational equations, two discussion questions, and three word problems.
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Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °
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## Topic review (newest first)
krassi_holmz
2006-03-08 17:43:49
I told you.
fgarb
2006-03-08 14:09:18
Looks right to me! Let us know if anything about the technique you used doesn't make sense.
RickyOswaldIOW
2006-03-08 10:27:33
I went back to this question and tried it out myself.
firstly I divide f(x) by one of the factors (x + 1). Then I take the quotient and divide that by the other factor (x - 1). I am now left with the quadratic 3x^2 + 7x + 4 which I factorise into (3x + 4) and (x + 1).
Thus the factors are:
(3x + 4)(x - 1)(x + 1)(x + 1) == (3x + 4)(x - 1)(x + 1)^2
As always, thanks for the help
RickyOswaldIOW
2006-03-08 08:13:22
I cannot divide 3x^4 + 7x^3 + x^2 - 7x - 4 by x^2 - 1 directly (not using my method of long division at least). If I divide by (x + 1) I am left with a cubic expression 3x^3 + 4x^2 - 3x - 4. Do I need to divide this by the other factor (x - 1)?
mathsyperson
2006-03-07 09:44:23
Using the difference of 2 squares rule, (x² - 1) = (x+1)(x-1).
Those two terms are both included in f(x), so dividing cancels them out.
f(x)/(x² - 1) = (4+3x)(x+1)
RickyOswaldIOW
2006-03-07 09:42:23
how do I divide f(x) by x^2-1?
krassi_holmz
2006-03-07 05:53:19
Good. but if a=7 and b=1 we get:
f(x)= (x^2-1)(4+3x)(1+x)= (x-1)(4+3x)(1+x)^2,
which is divisible by x^2-1
Well done ricky and fgarb!
RickyOswaldIOW
2006-03-07 05:43:48
long division of x^2 - 1/f(x) just like you said krassi?
I cannot seem to make this work
x^2 - 1 / 3x^4 + 7x^3 (just to start)
so I divide 3x^4 by x^2 to give me 3x^2.
I then multiply 3x^2 by x^2 and then by -1 to give me 3x^4 - 3x^2.
I place this underneath 3x^4 + 7x^3 and subtract it, 3x^4 - 3x^4 = 0 as I would expect but I cannot subtract -3x^2 from 7x^3...
RickyOswaldIOW
2006-03-07 05:40:54
Do you know what is the polynomial division?
Nope, no idea
I just looked at what I could do to (x - 1) to make (x^2 - 1). I only got (x - 1) in the first place beause I misread the book! So that was luck, then fgarb pointed out that it was one of the factors anyway.
f(1) = 3(1)^4 + a(1)^3 + b(1)^2 - 7(1) - 4 = 0
a + b = 8
f(-1) = 3(-1)^4 + a(-1)^3 + b(-1)^2 - 7(-1) - 4 = 0
-a + b = -6
Thus: a = 7 and b = 1.
How do I go about factorising f(x) now? Do I use the same long division method I have been using on cubic polynomials?
krassi_holmz
2006-03-07 05:28:34
Good!
Well done!!
"Given that (x² - 1) is a factor of the polynomial f(x), where f(x) = 3x^4 + ax^3 + bx^2 - 7x - 4, find the values of a and b and hence factorise f(x) completely."
Do you know what is the polynomial division?
It may help you to reduce the power of f(x).
Just divide f(x) by (x^2-1) and then the remainder which you will get must be divisible to (x^2-1).
RickyOswaldIOW
2006-03-07 05:08:14
(x^2 - 1) / (x - 1) = x + 1!
RickyOswaldIOW
2006-03-07 04:57:51
I do indeed know how to find a and b if I have two factors but I really have no idea how to get the second factor. What must I do to x - 1 to make x^2 - 1?
Maybe I must simply guess the other factors and put them into f(x) and test different values of a and b from f(1) till I find another that makes f(x) = 0?
(x^2 - 1) / (x - 1) = ???
fgarb
2006-03-06 14:21:24
One of the factors of x^2-1 is what you were using before: x-1. You should be able to figure out the other one either by division or by trial and error, the answer isn't complicated.
Then, you know that f(x) has to be divisible by both x-1 and the other factor that you found. See if you can use both of those conditions to find a and b. Good luck!
RickyOswaldIOW
2006-03-06 13:55:26
Try breaking (x^2-1) up into its constituent factors and applying them part by part
How do I do this? I've never seen such a factor before.
fgarb
2006-03-06 10:39:49
Ah, that would explain it then. Saying something has (x^2 - 1) as a factor is actually saying it has two different factors - that is, anything that is a factor of (x^2-1) must also be a factor of the equation you're trying to solve for. Try breaking (x^2-1) up into its constituent factors and applying them part by part and see if that gets you anywhere. Good luck, and feel free to ask for more help if you get stuck with that.
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Type your value in a box, click your mouse anywhere on the page, or press tab key.
Power of 10 notation can be used as follows: 3.2×104 is entered 3.2e4
The natural unit for measuring angles is the radian, which is derived from the formula for
the circumference c of a circle: c = 2 π × r, where r is the radius. If r = 1, then c = 2 π,
so the radian is defined such that a circle has 2 π radians. It also has 360 degrees.
1 radian = 360 / 2 π = 180 / π =57.295779513082320876798154814106 degrees
.
Angle Degrees Part of Pi in numbers 90° π/2 1.570796327 60° π/3 1.047197551 45° π/4 0.785398163 30° π/6 0.523598775
Formulas
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# 'for' loop pattern
Page 1 of 1
## 1 Replies - 1694 Views - Last Post: 12 June 2012 - 07:39 AMRate Topic: //<![CDATA[ rating = new ipb.rating( 'topic_rate_', { url: 'http://www.dreamincode.net/forums/index.php?app=forums&module=ajax§ion=topics&do=rateTopic&t=282545&s=c894bcb8291781bcb913de6ad93d8716&md5check=' + ipb.vars['secure_hash'], cur_rating: 0, rated: 0, allow_rate: 0, multi_rate: 1, show_rate_text: true } ); //]]>
### #1 dreamincodedot
• New D.I.C Head
Reputation: 0
• Posts: 2
• Joined: 12-June 12
# 'for' loop pattern
Posted 12 June 2012 - 07:23 AM
This is all that I've gotten so far..
for i in range (1, 10, 1):
print('#' * i)
for i in range (10, 0, -1):
print('#' * i)
I can't seeem to get the other side.
Is This A Good Question/Topic? 0
## Replies To: 'for' loop pattern
### #2 sepp2k
• D.I.C Lover
Reputation: 2277
• Posts: 3,507
• Joined: 21-June 11
## Re: 'for' loop pattern
Posted 12 June 2012 - 07:39 AM
There's two things, you need to realize:
1. For each line the number of #s on the left side is equal to the number of #s on the right side.
2. The number of #s on the left + the number of #s on the right + the number of spaces between them = the width of the shape (which from looking at the picture seems to be 12).
So if you know the width and you know how many #s you printed on the left side, you can calculate how many spaces to print before printing the right side.
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https://stats.stackexchange.com/questions/381789/what-is-the-difference-between-dice-loss-vs-jaccard-loss-in-semantic-segmentatio
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# What is the difference between dice loss vs jaccard loss in semantic segmentation task?
What is the difference between dice loss vs jaccard loss in semantic segmentation task?
Dice loss:
Dice = (2*|X & Y|)/ (|X|+ |Y|) = 2*sum(|A*B|)/(sum(A^2)+sum(B^2))
def dice_coef(y_true, y_pred):
y_true_f = K.flatten(y_true)
y_pred_f = K.flatten(y_pred)
intersection = K.sum(y_true_f * y_pred_f)
dice_coef = (2. * intersection + smooth) / (K.sum(y_true_f) + K.sum(y_pred_f) + smooth)
return 1.0-dice_coef
Not sure why Dice = (2*|X & Y|)/ (|X|+ |Y|) = 2*sum(|A*B|)/(sum(A^2)+sum(B^2)) ?
And seems implementation differ:
Jaccard loss:
Jaccard = (|X & Y|)/ (|X|+ |Y| - |X & Y|) = sum(|A*B|)/(sum(|A|)+sum(|B|)-sum(|A*B|))
def jaccard_loss(y_true, y_pred):
intersection = K.sum(K.abs(y_true * y_pred), axis=-1)
sum_ = K.sum(K.abs(y_true) + K.abs(y_pred), axis=-1)
jac = (intersection + smooth) / (sum_ - intersection + smooth)
return (1 - jac) * smooth
https://github.com/keras-team/keras-contrib/blob/master/keras_contrib/losses/jaccard.py
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http://puzzle.queryhome.com/376/metre-stick-broken-pieces-random-length-shorter-piece-average
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# A 5-metre stick is broken into two pieces at random. What is the length of the shorter piece, on average?
+1 vote
525 views
posted Apr 13, 2014
## 2 Solutions
1.25
because smaller stick will be between 0.1 to 2.4 size,
on average it will be 1.25
solution Apr 17, 2014
+1 vote
Expected length of smaller part will be 1/2 of 1/2*original length i.e. 5/4 m = 1.25m
solution Dec 19, 2014
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Nine unit circles are packed into a square, tangent to their neighbors and to the square. What is the length of the longest smooth path connecting two opposite corners of the square?
Assumptions:
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# MATH
posted by .
Determine how much would be accumulated at the end of 10 years at 7.5% interest if Woody saved \$30 every month.
• MATH -
I will assume that 7.5% per annum is compounded monthly.
so i = .075/12 = .00625
amount = 30 ( 1.00625^120 - 1)/.00628 = 5337.91
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Platinum 38 cubic meter to carats (Metric) of platinum converter
# platinum conversion
## Amount: 38 cubic meters (m3) of platinum volume Equals: 4,079,680,000.00 carats (Metric) (ct) in platinum mass
Calculate carats (Metric) of platinum per 38 cubic meters unit. The platinum converter.
TOGGLE : from carats (Metric) into cubic meters in the other way around.
### Enter a New cubic meter Amount of platinum to Convert From
* Enter whole numbers, decimals or fractions (ie: 6, 5.33, 17 3/8)
## platinum from cubic meter to carat (Metric) Conversion Results :
Amount : 38 cubic meters (m3) of platinum
Equals: 4,079,680,000.00 carats (Metric) (ct) in platinum
Fractions: 4,079,680,000.00 carats (Metric) (ct) in platinum
CONVERT : between other platinum measuring units - complete list.
## Platinum Amounts (solid platinum)
Here the calculator is for platinum amounts (solid platinum volume; dense, precious, gray to white metal rare in abundance on the planet earth. Its annual production is only a very few hundred tons. It is a very highly valuable metal. Platinum performs real well in resisting corrosion. Not only beautiful jewellery is made out of platinum, this metal enjoys quite a wide variety of uses. For instance in electronics, chemical industries and also in chemotherapy applications against certain cancers. Traders invest money in platinum on commodity markets, in commodity future trading as this material is also one of the major precious commodity metals. Thinking of going into investing in stocks? It would be a wise idea to start learning at least basics at a commodity trading school first, to get used to the markets, then start with small investments. Only after sell and buy platinum.)
Is it possible to manage numerous units calculations, in relation to how heavy other volumes of platinum are, all on one page? The all in one Pt multiunit calculation tool makes it possible to manage just that.
Convert platinum measuring units between cubic meter (m3) and carats (Metric) (ct) of platinum but in the other direction from carats (Metric) into cubic meters.
conversion result for platinum: From Symbol Equals Result To Symbol 1 cubic meter m3 = 107,360,000.00 carats (Metric) ct
# Precious metals: platinum conversion
This online platinum from m3 into ct (precious metal) converter is a handy tool not just for certified or experienced professionals. It can help when selling scrap metals for recycling.
## Other applications of this platinum calculator are ...
With the above mentioned units calculating service it provides, this platinum converter proved to be useful also as a teaching tool:
1. in practicing cubic meters and carats (Metric) ( m3 vs. ct ) exchange.
2. for conversion factors training exercises with converting mass/weights units vs. liquid/fluid volume units measures.
3. work with platinum's density values including other physical properties this metal has.
International unit symbols for these two platinum measurements are:
Abbreviation or prefix ( abbr. short brevis ), unit symbol, for cubic meter is: m3
Abbreviation or prefix ( abbr. ) brevis - short unit symbol for carat (Metric) is: ct
### One cubic meter of platinum converted to carat (Metric) equals to 107,360,000.00 ct
How many carats (Metric) of platinum are in 1 cubic meter? The answer is: The change of 1 m3 ( cubic meter ) unit of a platinum amount equals = to 107,360,000.00 ct ( carat (Metric) ) as the equivalent measure for the same platinum type.
In principle with any measuring task, switched on professional people always ensure, and their success depends on, they get the most precise conversion results everywhere and every-time. Not only whenever possible, it's always so. Often having only a good idea ( or more ideas ) might not be perfect nor good enough solutions. Subjects of high economic value such as stocks, foreign exchange market and various units in precious metals trading, money, financing ( to list just several of all kinds of investments ), are way too important. Different matters seek an accurate financial advice first, with a plan. Especially precise prices-versus-sizes of platinum can have a crucial/pivotal role in investments. If there is an exact known measure in m3 - cubic meters for platinum amount, the rule is that the cubic meter number gets converted into ct - carats (Metric) or any other unit of platinum absolutely exactly. It's like an insurance for a trader or investor who is buying. And a saving calculator for having a peace of mind by knowing more about the quantity of e.g. how much industrial commodities is being bought well before it is payed for. It is also a part of savings to my superannuation funds. "Super funds" as we call them in this country.
Conversion for how many carats (Metric) ( ct ) of platinum are contained in a cubic meter ( 1 m3 ). Or, how much in carats (Metric) of platinum is in 1 cubic meter? To link to this platinum - cubic meter to carats (Metric) online precious metal converter for the answer, simply cut and paste the following.
The link to this tool will appear as: platinum from cubic meter (m3) to carats (Metric) (ct) metal conversion.
I've done my best to build this site for you- Please send feedback to let me know how you enjoyed visiting.
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https://www.numwords.com/words-to-number/en/1379
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NumWords.com
How to write One thousand three hundred seventy-nine in numbers in English?
We can write One thousand three hundred seventy-nine equal to 1379 in numbers in English
< One thousand three hundred seventy-eight :||: One thousand three hundred eighty >
Two thousand seven hundred fifty-eight = 2758 = 1379 × 2
Four thousand one hundred thirty-seven = 4137 = 1379 × 3
Five thousand five hundred sixteen = 5516 = 1379 × 4
Six thousand eight hundred ninety-five = 6895 = 1379 × 5
Eight thousand two hundred seventy-four = 8274 = 1379 × 6
Nine thousand six hundred fifty-three = 9653 = 1379 × 7
Eleven thousand thirty-two = 11032 = 1379 × 8
Twelve thousand four hundred eleven = 12411 = 1379 × 9
Thirteen thousand seven hundred ninety = 13790 = 1379 × 10
Fifteen thousand one hundred sixty-nine = 15169 = 1379 × 11
Sixteen thousand five hundred forty-eight = 16548 = 1379 × 12
Seventeen thousand nine hundred twenty-seven = 17927 = 1379 × 13
Nineteen thousand three hundred six = 19306 = 1379 × 14
Twenty thousand six hundred eighty-five = 20685 = 1379 × 15
Twenty-two thousand sixty-four = 22064 = 1379 × 16
Twenty-three thousand four hundred forty-three = 23443 = 1379 × 17
Twenty-four thousand eight hundred twenty-two = 24822 = 1379 × 18
Twenty-six thousand two hundred one = 26201 = 1379 × 19
Twenty-seven thousand five hundred eighty = 27580 = 1379 × 20
Sitemap
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http://mathhelpforum.com/advanced-statistics/108046-subtracting-two-normal-distributions.html
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# Thread: Subtracting two normal distributions
1. ## Subtracting two normal distributions
I have two distributions X~N(.65,.0002) and Y~N(.59,.0003) if I want the mean and SD do I just subtract them? and is X-Y also a normal distribution?
2. Hello,
If they're independent, X-Y will indeed be a normal distribution.
Now some properties... :
$\displaystyle \mathbb{E}[aX+bY]=a\mathbb{E}[X]+b\mathbb{E}[Y]$
$\displaystyle Var(aX+bY)=a^2Var(X)+b^2Var(Y)$ if they're independent. If not, you'll have to introduce the covariance (check your lessons)
So Var(X-Y)=Var(X)+Var(Y), if they're independent. Note that the variance is always positive !
3. Yes X and Y are independent. So would it just be $\displaystyle u_{x-y}$ = .06 and $\displaystyle SD_{x-y}$ = |.0002-.0003|? To add more information to this, X is the diameter of a metal pipe and the Y is the diameter of a pipe clamp so the X must fit inside Y. I'm also trying to figure out the probability that the pipe will fit inside the clamp.
,
,
,
,
### subtracting two normal distributions
Click on a term to search for related topics.
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Class 12
MATHS
परिशिष्ट
# Delta ABC में सिद्ध कीजिए कि : <br> (a sin(B-C))/(b^(2)-c^(2))=(bsin(C-A))/(c^(2)-a^(2))=(csin(A-B))/(a^(2)-b^(2))
Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.
Updated On: 1-9-2020
Apne doubts clear karein ab Whatsapp par bhi. Try it now.
Watch 1000+ concepts & tricky questions explained!
19.4 K+
900+
Text Solution
(a sin(B-C))/(b^(2)-c^(2))=(bsin(C-A))/(c^(2)-a^(2))=(csin(A-B))/(a^(2)-b^(2))
127303395
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05/23/18
05/22/18
#### balance the equation also what type of reaction is this
_Ca(No3)2+_NaOH=__Ca(OH)2+__NaNO3” __Mg(No3)2+__K3PO4=__Mg3(PO4)2+__KNO3
05/20/18
#### Marisa has taken 5 math tests and scored the following, 83,88,95,83,92. what does she need to score 6th math test in order to have a test average of 90 overall?
so i have almost solved but i am stuck i add them all up so it looks like 83+88+92+95+x=90 all over 6 then i got 441+x 6 but idk what to do now because the correct... more
05/18/18
#### A triangle has an area of 120 sq units. Need to find the base and the height.
120=1/2(3x+9)(x=5)
05/14/18
#### The equation (x^2+y^2)^2=2(x^3+y^3) graphs the bean curve. Approximate the value of y at x=1/2 by using the linear approximation...
The equation (x^2+y^2)^2=2(x^3+y^3) graphs the bean curve. Approximate the value of y at x=1/2 by using the linear approximation to the curve at the point (1,1). Possible answers... more
05/14/18
#### Let P be the population of some geographical area. If the population of the area changes over time based on...
Let P be the population of some geographical area. If the population of the area changes over time based on the following criteria, which of them signals exponential growth of the... more
05/14/18
#### The graph of x^4=4(x^2-y^2) has how many horizontal tangents?
Possible answers include: A. 1 B. 2 C. 4 D. 5
Help Asap Please Calculus Math Help
05/14/18
#### Consider the differential equation dy/dx=x^2+4y-1. If y=f(x) is the solution to the differential equation, at what point does f have a relative maximum?
Possible answers include: A. (3, -2) B. (0, 1/4) C. (-2, -1) D. (-5,-6)
05/10/18
#### Chemistry Question
Calculate the ratio of moles of acid to moles of base to create a 750 mL buffer solution with pH = 4.2 using acetic acid, CH3OOH, and acetate CH3COO. The ka for acetic acid is 1.8 x 10^(-5)
05/10/18
#### Chemistry Question
Consider an aqueous solution of ammonia, NH3, a base. If the resulting pH of this solution is 9.5, then what is the concentration of this ammonia solution? Note that the Ka for ammonium, NH4, is... more
Help Asap Please Maths Word Problem Question
05/07/18
Tuyen walked 2.0 mi due north 5 mi due west 3.0 mi due south, and 1.0 mi due east. How far from starting point?
05/04/18
05/02/18
#### 2cos2x(sin2x) + sin2x = 0?
I need help solving this problem, Solve the following trigonometric equation on the interval [ 0, 2pi ) : 2cos2x(sin2x) + sin2x = 0?
05/02/18
The volume of a sandbox is 48ft^3. The height of the sandbox is 2 feet. The width is w feet and the length is w+2 feet. Use the formula V=lwh to find the volume of w
05/01/18
#### help asap math due
a class used cars and buses to go on a field trip. they used 8 vehicles to go on the trip. each car can hold 3 students and each bus holds 30 students. if 78 students were on the trip how many cars... more
Help Asap Please Test Preparation Trigonometry
04/30/18
#### Find the height of the mountain to the nearest foot.
Bob is driving along a straight and level road straight toward a mountain. At some point on his trip, hemeasures the angle of elevation to the top of the mountain and finds it to be 24° 10′. He... more
04/30/18
#### Slope intercept
write an equation for the line that passes through (2,4) and (5,-2)
Help Asap Please Algebra 2 Algebra
04/30/18
#### Hyperbolas question on constant difference
Identify the constant difference for a hyperbola with foci (0, −5) and (0, 5) and a point on the hyperbola (0, 3).
Help Asap Please Test Preparation Trigonometry
04/29/18
#### 8(cos 90° + i sin 90°)/3(cos 30° + i sin 30°)
Find the following quotient, and write the quotient in rectangular form, using exact values. 8(cos 90° + i sin 90°)/3(cos 30° + i sin 30°)
Help Asap Please Test Preparation Trigonometry
04/29/18
#### Express the complex number in polar form
Express the complex number in polar form and use DeMoivreʹs Theorem to find the given power. Write your answer inrectangular form. Give your answer in exact form unless otherwise specified. 29) (-... more
Help Asap Please Test Preparation Trigonometry
04/29/18
#### Find the product
Find the product. Write the product in rectangular form, using exact values. [4 cis 300°] [5 cis 120°]
Help Asap Please Test Preparation Trigonometry
04/29/18
#### Find the angle between the given vectors to the nearest tenth of a degree.
Find the angle between the given vectors to the nearest tenth of a degree.a = 7i + 2j, b = -6i + 6j
## Still looking for help? Get the right answer, fast.
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# 9th Physics Chapter#1 Test
##### Citation preview
EDUCATIONAL WAVE PAKISTAN
Name
Class
Roll #
Date
9th
TEST CODE
Time Allowed
Max.Marks
25
Obtained Marks
1
Question#1:Choose the Correct option.
: (1
1)Which of the following is base quantity.
Momentum
D
Torque
C
Volume
B
D
100ug
C
2mg
B
100cm 3
D
10cm 3
C
B
D
4
C
3
B
0.01g
A
D
Heat
1cm 3
A
2
A
(5
5)Study of internal structure of earth is called: Sound
A
210.0(4
4)Number of significant figures in 210.0 are: 5
(3
3)1 Litre volume= 1000cm 3
Mass
(2
2)Which of the following is smallest quantity. 5000ng
45 Min.
C
Geo Physics
B
Atomic Physics A
Question#2: Answer the following Short Questions. 1)Define Plasma physics and Geo physics. 2)What is difference b/w base and derived quantities? 3)What are SI units?Explain their role in development of science.
2 (1 (2 SI SI(3
4)What is meant by vernier constant?
(4
5)Estimate 16 years age in seconds?
16(5
6)What is meant by zero error and zero correction?
(6
7)Define Physics? 8)Define nuclear physics and atomic physics. 9)What is meant by pitch of screw gauge? 10)Why the measurements taken by screw gauge are considered more reliable than that of vernier caliper?
(7 (8
(9 (10
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# Tagged Questions
The tag has no usage guidance.
1answer
199 views
### How does the expansion of the universe not violate causality?
It is often said that faster than light travel would violate causality. However, because the universe is expanding, there are actually distant stars that move away from us at a speed greater than the ...
4answers
2k views
### Can causality be violated?
A common justification for prohibiting many unusual phenomena such as faster than light travel is that if they were possible, causality would be violated. Let's define causality as: You cannot ...
5answers
4k views
### How does “warp drive” not violate Special Relativity causality constraints?
I'm talking about this nonsense: http://www.washingtonpost.com/news/post-nation/wp/2014/06/11/this-is-the-amazing-design-for-nasas-star-trek-style-space-ship-the-ixs-enterprise/ Now, I'm aware that ...
1answer
456 views
### Does negative energy density (i.e. weak energy condition violation) create closed timelike curves?
I remember reading something about Stephen Hawking denying the fact you can't make CTC's (Closed Timelike Curves) without weak energy condition violation. If this is true, where do the light cones ...
0answers
37 views
### Virtual Particles and Causation [duplicate]
Sometimes when people debate what type of cause a universe with a beginning may have Virtual Particles has been used as an example of a thing that can arise without a cause. So my question would be ...
4answers
363 views
### Principle of locality
Why does the principle of locality have so such great importance in physics that theory should be consistent with it?
0answers
115 views
### Does causality alone resolve the mathematical ambiguity of expressing physical systems?
Newton's 2nd law of motion is most often written in the differential form $\sum F = {dp \over dt}$ but can also be expressed in an integral form $p = \int\sum F dt$ Each form of expressing ...
2answers
2k views
### What does “causally connected” or “causes” really mean?
In a different thread, a user stated the following about events preceding or following other events: However, if the two events are causally connected ("event A causes event B"), the causal order ...
3answers
2k views
### Time travel outside of light cone without causality violation
If one is able to travel into the past but at a spatial distance that puts him outside of his own past light cone would this be considered a causality violating trip? Looking at a Minkoski diagram, it ...
2answers
347 views
### Feynman's $i \epsilon$ prescription in loop expansion
I have some questions about the $i\epsilon$ factor in Feynman diagrams. First, what is the physical meaning of $i\epsilon$ in loop amplitudes. Second, how does it ensures unitarity? And third, Dyson ...
1answer
192 views
### Are there any causeless phenomena from the mainstream physical viewpoint? [closed]
EDIT: The orginal version did not produce any answers about physics. I know what life is, I have studied that for decades. I wanted to hear how the border between matter and spirit looks from the ...
1answer
209 views
### Superluminal travel without time travel
Can superluminal travel or communication be possible without leading to the possibility of closed timelike curves (CTCs) and causality violations? I have seen conflicting opinions regarding this ...
0answers
175 views
### Big Bang, Heat Death, and cause and effect
If the Universe has two 'end points', one being the Big Bang, and the other being heat death, is there anything in the laws of physics which forbid a random fluctuation in the heat death state from ...
2answers
439 views
3answers
2k views
### Why is causality preserved in special relativity?
PART 1: I was reading the article Relativity of simultaneity Wikipedia. I couldn't understand this line: "if the two events are causally connected ("event A causes event B"), the causal order is ...
2answers
583 views
### What would happen if some signal could move faster than light?
The two postulates of STR doesn't say that any signal cannot move faster than light. It also doesn't assert that any signal except light cannot have velocity equals to that of light. So at the very ...
2answers
227 views
### On the distinction of past and future: could one theoretically reverse direction of particles and cause time to appear to go backwards?
Based on my understanding of physics after seeing The Distinction of Past and Future on Project Tuva, there is no distinction between past and future on a fundamental level- all particle interactions ...
0answers
111 views
### Is there any place for teleology in physics? [closed]
Most physicists absolutely hate the idea of teleology. They take it as an unquestionable article of faith that causality only runs in one direction, that is, from the past to the future. Is there any ...
1answer
116 views
### Ensuring globally hyperbolic geodesically-complete spacetimes
Let's say we have an incomplete spacetime A that is globally hyperbolic, does there necessary exist a globally hyperbolic completion? My guess is no, in which case what further restrictions can be ...
4answers
3k views
### Space-like and time-like: where do the names come from?
Space-like separated events are events that, in a well-chosen reference frame, can take place at the same time but never happen at the same location. On the other hand for time-like events, one can ...
0answers
142 views
### Why can apparent horizon be computed based on its local geometry?
Why can apparent horizon be computed based on its local geometry? In the paper titled Black Holes, Geometric Flows, and the Penrose Inequality in General Relativity by Hubert L. Bray, has been written:...
1answer
126 views
### causal sketches [closed]
I don't have much of an idea of how to draw causal sketches. I know that you need to work out the gradient of the light cones, which can be done using a given metric and using null vectors. But how do ...
1answer
241 views
### Observables still commute even if fields only anti-commute
In Peskin & Schroeder page 56, after introducing anti commutation relations for the fields instead of commutation relations (in order to fix the negative energy problem as well as to have proper ...
1answer
156 views
### How to make a black hole?
Many Physics discussions I have often conclude with: Well you will then form a black hole... My questions are: Is there a general recipe for making a black hole? If not, then can you list the ...
2answers
502 views
### Virtual photons as force mediators in QED - really?
If the photon is the force vector for EM interactions, e.g. electrons, how does each electron 'know' where the other one is so that it can send it a photon? I've thought about this for a while. I ...
0answers
108 views
### What is wrong in following arguments about connection of local gauge invariance and causality?
There is a question and corresponding downvoting of my answer, so I decided to ask this question. There is my answer on it: "...The most theories of free fields are invariant under global gauge ...
2answers
422 views
### Why does the Dopfer EPR experiment require coincidence counting?
Dopfer Momentum-EPR experiment (1998) seems to provide a interesting tweak in the EPR experiment. To read more details on this experiment, see: Page 3 (labelled S290) of 'Experiment and the ...
1answer
588 views
### Does cosmic censorship rule out stable toroidal black holes? How?
I'm having a hard time understanding what the arguments against stable toroidal black holes are saying. For many of these, I can't figure out if they're talking about: A non-rotating toroidal event ...
1answer
238 views
### Quantization surface in QFT
What does the Quantization Surface mean here? Reference: H. Latal W. Schweiger (Eds.) - Methods of Quantization
5answers
1k views
### Light-like Interval
In SR, the interval $I$ between two spacetime events is called light-like if $I=0$. Griffiths in his Introduction to Electrodynamics book says that [page 503], If $I=0$ we call the interval light-...
3answers
1k views
### Are random quantum phenomena happening without a cause?
In everyday life, most of us assumes every event and object has a cause in some sense. I am wondering if the same is true for quantum physics. Does the random nature of quantum phenomena mean they ...
0answers
455 views
### Quantum Entanglement and Causality [duplicate]
How does Quantum Entanglement not violate the principle of relativity? Alice and Bob are working on an entangled system of electrons which is spaced long apart. Now if Alice measures one electron to ...
2answers
937 views
### Can special relativity distort the relative order in which events occur?
Pretend you are throwing darts at a dart board. You throw dart $d_1$ at time $t_1$. After you throw your first dart, you throw your second dart $d_2$ at time $t_2$. Given that $t_2 > t_1$ in a ...
1answer
433 views
### What does “all future lies within the event horizon” mean?
I was trying to find an answer as to why light does not escape black holes and I stumbled upon this Phys.SE question. In the answer it said that: "Since all future lies within the event horizon, ...
1answer
413 views
### Retarded potential in gravitational field?
Is there a retarded potential concept in gravitational field similar to electromagnetic radiation?
2answers
931 views
### What's wrong with this QFT thought experiment?
In quantum field theory, the propagator $D(x-y)$ doesn't vanish for space-like separation. In Zee's book, he claims that this means a particle can leak out of the light-cone. Feynman also gives this ...
2answers
136 views
### Which causal structures are absent from any “nice” patch of Minkowski space?
Which "causal separation structures" (or "interval structures") can not be found among the events in "any nice patch ($P$) of Minkowski space"?, where "causal separation structure" ($s$) should be ...
1answer
146 views
### Can the vanishing of the Riemann tensor be determined from causal relations?
Given a Lorentzian manifold and metric tensor, "$( M, g )$", the corresponding causal relations between its elements (events) may be derived; i.e. for every pair (in general) of distinct events in set ...
1answer
123 views
### Is this hypo-theoretical model of future prediction feasible? [closed]
First let me state that I am not, nor ever have I been, a physics student. I am working on an idea for a book I'm writing. This is a thought experiment that posits the existence of a computer system ...
1answer
188 views
### Causal structure of the inflationary multiverse
In the multiverse as it is described by eternal inflation, it is not clear to me what is its causal structure and in particular if the bubble-universes are causally connected. We start from a de-...
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Why the p-value is 0, or how to keep the precision for small values?
2
2
Entering edit mode
3.7 years ago
Star ▴ 60
Hi all,
I am trying to calculate p-value from z-score, but p-value is coming as 0.
2*pnorm(-abs(37.9353259515))
[1] 0
However, if I get a smaller z-score, I get the accurate p-value.
2*pnorm(-abs(31.4730702149))
[1] 2.029847e-217
Can anyone please suggest a way around to get accurate p-value in 1st case instead of 0?
Cheers!
R • 2.1k views
6
Entering edit mode
3.7 years ago
Back in the days when computers didn't have the oomph they have today, you'd see a lot of -log10(value) calculations used to report very small numbers. Using calculations on or presenting log-transformed values may be an alternative, until you get to a point where you need to report high-precision results with Rmfpr or similar, depending on your programming environment.
library(Rmpfr)
.N <- function(.) mpfr(., precBits = 200)
2 * pnorm(-abs(.N(c(31.4730702149, 37.9353259515))))
# 2 'mpfr' numbers of precision 200 bits
# [1] 2.0298465195466694380461698263104467695985404891415345411446557e-217
# [2] 6.7359508146380783828098958784721518979443306021620032358869696e-315
As to the "why" that explains R's native behavior, I'd suspect that it uses double-precision floating point values internally. This gives you a working limit on measurements with values ranging from approximately ±5e−324 to ±5e324 (subnormal) or ±2.23e-308 to ±2.23e308 (normal).
Given your example, I'd say R is probably using the "normal" range, which accounts for a binary error factor at the edges ("epsilon") that lets you do arithmetic more easily and accurately.
Long answer, short: If you need to work with numerical values outside this range — and there can be cases when data need to be ranked with scores of this type — either log-transform your numbers or use a third-party library to get 128-bit or greater precision. 128-bit precision will support values ranging from approximately ±6.48e−4966 to ±6.48e4966 (subnormal), or ±1.19e-4932 to ±1.19e4932 (normal).
I'm certain that the more precision you require from third-party libraries, the longer it will take to do calculations. Performance is already an issue in R, so that is something to keep in mind. A log-transform is very probably cheap and easy, comparatively.
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Entering edit mode
3.7 years ago
You've reached the limits of machine precision. 0 and 1e-217 (1 preceded by 216 zeros) are not meaningfully different. I don't see how adding more zeros will help you. This looks to me like a case of the XY problem.
0
Entering edit mode
(This is probably good enough to be an answer rather than a comment.)
0
Entering edit mode
Is it architecture-dependent? I get the following on:
• Scientific Linux release 7.4 (Nitrogen)
• Kernel name is : Linux 3.10.0-514.6.1.el7.x86_64 x86_64
• CPU Type : Intel(R) Xeon(R) CPU E5-2680 v4 @ 2.40GHz
.
.Machine$double.xmin [1] 2.225074e-308 options(scipen=999) .Machine$double.xmin
[1] 0.00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000002225074
0
Entering edit mode
This is a complicated thing and I am not sure I understand all of it. This has to do with floating point arithmetic. As far as I know, machine precision depends on the numeric type (i.e. whether using float or double or long), on the programming language (or the compiler) and also the OS. Then I think most values in use are actually defined by some IEEE standard but not every application has to use them. To complicate matters, there are different "types" of machine precision. Strictly speaking machine precision is the smallest number eps such that 1+eps != 1 (i.e. .Machine$double.eps). Then there's the smallest non-zero number (.Machine$double.xmin). In R, .Machine gives you what the machine is capable of but that doesn't mean this is what's used by R or even some packages.
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# For any rng $R$, can we attach a unity?
Let $R$ be an rng. (There may be no unity)
Then, does there always exist a ring(with unity) $A$ such that $R$ is a subrng of $A$?
• possible duplicate of Is it possible to extend a commutative ring to have a unity? May 27, 2015 at 18:37
• @bburGsamohT: the question you linked refers to commutative rings; this one is more general, referring as it does to general (not necessarily commutative) rngs. Cheers! May 27, 2015 at 22:37
• @RobertLewis well, that difference is completely trivial considering that it makes no difference in the given answers. If the two questions answers were traded, then this should be closed despite the wording. However, the answers here are a bit better written, and this subsumes that one, so the other should be closed as a dupe now. May 28, 2015 at 3:09
One can always find a ring with unity containing a ring (without unity). There are many different ways to do that ; let's look at one of them. Let $R$ denote the ring without unity and consider $A=\mathbb{Z}\times R$ with the canonical addition and the following multiplication
$$(m,a)\cdot (n,b)=(mn, na+mb+ab)$$
$A$ with these two operations is a ring and $\left(1,0_R\right)$ is a unity for that ring. And $R\to A$ $a\to (0,a)$ is a canonical injection of $R$ in $A$
• To give some explanation (for OP and future others) for why this definition of multiplication makes sense: if we informally considered this as a sum of two rings, then we'd want to evaluate $(m+a)(n+b)$ the obvious thing to do is $mn + m.b + n.a + ab$. $mn\in\Bbb Z$ but the other three terms are in $R$ hence the reason for collecting them the way we do. May 27, 2015 at 18:49
• It's OK if the starting ring $R$ happens to contain a unity, say $e$.It won't conflict via the embedding with the element $(1,0_R)$ since the image of $e$ is $(0,e)$ and the latter is not an identity for the full ring on $\mathbb{Z} \times R.$. May 27, 2015 at 18:50
• @Cameron Williams you are perfectly right. In fact it is a general construction based on the direct sum $\mathbb{Z}\oplus R$ which in our case is isomorphic to the direct product. May 27, 2015 at 18:52
• In regard to coffemath's comment, we see that $(0, e)$ is in fact a non-unit idempotent in $\Bbb Z \times R$: $(0, e) \cdot (0, e) = (0, e^2) = (0, e)$, since $e$ is the unit of $R$ in this case. As regard to generality, it appears to me that if $S$ is unital, the set $S \times R$ with componentwise addition and multiplication $(s_1, r_1) \cdot (s_2, r_2) = (s_1 s_2, s_1 r_2 + s_2 r_1 + r_1 r_2)$ will satisfy all the axioms for a unital ring, with $1_{S \times R} = (1, 0)$. Cheers! May 27, 2015 at 19:01
For future readers, the above construction can be greatly generalized.
Let $$R$$ be a rng (may have no unity) and $$(M,+,•)$$ be an $$R$$-module.
Define $$S=\mathbb{Z}\times R$$ and define operations as $$(n,a)+(m,b)=(n+m,a+b)$$ and $$(n,a)(m,b)=(nm,ma+nb+ab)$$.
Then, $$R$$ is a subrng of $$S$$ with respect to an embedding $$r\mapsto (0,r)$$.
Now, define $$(n,a)\ast x = a•x + nx$$ where $$x\in M$$.
Then $$\ast$$ indeed induces $$M$$ to be an $$S$$-module and $$•$$ is the restriction of $$ast$$ on $$R\times M$$.
To sum up, the following is true:
Let $$R$$ be an rng and $$(M,+,•)$$ be an $$R$$-module.
Then, there exists a ring $$S$$ and an operation $$\ast:S\times M\rightarrow M$$ such that $$(M,+,\ast)$$ is an $$S$$-module and $$\ast\upharpoonright (R\times M)=•$$.
This means that, defining module on rngs (may be without unity) is equivalent to defining module on rings (with unity).
So, by defining modules on rings with unity, one can handle both two cases.
Moreover, here is one application of the original one : "every rng is a subrng of a ring with unit"
Let $$R$$ be an rng and consider a polynomial rng $$R[X]$$.
Since $$R$$ may not have a unity, you cannot formally define $$X$$ as an element of $$R[X]$$.
However, by extending $$R$$ to a ring $$S$$, since $$R[X]$$ is a subrng of $$S[X]$$, you can consider $$X$$ as an actual object!
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# Nanometer to Decimeter conversion
Result
dm
Click Result to Copy
nm=dm
How did we calculate?
To Calculate, we took the value you submitted and divided it by 100,000,000 to get the result.
(/100,000,000)
Our nanometer to decimeter(nm to dm) conversion tool is a free converter that enables you to convert from nanometer to decimeter easily.
## How to Convert nanometer to decimeter
To convert a nanometer measurement(nm) to a decimeter measurement(dm), divide the length by the conversion ratio. Since one decimeter is equal to 100,000,000 nanometers, you can use this simple formula to convert:
### What is the formula to convert from nanometer to decimeter?
decimeter=nm / 100,000,000
### Convert 5 nanometers to Decimeter
5 nm = (5 / 100,000,000) = 5e-8dm
### Convert 10 nanometers to Decimeter
10 nm = (10 / 100,000,000) = 1e-7 dm
### Convert 100 nanometers to Decimeter
100 nm = (100 / 100,000,000) = 0.000001dm
## Nanometer
### What is a Nanometer?
A micrometer is a metric unit for measuring length equal to 0.001 mm, or about 0.000039 inches. Its symbol is μm. The micrometer is commonly employed to measure the thickness or diameter of microscopic objects, such as microorganisms and colloidal particles.
A micrometer can be abbreviated as µm; for example, 1 Micrometer can be written as 1µm.
### What is the Nanometer used for?
Nanometers are used to measure the smallest things, usually the size of an atom or molecule. Typically, the size of transistors on a semiconductor-based processor is calculated in nanometers.
## Decimeter
### What is an Decimeter?
A decimeter is a unit of length in the metric system. The term “Deci” means one-tenth, and therefore decimetre means one-tenth of a meter. Since a meter is 100 cm, one-tenth of 100 cm is 10 cm. Thus one Decimeter measures 10 cm.
Decimeter can be abbreviated as dm; for example, 1 Decimeter can be written as 1dm.
### What is the Decimeter used for?
A decimeter is not widely used but is an important unit. In real life, we rarely find measurements written in decimetres. Since one meter is not a very long length, it is easier to use 0.1 m or 0.5 m when the length is shorter than a meter.
## How to use our Nanometer to Decimeters converter (nm to dm converter)
Follow these 3 simple steps to use our nanometer to decimeter converter
1. Input the unit of nanometers you wish to convert
2. Click on convert and watch this result display in the box below it
3. Click Reset to reset the nanometer value
## Nanometer to decimeter Conversion Table
nanometersdecimeters
nm dm
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Switch to:
Zoltek Companies, Inc. (NAS:ZOLT)
Beneish M-Score
0.00 (As of Today)
The zones of discrimination for M-Score is as such:
An M-Score of less than -2.22 suggests that the company is not an accounting manipulator.
An M-Score of greater than -2.22 signals that the company is likely an accounting manipulator.
Zoltek Companies, Inc. has a M-score of -2.94 suggests that the company is not a manipulator.
During the past 13 years, the highest Beneish M-Score of Zoltek Companies, Inc. was 0.00. The lowest was 0.00. And the median was 0.00.
Definition
The M-score was created by Professor Messod Beneish. Instead of measuring the bankruptcy risk (Z-Score) or business trend (F-Score), M-score can be used to detect the risk of earnings manipulation. This is the original research paper on M-score.
The M-Score Variables:
The M-score of Zoltek Companies, Inc. for today is based on a combination of the following eight different indices:
M = -4.84 + 0.92 * DSRI + 0.528 * GMI + 0.404 * AQI + 0.892 * SGI + 0.115 * DEPI = -4.84 + 0.92 * 1.114 + 0.528 * 1.2726 + 0.404 * 0.7937 + 0.892 * 0.7975 + 0.115 * 0.9543 - 0.172 * SGAI + 4.679 * TATA - 0.327 * LVGI - 0.172 * 1.3576 + 4.679 * -0.0963 - 0.327 * 0.7691 = -2.94
* All numbers are in millions except for per share data and ratio. All numbers are in their own currency.
This Year (Dec13) TTM: Last Year (Dec12) TTM: Accounts Receivable was \$26.2 Mil. Revenue was 35.123 + 40.958 + 30.31 + 33.306 = \$139.7 Mil. Gross Profit was 5.281 + 7.519 + 6.202 + 7.032 = \$26.0 Mil. Total Current Assets was \$146.9 Mil. Total Assets was \$356.1 Mil. Property, Plant and Equipment(Net PPE) was \$208.7 Mil. Depreciation, Depletion and Amortization(DDA) was \$18.5 Mil. Selling, General & Admin. Expense(SGA) was \$14.1 Mil. Total Current Liabilities was \$22.4 Mil. Long-Term Debt was \$17.6 Mil. Net Income was -2.264 + -0.15 + -0.897 + 3.306 = \$-0.0 Mil. Non Operating Income was -1.817 + -1.449 + -1.211 + 1.72 = \$-2.8 Mil. Cash Flow from Operations was 13.222 + 12.041 + 4.577 + 7.209 = \$37.0 Mil. Accounts Receivable was \$29.4 Mil. Revenue was 35.877 + 44.199 + 48.078 + 47.014 = \$175.2 Mil. Gross Profit was 9.077 + 10.22 + 11.278 + 10.968 = \$41.5 Mil. Total Current Assets was \$143.4 Mil. Total Assets was \$358.4 Mil. Property, Plant and Equipment(Net PPE) was \$214.6 Mil. Depreciation, Depletion and Amortization(DDA) was \$18.1 Mil. Selling, General & Admin. Expense(SGA) was \$13.0 Mil. Total Current Liabilities was \$24.1 Mil. Long-Term Debt was \$28.2 Mil.
1. DSRI = Days Sales in Receivables Index
Measured as the ratio of daysÂ’ sales in receivables in year t to year t-1.
A large increase in DSR could be indicative of revenue inflation.
DSRI = (Receivables_t / Revenue_t) / (Receivables_t-1 / Revenue_t-1) = (26.164 / 139.697) / (29.449 / 175.168) = 0.18729107 / 0.16811861 = 1.114
2. GMI = Gross Margin Index
Measured as the ratio of gross margin in year t-1 to gross margin in year t.
Gross margin has deteriorated when this index is above 1. A firm with poorer prospects is more likely to manipulate earnings.
GMI = GrossMargin_t-1 / GrossMargin_t = (GrossProfit_t-1 / Revenue_t-1) / (GrossProfit_t / Revenue_t) = (41.543 / 175.168) / (26.034 / 139.697) = 0.2371609 / 0.18636048 = 1.2726
3. AQI = Asset Quality Index
AQI is the ratio of asset quality in year t to year t-1.
Asset quality is measured as the ratio of non-current assets other than plant, property and equipment to total assets.
AQI = (1 - (CurrentAssets_t + PPE_t) / TotalAssets_t) / (1 - (CurrentAssets_t-1 + PPE_t-1) / TotalAssets_t-1) = (1 - (146.929 + 208.727) / 356.051) / (1 - (143.356 + 214.592) / 358.449) = 0.00110939 / 0.00139769 = 0.7937
4. SGI = Sales Growth Index
Ratio of sales in year t to sales in year t-1.
Sales growth is not itself a measure of manipulation. However, growth companies are likely to find themselves under pressure to manipulate in order to keep up appearances.
SGI = Sales_t / Sales_t-1 = Revenue_t / Revenue_t-1 = 139.697 / 175.168 = 0.7975
5. DEPI = Depreciation Index
Measured as the ratio of the rate of depreciation in year t-1 to the corresponding rate in year t.
DEPI greater than 1 indicates that assets are being depreciated at a slower rate. This suggests that the firm might be revising useful asset life assumptions upwards, or adopting a new method that is income friendly.
DEPI = (Depreciation_t-1 / (Depreciaton_t-1 + PPE_t-1)) / (Depreciation_t / (Depreciaton_t + PPE_t)) = (18.075 / (18.075 + 214.592)) / (18.497 / (18.497 + 208.727)) = 0.07768614 / 0.08140425 = 0.9543
6. SGAI = Sales, General and Administrative expenses Index
The ratio of SGA expenses in year t relative to year t-1.
SGA expenses index > 1 means that the company is becoming less efficient in generate sales.
SGAI = (SGA_t / Sales_t) / (SGA_t-1 /Sales_t-1) = (14.119 / 139.697) / (13.041 / 175.168) = 0.10106874 / 0.07444853 = 1.3576
7. LVGI = Leverage Index
The ratio of total debt to total assets in year t relative to yeat t-1.
An LVGI > 1 indicates an increase\$sgai= in leverage
LVGI = ((LTD_t + CurrentLiabilities_t) / TotalAssets_t) / ((LTD_t-1 + CurrentLiabilities_t-1) / TotalAssets_t-1) = ((17.597 + 22.374) / 356.051) / ((28.173 + 24.146) / 358.449) = 0.11226201 / 0.1459594 = 0.7691
8. TATA = Total Accruals to Total Assets
Total accruals calculated as the change in working capital accounts other than cash less depreciation.
TATA = (IncomefromContinuingOperations_t - CashFlowsfromOperations_t) / TotalAssets_t = (NetIncome_t - NonOperatingIncome_t - CashFlowsfromOperations_t) / TotalAssets_t = (-0.005 - -2.757 - 37.049) / 356.051 = -0.0963
An M-Score of less than -2.22 suggests that the company will not be a manipulator. An M-Score of greater than -2.22 signals that the company is likely to be a manipulator.
Zoltek Companies, Inc. has a M-score of -2.94 suggests that the company will not be a manipulator.
Related Terms
Historical Data
* All numbers are in millions except for per share data and ratio. All numbers are in their own currency.
Zoltek Companies, Inc. Annual Data
Sep04 Sep05 Sep06 Sep07 Sep08 Sep09 Sep10 Sep11 Sep12 Sep13 DSRI 1.0913 0.5961 0.9187 1.3494 0.9255 0.956 0.8078 1.1266 0.9634 1.1717 GMI 0.9866 3.3653 0.1915 0.8423 1.0417 1.2489 2.1356 0.8867 0.4764 1.1533 AQI 1.256 0.9428 0.5187 0.5871 0.2396 0.5129 0.6025 0.364 6.2185 0.9862 SGI 1.025 1.604 1.6678 1.6337 1.2302 0.7475 0.9258 1.1808 1.2284 0.7537 DEPI 0.7736 0.6827 0.8705 1.6942 1.6927 0.9017 0.9019 0.8682 1 0.9599 SGAI 0.2844 0.8311 1.3729 0.7468 1.4876 1.1383 0.8591 0.7497 0.7614 1.4332 LVGI 1.1497 0.8877 0.8598 0.4385 1.1525 0.6028 0.697 1.1323 1.4522 0.7908 TATA -0.0489 -0.0805 -0.0826 0.0094 -0.0129 -0.0409 -0.0964 -0.0079 0.019 -0.0381 M-score -2.46 -1.43 -3.00 -1.49 -2.74 -2.91 -2.62 -2.57 -0.50 -2.66
Zoltek Companies, Inc. Quarterly Data
Sep11 Dec11 Mar12 Jun12 Sep12 Dec12 Mar13 Jun13 Sep13 Dec13 DSRI 1.1266 1.4533 1.1573 0.9914 0.9634 0.8423 0.7676 0.833 1.1717 1.114 GMI 0.8868 0.627 0.4998 0.4472 0.4763 0.6813 0.8627 1.0049 1.1533 1.2726 AQI 0.364 0.3803 1.949 3.7131 6.2185 9.0988 2.3255 1.0229 0.9862 0.7937 SGI 1.1808 1.2522 1.2247 1.3264 1.2284 1.0561 0.9185 0.7756 0.7537 0.7975 DEPI 0.8682 0.8473 0.8539 0.8538 1 1.0322 0.9607 0.9815 0.9599 0.9543 SGAI 0.7497 0.776 0.8349 0.7558 0.7614 0.8928 1.028 1.2745 1.4332 1.3576 LVGI 1.1323 1.394 1.4183 1.5309 1.4522 1.287 0.9873 0.633 0.7908 0.7691 TATA -0.0079 0.0473 0.0495 0.0327 0.0189 0.0276 -0.0045 -0.0193 -0.0381 -0.0963 M-score -2.57 -2.17 -1.91 -1.39 -0.50 0.59 -2.33 -2.84 -2.66 -2.94
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Next: Kinematic analysis of ADCIGs Up: Biondi: Anisotropic ADCIGs Previous: Phase and group angles
# Angle Gathers by anisotropic downward-continuation migration
In this section I develop the theory for anisotropic ADCIGs from the plane-wave'' viewpoint. I assume that in the proximity of the reflection point the source wavefield and the receiver wavefield are plane waves and I derive the relationships between the propagation angles of these plane waves and the slopes computed in the prestack image. This assumption is not restrictive because the source and receiver wavefields can always be considered as the superposition of plane waves.
In anisotropic media, when the reflector is dipping with respect to the normal to the isotropic axis of symmetry (horizontal direction for VTI) the incident and reflected aperture angles differ. This difference is caused by the fact that, although the phase slowness is function of the propagation angle, Snell law requires that the components parallel to the reflector of the incident and reflected slowness vectors must match at the interface. However, we can still define an average'' aperture angle and average'' dip angle using the following relationships:
(8)
where the and are the phase angles of the incident and reflected plane waves, respectively.
cig-aniso-v3
Figure 1
Sketch representing the reflection of a plane wave from a planar reflector in an anisotropic medium. The angles marked in the figure are all phase angles. They are defined as follows: and are the propagation angles of the incident and reflected plane waves, and are the true aperture angles for the the incident and reflected plane waves, is the true geological dip angle, are the average aperture angle and the average dip angle.
Figure shows the geometric interpretation of these angles. Notice that the average dip angle is different from the true geological dip angle ,and that the average aperture angle is obviously different from the true aperture angles and .However, these five angles are related and, if needed, the true angles can be derived from the average angles, as shown in Appendix A.
Prestack images defined in the subsurface-offset domain are transformed into the angle domain by applying slant stacks. The transformation axis is thus the physical dip of the image along the subsurface offset; that is, .The dip angles can be similarly related to the midpoint dips in the image; that is, .Following the derivation of acoustic isotropic ADCIGs by Sava and Fomel (2003) and of converted-waves ADCIGs by Rosales and Rickett (2001), we can write the following relationships between the propagation angles and the derivative measured from the wavefield:
(9) (10) (11)
where and are the phase slownesses for the source and receiver wavefields, respectively. We obtain the expression for the offset dip by taking the ratio of equation 11 with equation 9, and similarly for the midpoint dips by taking the ratio of equation 10 with equation 9, and after some algebraic manipulations, we obtain the following expressions:
(12) (13)
In contrast with the equivalent relationships valid for isotropic media, these relationships depend on both the aperture angle and the dip angle .The expression for the offset dip (equation 9) simplifies into the known relationship valid in isotropic media when either the difference between the phase slownesses is zero, or the dip angle is zero. In VTI media this happens for flat geological dips. In a general TTI medium this condition is fulfilled when the geological dip is normal to the axis of symmetry.
Solving for and we obtain the following:
(14) (15)
where for convenience I substituted the symbol for the normalized slowness difference'' .
Substituting equation 15 in equation 14, and equation 14 into equation 15, we get the following two quadratic expressions that can be solved to estimate the angles as a function of the dips measured from the image:
(16) (17)
These are two independent quadratic equations in and that can be solved independently. If the normalized slowness difference'' between the slowness along the propagation directions of the source and receiver wavefields are known, we can directly compute and ,and then the true and .One important case in this category is when we image converted waves.
For anisotropic velocities, the slownesses depend on the propagation angles, and thus the normalized difference depends on the unknown and .In practice, these equations can be solved by a simple iterative process that starts by assuming the normalized difference'' to be equal to zero. In all numerical test I conducted this iterative process converges to the correct solution in only a few iterations, and thus is not computationally demanding.
If the anisotropic slowness function were spatially homogeneous, equations 16 and 17 could be solved iteratively in the Fourier domain, and the transformation to the average angles and could be computed exactly without the need of estimating the apparent reflector dip in the space domain. When the anisotropic parameters are a function of the spatial variables; that is, in the majority of the real situations, the solution of equations 16 and 17 requires the estimation of the local reflector dip in the space domain. If necessary, the reflectors' dip can be either extracted from the interpretation of the horizons of interest, or can be automatically estimated from the image by applying one of the several methods that have been presented in the literature (see for example Fomel (2002)). In practice, the estimation of the reflector dip is seldom necessary. The numerical and real-data examples shown below indicate that for practical values of the anisotropy parameters the dependency of the estimate from the dip angles can be safely ignored for small dips, and it is unlikely to constitute a problem for steep dips.
Next: Kinematic analysis of ADCIGs Up: Biondi: Anisotropic ADCIGs Previous: Phase and group angles
Stanford Exploration Project
11/1/2005
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Session - Maple Help
For the best experience, we recommend viewing online help using Google Chrome or Microsoft Edge.
DeepLearning
SessionObject
object for DeepLearning computation session
Description
• A Session is an object which encapsulates the context in which Tensor objects within a DataflowGraph are evaluated.
• There can be more than one Session for the same DataflowGraph, and each Session has its own state and resources.
Creating and Using Sessions
• To obtain the default computation session for the current graph, use the GetDefaultSession command.
• To obtain a new session, use the Session command.
Operations with Sessions
• The following functions can be performed with a Session.
Examples
Here we execute the same simple graph (which performs arithmetic with three Tensors) in the default computation session and in a new session.
> $\mathrm{with}\left(\mathrm{DeepLearning}\right):$
> $\mathrm{t1}≔\mathrm{Variable}\left(\left[1.5,2.0\right],\mathrm{datatype}={\mathrm{float}}_{8}\right)$
${\mathrm{t1}}{≔}\left[\begin{array}{c}{\mathrm{DeepLearning Variable}}\\ {\mathrm{Name: Variable:0}}\\ {\mathrm{Shape: \left[2\right]}}\\ {\mathrm{Data Type: float\left[8\right]}}\end{array}\right]$ (1)
> $\mathrm{t2}≔\mathrm{Variable}\left(\left[2.5,6.0\right],\mathrm{datatype}={\mathrm{float}}_{8}\right)$
${\mathrm{t2}}{≔}\left[\begin{array}{c}{\mathrm{DeepLearning Variable}}\\ {\mathrm{Name: Variable:0}}\\ {\mathrm{Shape: \left[2\right]}}\\ {\mathrm{Data Type: float\left[8\right]}}\end{array}\right]$ (2)
> $\mathrm{t3}≔\mathrm{Variable}\left(\left[7.2,4.2\right],\mathrm{datatype}={\mathrm{float}}_{8}\right)$
${\mathrm{t3}}{≔}\left[\begin{array}{c}{\mathrm{DeepLearning Variable}}\\ {\mathrm{Name: Variable:0}}\\ {\mathrm{Shape: \left[2\right]}}\\ {\mathrm{Data Type: float\left[8\right]}}\end{array}\right]$ (3)
Compatibility
• The DeepLearning[SessionObject] command was introduced in Maple 2018.
• For more information on Maple 2018 changes, see Updates in Maple 2018.
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## Thursday, November 03, 2016
Direction (Q. 1 -5) : Study the following graph carefully to answer the questions that follow: The ratio of imports to exports of two countries in different years
1. The ratio of imports and exports of country A in the year 2008 was what percent more than that for the country B in the year 2005 ?
2. In which of the following year was the value of exports less than the value of imports in the case of country A ?
1) 2006 2) 2012 3) 2011 4) 2005 5) None of these
3. If the imports of country A in 2011 and the exports of country B in 2009 were `15.6 lakhs and `16.2 lakhs respectively, What is the ratio of imports of country B in the year 2009 to exports of country A in 2011 ?
1) 2187 : 5000 2) 13 : 9 3) 9 : 13 4) 637 : 505 5) None of these
4. If the imports of country A in year 2007 and the exports of country B in 2012 were `7.4 lakhs and `5 lakhs respectively, the imports of country B in 2012 was what percent less than the exports of country A in 2007 ?
1) 47 2) 28 3) 32 4) Can't be determined 5) None of these
5. If the exports of country B in 2005 was `35 lakhs, what would be the value of imports in the same year for country A?
1) `210 lakhs 2) `189 lakhs 3) `58 lakhs 4)Can't be determined 5) None of these
Direction (Q. 6 -10) : Study the following pie–chart and answer the questions given below: There are six company which produces two items P and Q: Percent distribution of total production by the six company :
The following table shows the ratio of cost of production between item P and Q and Percent profit earned on the two items :
6. What is the total cost of the production of item P by companies C and E together ? (In `lakhs)
1) 7.5 2) 15 3) 30 4) 12 5) None of these
7. What is the total profit earned by company C on items P and Q together ? (In `lakhs)
1) `3.33 lakhs 2) `1.665 lakhs 3) `3 lakhs 4) Can't be determined 5) None of these
8. What is the ratio of the cost of the production of item P of company A to that of item Q of company B ?
1) 3 : 5 2) 9 : 13 3) 154 : 173 4) Can't be determined 5) None of these
9. Cost of production of item Q of company D is what percent more than the profit earned by company F on item P ?
1) 552.26 2) 996.87 3) 1134.56 4) 87.44 5) None of these
10. What is total profit earned by company B on item Q and the profit earned by company C on item P ? (in `lakhs)
1) 2.48 2) 1.08 3) 4.2 4) 3.56 5) None of these
5.4
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## Cousins in Binary Tree
In a binary tree, the root node is at depth 0, and children of each depth k node are at depth k+1. Two nodes of a binary tree are cousins if they have the same depth, but have different parents. We are given the root of a binary tree with unique values, and the values x and y of two different nodes in the tree. Return true if and only if the nodes corresponding to the values x and y are cousins.
by lek tin in "algorithm" access_time 2-min read
## Check if a String Is a Valid Sequence From Root to Leaves Path in a Binary Tree
Given a binary tree where each path going from the root to any leaf form a valid sequence, check if a given string is a valid sequence in such binary tree. We get the given string from the concatenation of an array of integers arr and the concatenation of all values of the nodes along a path results in a sequence in the given binary tree. Example 1: Input: root = [0,1,0,0,1,0,null,null,1,0,0], arr = [0,1,0,1] Output: true Explanation: The path 0 -> 1 -> 0 -> 1 is a valid sequence (green color in the figure).
by lek tin in "algorithm" access_time 2-min read
## Flatten Binary Tree to Linked List
Given a binary tree, flatten it to a linked list in-place. For example, given the following tree: 1 / \ 2 5 / \ \ 3 4 6 The flattened tree should look like: 1 \ 2 \ 3 \ 4 \ 5 \ 6 Solution # Definition for a binary tree node. # class TreeNode: # def __init__(self, x): # self.val = x # self.left = None # self.
by lek tin in "algorithm" access_time 1-min read
## Delete Nodes and Return Forest
Given the root of a binary tree, each node in the tree has a distinct value. After deleting all nodes with a value in to_delete, we are left with a forest (a disjoint union of trees). Return the roots of the trees in the remaining forest. You may return the result in any order. Example 1 Input: root = [1,2,3,4,5,6,7], to_delete = [3,5] Output: [[1,2,null,4],[6],[7]] Constraints 1 The number of nodes in the given tree is at most 1000.
by lek tin in "algorithm" access_time 1-min read
## Balanced Binary Tree
Given a binary tree, determine if it is height-balanced. For this problem, a height-balanced binary tree is defined as: a binary tree in which the left and right subtrees of every node differ in height by no more than 1. Example 1 Given the following tree [3,9,20,null,null,15,7]: 3 / \ 9 20 / \ 15 7 Return true. Example 2 Given the following tree [1,2,2,3,3,null,null,4,4]: 1 / \ 2 2 / \ 3 3 / \ 4 4 Return false.
by lek tin in "algorithm" access_time 2-min read
## Print Binary Tree
Print a binary tree in an m*n 2D string array following these rules: The row number m should be equal to the height of the given binary tree. The column number n should always be an odd number. The root node’s value (in string format) should be put in the exactly middle of the first row it can be put. The column and the row where the root node belongs will separate the rest space into two parts (left-bottom part and right-bottom part).
by lek tin in "algorithm" access_time 2-min read
## Construct Binary Tree From Preorder and Inorder Traversal
Given preorder and inorder traversal of a tree, construct the binary tree. Note You may assume that duplicates do not exist in the tree. For example, given: preorder = [3,9,20,15,7] inorder = [9,3,15,20,7] Return the following binary tree: 3 / \ 9 20 / \ 15 7 Hint Solution: /** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ class Solution { public TreeNode buildTree(int[] preorder, int[] inorder) { return helper(0, 0, inorder.
by lek tin in "algorithm" access_time 2-min read
## Binary Tree Inorder Traversal
Given a binary tree, return the inorder traversal of its nodes’ values. Example: Input: [1,null,2,3] 1 \ 2 / 3 Output: [1,3,2] Follow-up the Recursive solution is trivial, could you do it iteratively? Solution # Definition for a binary tree node. # class TreeNode: # def __init__(self, x): # self.val = x # self.left = None # self.right = None class Solution: def inorderTraversal(self, root): """ :type root: TreeNode :rtype: List[int] """ """Check root == None to reduce time on checking""" if root == None: return [] stack = [] result = [] current = root while (current!
by lek tin in "algorithm" access_time 1-min read
## Subtree With Maximum Average
Given a binary tree, find the subtree with maximum average. Return the root of the subtree. It’s guaranteed that there is only one subtree with maximum average. Example Given a binary tree: 1 / \ -5 11 / \ / \ 1 2 4 -2 return the node 11. Solution public class Solution { private class ResultType { public int sum, size; public ResultType(int sum, int size) { this.sum = sum; this.
by lek tin in "algorithm" access_time 2-min read
## Validate Subtree
Given two trees, T1 and T2, write an function to test whether T2 is a subtree of T1. Solution public class Subtree { public boolean isSubTree(TreeNode T1, TreeNode T2) { if (T2 == null) return true; if (T1 == null) return false; return (isSameTree(T1,T2) || isSubTree(T1.left, T2) || isSubTree(T1.right, T2)); } public boolean isSameTree(TreeNode T1, TreeNode T2) { if (T1 == null && T2 == null) return true; if (T1 == null || T2 == null) return false; if (T1.
by lek tin in "algorithm" access_time 1-min read
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Your browser does not support iframes.
Java Circuit Simulation Workaround ▾
Even after following the above instructions, loading applets may still show warning concerning “unsigned application” and “unknown publisher”. For Teahlab in particular, these warnings are due to the fact that we have opted not to pay a third party such as Verisign to sign our applets. Any warning that comes up when you try to run our applets should emphasize that our applets will always run with “limited access”, which is Oracle’s way of letting you know that teahlab doesn’t do anything on your computer except running the circuits you see: in other words, our applets are safe to run.
Sincerely,
The Teahlab Team
# MOORE FINITE STATE MACHINE: CONTROL CIRCUIT FOR AUTOMATIC DOOR by Isai Damier (Let's connect on twitter @isaidamier )
#### Problem Description for Control Circuit
Some time ago, I had a fictitious meeting with the representatives of a department store. They were planning to upgrade their chain of stores with automatic double doors that swing open, and so they hired me to build a circuit to control the automatic operation of the doors. They had two specific requirements. First, while opening, a door should not hit shoppers who happen to be standing behind the entrance. Second, when no one is around, the door should stay closed to save on air conditioning cost.
#### Requirement Analysis for Control Circuit
This problem is asking me to replace a doorman with a machine (a circuit to be exact). So instead of trying to come up with some fancy engineering solution, I go to a doorman and ask him to explain his work to me. He took a piece of napkin and a pencil and drew the picture in Figure 1 below.
Figure 1: Doorman Diagram
I asked him to explain the drawing and here is what he said: “ This is no rocket science. A door is either open or closed. So I draw two boxes and label one open and one closed. If the door is closed, I open it only when there are people in front of it but nobody to the REAR of it. So I draw an arrow from closed to open and label it front. In addition, the only time I close a door that's already open is when there is nobody around. So I draw another arrow from open to closed and label it neither. Under all other conditions, if the door is open it will stay open. And if the door is closed, it will stay closed. So I draw the two other arrows and label them with everything else. ”
Clearly this doorman has an eye for details. My work is done. The doorman's schematic is all I need to build the circuit. Nevertheless, a good inventor never reveals his sources. Hence, in my official report I changed what the doorman told me so to sound high-tech. First, I labeled the drawing state transition diagram to make it sound sophisticated (see Figure 2 below). Second, I revised the doorman's explanation and wrote the following.
To read the state transition diagram in Figure 2, imagine the door has two floor pads -- a front pad and a rear pad -- and that the circuit controlling the door is listening to weight on the pads. Therefore, Neither means there is nobody on either pad; Front means there is someone on the front pad only; Rear means there is someone on the rear pad only; and Both means there are people on both pads.
Figure 2: State Transition Diagram
Of course I thanked the doorman for his pains and bought him lunch. I am a nice guy.
#### Designing the Control Circuit
In reality, the state diagram in Figure 1 is the solution to the problem. Once you have a state diagram, the rest is just techniques. Indeed there are many software applications right now that can print a working circuit from a state diagram. Therefore, it's very important to practice drawing state diagrams. Take a pencil and paper and try to draw the doorman's schematic from the description. Do not continue until you get the drawing right. You must think like the doorman. Be the doorman.
From the state diagram, you are four steps away from a working circuit.
1. Translate the state diagram to a table
2. Make the table look like K-maps
3. Get the Boolean expressions from the K-maps
4. Draw the circuit from the Boolean equations.
Step 4 is not really worth mentioning. But I add it for the sake of completeness.
#### Digital System Implementation Procedure
A technique is like a magic trick: It's impressive the first time you see it; but once you get the secret, it becomes a joke. Our first technique is to translate the state diagram in Figure 1 into a table. From the state diagram we see that the door can be in two possible states (open; closed); and movement from state to state is governed by four conditions (neither; front; rear; both). To construct the table, we will label the rows by states (i.e. closed; open). And we will label the columns by conditions (i.e. neither; front; rear; both). Figure 3 below is our empty table.
State Conditions
Neither Front Rear Both
Closed
Open
Figure 3: Empty State Transition Table
Next we fill the table by following the arrows in the state diagram. For instance, the arrow that goes from closed to open is labeled front; which means if the door is closed and the condition is front then the door will go open. So in Figure 4, we write open in the cell where closed and front intersect.
State Conditions
Neither Front Rear Both
Closed Open
Open
Figure 4: State Transition Table with one cell
Figure 5 represents the completed state transition diagram. I add a few more labels to make the table easier to read. You should check Figure 5 against Figure 1 to make sure that the information match exactly.
Given state of door Shopper location
Neither Front Rear Both
What will happen to door (Next State)
Closed Closed Open Closed Closed
Open Closed Open Open Open
Figure 5: State Transition Table
Nearly all designers use this table. As a result, different people call the table by different names: state table, excitation table, flow table, state assigned table, etc. Whatever name you choose to call the table, make sure your audience knows what you are talking about.
Because the table represents the operation of a logic circuit, it makes sense to replace the states with 0s and 1s: Closed = 0; Open = 1. Hence, Figure 5 becomes Figure 6.
Given state of door Shopper location
Neither Front Rear Both
What will happen to door (Next State)
0 0 1 0 0
1 0 1 1 1
Figure 6: Excitation Table
Some designers go further. They call the front pad F, the rear pad R, the given state q, and the next state Q; hence, changing the labels of the table. And they change the pad signals to 0s and 1s so that Neither becomes RF = 00, Front becomes RF = 01; Rear becomes RF = 10; Both becomes RF = 11. Hence, Figure 6 becomes Figure 7.
Given state of door q Shopper Location
RF
00 01 10 11
What will happen to door (Next State Q)
0 0 1 0 0
1 0 1 1 1
Figure 7: State Assigned Table
I told you this table were famous!
Now for the cool trick: if we interchange column 10 with column 11, Figure 7 becomes a K-map! And from the K-map we can get the logic expression for Q, which we do next.
Given state of door q Shopper Location
RF
00 01 11 10
What will happen to door (Next State Q)
0 0
1
0 0
1 0
1
1
1
Figure 8: K-Map
Q = R' • F + q•R.
This is our Boolean function, and Q is the next state of the door.
#### Finite State Machines have Memory
Normally at this point we would say our work is done and implement the circuit directly from the Boolean expression. But if you try to build this circuit, you will find that q does not have a source. We know that signal F is coming from the front pad and that signal R is coming from the rear pad. But where is signal q coming from?
Intuitively, we know that q is whatever Q was previously. In other words, if we run the control circuit five times, then q(t1) = Q(t0), … q(t5) = Q(t4). This means the circuit must find a way to remember the value of Q, so it can use the value as q next time it has to run. To remember past information, digital systems use flipflops as storage units. Hence, we must input the value of Q into a flipflop. And when we need the value to use as q, simply get it from the output of the flipflop.
We create the circuit as below, linking Q to q using a D-flipflop.
The basic logic NOT gate circuits: an interactive inverter logic circuit where you see its function.
Circuit 1 — Play around with the circuit to see how it works
In reality, which flipflop you use depends on which flipflop you have available. I choose a D-flipflop above because it was convenient. Recall how a D-flipflop works: after each clock cycle, the output becomes the input. Since that's exactly what our control circuit needs to do with Q and q, the D-flipflop was an easy choice.
But what if you only have a JK flipflop? Then you would apply the rules of a JK flipflop to the excitation table in Figure 7 in order to get the input equations for J and K. The rules for a JK flipflop are simple, see Figure 9 below:
• If the flipflop is to go from state 0 to state 0, then J=0 and K=d (d means don't care; you choose whatever you want for K, 0 or 1).
• If the flipflop is to go from state 0 to state 1, then J=1 and K=d.
• If the flipflop is to go from state 1 to state 1, then J=d and K=0.
• If the flipflop is to go from state 1 to state 0, then J=d and K=1.
q Q J K
0 0 0 d
0 1 1 d
1 0 d 1
1 1 d 0
Figure 9: JK Flipflop Characteristic Table
Figure 10 is a typical way of changing an excitation table to accommodate the JK flipflop. We basically create an extra column for JK under each RF value and then evaluate JK using the characteristic table in Figure 9.
Present
State
Flipflop Inputs
RF = 00 RF = 01 RF = 11 RF = 10
q Q JK Q JK Q JK Q JK
0 0 0d 1 1d 0 0d 0 0d
1 0 d1 1 d0 1 d0 1 d0
Figure 10: Excitation Table with JK Flipflops
To get the Boolean expression for J, we assemble all the J values in one table, as in Figure 11 below. And since the table is already in K-map form, we simply extract the expression: so that J = R' • F.
q RF = 00 RF = 01 RF = 11 RF = 10
0 0
1
0 0
1 d
d
d d
Figure 11: Karnaugh map for J input of flip flop
Doing similarly for K in Figure 12, we get K = R' • F'.
q RF = 00 RF = 01 RF = 11 RF = 10
0
d
d d d
1
1
0 0 0
Figure 12: Karnaugh map for K input of flip flop
We show the JK version of the circuit below. Notice how using a JK flipflop obviates the need to get feedback data from the state variable q. That often happens; sometimes using the right flipflop can reduce the complexity of a circuit.
Circuit 2 — Play around with the circuit to see how it works
#### Epilogue
While it is always possible to build sequential systems that do not depend on the passage of time (i.e. asynchronous systems), it is often easier to build machines that do indeed depend on some representation of time. We live in a temporal universe. For us time is always continual. Hence, in both the D-flipflop and the JK flipflop versions of the control circuit, we use a time signal to regulate the circuit. In reality the source of the time (CLK) signal can be an actual clock or any persistent system. We do not account for the CLK signal in the Boolean expressions because the CLK signal comes standard with the flipflops.
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Find an Affinity Problem
I - Equivalent (Alternative) Solutions:Is there more than one solution to a problem? Are there "equivalent algorithms" for solving a particular problem? What is the difference between an "equivalent solution" and "equivalent algorithm"?
Two algorithms are equivalent when they describe the SAME methodology, whereas equivalent solutions are based on getting the same result irrespective of the method.
Example 1: Sorting - there are many methods of sorting each of which provides the same answer for a given set of inputs, but some are better than others! Affinity solutions are important in exploring for a most efficient method.
Example 2: The first, obvious, solution to a problem may not always be the easiest or best solution. Even famous people such as Albert Einstein or John von Neumann did not always find the easiest alternative solutions to given problems. Review von Neumann's solution to the "Fly and Trains" problem.
More Examples: I need to join two pieces of wood together: Equivalent solutions: Glue Nails Screws I need to get from New York to Washington DC: Equivalent solutions: Drive a car Catch a train Fly a plane
II - Equivalent Problems: Can a problem in one area be equivalent to an already solved problem in another area?
The Human Genome problem may have solutions in Syntactic Analysis or Code Breaking The "Four color problem": [Theorem] If a plane is divided into connected regions that are to be colored so that no two adjacent regions have the same color, it is never necessary to use more than four colors. Solutions to Jigsaw problems are also found in Syntactic Analysis
Last updated 2002/05/31
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# Question: What Is The Sum Of Any Two Old Number?
## What is the sum of any two odd number?
The sum of two odd numbers is always even.
The product of two or more odd numbers is always odd..
## What is the sum of first n even number?
Sum of first n even numbers = n * (n + 1).
## How do you find the sum of two odd numbers even?
The sum of two odd integers is even. Proof: If m and n are odd integers then there exists integers a,b such that m = 2a+1 and n = 2b+1. m + n = 2a+1+2b+1 = 2(a+b+1).
## What is the sum of any two number?
The result of adding two or more numbers. (because 2 + 4 + 3 = 9).
## What is the sum of three odd numbers?
Then the sum of three odd numbers can be written as: (2m−1)+(2n−1)+(2k−1) =2(m+n+k−1)−1. which, as we know is an odd number. Therefore the sum of three odd numbers is always an odd number.
## What is the odd number?
Any integer (not a fraction) that cannot be divided exactly by 2. The last digit is 1, 3, 5, 7 or 9. Example: −3, 1, 7 and 35 are all odd numbers.
## What is the sum of first 100 odd numbers?
Thus, the sum of the first 100 odd integers will be equal to 100^2 = 10,000.
## What is the sum of first 100 even numbers?
The number series 2, 4, 6, 8, 10, 12, . . . . , 200. Therefore, 10100 is the sum of first 100 even numbers.
## How do you find the sum of even numbers?
The sum of even natural numbers between 1 and 2n can be written as the sum of 2*i from i=1 to i=n. Factoring out the 2, it is equal to 2 times the sum of i from i=1 to n. Using the formula for the sum of consecutive numbers, we get that the sum of all even numbers from 1 to 2n equals n*(n+1).
## What is the sum of two odd numbers and one even number?
(b) The sum of two odd numbers and one even number is even. This is true. We know that sum of two odd numbers is always even. Adding one more even number will keep the result an even number.
## What is the smallest prime number?
A prime number is a whole number greater than 1 that can only be divided by itself and 1. The smallest prime numbers are 2, 3, 5, 7, 11, 13, 17, 19 and 23. The number 2 is the only even prime number.
## What is the smallest even number?
2Even numbers always end up with the last digit as 0, 2, 4, 6 or 8. Some examples of even numbers are 2, 4, 6, 8, 10, 12, 14, 16. These are even numbers as these numbers can easily be divided by 2. It should be noted that the smallest positive even natural number is 2.
## What is the sum of first 20 odd numbers?
The number series 1, 3, 5, 7, 9, . . . . , 39. Therefore, 400 is the sum of first 20 odd numbers.
## Is sum of two prime number is always even?
2, 3, 5, 7, 11, 13, 17, ….. To find, the sum of two prime numbers is always even = ? Every prime number is an odd number except 2, 2 is the even number. Hence, the sum of two prime numbers is “not always even”.
## What is the sum of even number?
Using the sum of n consecutive integers formula, we can calculate the sum of 1 through 50: 50 x 51/2 = 25 x 51 = 1275. And we multiply that by 2 (the number we originally factored out). 1275 x 2 = 2550. This approach will work for all even numbered series.
## What is the sum of odd numbers?
The total of any set of sequential odd numbers beginning with 1 is always equal to the square of the number of digits, added together. If 1,3,5,7,9,11,…, (2n-1) are the odd numbers, then; Sum of first odd number = 1. Sum of first two odd numbers = 1 + 3 = 4 (4 = 2 x 2).
## What is the sum of first 25 even numbers?
The number series 2, 4, 6, 8, 10, 12, . . . . , 50. Therefore, 650 is the sum of first 25 even numbers.
## Is 2 an odd number?
Any integer that cannot be divided exactly by 2 is an odd number. Odd numbers are in between the even numbers.
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Solving House Robber III in go. Please try yourself first to solve the problem and submit your implementation to LeetCode before looking into solution.
## Problem Description
The thief has found himself a new place for his thievery again. There is only one entrance to this area, called the "root." Besides the root, each house has one and only one parent house. After a tour, the smart thief realized that "all houses in this place forms a binary tree". It will automatically contact the police if two directly-linked houses were broken into on the same night.
Determine the maximum amount of money the thief can rob tonight without alerting the police.
Example 1:
``````Input: [3,2,3,null,3,null,1]
3
/ \
2 3
\ \
3 1
Output: 7
Explanation: Maximum amount of money the thief can rob = 3 + 3 + 1 = 7.``````
Example 2:
``````Input: [3,4,5,1,3,null,1]
3
/ \
4 5
/ \ \
1 3 1
Output: 9
Explanation: Maximum amount of money the thief can rob = 4 + 5 = 9.``````
See the full details of the problem House Robber III at LeetCode
Originally posted at: @github.com/halfrost/LeetCode-Go
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Math 241A Fall 2013 HOME WORK
TEXT: Conway "Functional Analysis" Edition 2
Please turn in ths HW the beginning of section ??
Please tell me what you think of the Homework. I NEED FEEDBACK!!
Lecture 1. Defs of Banach space and Banach Algebra.
Examples: C[0,1] and L^p
Ch 1 Hilbert Space.
p6 sec 1 ex 3 Show that H is a preHilbert space. You do not need to prove it is complete.
Chapter 1 p6 sec 1 ex 6 .
p11 sec 2 ex 1,2, 3
p13 sec 3 ex 5
p18 sec 4 ex 13, 19
p23 sec 5 ex 2, 3
Chapter 2
Operators on Hilbert space
sec 1 Great Examples of Operators ex 2
Section 2, Adjoints ex. 6, 11, 16
HW 1 Turn in the above on ??
Chapter 2
Section 1, ex. 1, 3, 5, 8
Section 3 Projections , ex. 6, 11
Section 4 Compact Operators, ex. 1, 2, 4, 5
Section 5, Diagonalizing Compact SelfAdj Operators, ex. 1
HW 2 Turn the above in on ??
Chapter 3 Banach Spaces
Bill already lectured on sec 1, 2 in the early part of the course while discussing Hilbert Space.
Section 1, ex. 3, 13
Section 2, ex. 5, 6
Section 3, ex. 1
Section 4, Quotient Spaces ex. 1, 6
Section 5, Linear Functionals , ex. 2
HW 3 Turn the above in on ??
Optional in 2013
Section 6, HB Theorem , READ IT
Section 7, Banach Limits, p. 83 SKIPPED
Section 9, Ordered Vector Spaces, p.88, ex. 4, 6, 7, 8, 9
Order1. As defined in the Banach Limit section 3.7 take
V to be l^\infty and S = sequences which posess a limit C= all nonnegative sequences in V.
What are (all of) the order units in S?
Ch 4 Section 3 Separating Hyperplanes Ex 2, 7 and Ex 10 with TVS replaced by Banach Space.
End Optional
Spectral Theory
If you have not done so yet do this:
S1. Suppose $M$ is multiplication by x on L^2[0,1, u] where u is a probability measure supported on [0,1]. For which such measures u is M a compact operator?
S2. Suppose A1 A2 A3 are commuting self adjoint operators on a seperable Hilbert space H.
Suppose the spectrum of Aj is contained in [-2,2] for each j.
Let R:= [-2,2]^3
The following is true and you may assume it:
If p(x) is positive on R, then p(A) is PsD. Here x =(x1,x2,x3).
PROBLEMs on cyclic vectors:
C0. Say exactly which symmetric matrices have a cyclic vector
C1. What is a natural notion of cyclic vector for A1 A2 A3
C2. Assuming that they have a cyclic vector, what is a natural spectral representation for A1 , A2, A3?
C3. Prove it
HW 4 Discuss the spectral theory problems above with the Prof in Week 10 of the quarter.
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Category
Forum
# How to Use A Looping to Split the Data In Matlab?
In MATLAB, you can use a loop to split data by creating a loop that iterates through your data and splits it based on a specific condition or delimiter. You can use a for loop or a while loop to iterate through your data, checking for the condition or delimiter that indicates where to split the data. Once the condition is met, you can use the split function or indexing to separate the data into separate arrays or variables. This can be useful for processing large datasets or working with data that needs to be split into different sections for analysis or manipulation. By using a loop to split the data, you can automate the process and make it more efficient.
## Best MATLAB Books to Read in 2024
1
Rating is 5 out of 5
MATLAB and Simulink Crash Course for Engineers
2
Rating is 4.9 out of 5
MATLAB for Engineers
3
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MATLAB: A Practical Introduction to Programming and Problem Solving
4
Rating is 4.7 out of 5
MATLAB For Dummies (For Dummies (Computer/Tech))
5
Rating is 4.6 out of 5
MATLAB: A Practical Introduction to Programming and Problem Solving
6
Rating is 4.5 out of 5
MATLAB and Simulink In-Depth: Model-based Design with Simulink and Stateflow, User Interface, Scripting, Simulation, Visualization and Debugging
7
Rating is 4.4 out of 5
Radar Systems Analysis and Design Using MATLAB
## What is the difference between a while loop and a for loop in MATLAB?
In MATLAB, both while loops and for loops are used to repeat a block of code multiple times. The main difference between them is in how they are used and configured:
1. While loop:
• A while loop will continue to execute a block of code as long as a specified condition is true.
• The condition is checked at the beginning of each iteration before the code block is executed.
• The loop will continue to run until the condition evaluates to false.
• While loops are typically used when the number of iterations is not known beforehand or when the number of iterations is dynamic based on some condition.
Example:
```1 2 3 4 5 ``` ```counter = 1; while counter <= 10 disp(counter); counter = counter + 1; end ```
1. For loop:
• A for loop will execute a block of code a specific number of times, usually based on the number of elements in an array or a specified range.
• The loop will iterate over a predetermined range of values and execute the block of code for each iteration.
• For loops are typically used when the number of iterations is known beforehand and the loop needs to iterate over a range of values.
Example:
```1 2 3 ``` ```for i = 1:10 disp(i); end ```
In summary, while loops are used for executing a block of code repeatedly based on a condition, while for loops are used when the number of iterations is known and fixed.
## How to break out of a loop in MATLAB?
To break out of a loop in MATLAB, you can use the `break` statement. The `break` statement allows you to exit a loop prematurely, without waiting for the loop condition to become false.
For example, in a `for` loop, you can use the `break` statement to exit the loop when a certain condition is met:
```1 2 3 4 5 6 ``` ```for i = 1:10 if i == 5 break; end disp(i); end ```
In this example, the loop will break when `i` is equal to 5, and the value of `i` will not be displayed.
You can also use the `break` statement in a `while` loop to exit the loop when a certain condition is met:
```1 2 3 4 5 6 7 8 ``` ```i = 1; while i <= 10 if i == 5 break; end disp(i); i = i + 1; end ```
In this example, the loop will break when `i` is equal to 5, and the value of `i` will not be displayed.
## How to use a loop to iterate over a cell array in MATLAB?
To iterate over a cell array in MATLAB, you can use a `for` loop or a `while` loop. Here is an example using a `for` loop:
```1 2 3 4 5 6 7 ``` ```% Create a cell array cellArray = {1, 'two', [3 4]}; % Iterate over the cell array using a for loop for i = 1:length(cellArray) disp(cellArray{i}); end ```
In this example, we create a cell array `cellArray` with three elements. We then use a `for` loop to iterate over each element of the cell array and display its contents using `disp()`.
You can also use nested loops to iterate over each element of a nested cell array, as shown in the example below:
``` 1 2 3 4 5 6 7 8 9 10 11 12 13 ``` ```% Create a nested cell array nestedCellArray = {1, {'two', 'three'}, {[4 5], [6 7]}}; % Iterate over the nested cell array using nested for loops for i = 1:length(nestedCellArray) if iscell(nestedCellArray{i}) for j = 1:length(nestedCellArray{i}) disp(nestedCellArray{i}{j}); end else disp(nestedCellArray{i}); end end ```
In this example, we create a nested cell array `nestedCellArray` with three elements, where the second and third elements are themselves cell arrays. We use nested `for` loops to iterate over each element of the nested cell arrays and display their contents.
## How to increment a loop variable in MATLAB?
In MATLAB, you can increment a loop variable by simply adding 1 to it within the loop. Here is an example:
```1 2 3 4 ``` ```for i = 1:10 disp(i); i = i + 1; % incrementing i by 1 end ```
In this example, the loop variable `i` starts at 1 and is incremented by 1 in each iteration of the loop. The output will be:
``` 1 2 3 4 5 6 7 8 9 10 ``` ```1 2 3 4 5 6 7 8 9 10 ```
## What is the significance of loop condition in MATLAB?
The loop condition in MATLAB is significant because it determines when the loop will stop executing. A loop condition is typically a logical expression that is evaluated at the start of each iteration of the loop. If the condition is true, the loop continues to execute, but if the condition is false, the loop stops executing and control passes to the next section of code.
The loop condition is crucial for controlling the repetition of a certain block of code. It allows the programmer to specify the number of times the loop should run or to define certain criteria under which the loop should stop running. Without a loop condition, the loop would continue indefinitely, leading to an infinite loop.
Additionally, using a loop condition can make the code more efficient by preventing unnecessary repetitions of the loop code. It allows the programmer to create flexible and dynamic loops that can adapt to changing conditions or data inputs.
## What is the role of loop iteration order in MATLAB?
The order of loop iteration in MATLAB determines the sequence in which the statements inside the loop are executed. By controlling the order of loop iteration, you can achieve different outcomes in your program. The iteration order can affect the speed, accuracy, and final result of your code, so it is important to consider it when designing your algorithm.
In MATLAB, the default loop iteration order is determined by the data structure being iterated over. For example, when iterating over a vector or matrix, the default order is column-wise iteration. However, you can customize the iteration order by using different loop constructs such as "for" loops, "while" loops, and nested loops.
By carefully choosing the iteration order, you can optimize the performance of your code and make it more readable and maintainable. It is important to understand the implications of loop iteration order in MATLAB in order to write efficient and effective code.
## Related Posts:
To interface MATLAB with other programming languages such as C/C++ and Python, you can use MATLAB's built-in features and tools. One common method is to use MATLAB's MEX functions, which allow you to create functions in C or C++ that can be called from...
To create and deploy standalone MATLAB applications, you can use MATLAB's Application Compiler tool. This tool allows you to package your MATLAB code and functions into a standalone executable that can be run on computers without MATLAB installed.To create...
To load struct fields from MATLAB to C++, you can use the MATLAB Engine API. This API allows you to exchange data between MATLAB and C++ code seamlessly.First, you need to initialize the MATLAB Engine in your C++ code. Then, you can use the mxArray data type t...
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# Mechanical Properties of Fluids - Online Test
Q1. The basic property of a fluid that makes it different from solids
Explaination / Solution:
fluids can flow due to unbalanced forces between the atoms of fluids.
Q2. Density is defined as
Explaination / Solution:
Density is defined as the compactness of substance.
Mathematically,
Density(D)=
Q3. The specific gravity of a material is
Explaination / Solution:
The specific gravity of an object is the ratio between the density of an object to a reference liquid. Usually, this reference liquid is water, which has a density of 1 g/mL or 1 g/cm3.
Water has a specific gravity equal to 1. Materials with a specific gravity less than 1 are less dense than water, and will float on the pure liquid; substances with a specific gravity more than 1 are more dense than water, and will sink.
Q4. Pressure p at any point in a fluid at rest is
Explaination / Solution:
It is important to note that it is valid only for a fluid at rest. In the case of a moving fluid, pressures in different directions could be different depending upon fluid accelerations in different directions.
Q5. According to Pascal’s Law the
Explaination / Solution:
According to Pascal's Law,
P-P= hdg
from above
Change in pressure is directly proportional to depth from the free surface.
At the same horizonatal line all point are at the same depth and have same value of acceleration due to gravity and denity of water as well.
Q6.
The pressure at a depth of h in a fluid of density at a place where the acceleration due to gravity is g and the pressure at h=0 is is given by
Explaination / Solution:
According to Pascal's Law,
Change in Pressure (P-P0) = hg
if h=0 then P-P0 = 0
and P=P0
Q7. The units of pressure in SI system is
Explaination / Solution:
The SI unit for pressure is the pascal (Pa), equal to one newton per square metre (N/m2, or kg. m. s−2). This name for the unit was added in 1971; before that, pressure in SI was expressed simply in newtons per square metre
Q8. Gauge pressure at a point is
Explaination / Solution:
Gauge pressure is zero-referenced against ambient air pressure, so it is equal to absolute pressure minus atmospheric pressure. Negative signs are usually omitted
The difference between absolute pressure and atmospheric pressure is what we call gauge pressure (). It can be calculated if we know the absolute and atmospheric pressures using this formula:
Q9. According to Pascal’s law for transmission of fluid pressure external pressure applied on any part of a fluid contained in a vessel is ___________________ in all directions
Explaination / Solution:
Pascal's principle is defined as a change in pressure at any point in an enclosed fluid at rest is transmitted undiminished to all points in the fluid.
This principle is stated mathematically as:
is the hydrostatic pressure (given in pascals in the SI system), or the difference in pressure at two points within a fluid column, due to the weight of the fluid.
Q10. In a hydraulic lift the force applied on the smaller cylinder of area A1 is F1. If the area of the larger cylinder is Athe maximum weight that can be lifted is
Explaination / Solution:
According to Pascal's Law,
Pressure applied to any point inside the liquid is trnasmiteed equally in all direction so,
Pressure applied on the smaller cylinder is equal to the pressure on the other cylinder,which is given by
So,
Maximum force on the other side is ,
F2 =
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# hum/176
In this week's video, Bill Goodykoontz discusses the effects of violence on audiences and gamers. How much responsibility do filmmakers, game manufacturers, etc. bear when it comes to depicting violence? How much responsibility do parents have in allowing children exposure to these sources?
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1. What kind of HELP do you need? You need to be specific when asking questions here.
If all you do is post your entire assignment, with no evidence of thinking on your part, nothing will happen since no one here will do your work for you. But if you are specific about what you don't understand about the assignment or exactly what help you need, someone might be able to assist you.
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One question I get asked a lot is: What is the most exciting new statistical feature in the 14.1 release? And they get a bit frustrated when I say: It depends.
But it does depend! SAS statistical software provides a broad array of capabilities that help users track disease outbreaks, predict cell phone plan choices, design clinical trials, improve health care utilization, plan agricultural experiments, create more effective web sites, and determine insurance premiums. And those are just a few examples. Which new feature makes your day depends on your particular area of statistical practice.
So, let me answer the question this way:
If I were a sample survey researcher, I would be excited about new techniques for dealing with missing data from surveys!
Nonresponse in surveys is a big problem. Even with government surveys conducted in-person by researchers, there are question that respondents don’t want to answer, such as topics like income or diet. Missing data matters because the results of the analysis may be biased if non-respondents are different from respondents, and the results may also be less precise.
One way to deal with missing data is to impute them — in other words, replace them with observed values from the same question. And in SAS/STAT 14.1 this is what the new SURVEYIMPUTE procedure does. Once you produce a data set with the imputed values, you can analyze it with the usual survey data analysis procedures, incorporating weights that account for the imputation.
Imputation methods in PROC SURVEYIMPUTE include single and multiple hot-deck imputation and fully efficient fractional imputation (FEFI). Donor selection techniques include simple random selection with or without replacement, probability proportional to weights selection, and approximate Bayesian bootstrap selection.
If I were a data scientist, I would be thrilled with new methods for fitting generalized additive models!
These models are useful because you can include covariate effects that enter the model in a complex, nonlinear fashion, without specifying a parametric form as in standard regression. The new GAMPL procedure, which relies on penalized likelihood estimation, is appropriate for problems that can have hundreds or even thousands of covariates. The GAMPL procedure is also a high-performance procedure that can operate in a distributed computing environment when you have a SAS High Performance Statistics license.
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I would be excited to see the graphical displays that HPSPLIT now produces for classification and regression trees, now specified with a familiar modeling syntax. These techniques, which have been a mainstay of the data mining and machine learning communities, are well suited for complex ecological data.
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# Brute-force string generator
Posted on
Problem
I have created a brute-force algorithm class in C# and I was wondering if I could speed up the process of creating a new string. There are two methods inside the class: one returns the brute-force string and one returns the number of strings created.
``````public class BruteForce
{
public char[] CharList = new char[] { 'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z', 'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R', 'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z', '0', '1', '2', '3', '4', '5', '6', '7', '8', '9' };
public List<int> IntList = new List<int> { 0 };
public string GenerateString()
{
int Number = IntList.Count - 1;
if (IntPasswordsGenerated == 1) return CharList[0].ToString();
do
{
IntList[Number]++;
if (IntList[Number] == CharList.Length && Number == 0)
{
IntList[Number] = 0;
break;
}
else if (IntList[Number] == CharList.Length)
{
IntList[Number] = 0;
Number--;
continue;
}
else
{
break;
}
} while (true);
string BruteForceString = "";
foreach (int CurrentInt in IntList)
BruteForceString += CharList[CurrentInt];
return BruteForceString;
}
{
}
}
``````
Solution
What you have implemented is a back-to-basics number-system in base 62 Your `IntList` List is a mechanism of having the equivalent of “units”, “tens”, “hundreds”, except you are in base 62 so it’s “units”, “sixty-twos”, …..
When you increment a value, and it overflows, you then set it back to zero, and increment the next column instead.
In addition, you have a special case for the first time around, where you have to bypass the logic to ensure you have the right initial value returned.
So, there’s a trick for the first time around that’s easy to implement. The trick is to think of the array as holding the “next” value to return. Your code then becomes:
`````` string BruteForceString = "";
foreach (int CurrentInt in IntList)
BruteForceString += CharList[CurrentInt];
// Now calculate the **next** password
......
return BruteForceString;
``````
Note how your initial array already contains the “next” password when initialized. Then, you can convert that to a string, and then increment the array to be ready for the next call.
But, I want to suggest that you are doing it all horribly wrong… 😉
What you should be doing is using a simple `long` value to store your next value, and then using integer division and modulo to work your base-62 system.
Additionally, your variable names are quite horrible…. but that’s OK, we will get rid of them all except the worst….
``````private ulong IntPasswordsGenerated = 0;
public string GenerateString()
{
ulong base = CharList.Length;
// get the current value, increment the next.
do
{
current /= base;
} while (current != 0);
}
``````
The magic is almost entirely in this line here:
`````` ulong current = IntPasswordsGenerated++;
``````
hat takes a copy of the “next” password, and then afterwards increments what the next password will be.
Using division and modulo is likely a lot faster than the multiple loops required to check and validate the multiple array elements in your custom number system.
Finally, by taking a copy of the next password, and by keeping the password as a single `ulong` instance variable, it becomes relatively simple to lock on a small section of code and then make your method able to run in multiple threads… allowing you to brute force more passwords at once though parallel processing. All you need to do that is drop the `ulong` to a long, and then use the `Interlock.increment()` method to do a thread-safe, atomic increment (assuming you have a 64-bit machine).
You are always writing the type of variables inside the variables’ names, for example: `IntPasswordsGenerated` Remove them as the compiler already takes care of types for You.
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http://mathhelpforum.com/differential-equations/97326-solved-re-writing-ode.html
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1. ## [SOLVED] Re-writing ODE
im stuck on a problem. Show that the ODE $\displaystyle dy/dx=\frac{2x+e^y}{1+x^2}$, can be written in the form $\displaystyle (1+2xe^{-y})dx-(1+x^2)e^{-y}dy=0$
Im stuck and i cant figure out what to do or what method to use.
A nudge in the right direction would be good
EDIT: sorry about that, i fixed it up.
Sorry fixed the typo.. :S
2. Hello,
Is it $\displaystyle \frac{2x+e^y}{1+x}$ ?
Is it $\displaystyle 2x+\frac{e^y}{1+x}$ ?
In either case are you sure there's not a typo somewhere ? Because I really can't see where 1+x² comes from :s
3. fixed.
4. That's better
$\displaystyle \frac{dy}{dx}=\frac{2x+e^y}{1+x^2} \Rightarrow (1+x^2) dy=(2x+e^y) dx$
Do you agree ? (that's just cross multiplying)
Then you may ask... "How did they get there ?"
Notice that there are $\displaystyle e^{-y}$ and where there was $\displaystyle e^y$, there's 1.
So what you have to think is "I may have to divide the whole equation by $\displaystyle e^y$"
See what this gives :
$\displaystyle \frac{1+x^2}{e^y} \cdot dy=\frac{2x+e^y}{e^y} \cdot dx$
(1/e^y = e^(-y))
$\displaystyle e^{-y} (1+x^2) dy=(2xe^{-y}+1) dx$
Looks better to you ?
5. wow thanks a tonne! yea i did the whole cross multiplying thing already, i just got confused when i saw $\displaystyle {e^-y}$ and that 1.
thanks for that ill take it from there, ill let you know how i go!
6. yea thats the nudge i needed. thanks again, i didnt have to really do anything after that, aside from from minus one side to the other.
thanks again!
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Accessibility statement
# Generalised Linear Models - MAT00103H
« Back to module search
• Department: Mathematics
• Credit value: 20 credits
• Credit level: H
• Academic year of delivery: 2023-24
• See module specification for other years: 2024-25
## Module summary
This module introduces students to setting up models, their estimation and comparisons for diverse response and explanatory variables, departing from the traditional assumptions of normality and linearity.
• None
## Module will run
Occurrence Teaching period
A Semester 1 2023-24
## Module aims
This module introduces students to setting up models, their estimation and comparisons for diverse response and explanatory variables, departing from the traditional assumptions of normality and linearity.
## Module learning outcomes
By the end of the module, students will be able to:
1. Demonstrate the unifying role of exponential families when studying the association between response and explanatory variables measured in diverse scales.
2. Perform maximum likelihood based inference for GLMs, including in the context of logistic and Poisson regression.
3. Provide descriptive statistics and graphical summaries of information contained in data from survival experiments in different types of studies.
4. Use estimation and hypothesis testing for inference for survival data.
5. Use the statistical programme R to perform data analysis in the GLM and survival analysis contexts.
## Module content
Data that motivates this module may arise from a variety of fields from medicine to insurance and engineering, each with their characteristic challenges departing from the standard modelling tools. This module will extend your modelling skills from the limited umbrella of linear models to now encompass response data that may be following distributions (such as Binomial, Poisson, Gamma) which are part of the exponential family of distributions and may be non-linearly connected to a set of covariates. This module will also introduce you to modelling survival data, where observations are measured as time to event and often subjected to censoring, e.g. event times are not known completely.
## Indicative assessment
Closed/in-person Exam (Centrally scheduled) 100
None
### Indicative reassessment
Closed/in-person Exam (Centrally scheduled) 100
## Module feedback
Current Department policy on feedback is available in the student handbook. Coursework and examinations will be marked and returned in accordance with this policy.
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MATLAB Answers
# Fill NaN cell with mean of eight surrounding cells in grid data in matlab
24 views (last 30 days)
Shakir Hussain on 21 Apr 2018
Edited: Jan on 25 Apr 2018
I am trying to fill the NaN values of grid data with 8 surrounding values but could not understand, what is going wrong here in my matlab code. The data is 752*891*11 %%% (11 years precipitation of 752*891 cells).
for i = 2 : 752
for j = 2 : 891
for k = 1 : 11
if isnan(data(i,j,k)) == 1
data(i,j,k) = nanmean(nanmean(data(i-2:i+2,j-2:j+2,k-2:k+2)));
end
end
end
end
Thanks in advance for help
#### 0 Comments
Sign in to comment.
### Accepted Answer
Jan on 21 Apr 2018
Edited: Jan on 25 Apr 2018
If you want to replace a scalar based on a 3D array, you either need 3 nanmean calls:
```nanmean(nanmean(nanmean(data(i-1:i+1, j-1:j+1, k-1:k+1))))
```
with i-1:i+1, instead of i-2:i+2. This would be easier:
```if isnan(data(i,j,k)) % "== 1" is not needed
tmp = data(i-1:i+1, j-1:j+1, k-1:k+1); % 3D block
data(i,j,k) = nanmean(tmp(:)); % Make it a vector
end
```
But this is a 3x3x3 neighborhood with 27 elements, not 8. I assume you mean:
```if isnan(data(i,j,k)) % "== 1" is not needed
tmp1 = data(i-1:i+1, j, k);
tmp2 = data(i:i, j-1:j+1, k);
tmp3 = data(i, j, k-1:k+1);
tmp = [tmp1(:); tmp2(:); tmp3(:)]; % Create a vector
data(i,j,k) = nanmean(tmp);
end
```
Because the center point data(i,j,k) is included 3 times, but ignored by nanmean, you have 6 neighbors now, not 8.
So please explain again, what you exactly want.
#### 7 Comments
Show 4 older comments
Shakir Hussain on 24 Apr 2018
The above four preference is for single year and wants to repeat the same way for 11 years. We have both positive and negative values of temperature and your above given code is pop up with (Subscript indices must either be real positive integers or logicals.) error.
Jan on 24 Apr 2018
When you access the index i-2:i+2, the loop for i must start at 3, not at 2, because 0 is not a valid index. And the loop must stop at size(data, 1) - 2. Equivalently for j.
Please post a copy of the complete message, if you mention an error in the forum.
Shakir Hussain on 25 Apr 2018
Thank you jan The problem has solved
Sign in to comment.
### More Answers (1)
Walter Roberson on 22 Apr 2018
```means = conv2(YourMatrix, [1 1 1;1 0 1;1 1 1]/8,'same') ;
mask = isnan(YourMatrix);
YourMatrix(mask) = means(mask);
```
No loops needed.
Note: you would need a little adjustment to handle a nan on the edge of the matrix, to calculate the means properly.
#### 4 Comments
Show 1 older comment
Walter Roberson on 22 Apr 2018
Ah, good point, the 0 * nan at the point itself makes the conv2() result into nan.
Jan on 22 Apr 2018
A solution to use conv2 with NaNs:
```nanX = isnan(X);
X(nanX) = 0;
mask = [1 1 1; 1 0 1; 1 1 1];
means = conv2(X, mask, 'same') ./ ...
conv2(~nanX, mask, 'same');
X(nanX) = means(nanX);
```
But I'm not sure, how this can be applied to the OP's problem. A neighborhood of 6 or 10(?) elements is wanted.
Image Analyst on 23 Apr 2018
Just change the mask shape and elements to be whatever is wanted.
Sign in to comment.
Sign in to answer this question.
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## CHAPTER OBJECTIVES
• Describe statistical tests used to evaluate differences among groups
• Define statistically dependent (i.e., dependent or paired observations) samples and state how these affect the choice of a statistical test
• Explain how the scale of measurement for the dependent variable influences the choice of a statistical test
• Define nonparametric statistical methods and identify common nonparametric tests
• Identify statistical tests to evaluate relationships between two continuous variables and between two ordinal variables
## KEY TERMINOLOGY
• Analysis of variance (ANOVA)
• Bivariate analyses
• Chi-square test of homogeneity
• Independent groups t-test
• Kruskal-Wallis test
• Negative relationship
• Nonparametric methods
• Paired t-test
• Parametric methods
• Pearson correlation coefficient
• Positive relationship
• Scatterplot
• Sign test
• Spearman rank correlation coefficient
• Wilcoxon-Mann-Whitney test
## INTRODUCTION
One of the most common research designs encountered in biomedicine involves the comparison of outcomes in two groups. Typically, one group of patients will receive an experimental treatment and a second group receives a comparison treatment that may be a placebo, a second type of treatment, or usual care. Within pharmacy, the classic two-group study is a study comparing a drug treatment to placebo. The outcomes from each group are compared using a statistical test and conclusions are made regarding the efficacy of treatment based on the results of the statistical test.
The purpose of this chapter is to describe the most commonly used bivariate analyses,a including those used to compare groups, and to illustrate these techniques using small datasets. The chapter begins by describing the process of statistically comparing averages (or means) and then proportions between two independent groups. The situation of nonindependent groups, such as when a group of patients is measured at two different time points, is then discussed. Comparisons of the rank order of responses in two or more groups are also described, as is the calculation of the correlation between two variables. These statistical tests are described in the context of a case scenario. The chapter ends by reviewing the use and assumptions of the described statistical tests for comparing groups as shown with a flowchart.
aBivariate analysis is a term used to denote an analysis with just two variables. Some use the term univariate analysis to describe analysis of a single dependent variable, even though there may be multiple independent variables. Using this definition, some of the techniques discussed in this chapter, such as ANOVA and the t-test, may also be considered univariate techniques.
## CASE SCENARIO
Consider the situation of a pharmacist-run community health center where services are provided to a substantial number of patients with Type II diabetes. Community health center pharmacists have worked with the local YWCA to provide exercise classes and make gym facilities available to their patients. They know that exercise should have a positive effect on patients with diabetes; that is, exercise increases the impact ...
### Pop-up div Successfully Displayed
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Show that OD is perpendicular to OA.
angle AOB + angle BOC = angle COD
28 + 17 = 45
angle AOB + angle COD = 90
Therefore, OD is perpendicular to OA.
2. Originally Posted by stellina_91
Show that OD is perpendicular to OA.
angle AOB + angle BOC = angle COD
28 + 17 = 45
angle AOB + angle COD = 90
Therefore, OD is perpendicular to OA.
You are right.
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# Investigation of Algorithms II .
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Category: Animals / Pets
Description
2. Nuts and bolts. Before we endeavor to investigate a calculation, we have to characterize two things:How we measure the span of the inputHow we measure the time (or space) requirementsOnce we have done this, we discover a comparison that depicts the time (or space) necessities as far as the extent of the inputWe disentangle the mathematical statement by disposing of constants and tossing everything except the quickest developing term.
Transcripts
Slide 1
Examination of Algorithms II
Slide 2
Basics Before we endeavor to investigate a calculation, we have to characterize two things: How we measure the extent of the info How we measure the time (or space) prerequisites Once we have done this, we discover a condition that portrays the time (or space) necessities as far as the span of the information We disentangle the condition by disposing of constants and disposing of everything except the quickest developing term
Slide 3
Size of the info Usually it\'s very simple to characterize the span of the info If we are sorting an exhibit, it\'s the span of the cluster If we are figuring n! , the number n is the "size" of the issue Sometimes more than one number is required If we are attempting to pack objects into boxes, the outcomes may rely on upon both the quantity of articles and the quantity of boxes Sometimes it\'s difficult to characterize "size of the information" Consider: f(n) = if n is 1, then 1; else if n is even, then f(n/2); else f(3*n + 1) The conspicuous measure of size, n , is not really a decent measure To see this, process f(7) and f(8)
Slide 4
Measuring prerequisites If we need to know how much time or space a calculation takes, we can do exact tests — run the calculation over various sizes of info, and measure the outcomes This is not investigation However, experimental testing is valuable as a keep an eye on examination Analysis implies making sense of the time or space necessities Measuring space is typically clear Look at the sizes of the information structures Measuring time is generally done by tallying trademark operations Characteristic operation is a troublesome term to characterize In any calculation, there is some code that is executed the most circumstances This is in a deepest circle, or a most profound recursion This code requires "consistent time" (time limited by a steady) Example: Counting the correlations required in a cluster seek
Slide 5
Big-O and companions Informal definitions: Given an unpredictability work f(n) , ( f(n) ) is the arrangement of many-sided quality capacities that are lower limits on f(n) O( f(n) ) is the arrangement of multifaceted nature works that are upper limits on f(n) ( f(n) ) is the arrangement of many-sided quality capacities that, given the right constants, accurately portrays f(n) Example: If f(n) = 17x 3 + 4x – 12 , then ( f(n) ) contains 1 , x , x 2 , log x , x log x , and so forth. O ( f(n) ) contains x 4 , x 5 , 2 x , and so forth ( f(n) ) contains x 3
Slide 6
Formal meaning of Big-O * A capacity f(n) is O(g(n)) if there exist positive constants c and N to such an extent that, for all n > N , 0 < f(n) < cg(n) That is, if n is sufficiently enormous (bigger than N — we couldn\'t care less about little issues), then cg(n) will be greater than f(n) Example: 5x 2 + 6 is O(n 3 ) in light of the fact that 0 < 5n 2 + 6 < 2n 3 at whatever point n > 3 ( c = 2 , N = 3 ) We could similarly too utilize c = 1 , N = 6 , or c = 50 , N = 50 obviously, 5x 2 + 6 is likewise O(n 4 ) , O(2 n ) , and even O(n 2 )
Slide 7
Formal meaning of Big- * A capacity f(n) is (g(n)) if there exist positive constants c and N to such an extent that, for all n > N , 0 < cg(n) < f(n) That is, if n is sufficiently huge (bigger than N — we couldn\'t care less about little issues), then cg(n) will be littler than f(n) Example: 5x 2 + 6 is (n) since 0 < 20n < 5n 2 + 6 at whatever point n > 4 ( c=20 , N=4 ) We could similarly too utilize c = 50 , N = 50 obviously, 5x 2 + 6 is additionally O(log n) , O( n ) , and even O(n 2 )
Slide 8
Formal meaning of Big- * A capacity f(n) is (g(n)) if there exist positive constants c 1 and c 2 and N to such an extent that, for all n > N , 0 < c 1 g(n) < f(n) < c 2 g(n) That is, if n is sufficiently huge (bigger than N ), then c 1 g(n) will be littler than f(n) and c 2 g(n) will be bigger than f(n) as it were, is the "best" many-sided quality of f(n) Example: 5x 2 + 6 is (n 2 ) on the grounds that n 2 < 5n 2 + 6 < 6n 2 at whatever point n > 5 ( c 1 = 1 , c 2 = 6 )
Slide 9
cg(n) f(n) is O(g(n)) f(n) is (g(n)) f(n) f(n) cg(n) N c 1 g(n) f(n) is (g(n)) f(n) c 2 g(n) N Graphs Points to notice: What occurs close to the starting ( n < N ) is not vital cg(n) dependably goes through 0 , but rather f(n) may not (why?) In the third chart, c 1 g(n) and c 2 g(n) have the same "shape" (why?)
Slide 10
Informal audit For any capacity f(n) , and sufficiently huge estimations of n , f(n) = O(g(n)) if cg(n) is more noteworthy than f(n) , f(n) = theta(g(n)) if c 1 g(n) is more prominent than f(n) and c 2 g(n) is not as much as f(n) , f(n) = omega(g(n)) if cg(n) is not as much as f(n) , ...for appropriately picked estimations of c , c 1 , and c 2
Slide 11
The End The formal definitions were taken, with some slight alterations, from Introduction to Algorithms, by Thomas H. Cormen, Charles E. Leiserson, Donald L. Rivest, and Clifford Stein
Recommended
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# U.S. inflation rate in 1874: -5.00%
## Inflation Calculator
\$
### Inflation in 1874 and Its Effect on Dollar Value
Purchasing power increased by 5.00% in 1874 compared to 1873. On average, you would have to spend 5.00% less money in 1874 than in 1873 for the same item. This is an example of deflation.
In other words, \$1 in 1873 is equivalent in purchasing power to about \$0.95 in 1874.
The 1873 inflation rate was -1.64%. The inflation rate in 1874 was -5.00%. The 1874 inflation rate is lower compared to the average inflation rate of 2.16% per year between 1874 and 2020.
Inflation rate is calculated by change in the consumer price index (CPI). The CPI in 1874 was 11.40. It was 12.00 in the previous year, 1873. The difference in CPI between the years is used by the Bureau of Labor Statistics to officially determine inflation. Because the 1874 CPI is less than 1873 CPI, negative inflation (also known as deflation) has occurred.
Average inflation rate -5.00% Converted amount (\$1 base) \$0.95 Price difference (\$1 base) \$-0.05 CPI in 1873 12.000 CPI in 1874 11.400 Inflation in 1873 -1.64% Inflation in 1874 -5.00%
USD Inflation since 1635
Annual Rate, the Bureau of Labor Statistics CPI
### Inflation by Country
Inflation can also vary widely by country. For comparison, in the UK £1.00 in 1873 would be equivalent to £0.96 in 1874, an absolute change of £-0.04 and a cumulative change of -3.85%.
Compare these numbers to the US's overall absolute change of \$-0.05 and total percent change of -5.00%.
### Inflation by Spending Category
CPI is the weighted combination of many categories of spending that are tracked by the government. This chart shows the average rate of inflation for select CPI categories between 1873 and 1874.
Compare these values to the overall average of -5.00% per year:
Category Avg Inflation (%) Total Inflation (%) \$1 in 1873 → 1874
Food and beverages 0.00 0.00 1.00
Housing 0.00 0.00 1.00
Apparel 0.00 0.00 1.00
Transportation 0.00 0.00 1.00
Medical care 0.00 0.00 1.00
Recreation 0.00 0.00 1.00
Education and communication 0.00 0.00 1.00
Other goods and services 0.00 0.00 1.00
For all these visualizations, it's important to note that not all categories may have been tracked since 1873. This table and charts use the earliest available data for each category.
### How to Calculate Inflation Rate for \$1, 1873 to 1874
This inflation calculator uses the following inflation rate formula:
CPI in 1874CPI in 1873
×
1873 USD value
=
1874 USD value
Then plug in historical CPI values. The U.S. CPI was 12 in the year 1873 and 11.4 in 1874:
11.412
×
\$1
=
\$0.95
\$1 in 1873 has the same "purchasing power" or "buying power" as \$0.95 in 1874.
To get the total inflation rate for the 1 years between 1873 and 1874, we use the following formula:
CPI in 1874 - CPI in 1873CPI in 1873
×
100
=
Cumulative inflation rate (1 years)
Plugging in the values to this equation, we get:
11.4 - 1212
×
100
=
-5%
### Comparison to S&P 500 Index
To help put this inflation into perspective, if we had invested \$1 in the S&P 500 index in 1873, our investment would be nominally worth approximately \$1.02 in 1874. This is a return on investment of 2.08%, with an absolute return of \$0.02 on top of the original \$1.
These numbers are not inflation adjusted, so they are considered nominal. In order to evaluate the real return on our investment, we must calculate the return with inflation taken into account.
The compounding effect of inflation would account for -5.26% of returns (\$-0.05) during this period. This means the inflation-adjusted real return of our \$1 investment is \$0.07. You may also want to account for capital gains tax, which would take your real return down to around \$0 for most people.
Investment in S&P 500 Index, 1873-1874
Original Amount Final Amount Change
Nominal \$1 \$1.02 2.08%
Real
\$1 \$1.07 7.46%
Information displayed above may differ slightly from other S&P 500 calculators. Minor discrepancies can occur because we use the latest CPI data for inflation, annualized inflation numbers for previous years, and we compute S&P price and dividends from January of 1873 to latest available data for 1874 using average monthly close price.
For more details on the S&P 500 between 1873 and 1874, see the stock market returns calculator.
Politics and news often influence economic performance. Here's what was happening at the time:
• The first Battle of the Stronghold takes place where the US Army is defeated by a group of Modoc warriors.
• John Morseby discovers the site of Port Moresby, Papua New Guinea and claims it for Britain.
• Sultan Bargash closes the slave market of Zanzibar.
### Data Source & Citation
Raw data for these calculations comes from the Bureau of Labor Statistics' (CPI), established in 1913. Inflation data from 1665 to 1912 is sourced from a historical study conducted by political science professor Robert Sahr at Oregon State University.
You may use the following MLA citation for this page: “Inflation Rate in 1874 | Inflation Calculator.” Official Inflation Data, Alioth Finance, 27 Mar. 2020, https://www.officialdata.org/inflation-rate-in-1874.
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19 years, 5 days
## yes, but what's wrong with the sum?...
Thanks, Joe. I thought there would be a way to do it in a direct way, by summing c(j) over j from 0 to l-1. What's wrong with the sum here?
## yes, but what's wrong with the sum?...
Thanks, Joe. I thought there would be a way to do it in a direct way, by summing c(j) over j from 0 to l-1. What's wrong with the sum here?
## distribution of relative errors of appro...
Sorry for the subs question. I was trying to compute the probabilities for the values that relative error M (in approximating A with F) can take on. The minimum of such values is zero, the maximum is 2 and there is a finite number of them in between. I failed to interpret this
f := `assuming`([simplify(Probability(M = x))], [x >= 0]); pr := simplify(subs(p = 1/3, %)); plot(pr, x = -1 .. 3)
so I asked Maple whether M could be described by a means of a generalised density, using the Dirac delta.
Maple returned an expression involving the Dirac delta function, but I failed to find out how to make use of it or whether it is valid.
Perhaps I should generalise my question: what would be the standard approach to describe the (absolute) relative error M probabilistically? How to plot this description?
## distribution of relative errors of appro...
Sorry for the subs question. I was trying to compute the probabilities for the values that relative error M (in approximating A with F) can take on. The minimum of such values is zero, the maximum is 2 and there is a finite number of them in between. I failed to interpret this
f := `assuming`([simplify(Probability(M = x))], [x >= 0]); pr := simplify(subs(p = 1/3, %)); plot(pr, x = -1 .. 3)
so I asked Maple whether M could be described by a means of a generalised density, using the Dirac delta.
Maple returned an expression involving the Dirac delta function, but I failed to find out how to make use of it or whether it is valid.
Perhaps I should generalise my question: what would be the standard approach to describe the (absolute) relative error M probabilistically? How to plot this description?
## I think Maple should do it automatically...
I think this conversion should be done automatically by Maple, perhaps in future versions. If the series converges, I'd like to know it direct from Maple, not from other sources.
## I think Maple should do it automatically...
I think this conversion should be done automatically by Maple, perhaps in future versions. If the series converges, I'd like to know it direct from Maple, not from other sources.
## yes, correct but.......
yes, that the correct result, but what I want is a closed form expression involving h in place of numbers X and Y in Xalpha +Yalpha^2. Your proc does what mine proc labelled A does, but what I want is an expression in alpha and h only, without sums. The triple sum expression is a minimal example that causes problems.I don't know why.
The expression I'd like maple to provide for me can be obtained manually, by looking up the coefficient sequences at Sloan's OEIS, but I think there must be a way to do it automatically with maple. Is there such a way?
## yes, correct but.......
yes, that the correct result, but what I want is a closed form expression involving h in place of numbers X and Y in Xalpha +Yalpha^2. Your proc does what mine proc labelled A does, but what I want is an expression in alpha and h only, without sums. The triple sum expression is a minimal example that causes problems.I don't know why.
The expression I'd like maple to provide for me can be obtained manually, by looking up the coefficient sequences at Sloan's OEIS, but I think there must be a way to do it automatically with maple. Is there such a way?
## Any general strategy...?...
That's the kind of proof I wamted. For some reason, simplify(eq,trig) with 13.00 considers cos(theta)*(4*cos(theta)^2-3) simpler than cos(3*theta) and report accordingly. Any general strategy that helps not overlook useful simplifications?
## Any general strategy...?...
That's the kind of proof I wamted. For some reason, simplify(eq,trig) with 13.00 considers cos(theta)*(4*cos(theta)^2-3) simpler than cos(3*theta) and report accordingly. Any general strategy that helps not overlook useful simplifications?
## yes, non-maple type...
Yes, a simple typo. Thanks. Should read:
3) one can show, rather generally, that inequality S<sigma^2/mu^2 is equivalent to m1>m2.
## yes, non-maple type...
Yes, a simple typo. Thanks. Should read:
3) one can show, rather generally, that inequality S<sigma^2/mu^2 is equivalent to m1>m2.
## wiki for Maple...
Recently, we had to think of creating a forecasting wiki. We didn't consider the wikibook line at all. Eventually we chose the mediawiki engine (of so many others) and http://www.siteground.com/wiki-hosting.htm (of so many others). It would be excellent if we have allmaple.info or the like wiki website in the visible future. It is best to separate from wikipedia. NB: It is not an ad or propaganda, I just did some deep research for our own needs.
## Yes, but 0.5 is so special...
Thanks for the discussion, Joe. The relavant maple help page does not seem to distinguish between the Pascal and Polya cases of NBD, giving the condition of x>0 without specifying the set of numbers to which it applies. Even with an integer-valued x, it fails to work properly for p<>0.5; try [seq(print(tmp(i, .7)), i = 1 .. 4)], for example. Can the problem be cured locally, without stats package?
## Thanks to all - very...
Thanks to all - very interesting comments and solutions. I've re-coded my original example taking into account your advice - F_4 is clearly best and my F_5 similar. It seems seq and explicit loop structures are similarly efficient for this example? Is there anything else in the code for F_5 or F_4 that slows it down? For example, I read it somewhere that floor, round, ceil and the like are very slow and must be avoided. Thanks. The current performance is as below: F[5] := proc (n) local S, H, s; S := Statistics:-Sample(Statistics:-RandomVariable((':-Poisson')(1))); s := proc (x) options operator, arrow; floor(convert(S(x), `+`)) end proc; H := proc (g) options operator, arrow; map(proc (t) options operator, arrow; `if`(t = 0, 0, s(g)) end proc, g) end proc; [seq(H(floor(S(1)[1])), i = 1 .. n)] end proc testt := proc (n::posint) #F[1]:=FFFF (original);F[2]:=MyF (I'm not sure I understand)[];F[3]:=FF1(scott);F[4]:=FFFF[2] #(acer[]);F[5]:=new: local t, T, i; for i to 5 do t := time(); F[i](n); T[i] := time()-t end do; [seq(F[i] = T[i], i = 1 .. 5)] end proc: > testt(100); testt(100); testt(100); testt(100); testt(100); print(`output redirected...`); # input placeholder [F[1] = 0.547, F[2] = 0.109, F[3] = 0.188, F[4] = 0.015, F[5] = 0.016] [F[1] = 0.219, F[2] = 0.062, F[3] = 0.188, F[4] = 0.016, F[5] = 0.015] [F[1] = 0.188, F[2] = 0.078, F[3] = 0.156, F[4] = 0.016, F[5] = 0.016] [F[1] = 0.203, F[2] = 0.062, F[3] = 0.188, F[4] = 0.016, F[5] = 0.015] [F[1] = 0.219, F[2] = 0.062, F[3] = 0.172, F[4] = 0.016, F[5] = 0.015] > testt(2000); testt(2000); testt(2000); print(`output redirected...`); # input placeholder [F[1] = 4.329, F[2] = 1.781, F[3] = 4.500, F[4] = 0.406, F[5] = 0.344] [F[1] = 4.625, F[2] = 1.781, F[3] = 4.703, F[4] = 0.360, F[5] = 0.437] [F[1] = 4.734, F[2] = 2.016, F[3] = 4.937, F[4] = 0.438, F[5] = 0.375]
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cleveland_shape_pal: Shape palette from Cleveland "Elements of Graphing Data"... In ggthemes: Extra Themes, Scales and Geoms for 'ggplot2'
Description
Shape palettes for overlapping and non-overlapping points.
Usage
`1` ```cleveland_shape_pal(overlap = TRUE) ```
Arguments
`overlap` `logical` Use the scale for overlapping points?
Note
In the Elements of Graphing Data, W.S. Cleveland suggests two shape palettes for scatter plots: one for overlapping data and another for non-overlapping data. The symbols for overlapping data relies on pattern discrimination, while the symbols for non-overlapping data vary the amount of fill. This palette attempts to create these palettes. However, I found that these were hard to replicate. Using the R shapes and unicode fonts: the symbols can vary in size, they are dependent of the fonts used, and there does not exist a unicode symbol for a circle with a vertical line. If someone can improve this palette, please let me know.
Following Tremmel (1995), I replace the circle with a vertical line with an encircled plus sign.
The palette `cleveland_shape_pal()` supports up to five values.
References
Cleveland WS. The Elements of Graphing Data. Revised Edition. Hobart Press, Summit, NJ, 1994, pp. 154-164, 234-239.
Tremmel, Lothar, (1995) "The Visual Separability of Plotting Symbols in Scatterplots", Journal of Computational and Graphical Statistics, https://www.jstor.org/stable/1390760
Other shapes: `circlefill_shape_pal()`, `scale_shape_circlefill()`, `scale_shape_cleveland()`, `scale_shape_tremmel()`, `tremmel_shape_pal()`
Examples
``` 1 2 3 4 5 6 7 8 9 10 11 12``` ```### (discrete). library("ggplot2") p <- ggplot(mtcars) + geom_point(aes(x = wt, y = mpg, shape = factor(gear))) + facet_wrap(~am) + theme_bw() # overlapping symbol palette p + scale_shape_cleveland() # non-overlapping symbol palette p + scale_shape_cleveland(overlap = FALSE) ```
Example output
```
```
ggthemes documentation built on Jan. 20, 2021, 5:05 p.m.
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# Conventions: Double RKCB
#### ByEddie Kantar
Oct 8, 2015
Source: Bridge Guys
This conventional variation of the Six Ace Blackwood convention was devised by Mr. Edwin Kantar, and published in his book Roman Keycard Blackwood.
In one or several of these publications Mr. Edwin Kantar addresses the conventional method designated as Six Ace Blackwood, which he has varied and modified to meet the same requirements in a revised manner, but which continues to have similarities to the Six-Ace Blackwood conventional method known as 6A-RKCB, the origin of which is unknown.
The concept behind this variation is the same, namely first to discover that the partnership possesses two suits of eight cards or more (so-called double suit agreement), so that there are two combined suits and therefore two suit fits. Once this feature has been discovered, then either partner of the partnership can initiate the artificial 4 No Trump bid to determine whether or not slam is a possibility.
The partnership attempts to discover the number of Aces, which are considered Key Cards, and also whether the partnership possesses both Kings of the two suits, also known as Key Cards. The official designation of Six Ace Keycard is somewhat a misnomer, or a name wrongly or unsuitably applied, since Kings are not Aces.
As soon as the partnership has established the trump suit or inferred the trump suit, then one partner will initiate the 4 No Trump bid to ask for Aces and Kings. The responses of the partner are shown below and these responses are the responses suggested by Mr. Edwin Kantar in his publication(s).
5: Shows 1 or 4 Keycards. In this response no information about either Queen of the two suit fits is included. 5: Shows 0 or 3 Keycards. In this response no information about either Queen of the two suit fits is included. 5: Shows 2 Keycards. This response also shows that the partner holds no Queen in the two suit fits. 5: Shows 2 Keycards and possession of the lower-ranking Queen of the two suit fits. 5NT: Shows 2 Keycards and possession of the higher-ranking Queen of the two suit fits. 6: Promises both Queens of the two suit fits.
According to the modifications of Mr. Edwin Kantar there is not any possibility to show a void such as a distribution of 6-4-3-0. This is owing to a lack of bidding space. This is also the case with the version known as Six Ace Roman Keycard Blackwood or, as it is more commonly known, Six Ace Blackwood.
Following the two responses of 5 Clubs and 5 Diamonds there might be instances, in which it would be preferable to know whether or not there are no Queens, 1 Queen, or even both Queens are held in the agreed two suits. Mr. Edwin Kantar suggests the following guidelines to discover whether or not any Queen is held in the two suit fits.
If the response is 5 Clubs, then 5 Diamonds is the Queen-asking bid.
5: Shows an absence of both Queens. 5: Promises the Queen of the lower-ranking suit of the two suit fits. 5NT: Promises the Queen of the higher-ranking suit of the two suit fits. 6: Promises both Queen of the two suit fits.
If Diamonds is one of the agreed two suits, then the next higher-ranking suit, which is not one of the agreed two suits, becomes the Queen-asking bid. In this instance the steps will simply be identical in significance although different in rank.
If the response is 5 Diamonds, then the Queen-asking partner employs the following guidelines:
5: This is the correct bid if Hearts is not one of the agreed two suits. If Hearts is one of the agreed two suits, then Spades becomes the Queen-asking bid and the ensuing steps according to the guidelines show an absence or possession of the Queen(s) of the agreed two suits. If Spades is the second agreed suit, then 5 No Trump becomes the Queen-asking bid, and the steps are adjusted accordingly. This particular bidding sequence, however, requires that the partnership hold the values to safely play a small slam contract. (If these values are not present, then the partnership should agree to a sign-off bid.) 5: This is the correct bid if Spades is not one of the agreed two suits. 6: This is the correct bid if Spades and Hearts (both Major suits) are the agreed two suits.
Once the Queen-asking bid has been initiated, then the answering partner responds in steps according to the following guidelines:
First Step: Shows the absence of any Queen of the agreed two suits. Second Step: Shows the lower-ranking Queen of the agreed two suits. Third Step: Shows the higher-ranking Queen of the agreed two suits. Fourth Step: Shows both Queens in the agreed two suits.
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# Delaunay Triangulations
## Presentation on theme: "Delaunay Triangulations"— Presentation transcript:
Delaunay Triangulations
COSC 6114 Prof. Andy Mirzaian Voronoi Diagrams & Delaunay Triangulations
Voronoi Diagram & Delaunay Triangualtion
Algorithms Divide-&-Conquer Plane Sweep Lifting into d+1 dimensions Edge-Flip Randomized Incremental Construction Applications Proximity space partitioning and the post office problem Height Interpolation Euclidean: Minimum Spanning Tree, Traveling Salesman Problem, Minimum Weight Triangulation, Relative Neighborhood Graph, Gabriel Graph. Extensions Higher Order Voronoi Diagrams Generalized metrics - Robot Motion Planning
References: [M. de Berge et al] chapters 7, 9, 13
[Preparata-Shamos’85] chapters 5, 6 [O’Rourke’98] chapter 5 [Edelsbrunner’87] chapter 13 AAW Lecture Notes 16, 17, 18, 19
Introduction
Voronoi Diagram & Delaunay Triangulation
P = { p1, p2, … , pn} a set of n points in the plane.
Voronoi Diagram & Delaunay Triangulation
Nearest site proximity partitioning of the plane Voronoi(P): # regions = n, # edges 3n-6, # vertices 2n-5.
Voronoi Diagram & Delaunay Triangulation
Delaunay Triangulation = Dual of the Voronoi Diagram. DT(P): # vertices = n, # edges 3n-6, # triangles 2n-5.
Voronoi Diagram & Delaunay Triangulation
Delaunay triangles have the “empty circle” property.
Voronoi Diagram & Delaunay Triangulation
Voronoi Diagram pi Voronoi Region of pi: Voronoi Diagram of P:
P = { p1, p2, … , pn} a set of n points in the plane. Assume: no 3 points collinear, no 4 points cocircular. PB(pi, pj) perpendicular bisector of pipj. pi pj H(pi, pj) half-plane Voronoi Region of pi: pi Voronoi Diagram of P:
Voronoi Diagram Properties
Each Voronoi region V(pi) is a convex polygon (possibly unbounded). V(pi) is unbounded pi is on the boundary of CH(P). Consider a Voronoi vertex v = V(pi) V(pj) V(pk). Let C(v) = the circle centered at v passing through pi, pj, pk. C(v) is circumcircle of Delaunay Triangle (pi, pj, pk). C(v) is an empty circle, i.e., its interior contains no other sites of P. pj = a nearest neighbor of pi V(pi) V(pj) is a Voronoi edge (pi, pj) is a Delaunay edge. more later …
Delaunay Triangulation Properties
DT(P) is straight-line dual of VD(P). DT(P) is a triangulation of P, i.e., each bounded face is a triangle (if P is in general position). (pi, pj) is a Delaunay edge an empty circle passing through pi and pj. Each triangular face of DT(P) is dual of a Voronoi vertex of VD(P). Each edge of DT(P) corresponds to an edge of VD(P). Each node of DT(P), a site, corresponds to a Voronoi region of VD(P). Boundary of DT(P) is CH(P). Interior of each triangle in DT(P) is empty, i.e., contains no point of P. more later …
ALGORITHMS
A brute-force VD Algorithm
P = { p1, p2, … , pn} a set of n points in the plane. Assume: no 3 points collinear, no 4 points cocircular. intersection of n-1 half-planes Voronoi Region of pi: Voronoi Diagram of P: Voronoi region of each site can be computed in O(n log n) time. There are n such Voronoi regions to compute. Total time O(n2 log n).
Divide-&-Conquer Algorithm
M. I. Shamos, D. Hoey [1975], “Closest Point Problems,” FOCS, D.T. Lee [1978], “Proximity and reachability in the plane,” Tech Report No, 831, Coordinated Sci. Lab., Univ. of Illinois at Urbana. D.T. Lee [1980], “Two dimensional Voronoi Diagram in the Lp metric,” JACM 27, The first O(n log n) time algorithm to construct the Voronoi Diagram of n point sites in the plane.
ALGORITHM Construct Voronoi Diagram (P)
INPUT: P = { p1, p2, … , pn} sorted on x-axis. OUTPUT: CH(P) and DCEL of VD(P). [BASIS]: if n1 then return the obvious answer. [DIVIDE]: Let m n/2 Split P on the median x-coordinate into L = { p1, … , pm} & R = { pm+1, … , pn}. [RECUR]: (a) Recursively compute CH(L) and VD(L). (b) Recursively compute CH(R) and VD(R). [MERGE]: (a) Compute Upper & Lower Bridges of CH(L) and CH(R) & obtain CH(P). (b) Compute the y-monotone dividing chain C between VD(L) & VD(R). (c) VD(P) [C] [VD(L) to the left of C] [VD(R) to the right of C]. (d) return CH(P) & VD(P). END. O(1) O(n) T(n/2) T(n/2) O(n) T(n) = 2 T(n/2) + O(n) = O( n log n).
P = { p1, p2, … , pn} a set of n points in the plane.
VD(P) = [C] [VD(L) to the left of C] [VD(R) to the right of C] .
VD(L) and CH(L)
VD(R) and CH(R)
Upper & Lower bridges between CH(L) and CH(R) & two end-rays of chain C.
Construct chain C. (1,5) (3,5) 5 8 1 3 6 (3,6) (4,6) 9 4 (4,7) 7 2 (2,7)
Construct chain C. 5 8 1 3 6 9 4 7 2
Crop VD(L) & VD(R) at C. 5 8 1 3 6 9 4 7 2
VD(P) and CH(P) 5 8 1 3 6 9 4 7 2
Fortune’s Algorithm O(n log n) time algorithm by plane-sweep.
Steve Fortune [1987], “A Sweepline algorithm for Voronoi Diagrams,” Algorithmica, Guibas, Stolfi [1987], “Ruler, Compass and computer: The design and analysis of geometric algorithms,” Proc. of the NATO Advanced Science Institute, series F, vol. 40: Theoretical Foundations of Computer Graphics and CAD, O(n log n) time algorithm by plane-sweep. See AAW animation. Generalization: VD of line-segments and circles.
The Waive Propagation View
Simultaneously drop pebbles on calm lake at n sites. Watch the intersection of expanding waves.
Time as 3rd dimension z=time p y x a a y x
apex of the cone y x All sites have identical opaque cones.
Time as 3rd dimension z x y base plane
q base plane All sites have identical opaque cones. cone(p) cone(q) = vertical hyperbola h(p,q). Vertical projection of h(p,q) on the xy base plane is PB(p,q).
Time as 3rd dimension z x y base plane
Visible intersection of the cones viewed upward from z = - is VD(P).
Conic Sections: Focus-Directrix
Eccentricity constant: focus f w h directrix l 0 = e point (focus) 0 < e < 1 ellipse e = 1 parabola e > 1 hyperbola
Sweep Plane & Sweep Line
base plane sweep line a z y x
Sweep Plane & Cone Intersection
v sweep line l a base plane p a u w y x Vertical projection of intersection of cone(p) & the sweep plane on the base plane is a parabola with focus p and directrix l.
Parabolic Evolution l p l focus p y x
Parabolic Evolution l p l focus p y x
Time snapshots of moving parabola associated with site p
3 2 1 p y x sweep line 1 2 3
The parabolic front Sweep plane opaque. So we don’t see future events.
Any part of a parabola inside another one is invisible, since a point (x,y) is inside a parabola iff at that point the cone of the parabola is below the sweep plane. Parabolic Front = visible portions of parabola; those that are on the boundary of the union of the cones past the sweep. Parabolic Front is a y-monotone piecewise-parabolic chain. (Any horizontal line intersects the Front in exactly one point.) Each parabolic arc of the Front is in some Voronoi region. Each “break” between 2 consecutive parabolic arcs lies on a Voronoi edge. sweep line
Evolution of the parabolic front
The breakpoints of the parabolic front trace out every Voronoi edge as the sweep line moves from x = - to x = + . Every point of every Voronoi edge is a breakpoint of the parabolic front at some time during the sweep. Proof: (a) Fig 1: Event w: Cu is an empty circle. (b) Fig 2: At event w point u must be a breakpoint of the par. front Otherwise: Some parabola Z covers u at v Focus of Z is on Cv and Cv is inside Cu Focus of Z is inside Cu Cu is not an empty circle a contradiction. Fig 1. p u w sweep line Cu q Z p Fig 2. v u w Cv Cu sweep line q
The Discrete Events SITE EVENT: Insert into the Parabolic Front.
CIRCLE EVENT: Delete from the Parabolic Front.
SITE EVENT A new parabolic arc is inserted into the front when sweep line hits a new site. p p p s s s q q q 3 1 2
SITE EVENT A new parabolic arc is inserted into the front when sweep line hits a new site. p p p s s s q q q A parabola cannot appear on the front by breaking through from behind. The following are impossible: b b b b a a a a g g t+Dt t t+Dt t
CIRCLE EVENT Circle event w causes parabolic arc b to disappear.
a and g cannot belong to the same parabola. p p p b a a a q u w q u q b w w g g g s s s
DATA STRUCTURES (T & Q) T:
T: [SWEEP STATUS: a balanced search tree] maintains a description of the current parabolic front. Leaves: arcs of the parabolic front in y-monotone order. Internal nodes: the break points. T: sweep direc. x A C D B y Par(C) Par(D) Par(A) Par(B) Operations: (a) insert/delete an arc. (b) locate an arc intersecting a given horizontal line (for site event). (c) locate the arcs immediately above/below a given arc (for circle event). We also hang from this the part of the Voronoi Diagram swept so far. - Each leaf points to the corresponding site. - Each internal node points to the corresponding Voronoi edge.
DATA STRUCTURES (T & Q) Q: [SWEEP SCHEDULE: a priority queue] schedule of future events: all future site-events & some circle-events, i.e., those corresponding to 3 consecutive arcs of the current parabolic front as represented by T. The others will be discovered & added to the sweep schedule before the sweep lines advances past them. Conversely, not every 3 consecutive arcs of the current front specify a circle-event. Some arcs may drop out too early.
Event Processing & Scheduling
Event-driven simulation loop: At each iteration remove the next event (with min x-coordinate) from Q & simulate the effect of the sweep-line advancing past that event point.
Event Processing & Scheduling
Event-driven simulation loop: At each iteration remove the next event (with min x-coordinate) from Q & simulate the effect of the sweep-line advancing past that event point. death(a) : pointing to a circle-event in Q as the meeting point of the Voronoi edges. (If the edges are diverging, then death(a) = nil.) Remove circle-event death(a) if: (a) a is split in two by a site-event, or (b) whenever one of the two arcs adjacent to a is deleted by a circle-event. a
Event Processing & Scheduling
Event-driven simulation loop: At each iteration remove the next event (with min x-coordinate) from Q & simulate the effect of the sweep-line advancing past that event point. A circle-event update: each parabolic arc b (leaf of T) points to the earliest circle-event, death(b), in Q that would cause deletion of b at the corresponding Voronoi vertex. a spurious circle-event a death(b’) death(b) b’ v b s s b’’ death(b’’) g g
Event Processing & Scheduling
Event-driven simulation loop: At each iteration remove the next event (with min x-coordinate) from Q & simulate the effect of the sweep-line advancing past that event point. (a,g,d) do not define a circle-event: (a,c,d) is not a circle-event now, it is past the current sweep position. a b a b c d g d
ANALYSIS |T| = O(n) : the front always has O(n) parabolic arcs, since splits occur at most n times by site events Also by Davenport-Schinzel: … a … b … a … b … is impossible [At most 2n-1 parabolic arcs in T.] |Q| = O(n) : there are at most n site-events and O(n) triples of consecutive arcs on the parabolic front to define circle-events. Total # events = O(n), Time per event processing = O(log n). THEOREM: Fortune’s algorithm computes Voronoi Diagram of n sites in the plane using optimal O(n log n) time and O(n) space.
Delaunay Triangulation
Terrain Height Interpolation
A perspective view of a terrain. A topographical map of a terrain.
Terrain Height Interpolation
A perspective view of a terrain. A topographical map of a terrain. Terrain: A 2D surface in 3D such that each vertical line intersects it in at most one point. f : 2 f(p) = height of point p in the domain A of the terrain. Method: Take a finite sample set P A. Compute f(P), and interpolate on A. f P A
Triangulations of Planar Point Sets
P = {p1, p2, … , pn } 2. A triangulation of P is a maximal planar straight-line subdivision with vertex set P. THEOREM: Let P be a set of n points, not all collinear, in the plane Suppose h points of P are on its convex-hull boundary. Then any triangulation of P has 3n-h-3 edges and 2n-h-2 triangles. Proof: m = # triangles 3m + h = 2E (each triangle has 3 edges; each edge incident to 2 faces) Euler: n – E + (m+1) = 2 m = 2n - h - 2, E = 3n – h – 3.
Delaunay Graph: Dual of Voronoi Diagram
Delaunay Graph DG(P) as dual of Voronoi Diagram VD(P).
Delaunay Graph: Dual of Voronoi Diagram
Delaunay Graph DG(P) as strainght-line dual of Voronoi Diagram VD(P).
Delaunay Graph is a Triangulation
Alternative Definition of Delaunay Graph: A triangle D(pi , pj , pk) is a Delaunay triangle iff the circumscribing circle C(pi , pj , pk) is empty. Line segment (pi, pj) is a Delaunay edge iff there is an empty circle passing through pi and pj, and no other point in P. THEOREM: Delaunay Graph of P is a straight-line plane graph, & a triangulation of P. Proof: Follows from the following Lemmas.
Delaunay Graph is a Triangulation
LEMMA 1: Every edge of CH(P) is a Delaunay edge. Proof: Consider a sufficiently large circle that passes through the 2 ends of CH edge e, and whose center is separated from CH(P) by the line aff(e). e
Delaunay Graph is a Triangulation
LEMMA 2: No two Delaunay triangles overlap. Proof: Consider circumscribing circles of two such triangles. Line L separates the two triangles. L empty area
Delaunay Graph is a Triangulation
LEMMA 3: pi & pj are Voronoi neighbors (pi , pj) is a Delaunay edge. Proof: Consider the circle that passes through pi & pj and whose center is in the relative interior of the common Voronoi edge between V(pi) & V(pj). pj V(pj) pi V(pi)
Delaunay Graph is a Triangulation
LEMMA 4: If pj and pk are two (rotationally) successive Voronoi neighbors of pi & pjpipk < 180, then D(pi , pj , pk) is a Delaunay triangle. Proof: pj & pk must also be Voronoi neighbors Now apply Lemma 3 to (pi , pj), (pi , pk), (pj , pk).
Delaunay Graph is a Triangulation
LEMMA 4: If pj and pk are two (rotationally) successive Voronoi neighbors of pi & pjpipk < 180, then D(pi , pj , pk) is a Delaunay triangle. Proof: pj & pk must also be Voronoi neighbors Now apply Lemma 3 to (pi , pj), (pi , pk), (pj , pk). COROLLARY 5: For each pi P, the Delaunay triangles incident to pi completely cover a small open neighborhood of pi inside CH(P). CH(P) pi pi
Delaunay Graph is a Triangulation
LEMMA 6: Every point inside CH(P) is covered by some Delaunay triangle in DG(P). Proof: Let q be an arbitrary point in CH(P). Let (pi , pj) be the Delaunay edge immediately below q. ((pi , pj) exists because all convex-hull edges are Delaunay by Lemma 1.) From Corollary 5 let D(pi,pj,pk) be the next Delaunay triangle incident to pi as in the Figure below. Then, either q D(pi,pj,pk), or the choice of (pi , pj) is contradicted. pk pk pk q q q pi pj pi pj pi pj The THEOREM follows from Lemmas 2-6. We now use DT(P) to denote the Delaunay triangulation of P.
Angles in Delaunay Triangulation
DEFINITION: T = an arbitrary triangulation (with m triangles) of point set P. a1, a2, …, a3m = the angles of triangles in T, sorted in increasing order. A(T ) = (a1 , a2 , … , a3m) is called the angle-vector of T. THEOREM: DT(P) is the unique triangulation of P that lexicographically maximizes A(T ). Proof: Later. COROLLARY: DT(P) maximizes the smallest angle. Useful for terrain approximation by triangulation & linear interpolation. Small angles (long skinny triangles) cause large approximation errors.
DT & VD via CH K.Q. Brown [1979], “Voronoi diagrams from convex hulls,” IPL K.Q. Brown [1980], “Geometric transforms for fast geometric algorithms,” PhD. Thesis, CMU-CS Guibas, Stolfi [1987], “Ruler, Compass and computer: The design and analysis of geometric algorithms,” Proc. of the NATO Advanced Science Institute, series F, vol. 40: Theoretical Foundations of Computer Graphics and CAD, Guibas, Stolfi [1985], “Primitives for the manipulation of general subdivisions and the computation of Voronoi diagrams,” ACM Trans. Graphics 4(2), [Edelsbrunner’87] pp: Aurenhammer [1987], “Power diagrams: properties, algorithms, and applications,” SIAM J. Computing 16,
Paraboloid of Revolution L l: (x,y) (x,y, x2+y2)
DT in d CH in d+1 z=l(x,y) Paraboloid of Revolution L l: (x,y) (x,y, x2+y2) y x
DT in 2 CH in 3 SUMMARY: Consider a plane P in 3 and the paraboloid of revolution L. Projection of PL down to 2 is a circle C. Every point of L below P projects down to interior of C. Every point of L above P projects down to exterior of C. 2D (“Nearest-Point” and “Farthest-Point”) Delaunay Triangulation algorithm via 3D-convex-hull in O(n log n) time.
DT in d CH in d+1 z CH((P)) y DT(P) x
DT in d CH in d+1 P = {p1, p2, … , pn } d.
l(P) = {l(p1), l(p2), …, l(pn)} d+1. Assume P is in general position. THEOREM: Projection of lower convex-hull of l(P) down to d is the “nearest-point” DT(P). Projection of upper convex-hull of l(P) down to d is the “farthest-point” DT(P).
VD in d Half-Spaces in d+1
z=l(x,y) P(p) Paraboloid of Revolution L l: (x,y) (x,y, x2+y2) l(p) y p x
VD in d Half-Spaces in d+1
Observation: dist(s,t) = dist(p,r)2 . t d2 L: y = x2 s A d=1 example: 1 r p d P(p)
VD in d Half-Spaces in d+1
A d=1 example: L: y = x2 P(q) P(p) 1 q p
VD in d Half-Spaces in d+1
P(p) P(q) l(q) l(p) q PB(p,q) p d
VD in d Half-Spaces in d+1
The projection of P(p)P(q) down to d is the perpendicular-bisector of p & q. P(p) is above P(q) at points of d that are closer to q. P+(p) = the closed half-space in d above P(p). THEOREM: Voronoi Diagram in d is obtained by projecting down the intersection of half-spaces P+(p) , i=1..n, in d+1. Duality Reminder: Intersection of half-spaces Convex-Hull of Points.
VD in d Half-Spaces in d+1
LEMMA: The projection down to d of the intersection of any non-vertical hyper-plane in d+1 with the paraboloid of revolution L : xd+1 = x12 + x22 + … + xd2 is a sphere in d. Proof:
P 2 l(P) 3 COROLLARY: Consider 4 points pi = (xi , yi), i=1..4 in 2 , and their immage l(pi) = (xi , yi , xi2 + yi2) in 3. pi are cocircular in 2 l(pi) are coplanar in 3. Point D=p4 is inside the circumcircle of A=p1, B=p2, C=p3 a1 + a3 < 180 (a2 + a4 > 180) l(D) is below the plane passing through l(A), l(B), l(C). D(A,B,C,D) and D(A,B,C) have the same sign. i.e., B a3 a4 D C a2 a1 A Incircle(A,B,C,D)
A simple O(n2) time DT Algorithm
Step 1: Let T be an arbitrary triangulation of P [e.g., use sweep in O(n log n) time] Step 2: while T has a quadrangle of the form below with A + B > 180 do flip diagonal CD (i.e., replace it with diagonal AB). [O(n2) iterations] C C flip CD B B A A Fact 1: 6 sorted angles are lexicographically increased D D C B A don’t flip CD D
A snapshot of the Algorithm
A snapshot of the Algorithm
FLIP e1
A snapshot of the Algorithm
A snapshot of the Algorithm
FLIP e2
A snapshot of the Algorithm
A snapshot of the Algorithm
FLIP e3
A snapshot of the Algorithm
A snapshot of the Algorithm
FACTS FACT 1: The flip operation cd ab is like attaching the tetrahedron (a,b,c,d) to the “lifted convex-hull” in 3D constructed so far [so its volume increases]. Proof: (1) There are at most O(n2) edges in the tetrahedronization in 3D After an edge is flipped out, it cannot reappear. O(n2) flips at most (2) Euler: n nodes, e edges, f faces, t tetrahedra, h hull vertices: n – e + f– t = 1 & 4t + (2h –4) = 2f t = e – (n + h – 3) = O(n2).
FACTS The O(n2) bound on the number of flips is tight:
flip “inside out”
FACTS FACT 2: Let Cabc denote the circumcircle of triangle (a,b,c). a
FACTS FACT 3: A flip lexicographically increases the sorted sequence of 6 angles of the pairs of triangles involved before and after the flip. COROLLARY 4: Each flip lexicographically increases the angle-vector of the triangulation. THEOREM: DT(P) is the unique triangulation of P that lexicographically maximizes A(T ).
Randomized Incremental Construction
Guibas, Knuth, Sharir [1990], “Randomized Incremental Construction of Delaunay and Voronoi Diagrams,” Proc. 17th Annual Int’l Coll. Automata & Lang. Progr. (CALP’90), LNCS 443, Springer-Verlag, Edelsbrunner, Shah [1992], “Incremental Topological Flipping works for regular Triangulations,” CG,
Randomized Incremental DT Algorithm
Randomly permute the given sites, say p1, p2, … , pn. Construct Delaunay Triangulation of p1, p2, p3. for i 4 .. n do (a) Find triangle (or outer-face) of DT that contains pi (b) Add pi to that face and triangulate it (c) Perform repeated “Incircle-Test” and do edge-flips. return resulting DT. end Inserting a new site pi : flip pi
Step 3(c): Use the “Histroy DAG”
Maintain all versions of the DT on top of one another, plus appropriate links between overlapping “old” & “new” triangles. This hierarchical structure will be used to perform the point-location step 3(a) also. No more preprocessing needed for nearest-neighbor queries. The Voronoi Diagram can also be constructed in a similar randomized manner with a hierarchical representation of all the past history of insertions.
ALGORITHM Randomized Incremental DT Construction
Initialize DT to “bounding-box” triangle W1 W2 W3 (3 points at ). for k 1 .. n do select a random site P not yet selected. locate triangle ABC that contains P. replace ABC by PAB, PBC, PCA X A ; progress true repeat Y 3rd vertex of triangle left of PX if X & Y are not both some Wi (at ) then do find Z P s.t. XYZ is a current triangle if Incircle(P,X,Y,Z) then do flip triangles PXY & ZXY to PXZ & PZY progress false end if if progress then X Y until X = A and progress end B C P A Y P Z X expected time = O(n log n) expected space = O(n)
Triangulation History DAG split D1 flip rs flip rt D1 D2 D3 s p D2 r
History DAG will be used for Nearest Site Point Location query answering.
Each node in History DAG has out-degree at most 3. When inserting a new site, all Delaunay edges created due to edge-flips are incident to the new site. pi pi
THEOREM 1: The randomized Delaunay Triangulation algorithm satisfies the following properties on all inputs of size n: The expected total number of structural changes (i.e., edge-flips …) that happen to the diagram is only O(n). [The worst-case is Q(n2).] Using the History DAG, the amortized expected cost per site insertion is O(log n), i.e., the overall expected time is O(n log n). Proof: See next slides. THEOREM 2: Suppose the Delaunay diagram of m sites is already computed and n additional sites are to be added. If the new sites are randomly and incrememntally inserted into the diagram, then the total number of structural changes has expected value O(m log n + n) [and this bound is tight].
Worst-case Q(n2) edge-flips:
insert next y=1 x insert first Note: In any triangulation of the n sites: # triangles = n – h – 2 2n # edges = n –h – 3n where h is the # hull vertices.
LEMMA 1: Expected # triangles created by the RIA is at most 9n + 1 = O(n).
Proof: Pk = { p1, p2, …, pk } the first k points (random). Tk = Delaunay Triangulation of Pk {W1, W2, W3}. # edges of Tk 3(k+3) –6 = 3k + 3. # edges of Tk excluding triangle (W1, W2, W3) 3k. E[degree of pk in Tk ] (2 3k) / k = 6. Backwards analysis: pk can be any of the points in Pk with equal probability. # triangles created at iteration k 2 [degree of pk in Tk] – 3. E[# triangles created in iteration k] 2 6 –3 = 9. E[# triangles created in all iter.] 9n. [by linearity of expectation] (The one extra in 9n+1 is the initial “bounding triangle (W1, W2, W3).
Random-Sampling technique of Clarkson-Shor.
Definition: Consider a triangle D = (pi , pj , pk) C(D) = circumcircle of triangle D. X(D) = set of points in P that are in the interior of C(D). width(D) = || X(D) ||. pj pi width = 0 D is a Delaunay triangle. pk
Expected time = point locations + History DAG construction
Expected time excluding point-location = expected size of the History DAG = expected # triangles created by RIA = O(9n + 1) = O(n). Expected time for all point-locations = O( n + D width(D) ) [where D is a Delaunay triangle at some iteration] Why? Point-location for insertion of site p: Starting with (W1, W2, W3) in the History DAG, trace the path of all triangles that contain p. Charge the cost of each visited triangle node to either (W1, W2, W3) or a triangle D that was Delaunay at some time. The total charge will be as above. (See next slide.)
Expected time for all point-locations = O( n + D width(D) )
Expected time for all point-locations = O( n + D width(D) ) [where D is a Delaunay triangle at the end of some iteration] Width(D) = W D will be charged by at most W points, at most once for each. Why? Suppose point-location of p visits a triangle D. So, p is inside D, hence p X(D). (i) If D is Delaunay at the time of its creation, then charge that visit to D. (ii) If D is not Delaunay at the time of its creation, then charge that visit to its neighbor D’. (D’ must have been Delaunay.) p D p X(D’). D D’
LEMMA 2: E[ D width(D) ] = O(n log n).
Proof: Pk = { p1, p2, …, pk } (random). Tk = DT of Pk {W1, W2, W3}. [pk is incident to D.] Definition: for a point q P - Pk : R(Pk,q) = # triangles D Tk s.t. q X(D). R(Pk,q, pk) = # triangles D Tk s.t. q X(D), and pk is a vertex of D.
THEOREM 3: The randomized incremental algorithm calculates the Delaunay Triangulation of a set of n points in the plane in expected time O(n log n) & expected space O(n). Remark: This expected time is optimum, since there is a linear time reduction from sorting to DT (i.e., xi (xi, xi2) ) and expected time for sorting in the algebraic decision-tree model of computation is W(n log n).
Generalizations & Applications
RIA for 3D Convex Hull Incremental construction of 3D Convex Hulls: Generalize the following idea to CH of any set P of n points in 3D. THEOREM 4: The expected overall # of faces that appear during the randomized incremental construction of the lower-CH of a set P of n points in 3D is O(n). Proof: width(XYZ) = # points of P that lie below the plane of XYZ The rest of the proof is again by the random-sampling technique of Clarkson-Shor. THEOREM 5: The randomized incremental construction of 3D convex hull takes optimal expected time O(n log n) and expected space O(n). [Worst-case could be Q(n2) time and space.]
RIA for Voronoi Diagram
IDEA: radial triangulation of Voronoi regions: Now apply previous ideas on these type of triangles. Each radial triangle is determined by at most 4 sites. No need for further point-location preprocessing for nearest-neighbors queries. The constructed structure will have that. THEOREM 6: The expected size of the structure is O(n). The structure can be built in expected O(n log n) time. The above point-location structure can be used to locate the final radial triangle that contains an arbitrary query point q in expected O(log2 n) time.
The Post Office Problem
PROBLEM: Preprocess a given set P of n points in the plane for: Nearest Neighbor Query: Given a query point q, determine which point in P is nearest to q. Shamos [1976]: Slab Method: Query Time: O(log n) Preprocessing Time: O(n2) Space: O(n2) Kirkpatrick [1983]: Triangulation refinement method for planar point location: Query Time: O(log n) Preprocessing Time: O(n log n) Space: O(n) Construct Voronoi Diagram. Each Voronoi region is convex, hence monotone. Triangulate the Voronoi regions in O(n) time. Then apply Kirkpatrick’s method. Andoni, Indyk [2006] “Near-optimal hashing algorithms for approximate nearest neighbor in high dimensions,” FOCS’06.
Largest Empty Circle Problem
PROBLEM: Determine the largest empty circle with center in CH(P). O(n) Candidate centers. All can be found in O(n) time (after VD(P) is given): (1) Voronoi vertex inside CH(P), (2) Intersection of a Voronoi edge and an edge of CH(P).
Subgraphs of Delaunay Triangulation
Gabriel Graph Relative Neighborhood Graph Euclidean Minimum Spanning Tree Nearest Neighbor Graph
NNG EMST RNG GG DT Delaunay Triangulation:
p q Delaunay Triangulation: (p,q) is a DT edge empty circle through p and q. p q Gabriel Graph: (p,q) is a GG edge empty circle with diameter (p,q), (i.e., (p,q) intersects its dual Voronoi edge). p q lune(p,q) Relative Neighborhood Graph: (p,q) is an RNG edge rP-{p,q}: (p,q) is NOT the longest edge of triangle (p,q,r) (i.e., d(p,q) max{d(p,r), d(q,r)}) (i.e., lune(p,q) is empty). Euclidean Minimum Spanning Tree: (p,q) is in EMST cycles: (p,q) is NOT the longest edge of the cycle. Nearest Neighbor Graph: (p,q) is a directed edge in NNG rP-{p,q}: d(p,q) d(p,r).
NNG EMST RNG GG DT Delaunay Triangulation
NNG EMST RNG GG DT Gabriel Graph:
NNG EMST RNG GG DT Gabriel Graph:
NNG EMST RNG GG DT Relative Neighborhood Graph:
NNG EMST RNG GG DT Euclidean Minimum Spanning Tree:
NNG EMST RNG GG DT Nearest Neighbor Graph:
Delaunay Triangulation
Euclidean Minimum Spanning Tree
General (m edge, n vertex graph) MST algorithms (See also AAW): Kruskal or Prim O(m log n) or O(m + n log n) time. Yao or Cheriton-Tarjan: O(m log log n) time Chazelle: O(m a(m,n)) time. EMST requires W(n log n) time in the worst-case. [Linear time reduction from the Closest Pair Problem.] EMST in O(n log n) time: (1) Compute DT in O(n log n) time (# edges in DT 3n –6). (2) Apply Prim or Kruskal MST algorithm to DT. Next we will show EMST can be obtained from DT in only O(n) time.
Euclidean Minimum Spanning Tree
D. Cheriton, R.E. Tarjan [1976] “Finding minimum spanning trees,” SIAM J. Comp. 5(4), Also appears in §6.1 of [Preparata-Shamos’85]. Cheriton-Tarjan’s MST algorithm works on general graphs When applied to a planar graph with n vertices and arbitrary edge-weights, it takes only O(n) time. The following graph operations preserve planarity: (a) vertex or edge removal, (b) edge contraction (shrink the edge & identify its two ends): e
Cheriton-Tarjan: MST algorithm (overview)
Input: edge-weighted graph G=(V,E) Q (* queue of sub-trees *) for v V do enqueue (v, Q) (* n single-node trees in Q *) while |Q| 2 do - let T1 be the tree at the front of Q - find edge (u,v) E with minimum weight s.t. u T1 and v T1 - let T2 be the tree (in Q) that contains v - T MERGE (T1 , T2) by adding edge (u,v) - remove T1 and T2 from Q - add T to the end of Q - CLEAN-UP after each stage (see next slide) end
Cheriton-Tarjan Invariants: (a) stage numbers of trees in Q form a non-decreasing sequence. (b) stage(T)=j implies T has at least 2j nodes. So, stage(T) log |T|. (c) after completion of stage j (i.e., the first time stage(T) > j, TQ) there are n/(2j) trees in Q. CLEAN-UP: After the completion of each stage do “clean-up”, i.e., shrink G to G*, where G* is G with each edge in the same tree contracted, i.e., each tree in Q is contracted to a single node, with only those edges (u,v)G* , uT, vT’, That are shortest incident edges between disjoint trees T, T’.
Cheriton-Tarjan: Algorithm
PROCEDURE MST of a Graph G=(V,E) Q (* initialize queue *) for v V do stage(v) 0 ; enqueue (v, Q) j 1 while |Q| 2 do - let T1 be the tree at the front of Q - if stage(T1) = j then CLEAN-UP; j j+1 - (u,v) shortest edge, s.t. u T1 and v T1 - let T2 be the tree (in Q) that contains v - T MERGE (T1 , T2) by adding edge (u,v) - stage(T) 1 + min{ stage(T1), stage(T2)} - remove T1 and T2 from Q - add T to the end of Q end
Cheriton-Tarjan: Analysis
FACTS: (a) A planar graph with m vertices has O(m) edges. (b) Shrunken version of a planar graph is also planar (c) CLEAN-UP of stage j takes O(n/2j) time. (d) # stages log n (e) During stage j each of the < 3n/ 2j edges of G* are checked at most twice (once from each end). So, stage j takes O(n/ 2j) time. (f) Cheriton-Tarjan’s algorithm on planar graphs takes:
THEOREM: The MST of any weighted connected planar graph with n vertices can be computed in optimal O(n) time. COROLLARY: Given DT(P) of a set P of n points in the plane, the following can be constructed in O(n) time: (a) GG(P), (b) RNG(P), (c) EMST(P), (d) NNG(P). Proof: (a) & (d): obvious. (c): use Cheriton-Tarjan on DT(P). (b): see Exercise.
Traveling Salesman Problem (TSP)
Input: An nn positive distance matrix D=(dij), i,j = 1..n, where dij is the travel distance from city i to city j. Output: A traveling salesman tour T. T starts from the home city (say, city 1) and visits each city exactly once and returns to home city. Goal: minimize total distance traveled on tour T: TOPT (1, 3, 4, 2) ((1, 3), (3,4), (4,2), (2,1)) C(TOPT) = d13 + d34 + d42 + d = = 15
Some Classes of TSP General TSP : distance matrix D is arbitrary
Metric-TSP: D satisfies the metric axioms Euclidean-TSP: n cities as n points in the plane with Euclidean inter-city distances These are all NP-hard. Related Problems: Minimum Spanning Tree Hamiltonian Cycle Graph Matching Eulerian Graphs
Hamiltonian Cycle Problem (HCP)
HCP: Given a graph G(V,E), does G have a Hamiltonian Cycle (HC)? HC is any simple spanning cycle of G, i.e., a cycle that goes through each vertex exactly once. HCP is known to be NP-hard. Non-Hamiltonian (Peterson graph) Hamiltonian (skeleton of dodecahedron)
A perfect matching M of weight 5+2+3+6=16 in graph G.
Graph Matching Definition: A matching M in a graph G(V,E) is a subset of the edges of G such that no two edges in M are incident to a common vertex. Weight of M is the sum of its edge weights. A perfect matching is one in which every vertex is matched. 7 5 3 8 9 6 1 3 4 2 7 A perfect matching M of weight =16 in graph G. FACT: Minimum weight maximum cardinality matching can be obtained in polynomial time [Jack Edmonds 1965].
Min Weight Euclidean Matching
Given 2n points in the Euclidean plane, match them in pairs with n matching edges of minimum total length. O(n3) on general weighted graphs: Jack Edmonds [1965] O(n2.5 log4 n): P.M. Vaidya [1989], SICOMP 18(6), Mirzaian [1993], “Minimum weight Euclidean matching and weighted relative neighborhood graphs,” WADS,
Eulerian Graph Definition: A graph G(V,E) is Eulerian if it has an Euler walk Euler Walk is a cycle that goes through each edge of G exactly once. An Eulerian graph G An Euler walk of G FACT 1: A graph G is Eulerian if and only if (a) G is connected and (b) every vertex of G has even degree. FACT 2: An Euler walk of an Eulerian graph can be found in linear time.
Reduction: Let n = |V|. Define the TSP distance matrix as:
General TSP THEOREM: Let r >1 be any constant r-approximation of general TSP is also NP-hard. [So, there is no polynomial-time r-approximation algorithm for general TSP, unless P=NP.] Proof: Reduction from Hamiltonian Cycle Problem (HCP). HCP: Given G(V,E), is G Hamiltonian? Reduction: Let n = |V|. Define the TSP distance matrix as: G is Hamiltonian C(TOPT) = n. G is not Hamiltonian C(TOPT) n + rn (1+r) n > r n. Suppose there were a poly-time r-approx TSP algorithm producing a tour T. C(T)/C(TOPT) r G has HC if and only if C(T) = n. This would provide a poly-time algorithm for the HCP!
Metric & Euclidean TSP Euclidean Traveling Salesman Problem (ETSP):
Metric Traveling Salesman Problem (metric-TSP): special case of general TSP (distance matrix is metric) NP-hard 2-approximation: Rosenkrants-Stearns-Lewis [1974] 1.5-approximation: Christofides [1976] Euclidean Traveling Salesman Problem (ETSP): special case of Metric-TSP (with Euclidean metric) PTAS: Sanjeev Arora [1996] [CSE4101/5101 Lecture Note 10].
2-approximation of metric-TSP
Rosenkrants-Stearns-Lewis [1974] (See also AAW) C(T) 2 C(TOPT) Minimum Spanning Tree (MST) Euler walk around double-MST Bypass repeated nodes on the Euler walk. FACT 1: Triangle inequality implies bypassing nodes cannot increase length of walk. FACT 2: C(T) 2 C(MST) 2 C(TOPT) .
r = 2 is tight (even for Euclidean instances)
Tour T Tour TOPT Euclidean Instance Euler walk of double MST MST
1.5-approximation for metric-TSP
Christofides [1976]: C(T) 1.5 C(TOPT) MST Odd degree nodes in MST M = Minimum weight perfect matching on odd-degree nodes E = MST + M is Eulerian Find an Euler walk of E. Bypass repeated nodes on the Euler walk to get a TSP tour T. FACTS: Any graph has even # of odd degree nodes. (Hence, M exists.) C(MST) C(TOPT) C(M) 0.5 C(TOPT) (See next slide.) C(E) = C(MST) + C(M) 1.5 C(TOPT) C(T) C(E) 1.5 C(TOPT)
C(M) 0.5 C(TOPT) Odd degree node in MST TOPT
C(M) 0.5 C(TOPT) C(M) C(M1) Odd degree node in MST M1 TOPT
C(M) 0.5 C(TOPT) C(M) C(M2) Odd degree node in MST M2 TOPT
C(M) 0.5 C(TOPT) C(M) C(M1) C(M) C(M2)
2 C(M) C(M1) + C(M2) C(TOPT) Odd degree node in MST M1 M2 TOPT
r = 1.5 is tight (even for Euclidean instances)
MST: MST + M: Tour T: Tour TOPT :
Open Problem Weighted Relative Neighborhood Graph (WRNG): RNG of a set of circular disks This arises in EMWM computation [Mirzaian’93] weighted distance (triangle inequality not satisfied): FACT: WRNG WGG WDT. WDT and WGG can be computed in O(n log n) time. Open Problem: Can WRNG be computed in O(n log n) time?
Minimum Weight Triangulation
Problem: Given a set P of n points in the Euclidean plane, find a triangulation of P with minimum total edge length. MWT is NP-hard: [LN16: Mulzer-Rote, SoCG’06] See LN17 & LN18 also. Delaunay Triangulation can give W(n) approximation ratio: e DT MWT Greedy Approximation: W( 𝑛 ) approx. ratio: - Start the triangulation T with no edges Repeatedly select the shortest edge e (from unselected edges that do not cross any selected edges) and add e to T.
In general, dynamic programming can be used for triangulation of point sets or polygons.
Maximize the minimum angle: Delaunay Triangulation , O(n log n) Minimize the maximum angle: SoCG’89-90, O(n2 log n) Minimize the longest edge: Use RNG Minimize total edge length: MWT [LN16,17,18]
Extensions of Voronoi Diagrams
Voronoi Diagram of line-segments, circles, … Medial Axis Order k Voronoi Diagram Farthest Point Voronoi Diagram (order n-1 VD) Weighted Voronoi Diagram & Power Diagrams Generalized metric (e.g., Lp metric)
VD of line-segments & circles
parabola par line line par line par line hyperbola
Medial Axis MA: Locus of centers of empty circles that touch 2 or more data points. [Blum’76] Applications: pattern recognition, medical CAT scan imaging, … MA of polygons in linear time: convex polygons: [Aggarwal et al. 1989] simple polygons: [Chin-Snoeyink-Wang, Discrete Comp. Geom.1999]
Higher Order VD [PrS85] § 6.3 [Ede87] § 13.3-13.5 [ORo98] § 6.6.
Aurenhammer [1987], “Power diagrams: properties, algorithms, and applications,” SIAM J. Computing 16, D.T. Lee [1982], “On k-Nearest Neighbors Voronoi Diagram in the Plane,” IEEE Trans. Computers, C-31,
Order k Voronoi Diagram
Given a set of n sites in space, partition the space into regions where any two points belong to the same region iff they have the same set of k nearest sites. 5,8 5,6 2,5 5 6,8 Example: Order 2 Voronoi Diagram Some Voronoi regions of order 2 are empty, e.g. (5,7). 2 4,5 6 2,4 8 1,2 4,6 1,4 4 1 6,7 3,4 4,7 7,8 3 7 1,3 3,7
Order k Voronoi Diagram
H(pi, pj) half-space pi pj Voronoi region of T (a convex polyhedron, possibly empty) Order k Voronoi Diagram of P: possible subsets of size k. Most have empty Voronoi regions. In 2D only O(k(n-k)) of them are non-empty [D.T. Lee’82].
VDk in d k-belt of arrangement of hyperplanes in d+1
k-belt = region between level k and level (k+1) in the arrangement. Level 2 2-belt Level 3
Example: VD2 in 1 y y=x2 1,2 x p3 p1 p2 p4 2,3 3,4
ALGORITHM Order-k VD of P 2
For each point p P do - lift p onto L: z = x2 + y2, call it l(p) - P(p) plane tangent to L at l(p) Construct k-belt of the arrangement of the planes P(p), p P, in 3. Project down this k-belt onto the base plane 2. This is the k-th order Voronoi Diagram of P. FACT 1: In O(n3) time we can compute all k-levels of the arrangement and find the k-th order VD. Improvement by [D.T. Lee 1982]: FACT 2: [Preparata-Shamos’85]: k-th order VD of n points in the plane can be obtained in time O(min { k2 , (n-k)2 } n log n). COROLLARY 3: Complexity of the k nearest neighbors query problem is: O(k + log n) Query Time O(k2 n log n) Preprocessing Time O(kn) Space.
VD in Lp metric L metric: || p1 – p2 || = max { | x1 – x2 | , |y1 – y2 | } p1 p2 “unit circle” PB L1 metric: || p1 – p2 ||1 = | x1 – x2 | + |y1 – y2 | p1 p2 “unit circle” PB
Largest empty square in O(n log n) time.
VD in Lp metric Voronoi Diagram in L metric in O(n log n) time & O(n) space. Largest empty square in O(n log n) time.
Robot Motion Planning B A
obstacle A robot Assume robot moves by translation only (no rotation).
Robot Motion Planning Approach: Voronoi Diagram
Minkowski Sum AB = { p + q | p A, q B } = p-q p AN -q diameter(A) = AAN q A
Exercises
(a) Draw the Voronoi diagram of 10 points all on a line
(a) Draw the Voronoi diagram of 10 points all on a line. (b) Draw separately the Voronoi diagram of 10 points all on a circle. (c) Draw the Voronoi diagram of 10 random points in the plane using the L1 metric instead of the L metric. Show an example of a Voronoi cell that is not convex. Let P and Q be two point sets in the plane, and let p P and q Q be the two points from these sets that minimize dist(p,q). True or false: line-segment (p,q) is an edge of DT(P Q). Explain. Dynamic maintenance of Voronoi Diagrams: We are given the Voronoi Diagram VD(P) of a set P of n points in general position in the plane. (a) Show how to delete a given point p P, thereby constructing the Voronoi Diagram of P-{p} Your method should take O(k log k) time where k is the number of edges on the boundary of the Voronoi region corresponding to p. (b) Optional: Can you improve the running time of your algorithm for part (a) to O(k)? (c) We are given a point qP and a Delaunay triangle in DT(P) whose circumscribing circle contains q Show how to insert q, thereby constructing the Voronoi Diagram of P{q}. Your method should take O(k) time, where k is size of the combinatorial “change” in the VD. [Note: DT(P) is not given.] Modified Delaunay Triangulation: We are given the Delaunay Triangulation of a set P of n point in the plane. Suppose the x-coordinate of each point in P with positive x-coordinate is increased by 1. There are no other changes to the coordinates of the points in P. How fast can you algorithmically update the Delaunay Triangulation of the modified P? Describe and justify your answer in detail. Let S be a set of n segments in the Euclidean plane. The distance between two segments s1, s2 is defined as d(s1, s2) = min { d(p, q) | p s1, q s2 }. Describe a randomized algorithm for computing a nearest (respectively, farthest) pair of segments in S whose expected running time is O(n log n).
Farthest Point Voronoi Diagram: The following problems have to do with the farthest-point Voronoi diagram. For a set S of n points in the plane, define the farthest-point Voronoi diagram of S, denoted FVD(S), as the union of farthest-point cells, where we define for each point pS, FV(p) to be the set of all points that have p as their farthest neighbor in S. (a) Give an example of a set S and a point pS such that FV(p) is empty. (b) Show that each non-empty region in a farthest-point Voronoi diagram is unbounded and associated with a point on the convex hull of S. (c) Show that FVD(S) in the plane is a tree. (d) Define the farthest-point Delaunay Triangulation FDT(S) as the graph-theoretic planar dual of FVD(S). Derive a relationship between FDT(S) and the convex hull of the 3-dimensional set of points defined by mapping each point (x,y) S to the point (x,y, x2 +y2). Use this fact to design a simple O(n log n) time algorithm for constructing FVD(S). (e) Prove that each vertex v of FVD(S) corresponds to a circle C determined uniquely by the sites defining v such that C contains all the points of S. (f) Show how to use FVD(S) to compute the smallest-radius circle that contains S. What is the running time of your method? Minkowski Sum: Let A and B be two convex polygons in the plane, with m and n vertices respectively. If we regard a point in the plane as a vector, then the Minkowski sum of A and B is defined as A B = {a + b a A , b B}. Show that A B is a convex polygon with at most m + n vertices. How fast can one compute A B?
Hausdorf Distance: The set distance between two sets of points A and B is defined as d(A,B) = min {d(p,q) | pA , qB } , where d(p,q) is the Euclidean distance between points p and q. The directed Hausdorf distance from a set of points A to a set of points B is H(A,B) = max {d(p,B) | pA } , and the Hausdorf distance between a set of points A and a set of points B is D(A,B) = max { H(A,B), H(B,A)}. [This metric is used to measure similarity between two sets of points.] Outline an efficient algorithm for computing the Hausdorf distance between two sets, A and B, each containing n points in the plane. What is the running time of your method? Voronoi Diagram in L metric: The L distance is dist(p, q) = max( |px − qx |, | py − qy | ). (a) Given a point q, describe the set of points that are at L distance w from q. (b) Given two points p and q, describe the set of points that are equidistant from p and q in the L distance [Hint: The bisector is a polygonal curve. As a function of the coordinates of p and q, indicate how many vertices this curve has and where these vertices are located. Give exact formula, and not just a high-level description.] (c) Argue that the Voronoi diagram for n sites in the L metric has O(n) line-segments and vertices [Hint: First show that each cell of the Voronoi diagram is connected. Take care in your application of Euler's formula, since some vertices may have degree two.] (d) Assuming that the L Voronoi diagram is given, describe a method to compute the largest empty axis-aligned square centered at a query point q. Your method should involve O(n) space and O(log n) query time. [You are not required to show how to build the data structure, only to argue its existence.]
Let P = {p1, p2, … , pn} be the vertices of a convex polygon, given in counterclockwise order. The purpose of this problem is to modify the randomized incremental Delaunay triangulation algorithm to run in O(n) expected time. Unlike the algorithm presented in class, we do not put all the sites within a large bounding triangle. We start with the triangle defined by three random sites from P. The remaining sites are inserted in random order; each new site is connected to the convex hull, and edge flips are performed until the circumcircle condition is satisfied. (See the figure below.) (a) Give a backwards analysis to show that the expected number of edge-flips performed with each insertion is O(1). (b) Whenever a new site is added, we need to determine where along the current convex hull it is to be added. Explain how to do this in O(1) expected time. You are allowed to preprocess the sites in O(n) time.
Given a set P of n points in the plane in general position (no 3 collinear & no 4 cocircular), the objective of this problem is to determine the pair of concentric circles C1 and C2 such that every point of P lies between these circles, and the area of the region between these two circles is minimized. [Note: There is a harder variant of this problem in which the goal is to minimize the width of this region.] Consider the subdivision S formed by overlaying NVD(P) and FVD(P), the nearest and farthest Voronoi diagrams of P, respectively. The vertices of S consist of the union of the vertices of the two Voronoi diagrams together with the points at which two edges, one from each diagram, intersect each other. (a) Prove that the center of the optimum concentric circles cannot lie in the interior of any face of S. (b) Prove that the center of the optimum concentric circles cannot lie in the relative interior of any edge of S. (c) Using (a) and (b), give an O(k + n log n)-time algorithm for solving this problem, where k is the number of vertices of S. (Assume NVD(P) and FVD(P) can be computed in O(n log n) time.) (d) Show that the problem can be solved in O(n) time by using the fact that linear programming with O(n) constraints and a constant number of variables can be solved in O(n) time. C1 C2
Determining whether a general graph has a spanning tree of degree at most three is NP-complete. In this problem, we consider this problem for planar triangulations. Given a connected graph G = (V,E), a spanning subgraph of G is any graph G’= (V,E’) with E’ E. A spanning tree is a connected acyclic spanning subgraph of G. Recall that a triangulation of a point set V is a PSLG whose vertex set is V, whose external face consists of the edges of the convex hull of V , and whose internal faces are triangles. Prove that any triangulation of a set of points in the plane (viewed as a planar graph) has a spanning tree of degree at most three. See Fig. (a). [Hint: First prove the following by induction on the number of vertices. Consider any PSLG triangulation whose external face is homeomorphic to a circle. Such a graph is characterized by three conditions: its internal faces are triangles, there are at least three vertices on its external face, and every vertex on its external face has exactly two neighbors on the external face. Prove that such a PSLG has a connected spanning subgraph of degree at most three such that every edge of the external face is in that subgraph. See Fig. (b).] (a) (b)
The Delaunay triangulation has an interesting monotonicity property with respect to ray shooting. To illustrate it, let us consider a triangulation T spanning a set P of n points in the plane. Let p be an arbitrary point in the plane. For any triangles ti and tj in T, we write ti <p tj if there exists a ray starting at p such that a traveller traversing along the ray from p passes through ti before tj (see Figure (a) below). If there is no cyclic sequence (t1 , t2 , ... , tk , t1 ) of triangles in T satisfying t1 <p t2 <p ... <p tk <p t1 , k ³ 2, we say that T is monotone with respect to ray shooting at p. If T is monomtone with respect to ray shooting at every point in the plane, we say T is ray-shoot monotone. Figure (b) below shows that not every triangulation is ray-shoot monotone. In that example t1 <p t2 because of ray r1, t2 <p t3 because of ray r2 , and t3 <p t1 because of ray r3. Thus, t1 <p t2 <p t3 <p t1 . Therefore, this triangulation is not monotone with respect to ray shooting at point p, and hence it is not ray-shoot monotone. Prove Delaunay triangulations are ray-shoot monotone. [Hint: use the lifting method.]
Computing geometric properties of a union of spheres is important to many applications in bioinformatics. The following problem is a 2-dimensional simplification of one of these problems. (a) You are given a set P of atoms representing a protein, which for our purposes will be represented by a collection of circles in the plane, all of equal radius ra . Such a protein lives in a solution of water. We will model a molecule of water by a circle of radius rb > ra . A water molecule cannot intersect the interior of any protein atom, but it can be tangent to one. We say that an atom a P is solvent-accessible if there exists a placement of a water molecule that is tangent to a, and the water molecule does not intersect any other atoms in P. In the figure below, all atoms are solvent-accessible except for three (as indicated). Given a protein molecule P of n atoms, devise an O(n log n) time algorithm to determine all solvent-inaccessible molecules of P. (b) Solve the problem when atoms in P are of non-uniform size, i.e., they are circles of varying radii.
Kinetic Delaunay Triangulation: We are given a set P = {p1 ,
Kinetic Delaunay Triangulation: We are given a set P = {p1 , ... , pn} of n stationary points and a moving point p0, all in the plane. Point p0 moves on a line trajectory with a given constant speed, i.e., at time t, this point is at location p0(t) = a × t + b, where a and b are given 2-vectors. Assume P is in general position (i.e., no three points collinear and no four points cocircular), and the trajectory line of p0 does not hit P. Consider the kinetic set Qt = P È {p0(t)} and its time varying Delaunay Triangulation DT(Qt ). We want to preprocess Qt into a compact data structure so that for any given query time t, we can efficiently extract DT(Qt ) from it. (a) Derive a bound on the combinatorial size of changes in DT(Qt ) as t goes from -¥ to +¥. (b) Design and analyze the preprocessing and query answering algorithms and the data structure. Relative Neighborhood Graph: Suppose we are given the Delaunay Triangulation DT(P) of a set P of n points, in general position, in the plane. We know that a Delaunay edge e is in RNG iff lune(e) is empty. Consider a Delaunay edge e that is not in RNG. Then, there must be some point pP (distinct from two ends of e) such that plune(e). We say such a site p kills edge e. (a) Show that if a Delaunay edge e is killed by a site p, then there is a sequence of one or more Delaunay edges (e1, e2, … , ek), all killed by p, e = ek , (p, e1) is a Delaunay triangle, and for i = 1..k-1, ei and ei are two edges of a (common) Delaunay triangle. (b) Start with the Delaunay triangulation and assume its edges are coloured blue. Consider a triangle incident to site p whose opposite side is a blue edge e killed by p. Update the triangulation by an edge-flip that (permanently) removes blue edge e and adds a new red edge incident to p. Is it possible to remove all Delaunay edges killed by p with repeated application of such an edge-flipping process? (c) Can we make the blue edge removals in part (b) permanent, while we continue applying the same process, restarting with different sites p? In other words, is it true that after the process in part (b) is sequentially applied to all sites p, the surviving blue edges form the RNG? (d) Show that given DT(P), RNG(P) can be computed in O(n) time. (e) What is the computational complexity of the Weighted RNG (WRNG)?
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# Tag Archives: von neumann regular
## A short survey of von Neumann regular rings
I’ve talked a lot about von Neumann regular rings on this blog, so I thought I’d write an informal short survey on them, collecting some facts we’ve already seen and many new ones. It should give you an idea of what von Neumann regular rings are. Most of the facts that I did not explicitly […]
## Roger Ming’s theorem on von Neumann regular rings
We say that an associative ring $A$ is von Neumann regular if for every $a\in A$ there exists a $x\in A$ such that $axa = a$. That is a rather strange condition, isn’t it? But, you can think of $x$ as a pseudoinverse to $a$. This weakening of inverses has a homological counterpart: if every […]
## A zero-dimensional ring that is not von Neumann regular
An associative ring $R$ is called von Neumann regular if for each $x\in R$ there exists a $y\in R$ such that $x = xyx$. Now let $R$ be a commutative ring. Its dimension is the supremum over lengths of chains of prime ideals in $R$. So for example, fields are zero dimensional because the only […]
## Injective and p-injective
An $R$-module $M$ is called injective if the functor $\Hom_R(-,M)$ is exact. The well-known Baer criterion states that an $R$-module $M$ is injective if and only if for every ideal $I$ of $R$, every map $I\to M$ can actually be extended to a map $R\to M$. For example, $\Q$ is an injective $\Z$-module. If every […]
## Local Rings of Weak Dimension Zero are Division
Let $R$ be a ring and $M$ be an $R$-module. The flat dimension of $M$ is the infimum over all lengths of flat resolutions of $M$. Usually, the flat dimension of $M$ is denoted by $\mathrm{fd}_R(M)$. For example, $\mathrm{fd}_{\mathbb{Z}}(\mathbb{Q}) = 0$. Since $\mathbb{Q}$ has projective dimension $1$, the flat dimension and projective dimension of a […]
## Basic Examples of the Tensor Product and Flatness
The tensor product is one of the most important constructions in mathematics, and here we shall see my favourite examples of the tensor product in action, hopefully to illuminate its properties for beginners. Proofs or references are provided, but since the emphasis is on examples, the proofs that are given are terse and details are […]
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# FW: Bogus mathematics, (was Re: L'état de PCT, c'est moi (was ...))
[From Rick Marken (2018.08.17.06:00)]
[From Bruce Abbott (2018.08.16.0930 EDT)]
Â
[Rick Marken 2018-08-15_18:27:22]
 [From Bruce Abbott (2018.08.15.1030 EDT)]
Â
BA: Pollick and Shapiro (1995) performed an analysis of the velocity-based formula… Their analysis was essentially identical to yours (both add affine velocity to the regression equation as a separate factor) but their conclusion was not. …Â they concluded that the power coefficient of 2/3 will appear across observations only if the affine velocity is constant across observations…
RM: Maoz et al. (2006) also found what we had found and came to the same mistaken conclusion about what it meant as did Pollick and Shapiro (1995). We addressed this in our rebuttal:" [their conclusion]** amounts to assuming that one’s theory of the cause of the power law is correct and any deviation of data from the theory is the fault of the data. **
Â
BA: The fact that you just restated this misconception is proof that you either did not read, did not understand, or have chosen to ignore, the analysis I presented that should be the subject of your reply. That analysis is a refutation of the position you take above in the second bolded section.
RM: I don’t understand what there is to refute. Both Pollick and Shapiro (1995) and Maoz et al. (2006) found exactly what we did: that the degree of deviation of the power coefficient found by regressing curvature on velocity will deviate from the power law value  (1/3 or 2/3 depending on how the variables are measured) in proportion to the covariance between curvature and the omitted variable, affine velocity. The deviation from 1/3 or 2/3 will be zero when the covariance between curvature and velocity is zero, which will happen when affine velocity is constant throughout the trajectory.Â
RM: Maybe you think that “only if the affine velocity is constant across observations…” means something different than what I think it means. I think it means “constant across observations throughout a trajectory”. Maybe you think it means “constant across observations of different trajectories”. Is that it? If so, that is not what Pollick and Shapiro (1995),  Maoz et al. (2006) and Marken & Shaffer (2018) mean by constant affine velocity.
Â
RM: But ignoring the mathematics for now, I’m still interested in knowing what you think is the correct application of PCT to the power law…
Â
BA: Ignoring the mathematics? Let’s not try to duck the issue by changing the subject. At this point there are only two appropriate responses to the analysis I laid out: (1) admit that you have made a serious error of mathematical logic, or (2), provide a mathematical proof that invalidates my refutation. To be clear, the serious error to which I refer is the following: That the formula for curvature implies that for a time series of V,C pairs, “the true (mathematical) power coefficient (beta) relating these variables is 2/3.�
Here’ the proof that the true power coefficient (beta) relating V (called v(t) and C (which is here measured as k(t), which is proportional to the reciprocal of the Gribble Ostry equatoin for curvature measure as R) is -1/3 (from Maoz et al. ,2006).Â
And their version of the omitted variable analysis is:Â
RM: The last sentence basically confirms our conclusion: that any trajectory, regardless of how it is produced, where log (affine velocity) and log (curvature) are uncorrelated, will precisely correspond to the power law.Â
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BA: To get you started along this path, the first thing I need from you is evidence that you have read the argument against your conclusion and have understood it well enough to restate it correctly in your own words. In your reply, insert restatement here:
RM: I’m sure that I can’t provide such evidence even if I do. I think we’ve reached the point where this is getting us nowhere. My goal in this whole long debate was simply to try to get people to ignore the power law and do some PCT research – research aimed at testing to determine what perceptual variable people control wen they move. I think it’s pretty clear that no one wants to do this; they want to study the power law. So go ahead. Do the research, publish it and I’ll see what I think. Maybe you are all geniuses and I’m just a self- deluded moron. That will be OK. I’m still pretty rich and very good looking. So I’ll get by;-)
BestÂ
Rick
···
Richard S. MarkenÂ
"Perfection is achieved not when you have nothing more to add, but when you
have nothing left to take away.�
                --Antoine de Saint-Exupery
[Rick Marken 2018-08-21_15:35:23]
[From Bruce Abbott (2018.08.21.0900 EDT)]
BA: To get you started along this path, the first thing I need from you is evidence that you have read the argument against your conclusion and have understood it well enough to restate it correctly in your own words. In your reply, insert restatement here:
RM: I’m sure that I can’t provide such evidence even if I do.
Â
BA: You can’t simply restate the argument I presented. Wow.
RM: Not that I can’t. I won’t. Perhaps you will understand how annoying you sound if I said the same thing to you: The first thing I need from you is evidence that you have read the argument against your conclusion and have understood it well enough to restate it correctly in your own words.Â
RM: I have seen absolutely no evidence that you understand my explanation of the power law – the explanation that I laid out so clearly in Marken & Shaffer (2017) and Marken and Shaffer (2018).
Â
RM: I think we’ve reached the point where this is getting us nowhere.
Â
BA: Well, yes, given that you refuse to cooperate, even to restate my position so that I know that you have read and understood it.
RM: Well, excuuuuuuuuse me for not cooperating with you. But my people haven’t had a very good experience with inquisitions so I guess I just don’t respond well to them.Â
Â
RM: My goal in this whole long debate was simply to try to get people to ignore the power law and do some PCT research – research aimed at testing to determine what perceptual variable people control wen they move.
Â
BA: Ignore the power law? The power law research is aimed at finding valid explanations for why tangential velocity often is a power-law function of curvature. Your “explanationâ€? – that it’s simply an artifact of the way curvature is calculated – is false, so the problem of explaining why such relationships emerge remains. Ignoring the power law has to be the worst method ever for developing such explanations!
RM: Then go ahead and start finding explanations. Publish papers that show how wrong I am and how right you are. But you are just wasting your time trying to get be to recant. I am obviously a hopeless case.Â
BA: The power law is not a “behavioral illusionâ€? – nobody is claiming that curvature is a stimulus, to wwhich tangential velocity is a response.Â
RM: So I’ve been told. Nevertheless, it is a behavioral illusion, just like selection by consequences is a behavioral illusion. A behavioral illusion exists when the behavior we see is not what it seems.
Â
BA: The relationship described as a power law probably emerges for many reasons, depending on the circumstances under which it is observed. For example, racecar drivers circulating around an oval track speed up on the straights and slow in the turns, thus conforming to the power law.Â
RM: It sounds like you are saying that in this situation curvature is a stimulus to which tangential velocity is a response. I thought inquisitors were taught to keep their story straight. Or was it that they are taught not to keep their story straight. That would work better, actually, wouldn’t it.
Â
RM: I think it’s pretty clear that no one wants to do this; they want to study the power law. So go ahead. Do the research, publish it and I’ll see what I think. Maybe you are all geniuses and I’m just a self- deluded moron.
Â
BA: Well, self-deluded for sure. You won’t even try to understand where your error lies. You act as if you think everyone else is a moron and you are the genius – the oonly one who can see that the emperor has no clothes.
RM: Then why not just ignore me and do your power law studies and show what’s really going on. And best of all, give your PCT explanation of the power law (if there is one) so you can get PCT out there for the public. You certainly don’t want me to be the only person publishing PCT stuff, do you?
Â
RM: That will be OK. I’m still pretty rich and very good looking. So I’ll get by;-)
Â
BA: Rich for sure – I bought your book, Controlling Peoplee!
RM: Well that’s a surprise. I thought Pope Boris had put my books on the Index.
BestÂ
Rick
Â
···
Â
Bruce
Â
Richard S. MarkenÂ
"Perfection is achieved not when you have nothing more to add, but when you
have nothing left to take away.�
                --Antoine de Saint-Exupery
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FIN 534 Week 11 Quiz 10 Click Here to Buy the Tutorial http://www.uophelp.com/FIN-534-/product-6359-FIN-534-Week-11-Quiz-10 For more course tutorials visit www.uophelp.com
Finance 534 week 11 quiz 10
Question 1
Suppose DeGraw Corporation, a U.S. exporter, sold a solar heating station to a Japanese customer at a price of 143.5 million yen, when the exchange rate was 140 yen per dollar. In order to close the sale, DeGraw agreed to make the bill payable in yen, thus agreeing to take some exchange rate risk for the transaction. The terms were net 6 months. If the yen fell against the dollar such that one dollar would buy 154.4 yen when the invoice was paid, what dollar amount would DeGraw actually receive after it exchanged yen for U.S. dollars?
Question 2
Suppose 144 yen could be purchased in the foreign exchange market for one U.S. dollar today. If the yen depreciates by 8.0% tomorrow, how many yen could one U.S. dollar buy tomorrow?
Question 3
Suppose one British pound can purchase 1.82 U.S. dollars today in the foreign exchange market, and currency forecasters predict that the U.S. dollar will depreciate by 12.0% against the pound over the next 30 days. How many dollars will a pound buy in 30 days? Answer
Question 4
Suppose 6 months ago a Swiss investor bought a 6-month U.S. Treasury bill at a price of \$9,708.74, with a maturity value of \$10,000. The exchange rate at that time was 1.420 Swiss francs per dollar. Today, at maturity, the exchange rate is 1.324 Swiss francs per dollar. What is the annualized rate of return to the Swiss investor?
Question 5
A box of candy costs 28.80 Swiss francs in Switzerland and \$20 in the United States. Assuming that purchasing power parity (PPP) holds, what is the current exchange rate?
Question 6
Suppose one year ago, Hein Company had inventory in Britain valued at 240,000 pounds. The exchange rate for dollars to pounds was 1£ = 2 U.S. dollars. This year the exchange rate is 1£ = 1.82 U.S. dollars. The inventory in Britain is still valued at 240,000 pounds. What is the gain or loss in inventory value in U.S. dollars as a result of the change in exchange rates?
Question 7
Which of the following is NOT a reason why companies move into international operations?
Question 8
If one U.S. dollar buys 1.64 Canadian dollars, how many U.S. dollars can you purchase for one Canadian dollar?
Question 9
If the inflation rate in the United States is greater than the inflation rate in Britain, other things held constant, the British pound will
Question 10
In 1985, a given Japanese imported automobile sold for 1,476,000 yen, or \$8,200. If the car still sold for the same amount of yen today but the current exchange rate is 144 yen per dollar, what would the car be selling for today in U.S. dollars?
Question 11
Suppose the exchange rate between U.S. dollars and Swiss francs is SF 1.41 = \$1.00, and the exchange rate between the U.S. dollar and the euro is \$1.00 = 1.64 euros. What is the cross-rate of Swiss francs to euros?
Question 12
Suppose hockey skates sell in Canada for 105 Canadian dollars, and 1 Canadian dollar equals 0.71 U.S. dollars. If purchasing power parity (PPP) holds, what is the price of hockey skates in the United States?
Question 13
If one Swiss franc can purchase \$0.71 U.S. dollars, how many Swiss francs can one U.S. dollar buy?
Question 14
2 out of 2 points
Suppose 90-day investments in Britain have a 6% annualized return and a 1.5% quarterly (90-day) return. In the U.S., 90-day investments of similar risk have a 4% annualized return and a 1% quarterly (90-day) return. In the 90-day forward market, 1 British pound equals \$1.65. If interest rate parity holds, what is the spot exchange rate?
•
Question 15
Which of the following statements is NOT CORRECT?
Fin 534 week 11 quiz 10
Fin 534 week 11 quiz 10
FIN 534 Week 11 Quiz 10 FIN 534 Week 11 DQ 2 FIN 534 Week 11 DQ 1 FIN 534 Week 10 Quiz 9 FIN 534 Week 10 DQ 2 FIN 534 Week 10 DQ 1 FIN 534 W...
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# Factors of 10
The factors of 10 are the numbers that divide the original uniformly. Factor pairs of the number 10 are the whole numbers, which produces the actual number when multiplied. For example, when we multiply 2 and 3, we get 6, i.e. 2 × 3 = 6. Here, 2 and 3 are the factors of 6. Similarly, we can find the factors of the given number, i.e. 10. However, we can also use the factorization method to get the factors of a number.
Factors of 10: 1, 2, 5 and 10
In the factorization method, first, consider the numbers 1 and 10 as factors of 10 and continue with finding the other pair of multiples of 10, which gives the results as an original number. To understand this method, read the article below to find the factors of 10 in pairs. Also, get the prime factors of 10 with the help of the division method.
## How to Find the Factors of 10?
Go through the following steps to find factors of 10.
Step 1: First, write the number 10
Step 2: Find the two numbers, which results in 10 under the multiplication, say 2 and 5, such that 2 × 5 = 10.
Step 3: We know that 2 and 5 are prime numbers with only two factors, i.e., one and the number itself.
The factors of 2 = 2 × 1
The factors of 5 = 5 x 1
So, we cannot further factorize them.
Step 4: Therefore, the factorization of 10 can be expressed as 10 = 2 × 5 × 1
Step 5: Finally, write down all the unique numbers which we can obtain from the above process.
Factors of 10 1, 2, 5, and 10
## Pair Factors of 10
To find the pair factors of 10, multiply the two numbers to get the original number as 10. We can write both positive and negative integers in pairs as shown below:
Positive pairs Negative pairs 1 × 10 = 10; (1, 10) (-1) × (-10) = 10; (-1, -10) 2 × 5 = 10; (2, 5) (-2) × (-5) = 10; (-2, -5) 5 × 2 = 10; (5, 2) (-5) × (-2) = 10; (-5, -2) 10 × 1 = 10; (10, 1) (-10) × (-1) = 10; (-10, -1)
Therefore, the positive pair factors of 10 are (1, 10), (2, 5), (5, 2) and (10, 1).
The negative pair factors of 10 are (-1, -10), (-2, -5), (-5, -2) and (-10, -1).
### Prime Factors of 10
As we know, 10 is a composite number, and it has prime factors. Now let us find the prime factors of 10.
• The first step is to divide the number 10 with the smallest prime number, i.e. 2.
10 ÷ 2 = 5
Now, divide 5 by 2.
5 ÷ 2 = 2.5
Factors should be whole numbers, so 2 cannot be the factor of 5. Hence, proceed with the next prime numbers, i.e. 3, 5
5 ÷ 3 = 1.6667
5 ÷ 5 = 1
• We have received number 1 at the end of the division process; thus, we cannot proceed further.
So, the prime factors of 10 are 2 and 5, where 2 and 5 are the prime numbers.
Prime factorisation of 10 is 2 × 5.
Some of the important facts about the factors of 10 are listed below:
• The number of factors of 10 is 4.
• The sum of all factors of 10 is equal to 18.
• The product of all factors of 10 is equal to square of 10 or 10 times of 10.
## Frequently Asked Questions on Factors of 10 – FAQs
### What are all the multiples of 10?
The multiples of 10 include: 10, 20, 30, 40, 50, 60, 70, 80, 90, 100, etc.,
### What are the prime factors of 10?
The prime factors of 10 are 2 and 5.
### What are factors of a number?
Factors are the numbers when multiplied together to get another number. A number can have two or more factors.
### What are all factors of 12?
All the factors of 12 are 1, 2, 3, 4, 6, and 12.
### What are factors of 100?
The factors of 100 include 1, 2, 4, 5, 10, 20, 25, 50, and 100.
### How many factors do 10 and 100 have?
The number 10 has 4 factors, such as 1, 2, 5 and 10.
The number 100 has 9 factors, such as 1, 2, 4, 5, 10, 20, 25, 50, and 100.
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# smooth submanifold contained in a subvariety of same dimension is real analytic
This theorem seems to usually be attributed to Malgrange in literature as it appeared in his book[1].
###### Theorem (Malgrange).
Suppose $M\subset{\mathbb{R}}^{N}$ is a connected smooth ($C^{\infty}$) submanifold and $V\subset{\mathbb{R}}^{N}$ is a real analytic subvariety of the same dimension as $M$, such that $M\subset V$. Then $M$ is a real analytic submanifold.
The condition that $M$ is smooth cannot be relaxed to $C^{k}$ for $k<\infty$. For example, note that in ${\mathbb{R}}^{2}$, the subvariety $y^{3}-x^{8}=0$, which is the graph of the $C^{1}$ function $y=\lvert x\rvert^{\frac{8}{3}}$, is not a real analytic submanifold.
## References
• 1 Bernard Malgrange. . Oxford University Press, 1966.
Title smooth submanifold contained in a subvariety of same dimension is real analytic SmoothSubmanifoldContainedInASubvarietyOfSameDimensionIsRealAnalytic 2013-03-22 17:41:16 2013-03-22 17:41:16 jirka (4157) jirka (4157) 5 jirka (4157) Theorem msc 14P99 RealAnalyticSubvariety
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`There are n people that are split into some unknown number of groups. Each person is labeled with a unique ID from 0 to n - 1.You are given an integer array groupSizes, where groupSizes[i] is the size of the group that person i is in. For example, if groupSizes[1] = 3, then person 1 must be in a group of size 3.Return a list of groups such that each person i is in a group of size groupSizes[i].Each person should appear in exactly one group, and every person must be in a group. If there are multiple answers, return any of them. It is guaranteed that there will be at least one valid solution for the given input.`
`There are n people that are split into some unknown number of groups. Each person is labeled with a unique ID from 0 to n - 1.`
`You are given an integer array groupSizes, where groupSizes[i] is the size of the group that person i is in. For example, if groupSizes[1] = 3, then person 1 must be in a group of size 3.`
`Return a list of groups such that each person i is in a group of size groupSizes[i].`
`Each person should appear in exactly one group, and every person must be in a group. If there are multiple answers, return any of them. It is guaranteed that there will be at least one valid solution for the given input.`
• groupSizes is an array of integers(numbers)
• Each person is an index, that’s the unique ID
• Each group’s size is stored in the groupSizes array
• groupSizes[Person’s Id] = group size of this Person
• Each person has ONLY ONE group
• Every person MUST BE in a group
• Each input(array of numbers) will always have AT LEAST 1 valid output
`let result = [], groups = {};`
`const groupThePeople = function(groupSizes) { let result = [], groups = {}; groupSizes.forEach((size, index) => { if (size in groups) { ... `
`... if (size in groups) { } else { groups[size] = [index] } ...`
`if (size in groups) { if (groups[size].length < size) { groups[size].push(index) } } else { groups[size] = [index] }`
`...if (groups[size].length === size) { }...`
`if (groups[size].length === size) { result.push(groups[size]); }`
`if (groups[size].length === size) { result.push(groups[size]); groups[size] = [];}`
`const groupThePeople = function(groupSizes) { let result = [], groups = {}; groupSizes.forEach((size, index) => { if (size in groups) { if (groups[size].length < size) { groups[size].push(index) } } else { groups[size] = [index] } if (groups[size].length === size) { result.push(groups[size]); groups[size] = []; } }) return result;};`
--
--
## More from Sam Lesser
Career Changer, Software Engineer & Web Developer
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Career Changer, Software Engineer & Web Developer
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Dark Mode On/Off
LAST UPDATED: AUGUST 9, 2021
# String slicing in Python to rotate a string
In this tutorial, you will learn to rotate a string in Python using slicing. Strings in Python are a sequence of characters. Slicing is the process of accessing a part or a subset of a given string. For a given string of size n, we have to rotate the string by d elements. Where d is smaller than or equal to the size of the string. A string can be rotated clockwise (right rotation) or anticlockwise (left rotation). We will perform both rotations and give output for both rotations in our program.
Look at the examples to understand the input and output format.
Input:
s="studytonight"
d=3
Output:
Left Rotation : dytonightstu
Right Rotation : ghtstudytoni
To solve this, we will be using the concept of slicing in Python. Slicing will be used to divide the string into two parts.
1. For Left rotation, we will store the characters from the index 0 to d in the first part and the remaining characters in the second part.
2. For Right rotation, we will store the characters from the index 0 to (n-d) where n is the size of the string and the remaining characters in the second part.
3. Now we will join the second and first part as second + first for both rotations.
## Algorithm
Look at the algorithm to understand the approach better.
Step 1- Define a function that will rotate the string
Step 2- In the function, declare variables for the first and second part of the string for both left and right rotation
Step 3- Use slicing to get the characters in the first and the second part as discussed above
Step 4- Print the rotated strings by joining the second and the first part
Step 5- Declare string and value of d
Step 6- Pass the string and d in the function
## Python Program
In this program, we have used len() method of the String class to get the length of the string. To join or concatenate the first and the second part of the string after slicing, we have used the concatenation (+) operator.
``````# Function to rotate string left and right by d
def rotate_string(s,d):
# slice string in two parts for left and right
Lfirst = s[0 : d]
Lsecond = s[d :]
Rfirst = s[0 : len(s)-d]
Rsecond = s[len(s)-d : ]
# now join two parts together
print ("Left Rotation : ", (Lsecond + Lfirst) )
print ("Right Rotation : ", (Rsecond + Rfirst))
string = 'Studytonight'
d=4
rotate_string(string,d) ``````
Left Rotation : ytonightStud
Right Rotation : ightStudyton
## Conclusion
In this tutorial, we have learned to write a program for rotating a string by d elements. We have performed both left and right rotation on a string using the slicing concept.
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# Fitting Reinforcement Learning model
Hi,
I am trying to adapt the reinforcement learning model example (Fitting a Reinforcement Learning Model to Behavioral Data with PyMC — PyMC example gallery) to a more complex model.
I am quite new to PyMC and I am having some problems writing a function to use with `scan`.
The task I am modelling is a 2 armed bandit task with independent rewards and punishments for each arm.
The model is a single learning rate Rescorla-Wagner type of model with a softmax decision rule parametrised by an inverse temperature.
At each trial the model chooses one of the two arms and it receives one of 4 different outcomes (reward, punishment, reward and punishment, nothing).
The model tracks 4 different beliefs (arm A reward, arm A punishment, arm B reward, arm B punishment) but only 2 of them are updated at each trial based on which arm is chosen.
``````import numpy as np
import jax.numpy as jnp
import arviz as az
import matplotlib.pyplot as plt
import seaborn as sns
import pymc as pm
import pytensor.tensor as pt
from pytensor import scan, function
import scipy
def simulate_model(params, beliefs, outcomes):
'''
params:
- learning rate
- inverse temperature
beliefs = [belief_A_rew, belief_A_pun, belief_B_rew, belief_B_pun]
outcomes = [reward_A, punish_A, reward_B, punish_B] and are [0/1]
'''
learning_rate = params[0]
inv_temp = params[1]
n_trials = outcomes.shape[0]
choices = np.zeros(n_trials)
for t in range(n_trials):
probA = np.exp(inv_temp * (beliefs[t, 0] - beliefs[t,1])) / (np.exp(inv_temp * (beliefs[t,0] - beliefs[t,1])) + np.exp(inv_temp* (beliefs[t,2] - beliefs[t,3])))
choices[t] = int((probA > np.random.rand()) == 0)
# choice is A
if choices[t] == 0:
beliefs[t+1,0] = beliefs[t,0] + learning_rate * (outcomes[t,0] - beliefs[t,0])
beliefs[t+1,1] = beliefs[t,1] + learning_rate * (outcomes[t,1] - beliefs[t,1])
beliefs[t+1,2] = beliefs[t,2] # don't update when not chosen
beliefs[t+1,3] = beliefs[t,3] # don't update when not chosen
else:
beliefs[t+1,0] = beliefs[t,0] # don't update when not chosen
beliefs[t+1,1] = beliefs[t,1] # don't update when not chosen
beliefs[t+1,2] = beliefs[t,2] + learning_rate * (outcomes[t,2] - beliefs[t,2])
beliefs[t+1,3] = beliefs[t,3] + learning_rate * (outcomes[t,3] - beliefs[t,3])
return choices, beliefs
``````
I am trying to follow the tutorial and I am trying to write the function to estimate the parameters using PyMC to then compare with the maximum likelihood results.
``````def update_model(choice, outcome, choice_probability, belief, learning_rate, inv_temp):
if choice == 0:
c = 0
lr = pt.as_tensor([learning_rate, learning_rate, 0, 0])
else:
c = 2
lr = pt.as_tensor([0, 0, learning_rate, learning_rate])
choice_probability = np.exp(inv_temp * (belief[c] - belief[c+1])) / (np.exp(inv_temp * (belief[0] - belief[1])) + np.exp(inv_temp* (belief[2] - belief[3])))
new_belief = belief + lr * (outcome - belief)
return choice_probability, new_belief
# choices is [1 x n_trials]
# outcomes is [4 x n_trials]
choices_pt = pt.as_tensor_variable(choices, dtype='int32')
outcomes_pt = pt.as_tensor_variable(outcomes, dtype='int32')
learning_rate = pt.scalar('learning_rate')
inv_temp = pt.scalar('inv_temp')
choice_probablities = pt.zeros(1, dtype='float64')
beliefs = pt.zeros((1,4), dtype='float64')
fn=update_1lr1t,
sequences=[choices_pt, outcomes_pt],
non_sequences=[learning_rate, inv_temp],
outputs_info=[choice_probablities, beliefs]
)
neg_loglike = -pt.sum(results[0])
pytensor_llik_td = pt.pytensor.function(
inputs=[learning_rate, inv_temp], outputs=neg_loglike, on_unused_input="ignore"
)
result = pytensor_llik_td(0.2, 2)
``````
The code works ok until I try to create the `pytensor.function` where I get the following error
TypeError: Inconsistency in the inner graph of scan ‘scan_fn’ : an input and an output are associated with the same recurrent state and should have compatible types but have type ‘Vector(float64, shape=(1,))’ and ‘Vector(float64, shape=(?,))’ respectively.
When I print the shapes of the tensors inside `update_model` I get
``````<Scalar(int32, shape=())> # choice
<Vector(int32, shape=(?,))> # outcome
<Vector(float64, shape=(1,))> # choice_probability
<Matrix(float64, shape=(1, 4))> # belief
``````
What is the correct way to pass a single row of `outcome` to `scan` at each iteration?
Any help is greatly appreciated!
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# Rotating objects in x y and z
1. Nov 28, 2008
### DaveC426913
My buddy, a Flash developer, wants to learn 3D rotation of objects from first principles, i.e. using trig. I know there are shrink-wrapped engines out there that will do this for you, but he wants to learn it. From me. Cuz I'm the math whiz.
I used to do this simply when I'd plot simple 3D graphs in "Basic classic" but now I'm stepping it up a notch.
I'll start off with a primitive object and rotation on one axis.
So... my primitive object is a line in the XYZ space. It extends from 0,0,0 to 0,0,100. From my observer PoV, where I can only see the XY plane, the line is seen end-on and will appear as a point.
Now I rotate my line about the origin by 5 degrees. I know that I've got to do sin(5), which is about .087(x100 = 8.7). So the line is now visible in my XY viewing plane connecting 0 , 0 and 8.7 , 0.
Good so far?
Ultimately, it seems to me, I'm going to end up with two equations, x=... and y=... where the right side will be a long string of trig functions with x,y and z angles and x,y and z distances as variables.
Are there more robust techniques I can use? My buddy talks about multiplying matrices, which I know nothing about (but who knows, maybe I'm a quick learn).
And then I've got to add perspective...
Last edited: Nov 28, 2008
2. Nov 28, 2008
### Staff: Mentor
First things first: It takes a minimum of three parameters to describe a rotation in three-space, not two. Pick up a book. Rotating the book by say 45 degrees about an axis normal to
• The cover of the book
• The top edge of the book
• The right edge of the book
• Some other non-principal axis
will yield different orientations. There are three rotational degrees of freedom in 3-space. For 4-space there are six rotational degrees of freedom. In generalized N-space, there are N*(N-1)/2 rotational degrees of freedom.
Rotations, in any dimension, have the following characteristics:
1. Leaving an object unrotated is a null rotation of the object. Mathematically, rotation has an identity element.
2. You can undo any rotation. Mathematically, every rotation has an inverse.
3. If you rotate an object in some way and then rotate in some other way, the result is still a rotation of the object. Mathematically, rotation is a closed operation.
4. Suppose you rotate an object 3 times: rotation a followed by rotation b followed by rotation c. Since rotation is closed, with can denote a•b to denote the end result of performing rotation a followed by rotation b, and b•c as the end result of performing rotation b followed by rotation c. So, what happens when you do rotations a then b then c? Is the result (a•b)•c or a•(b•c)? The answer is "yes". (a•b)•c = a•(b•c). Mathematically, rotation is transitive.
You can view rotation as a mathematical operator on the orientation of object. From above, there is an identity rotation, every rotation has an inverse, and rotation is closed and transitive. These are the attributes of a mathematical group. Rotation is such an important group that it has a name: SO(n), where n is the dimensionality of the space. That SO means "special orthogonal". I'll get to that later. You are interested in SO(3).
There are many different ways to represent rotations in 3-space. A few of them are
• Rotation matrices. The rotation group is called the special orthogonal group because a rotation in N-space can be described by a NxN matrix, but not just any random NxN matrix. Only NxN matrices (a) have a determinant of 1 ("special") and (b) whose inverse is equal to the transpose of the matrix ("orthogonal") describe a proper rotation, and all matrices that have these two characteristics describe a rotation. This is true for rotations in any dimension, not just three. This general characteristic is why the n-dimensional rotation group is called SO(n).
• Euler angles. This is a sequence of three rotations about three pre-specified axes. (Remember that it takes a minimum of three parameters to describe a rotation). Which pre-specified axes depends on who you talk to. Astronomers use a rotation about the z axis, followed by a second rotation about the rotated x axis, followed by a third rotation about the doubly-rotated z axis. In short, a z-x-z rotation sequence. Aerospace engineers use some permutation of rotations about the x, y, and z axes, such as an x-y-z rotation sequence or a z-y-x rotation sequence; there are six such permutations altogether. Euler angles, whatever form you choose to use, are evil.
• Eigen rotations.
• Unit quaternions.
• Rodrigues parameters.
• Modified Rodrigues parameters.
• ...
Please read this wikipedia article on rotation matrices, and then come back with questions.
On final note: Very few people make a proper distinction between rotation and transformation. They are distinct but related concepts. The distinction is easily visualizable in 2D. Imagine an opaque sheet of paper printed with X-Y grid lines (e.g., graph paper) and a transparent sheet of plastic similarly printed with X-Y grid lines. Mark some point on the graph paper and some point on the transparency; these points will serve as the origins of your graph paper and transparency reference frames. Now mark some other point on the graph paper with a dot. That point has some specific X and Y coordinates in the graph paper reference frame. Now put the plastic transparency on top of the graph paper such that the origins are collocated. Next, rotate the transparency with respect to the graph paper. Finally, determine the transparency frame X-Y coordinates of the point you marked on the graph paper.
The physical rotation of the transparency with respect to the opaque graph paper is just that: a physical rotation. You can describe this rotation in terms of a rotation matrix. The graph paper frame and transparency frame coordinates of that point you marked on the graph paper are related by a transformation matrix from the graph paper frame to the transparency frame. The rotation matrix from the graph paper frame to the transparency frame and the transformation matrix from the graph paper frame to the transparency frame are not the same thing. They are instead transposes of one another.
3. Nov 29, 2008
### DaveC426913
Where did I say two parameters? :grumpy: You misunderstood what I wrote.
I actually said six:
But ultimately, it is only displayed in two dimensions:
So, again, I'll have two equations (i,e, two outputs : x and y screen coords) that will have 6 parameters as input.
Much of the rest of what you say seems to be based on that initial misunderstanding.
Note that, ultimately, I'm representing objects in 2-space.
Last edited: Nov 29, 2008
4. Nov 29, 2008
### Staff: Mentor
Now you are the one misunderstanding. My post has four sections: (1) degrees of freedom in a rotation in N-space, (2) rotation forms a mathematical group, (3) there are many different ways to represent rotations in 3-space, and (4) difference between rotation and transformation. Only the first part, which comprises the first paragraph of my post, pertains to that original misunderstanding.
5. Nov 30, 2008
### DaveC426913
Well, I mean I don't really see how that will get me to my goal.
Oh. Ah. I see now. I said my guy wants to "learn" this; I realize that was disengenuous. I realize our goal is really to understand the specifiic steps to get us to a workable algorithm.
6. Nov 30, 2008
### Staff: Mentor
I'll start with 2D rotation and transformation matrices and extend this to 3D.
Physical rotation of a vector in 2D
Suppose you have a vector with cartesian coordinates (x,y) and want to rotate this by some angle theta. I'll use the conventional notations
• the x-axis is horizontal, positive to the right,
• the y-axis is vertical, positive upward,
• vectors are expressed as column vectors, and
• rotation is positive counterclockwise.
Expressed as a column vector, the vector from the origin to $(x,y)$ is
$$\boldsymbol{r} = \bmatrix x\\y\endbmatrix$$
The coordinates of the endpoint of the rotated vector are (derivation is an exercise left to the reader) $(x\cos\theta - y\sin\theta, x\sin\theta + y\cos\theta)$ or
$$\boldsymbol{r}' = \bmatrix x\cos\theta - y\sin\theta \\ x\sin\theta + y\cos\theta\endbmatrix = \bmatrix \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \endbmatrix \, \bmatrix x\\y\endbmatrix = \mathbf{R}(\theta) \, \boldsymbol{r}$$
where
$$\mathbf{R}(\theta) \equiv \bmatrix \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \endbmatrix$$
is the 2D matrix that physically rotates a column vector by an angle $\theta$.
Transformation of a vector in 2D
If you physically rotate the $\hat{\boldsymbol x}$ and $\hat{\boldsymbol x}$ unit vectors by some angle $\theta$ you will get a new coordinate system with unit vectors $\hat{\boldsymbol x}'$ and $\hat{\boldsymbol x}'$:
\aligned \hat{\boldsymbol x}' &= \mathbf{R}(\theta)\,\hat{\boldsymbol x} \\ \hat{\boldsymbol y}' &= \mathbf{R}(\theta)\,\hat{\boldsymbol y} \endaligned
Given some point with coordinates $(x,y)$ in the original coordinate system, the coordinates of that point in the rotated coordinate system are (derivation once again left to the reader) $(x',y') = (x\cos\theta + y\sin\theta, -x\sin\theta + y\cos\theta)$ or
$$\boldsymbol{r}' = \bmatrix x\cos\theta + y\sin\theta \\ -x\sin\theta + y\cos\theta\endbmatrix = \bmatrix \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \endbmatrix \, \bmatrix x\\y\endbmatrix = \mathbf{T}(\theta) \, \boldsymbol{r}$$
where
$$\mathbf{T}(\theta) \equiv \bmatrix \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \endbmatrix$$
is the 2D matrix that transforms a column vector to the frame physically rotated by an angle $\theta$ with respect to the original frame.
In this simple 2D world, the transformation and rotation matrices are obviously related by $\mathbf{T}(\theta) = \mathbf{R}(-\theta)$. A much better way to look at this is that the two are transposes of one another: $\mathbf{T}(\theta) = \mathbf{R}^T(\theta)$. The reason this is a much better way to look at things is that this transpose relationship holds in higher dimensions. Rotation and translation are conjugate operations.
Sequence of transformations in N-space
Suppose you have three reference frames in some N-space with a common origin but different orientations. I'll denote these as frames A, B, and C. Denote $\boldsymbol{r}_A$, $\boldsymbol{r}_B$, and $\boldsymbol{r}_C$ as the representations of some vector r in these three frames. These representations are related by the transformation matrices from one frame to another:
\aligned \boldsymbol{r}_B &= \mathb{T}_{A \to B} \boldsymbol{r}_A \\ \boldsymbol{r}_C &= \mathb{T}_{A \to C} \boldsymbol{r}_A \\ &= \mathb{T}_{B \to C} \boldsymbol{r}_B \endaligned
Combining the above,
\aligned \boldsymbol{r}_C &= \mathb{T}_{A \to C} \boldsymbol{r}_A \\ &= \mathb{T}_{B \to C} \boldsymbol{r}_B \\ &= \mathb{T}_{B \to C} (\mathb{T}_{A \to B} \boldsymbol{r}_A) \\ &= (\mathb{T}_{B \to C} \mathb{T}_{A \to B}) \boldsymbol{r}_A \endaligned
The final step uses the fact that matrix multiplication is transitive. Since the above must be true for any vector r,
$$\mathb{T}_{A \to C} = \mathb{T}_{B \to C}\,\mathb{T}_{A \to B}$$
With an even more rotations chained in a sequence,
$$\mathbf{T}_{F_1 \to F_n} = \mathbf{T}_{F_{n-1} \to F_n},\cdots\,\mathbf{T}_{F_1 \to F_2}$$
This result makes no assumptions of the dimensionality of the space. Transformation matrices for column vectors chain from right-to-left extends to all dimensions. Since rotation is the transpose of transformation, rotation matrices chain left-to-right.
The order of operations here is very important for any dimension higher than two. Two dimensional transformation (or rotation) matrices commute. For higher dimensions, multiplication of transformation (or rotation) matrices is not commutative.
Rotation/Transformation in 3-space
Rotation/transformation in 2-space is much easier to comprehend than in 3-space. Obviously, a rotation or transformation matrix for a vector in 3-space in 3-space is a 3x3 matrix. Constructing or making sense of a 2x2 rotation or transformation matrix is an easy task. One way to construct a 3x3 rotation or transformation matrix is to construct the rotation/transformation as a sequence of simpler rotations/transformations.
Euler's rotation theorem will help in this regard. Euler's rotation theorem says that any rotation in three space is a rotation can be expressed as a 2D rotation about some axis in 3-space. In particular, the transformation matrices corresponding to rotations about the x, y, and z axes are
\aligned \mathbf{T}_x(\theta) &= \bmatrix 1&0&0 \\ 0&\cos\theta&\sin\theta \\ 0&-\sin\theta&\cos\theta \endbmatrix \\ \mathbf{T}_y(\phi) &= \bmatrix -\sin\phi&0&\cos\phi \\ 0&1&0 \\ \cos\phi&0&\sin\phi \endbmatrix \\ \mathbf{T}_z(\psi) &= \bmatrix \cos\psi&\sin\psi&0 \\ -\sin\psi&\cos\psi&0 \\ 0&0&1 \endbmatrix \endaligned
While Euler angles are evil, they do serve a very useful purpose in the construction of transformation matrices.
Sources of confusion
One obvious source of confusion is the distinction between rotation and transformation. I harp on this because they are distinct concepts and because failing to distinguish between these concepts is the source of a lot of errors. In this post, I have use $\boldsymbol R$ to denote a rotation matrix and $\boldsymbol T$ to denote a transformation matrix to avoid confusion.
Another source of confusion is whether vectors are expressed as column or row vectors. While column vectors are used more widely than are row vectors, this is just convention. If you instead use a row vector convention,
• You will have to post-multiply a vector by a rotation or transformation matrix to form a rotated or transformed vector.
• The rotation matrix for a row vector is the transpose of the rotation matrix for a column vector; the same is true for transformation matrices.
• Rotation matrices for row vectors chain right-to-left and transformation matrices for row vectors chain left-to-right.
Last words
My wife has the last word. She wants me to stop playing and do some silly house chores.
7. Nov 30, 2008
### DaveC426913
Hm. Well, you lost me right about
here.
If I define a point in the 3D space at xyz, and I supply a rotation of xyz, will those equations spit out an xy coordinate for it in the viewing plane?
I'm sorry, I have great facility with conceptualizing mathematics but, without a formal post-secondary education, I'm afraid the representation on paper is beyond me. I guess I'm in over my head. :shy:
Last edited: Nov 30, 2008
8. Dec 2, 2008
### DaveC426913
I'm still at it.
My point at 0,0,100, if rotated by 5 degrees about the y axis, will end up at these coords:
$$x_g = z_{d} sin(y_a)$$
$$y_g = z_d sin(x_a)$$
where
- $$x_g$$ and $$y_g$$ are the viewing plane XY coords (i.e. the output),
- $$z_d$$ is the distance from origin in the z direction and
- $$x_a$$ and $$y_a$$ are the x and y angles of rotation, respectively.
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# Finding the volume of a cube based off the mass and density of a sphere contained inside it.
This is an extension to my last question, but since people didn't like that the problem involved atoms, I will change them to small spheres of cheese.
The mass of a sphere of cheese is $1.37\cdot 10^{-25}$ kg, and the density of the cheese is $8920 \mbox{kg/m}^3$.
Visualize a one cubic centimeter as formed by stacking up identical cubes, with one sphere of cheese at the center of each of these cubes. Determine the volume of each cube.
This seemed simple enough to me, I just solve for the volume of a sphere of cheese, then, since a cube holds exactly one sphere, the length of the sides of the cube would be equal to the diameter of the sphere.
But, when I calculate all of these things, the answer is coming up wrong on the website where I submit my homework (WebAssign).
Here are my calculations (please tell me if I'm doing something wrong!):
$$\rho = \frac{m}{V}$$ so $$V = \frac{m}{\rho}$$
$$V = \frac{1.37 \cdot 10^{-25} kg}{8920 kg/m^3}$$
$$V = 1.536 \cdot 10^{-29} m^3$$
The volume of a sphere is $V_{sphere} = \frac{4 \pi r^3}{3}$, so
$$1.536 \cdot 10^{29}m^3 = \frac{4 \pi r^3}{3}$$
$$3.667 \cdot 10^{-30}m^3 = r^3$$
$$\sqrt[3]{3.667 \cdot 10^{-30}m^3} = \sqrt[3]{r^3}$$
$$r = 1.542 \cdot 10^{-10}m$$
The diameter of a sphere is $d = 2r$, so
$$d = 1.542 \cdot 10^{-10}m \cdot 2$$
$$d = 3.084 \cdot 10^{-10}m$$
The volume of a cube is $V_{cube} = l^3$ and since a cube holds exactly one sphere, $l = d$, so
$$V = d^3$$ $$V = (3.084 \cdot 10^{-10}m)^3$$
$$V = 2.933 \cdot 10^{-29}m^3$$
$$V = 2.933 \cdot 10^{-23}cm^3$$
Where did I go wrong?
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[cairo] [PATCH pixman 11/13] pixman-filter: Gaussian fixes
Bill Spitzak spitzak at gmail.com
Thu Sep 11 19:12:28 PDT 2014
Simplified the function.
Expanded size slightly (from ~4.25 to 5) to make the cutoff less noticable.
The filter is truncated at a value of .001 instead of .006, this new
value is less than 1/2 of 1/255, rather than greater than it.
---
pixman/pixman-filter.c | 7 ++-----
1 file changed, 2 insertions(+), 5 deletions(-)
diff --git a/pixman/pixman-filter.c b/pixman/pixman-filter.c
index d763138..dcbed11 100644
--- a/pixman/pixman-filter.c
+++ b/pixman/pixman-filter.c
@@ -63,10 +63,7 @@ linear_kernel (double x)
static double
gaussian_kernel (double x)
{
-#define SQRT2 (1.4142135623730950488016887242096980785696718753769480)
-#define SIGMA (SQRT2 / 2.0)
-
- return exp (- x * x / (2 * SIGMA * SIGMA)) / (SIGMA * sqrt (2.0 * M_PI));
+ return exp (- x * x) / sqrt (M_PI);
}
static double
@@ -141,7 +138,7 @@ static const filter_info_t filters[] =
{ PIXMAN_KERNEL_BOX, box_kernel, 1.0 },
{ PIXMAN_KERNEL_LINEAR, linear_kernel, 2.0 },
{ PIXMAN_KERNEL_CUBIC, cubic_kernel, 4.0 },
- { PIXMAN_KERNEL_GAUSSIAN, gaussian_kernel, 6 * SIGMA },
+ { PIXMAN_KERNEL_GAUSSIAN, gaussian_kernel, 5.0 },
{ PIXMAN_KERNEL_LANCZOS2, lanczos2_kernel, 4.0 },
{ PIXMAN_KERNEL_LANCZOS3, lanczos3_kernel, 6.0 },
{ PIXMAN_KERNEL_LANCZOS3_STRETCHED, nice_kernel, 8.0 },
--
1.7.9.5
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# Why use vacuum permeability during derivation of $\vec{B}=\mu_0(\vec{H} + \vec{M})$
Why do we use $$\mu_0$$ during the derivation of $$\vec{B}=\mu_0(\vec{H} + \vec{M})$$ where $$\vec{M}$$ is magnetization?
The derivation given in Sadiku's Elements of Electromagnetics:
Let $$\vec{J_f}$$ be free volume current density, $$\vec{J_b}$$ be bound volume current density, \begin{align*} \nabla \, \times \left( \frac{\vec{B}}{\mu_0} \right) &= \vec{J_f} + \vec{J_b} = \vec{J} \\ &= \nabla \times \vec{H} \, + \nabla \times \vec{M} \\ &= \nabla \times (\vec{H} + \vec{M}) \\ \vec{B} &=\mu_0(\vec{H} + \vec{M}) \quad \blacksquare \end{align*}
I don't understand why we should use $$\mu_0$$ in the first place. Why don't we use $$\mu$$ instead? In free space, $$\vec{M} = 0$$ and \begin{align*} \nabla \times \vec{H} &= \vec{J_f} \\ \nabla \times \left( \frac{\vec{B}}{\mu_0} \right) &= \vec{J_f} \end{align*} then naturally we'd like to still have $$\nabla \times \vec{H} = \vec{J} = \vec{J_f} + \vec{J_b}$$ when $$\vec{M} \neq 0$$, so we could just change the $$\mu_0$$ to some constant $$\mu$$, so that \begin{align*} \nabla \times \vec{H} &= \vec{J} \\ \nabla \times \left( \frac{\vec{B}}{\mu} \right) &= \vec{J} = \vec{J_f} + \vec{J_b} \end{align*} but in the correct derivation, \begin{align*} \nabla \times \left( \frac{\vec{B}}{\mu_0} \right) &= \vec{J} = \vec{J_f} + \vec{J_b} \end{align*}
What is it that forces us to use $$\mu_0$$?
• My another question helped me to demystify this question. Basically, I was confused by the book defining $\oint {\bf H} \cdot d {\bf \ell} = I_{enc}$ while in fact it's not clearly explained in the text that $I_{enc} = I_{free}$. So after applying Stoke's theorem, I mistakenly treated $\nabla \times {\bf H} = {\bf J_f} + {\bf J_b}$ while it should be $\nabla \times {\bf H} = {\bf J_f}$ only, it's just definition and no point to add ${\bf J_b}$ to it. – endeneer May 19 '19 at 6:18
$${\bf B}=\mu_0({\bf H}+{\bf M})$$ is effectively just the definition of the auxiliary field $${\bf H}$$. $${\bf B}$$ is defined as the quantity that generates the velocity-dependent force in the Lorentz Force Law, and the magnetization $${\bf M}$$ is the magnetic moment per unit volume. $${\bf H}$$ is then defined in terms of the other two.
For the currents, the bound current $${\bf J}_{b}$$ is defined as the curl of $${\bf M}$$, and then the free current is whatever is left over, $${\bf J}_{f}={\bf J}-{\bf J}_{b}$$. There is thus no room to change the constant $$\mu_{0}$$ to something else.
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Home » Theory of Consumer Behavior » Law of Equi-Marginal Utility
# Law of Equi-Marginal Utility:
### (Equilibrium of the Consumer Through the Law of Equi-Marginal Utility):
Other Names of this Law:
### Law of Substitution OR Law of Maximum Satisfaction OR Law of Indifference OR Proportion Rule OR Gossen's Second Law.
In the cardinal utility analysis, the principle of equal marginal utility occupies an important place.
## Definition and Statement of Law of Equi-Marginal Utility:
The law of equi-marginal utility is simply an extension of law of diminishing marginal utility to two or more than two commodities. The law of equilibrium utility is known, by various names. It is named as the Law of Substitution, the Law of Maximum Satisfaction, the Law of Indifference, the Proportionate Rule and the Gossen’s Second Law.
In cardinal utility analysis, this law is stated by Lipsey in the following words:
“The household maximizing the utility will so allocate the expenditure between commodities that the utility of the last penny spent on each item is equal”.
As we know, every consumer has unlimited wants. However, the income this disposal at any time is limited. The consumer is, therefore, faced with a choice among many commodities that he can and would like to pay. He, therefore, consciously or unconsciously compress the satisfaction which he obtains from the purchase of the commodity and the price which he pays for it. If he thinks the utility of the commodity is greater or at-least equal to the loss of utility of money price, he buys that commodity.
As he buys more and more of that commodity, the utility of the successive units begins to diminish. He stops further purchase of the commodity at a point where the marginal utility of the commodity and its price are just equal. If he pushes the purchase further from his point of equilibrium, then the marginal utility of the commodity will be less than that of price and the household will be loser. A consumer will be in equilibrium with a single commodity symbolically:
MUx = Px
A prudent consumer in order to get the maximum satisfaction from his limited means compares not only the utility of a particular commodity and the price but also the utility of the other commodities which he can buy with his scarce resources. If he finds that a particular expenditure in one use is yielding less utility than that of other, he will tie to transfer a unit of expenditure from the commodity yielding less marginal utility. The consumer will reach his equilibrium position when it will not be possible for him to increase the total utility by uses. The position of equilibrium will be reached when the marginal utility of each good is in proportion to its price and the ratio of the prices of all goods is equal to the ratio of their marginal utilities.
The consumer will maximize total utility from his income when the utility from the last rupee spent on each good is the same. Algebraically, this is:
MUa / Pa = MUb / Pb = MUc = Pc = MUn = Pn
Here: (a), (b), (c)…. (n) are various goods consumed.
## Assumptions of Law of Equi-Marginal Utility:
The main assumptions of the law of equi-marginal utility are as under.
(i) Independent utilities. The marginal utilities of different commodities are independent of each other and diminish with more and more purchases.
(ii) Constant marginal utility of money. The marginal utility of money remains constant to the consumer as he spends more and more of it on the purchase of goods.
(iii) Utility is cardinally measurable.
(iv) Every consumer is rational in the purchase of goods.
## Example and Explanation of Law of Equi-Marginal Utility:
The doctrine of equi-marginal utility can be explained by taking an example. Suppose a person has \$5 with him whom he wishes to spend on two commodities, tea and cigarettes. The marginal utility derived from both these commodities is as under:
### Schedule:
Units of Money MU of Tea MU of Cigarettes 1 10 12 2 8 10 3 6 8 4 4 6 5 2 3 \$5 Total Utility = 30 Total Utility = 30
A rational consumer would like to get maximum satisfaction from \$5.00. He can spend money in three ways:
(i) \$5 may be spent on tea only.
(ii) \$5 may be utilized for the purchase of cigarettes only.
(iii) Some rupees may be spent on the purchase of tea and some on the purchase of cigarettes.
If the prudent consumer spends \$5 on the purchase of tea, he gets 30 utility. If he spends \$5 on the purchase of cigarettes, the total utility derived is 39 which are higher than tea. In order to make the best of the limited resources, he adjusts his expenditure.
(i) By spending \$4 on tea and \$1 on cigarettes, he gets 40 utility (10+8+6+4+12 = 40).
(ii) By spending \$3 on tea and \$2 on cigarettes, he derives 46 utility (10+8+6+12+10 = 46).
(iii) By spending \$2 on tea and \$3 on cigarettes, he gets 48 utility (10+8+12+10+8 = 48).
(iv) By spending \$1 on tea and \$4 on cigarettes, he gets 46 utility (10+12+10+8+6 = 46).
The sensible consumer will spend \$2 on tea and \$3 on cigarettes and will get maximum satisfaction. When he spends \$2 on tea and \$3 on cigarette, the marginal utilities derived from both these commodities is equal to 8. When the marginal utilities of the two commodities are equalizes, the total utility is then maximum, i.e., 48 as is clear from the schedule given above.
### Curve/Diagram of Law of Equi-Marginal Utility:
The law of equi-marginal utility can be explained with the help of diagrams.
In the figure 2.3 MU is the marginal utility curve for tea and KL of cigarettes. When a consumer spends OP amount (\$2) on tea and OC (\$3) on cigarettes, the marginal utility derived from the consumption of both the items (Tea and Cigarettes) is equal to 8 units (EP = NC). The consumer gets the maximum utility when he spends \$2 on tea and \$3 on cigarettes and by no other alternation in the expenditure.
We now assume that the consumer spends \$1 on tea (OC/ amount) and \$4 (OQ/) on cigarettes. If CQ/ more amounts are spent cigarettes, the added utility is equal to the area CQ/ N/N. On the other hand, the expenditure on tea falls from OP amount (\$2) to OC/ amount (\$1). There is a toss of utility equal to the area C/PEE. The loss is utility (tea) is greater than that The loss in utility (tea) is maximum satisfaction except the combination of expenditure of \$2 on tea and \$3 on cigarettes.
## Limitations/Exceptions of Law of Equi-Marginal Utility:
(i) Effect on fashions and customs: The law of equi-marginal utility may become inoperative if people forced by fashions and customs spend money on the purchase of those commodities which they clearly knows yield less utility but they cannot transfer the unit of money from the less advantageous uses to the more advantageous uses because they are forced by the customs of the country.
(ii) Ignorance or carelessness: Sometimes people due to their ignorance of price or carelessness to weigh the utility of the purchased commodity do not obtain the maximum advantage by equating the marginal utility in all the uses.
(iii) Indivisible units: If the unit of expenditure is not divisible, then again the law may become inoperative.
(iv) Freedom of choice: If there is no perfect freedom between various alternatives, the operation of law may be impeded.
## Importance of Law of Equi-Marginal Utility:
The law of equi-marginal utility is of great practical importance. The application of the principle of substitution extends over almost every field of economic enquiry. Every consumer consciously trying to get the maximum satisfaction from his limited resources acts upon this principle of substitution. Same is the case with the producer. In the field of exchange and in theory of distribution too, this law plays a vital role. In short, despite its limitation, the law of maximum satisfaction is meaningful general statement of how consumers behave.
In addition to its application to consumption, it applies equally to the theory of production and theory of distribution. In the theory of production, it is applied on the substitution of various factors of production to the point where marginal return from all the factors are equal. The government can also use this analysis for evaluation of its different economic prices.
The equal marginal rule also guides an individual in the spending of his saving on different types of assets. The law of equal marginal utility also guides an individual in the allocation of his time between work and leisure. In short, despite limitations the law of substitution is applied to all problems of allocation of scarce resources.
## Relevant Articles:
» Cardinal Utility Analysis » » » Law of Equi Marginal Utility » Derivation of the Demand Curve in Terms of Utility Analysis
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Development and Planning Economics Introduction to Development Economics Features of Developing Countries Economic Development and Economic Growth Theories of Under Development Theories of Economic Growth Agriculture and Economic Development Monetary Economics and Public Finance History of Money
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CBSE Class 10 Sample Paper Science Solution - For 2023 Boards
Class 10
Solutions to CBSE Sample Paper - Science Class 10
## Two students decided to investigate the effect of water and air on iron object under identical experimental conditions. They measured the mass of each object before placing it partially immersed in 10 ml of water. After a few days, the object were removed, dried and their masses were measured. The table shows their results.
Student Object Mass of Object before Rusting in g Mass of the coated object in g A Nail 3.0 3.15 B Thin plate 6.0 6.33
## (a) What might be the reason for the varied observations of the two students?
a)
• Rusting occurs in both A and B so there is an increase in mass due to the formation of rust .
• As the surface area of B is more , more of B is exposed to air and water.
• Hence, the extent of rusting in B is more.
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Help Computer Science Assignment Introduction I want to write a simple mathematical program for a finite-state computer. I am working on a long-term project, and I have some questions about the memory of an application. How does the memory of a finite- state computer work? It is a memory object. What is the memory of the computer? How can I apply the memory to a finite–state computer? I have some questions that I don’t have. 1. Is there a way to make the program work with a small memory? 2. Is there any other way to do it? I don’t have much experience with programming on finite- state computers. I’m in a lot of trouble right now (I’m new to programming). Is there any other program that could be used for this? What can I do to make the memory work with a large memory? If there are any other programs that could be useful for this, please give me some info about them. I’m new in programming (and I know I’m not the only one). I would like to write a program that will show me even the most basic information. Is that possible? If so, what’s the best language for this? Or should I just use Google’s search function? 3. If I want a simple program to show me even basic information, I would like to be able to do it in one line.
## Help With Programming Assignment
Can I do this in a simple form? Is it possible to do this in one line? 4. Is the memory a good idea? If I want to make a program which will show me basic information and be able to understand the complexity of the problem and how to solve it, I would be a good candidate. 4a. Is it possible to make a complex program with a little memory? The memory I have is the most important part of the program. My computer is a 64-bit machine with eight input/output sub-decorators. If the memory of that computer is an array, then a simple program would be very useful. Does the memory of this computer have any other advantages than the memory of its input sub-decompressor? A: If you want to make the same program work with two different processors, you can use the same memory for two different processor stacks. That makes more sense. For example, you could use 2 different processor stacks for the same computer. The memory of your disk can hold two identical disks. The memory of the disk can be used for two different program threads on the same core. The disk can hold four separate programs on the same device. If you have two different disk-based programs, you can write them on different disk-less systems.
## Get Programming Help
A program can be written in one of three ways: It can be written as a “plain program” (e.g., the “plain” program). It can have a “plain” function (e. g., “plain” functions) that can be written to disk. It can use the memory of one of the disk-less program threads (e. eg., “plain”, “plain”, or “plain”). It can read the memory of another disk-less system (eHelp Computer Science Assignment Monday, May 07, 2011 I’m thinking about having some class today that I can apply the Basic Basic Basic Basic (BBP) algorithm so that I can create a class that is just like my own class except it doesn’t have the entire class. I’m wondering if there are some classes that I can design that are similar? Here is what I have. The first class I have is an Object class, which is a class for interacting with objects. The other classes are Object and Object-Type, which I have to represent as classes, so I can have multiple classes that I want to work on.
## Software Engineering Assignments
The first object class I have contains an Object instance that is a class instance of Object and an Object instance class, which I can represent as Class. I have to have all classes that I have together as classes, and then it’s okay to have all the classes that I work on as more classes. So I have a class that’s just like the above class. The classes I have are basically the same except that I have the individual classes as classes. I have the classes I work on and I have a similar class that I can work with as classes. So I need a way that I can make a class that can make a more general class that I have, and that doesn’t have to be a class but I have to be able to create a class of that class that will work on more than one of the classes. So I’d do this in a couple of ways. First, I need to create a new class to represent the classes I have. I have a Class that I work with in this particular class. If I know how to represent the class, then I can create another class that will represent that class. I have another class that I work from on a set of classes that I know the class has. The classes that I need to represent to make this class work on more classes. I also need a way to have a more generic class that will be called after the class name.
## Do My Hw For Me
I need a way of creating a generic class that I’m able to work with for each class I have. This can be done using the constructor in a class. I don’t want to make a list of all the most common classes I work with. Second, I need a method that I can use to make a generic class. I need the class to be a subclass of the class object I want to represent. I need to be able create a method that takes a class, and create a generic class for the class that is the subclass. Third, I need this generic class to be able be used as a subclass for an object that I work in. I need a class that I actually want to work with when I need to work with another class, and then I need to have that generic class in the class that I am working with. This is my third method that I need. Fourth, I need the generic class to have a method that we can create for the class object in the class. I can create this method in a class, but I don’t have the class that it represents. I need it in a generic class, so I need to do it in a class that contains an object. Fifth, I need an implementation of the generic class that is provided with the classes that are represented.
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1
JEE Main 2019 (Online) 10th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
All the pairs (x, y) that satisfy the inequality
$${2^{\sqrt {{{\sin }^2}x - 2\sin x + 5} }}.{1 \over {{4^{{{\sin }^2}y}}}} \le 1$$
also satisfy the equation
A
sin x = |sin y|
B
sin x = 2sin y
C
2 sin x = sin y
D
2 |sin x | = 3 sin y
2
JEE Main 2019 (Online) 10th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
If $$\alpha$$ and $$\beta$$ are the roots of the quadratic equation,
x2 + x sin $$\theta$$ - 2 sin $$\theta$$ = 0, $$\theta \in \left( {0,{\pi \over 2}} \right)$$, then
$${{{\alpha ^{12}} + {\beta ^{12}}} \over {\left( {{\alpha ^{ - 12}} + {\beta ^{ - 12}}} \right).{{\left( {\alpha - \beta } \right)}^{24}}}}$$ is equal to :
A
$${{{2^{12}}} \over {{{\left( {\sin \theta - 8} \right)}^6}}}$$
B
$${{{2^6}} \over {{{\left( {\sin \theta + 4} \right)}^{12}}}}$$
C
$${{{2^{12}}} \over {{{\left( {\sin \theta + 8} \right)}^{12}}}}$$
D
$${{{2^{12}}} \over {{{\left( {\sin \theta - 4} \right)}^{12}}}}$$
3
JEE Main 2019 (Online) 9th April Evening Slot
MCQ (Single Correct Answer)
+4
-1
If m is chosen in the quadratic equation
(m2 + 1) x2 – 3x + (m2 + 1)2 = 0
such that the sum of its roots is greatest, then the absolute difference of the cubes of its roots is :-
A
$$4\sqrt 3$$
B
$$8\sqrt 3$$
C
$$8\sqrt 5$$
D
$$10\sqrt 5$$
4
JEE Main 2019 (Online) 9th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Let p, q $$\in$$ R. If 2 - $$\sqrt 3$$ is a root of the quadratic equation, x2 + px + q = 0, then :
A
p2 – 4q – 12 = 0
B
q2 – 4p – 16 = 0
C
q2 + 4p + 14 = 0
D
p2 – 4q + 12 = 0
EXAM MAP
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# Model theory - definiable sets
• Nov 16th 2010, 09:53 AM
Arczi1984
Model theory - definiable sets
Hi,
I'm newbie in logic and model theory. I've the following task:
By exhibiting suitable formulas, show that the set of even numbers is a $\displaystyle \sum_0^0$ set in $\displaystyle \mathbb{N}$.
Show the same for the set of prime numbers.
How can I solve this? Any advices?
I'll be grateful for help
• Nov 17th 2010, 01:44 PM
emakarov
What relations and functions are available in the language?
I understand that $\displaystyle \Sigma_0^0$ means a set definable by a formula with bounded quantifiers only. A number n is even iff there exists an m <= n such that 2m = n. A number n is prime iff $\displaystyle n\ne 1$ and for all m < n, if m divides n, then m = 1.
• Nov 19th 2010, 08:19 AM
Arczi1984
The structure of natural numbers is following:
$\displaystyle \mathbb{N}=(\omega, 0, 1, +, \cdot, <)$
One more definition:
"The quantifiers $\displaystyle \forall (x<y)$ and $\displaystyle \exists (x<y)$ are said to be bounded"
Your definitions of even and prime numbers are clear but how can I use this facts to show that mentioned sets are $\displaystyle \sum_0^0$ in $\displaystyle \mathbb{N}$? What formulas should I use?
• Nov 19th 2010, 09:12 AM
emakarov
Replace "there exists an x such that" with "$\displaystyle \exists x$", "and" with "$\displaystyle \land$", and "if P, then Q" with "$\displaystyle P\to Q$". Also, you can express 2 as 1 + 1, and "m divides n", since it is not in the language, has to be written using an existential quantifier: there exists k such that m * k = n.
To be honest, I don't see where your difficulty is. I wrote English sentences as close as possible to symbolic formulas, so a mechanical search/replace can turn one into the other.
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MEEN_315_Team_Report[up]
# MEEN_315_Team_Report[up] - MEEN 315 Team Report CARNOT...
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Unformatted text preview: MEEN 315 Team Report CARNOT ENGINE PROJECT ABSTRACT When discussing Carnot engines, we usually assume the engine to be ideal, meaning that the engine is in thermal equilibrium, in that case = and =, so that there is no external irreversibility. But in our project, we have a Carnot engine which is not in thermal equilibrium, therefore, must maintain a reasonable temperature difference between the two heat transfer media. Using these parameters we needed to find when the power output will be max and also when there will be max power. And we had to use EES to verify these results. In Parts A and B we had to derive the equations given in the team project paper. While in Parts C, D, and E, we had to use EES to solve for and in various different cases. INTRODUCTION In this project we had to assume that this was not an ideal Carnot engine, so we had to prove that the max power output occurs when = (eq 1) and that the maximum net power output occurs when = (eq 2) After deriving these equations we needed to verify those using EES. For Part C of the project, we needed to verify Parts A & B assuming ( = 1 (W/K)). In Part D, we had to find () and () assuming that the work output could be sold at 1.2 cents per joule and the energy input () cost 1 cent per joule. And in the last part of the project, Part E, we again needed to calculate for () and () but this time assuming that the work output could be sold at 1.2 cents per joule and the energy input () cost 1 cent per joule and the energy output () cost 0.1 cents per joule for post-processing. All of these graphs were created in EES and can be seen below. MATHEMATICAL MODEL AND RESULTS FOR PART (A) AND PART (B) To derive the required equations there were several steps taken to achieve the right results. Following are the steps taken to achieve these results....
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## This note was uploaded on 11/19/2011 for the course MEEN 315 taught by Professor Ramussen during the Fall '07 term at Texas A&M.
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# An Investigation to Determine the Water Potential of a potato tuber
Extracts from this document...
Introduction
An Investigation to Determine the Water Potential of a potato tuber Background This experiment involves the process of osmosis. Osmosis is the process by which water moves across a membrane is known as osmosis and is described as the diffusion of water through a selectively permeable membrane The direction and amount of movement of water is dependant on the water potential. The water potential is based on the number and concentration of water molecules a solution has. For example, distilled water has the highest water potential, as it is the purest state of water. A highly concentrated solute solution has a low water potential. The highest water potential value (which distilled water has) is 0. Any dissolved solute-solution has a negative water potential value. Water potential is measure in the units Pascal. Water moves from a region of high water potential to a region of low water potential. Therefore, it moves down the water potential gradient. In this experiment we will be trying to determine the water potential of a potato tuber. Osmosis in this respect works on a cellular level. The cells that we are looking at are large, thin-walled, and usually have a large central vacuole. They are often partially separated from each other. In areas not exposed to light, as in a potato, food storage in the form of starch grains is the main function (Where light is present, e.g. in a leaf, photosynthesis is the main function). The amount of water present in these cells results in them having certain water potentials. If we place these tissues in different concentration sucrose solution we can see the if water has gone in or out of the cells. If the environment is hypotonic (having a lower concentration of solute than the cell) the net movement of water was into the cell, it would become turgid, the cytoplasm and vacuole pushed up against the cell wall, making the cell stiffer and wider and also result in the whole cell to increase in mass. ...read more.
Middle
* The time that was spent in solution is also an important factor. The time that is spent in the solution also affects how much water has had time to enter or leave the cell which means the cell has not had time to reach the isotonic point. This factor will be taken care of by allowing the potato pieces to stay in the solutions for a few hours, which is more than enough time for them to reach their isotonic point. Any difference in time that the potato strips spend in the solutions will be held to a minimum by placing and removing them within a small space in time. * We also understand that the potatoes are not identical and have not received the same environmental factors when being grown. Since we are doing repeats in this experiment, there is not enough potato to make all of the strips so more than one potato has to be used. This factor cannot be helped except for me to choose the healthiest potatoes and makes sure they are of the same type of potato. Apparatus 1x Scalpel - used to cut the potato strips to equal length 1x Tile - used to protect the table when cutting the potato strips 1x Glass Rod - used to stir the solutions when mixing water and sucrose solution 1x 30cm Ruler - used for the approximate measure of the potato strips 1x Cork Borer - used to accurately cut out the potato strips from the potato. 2x large potatoes - used as a source of plant tissue 200ml of distilled water - used to make the solutions 8x 250ml beakers - used to contain the water and sucrose solution and the other different molar solutions 200ml of 1M sucrose solution - used to make the solutions 1x marker pen - used to mark the beakers so that the contents can be identified 2x 50ml measuring cylinders - accurately measure the amounts of liquids to make the different ...read more.
Conclusion
all gave that same answers. My conclusion of finding the water potential of the potato cell could have been helped if I had a greater range of molarities with smaller increments which would show more accurately mass changes and also allow me to plot a better graph. The amount of repeats was sufficient and I believe that any more repeats would just give the same answers. Also the potato pieces itself was not definitely from the same potato and was not exactly the same size, although I did try to cut them to 2cm each, this could have effected the amount of water gained or lost. Anomalies include that in the yellow repeat in the 0.2 molar solution. This reading is not consistent as it indicates that that potato strip has not gained or lost any mass, which is inconsistent with the other repeats which show a range of 2.58% - 3.40% mass changes. This anomaly could have been caused by the strip not being weighed correctly at the beginning of the experiment or at the end of the experiment. You'll notice that the graph that shows all the results from each experiment do not all give the same mass change. Human error may have caused this. When the potato strips were removed from the test tubes and dried I may well have dried some potato pieces more thoroughly than others may and so some would have more excess water, which would add to the mass. I think the conclusion reached is quite accurate. Although there was an anomalous result the rest lay on a smooth curve which intercepted the x-axis at a credible point. The exact value of this point may not be exactly accurate as it is calculated from a freehand curve of best fit. The best thing to do if I was to carry on this experiment would be to make a 0.24 molar solution of sucrose and place potato strips in it. If I was right then the potato strips should neither gain mass nor lose any mass. ...read more.
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# Related GCSE Life Processes & Cells essays
1. ## To determine the water potential of a potato tuber cell using varying salt solution.
5 star(s)
Pressure potentials are caused by resistance of the tissues to water flow7. To calculate water potential this formula is used: Water potential ? = solute potential ?s + pressure potential ?p The differences in pressure potential of potato cells placed in different concentrations of salt solution, will cause the cell to react in different ways.
2. ## An Experiment to determine Water Potential in Potato Tissue.
4 star(s)
The surface area is a variable because the larger the surface area the amount of osmosis increases. 3. The mass of the potatoes- To measure the mass of the potato I used an electronic weighing machine.
1. ## Aim To determine the water potential of a potato tuber cell
This will affect the accuracy of the result. For the improvement of this, I'm going to use one potato to produce 6 potato cylinders. Therefore each potato cylinder will have the same strength of cell wall. The method could be improved by using a more accurate and precise equipment. E.g.
2. ## The determination of the Water Potential of Potato Tuber Cells
Conclusion In my hypothesis I predicted that the potatoes in the solutions with the highest sucrose molarity would shrink and lose weight and that the potatoes in the weakest solutions will gain weight and increase in size.
1. ## 'Investigating how isotonic, hypertonic and hypotonic solutions affects the total mass of a potato ...
So the cell will stay the same size. * If the medium has a lower concentration of water then the cell will lose water by osmosis. The water crosses the cell membrane in both directions, but this time more water leaves the cell than enters it.
2. ## Investigate the water potential of potato tissue and compare this with the water potential ...
Apparatus Reason for use 2 x pipettes Allow a reasonable degree of accuracy and are easy to use 2 x 250cm3 measuring cylinders To enable a little more than the theoretical 192cm3 required in total of each substance for each tissue.
1. ## Investigating the cellular water potential of potato cells.
Solute molecules tend to be far too large to be able to pass through the small pores within certain cell membranes; therefore they are unable to diffuse from one of the membrane to the other. However, water molecules are small enough to do so, thus the membrane is said to
2. ## Investigation To Find the concentration of sucrose solution that has the same &quot;water potential&quot; ...
I have chosen to vary the concentration of the sugar solution. This will hopefully give me a varied set of results from which I hope to make a decent conclusion. If any of the non-variables below are not kept constant it would mean it would not be a fair test.
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# Using the Exploratory Modelling Workbench
Over the last 7 years, I have been working on the development of an open source toolkit for supporting decision-making under deep uncertainty. This toolkit is known as the exploratory modeling workbench. The motivation for this name is that in my opinion all model-based deep uncertainty approaches are forms of exploratory modeling as first introduced by Bankes (1993). The design of the workbench has undergone various changes over time, but it has started to stabilize in the fall of 2016. This summer, I published a paper detailing the workbench (Kwakkel, 2017). There is an in depth example in the paper, but in a series of blogs I want to showcase the funtionality in some more detail.
The workbench is readily available through pip, but it requires ipyparallel and mpld3 (both available through conda), SALib (via pip), and optionality platypus (pip install directly from github repo).
## Adapting the DPS example from Rhodium
As a starting point, I will use the Direct Policy Search example that is available for Rhodium (Quinn et al 2017). I will adapt this code to work with the workbench. In this way, I can explain the workbench, as well as highlight some of the main differences between the workbench and Rhodium.
<br /># A function for evaluating our cubic DPS. This is based on equation (12)
# from [1].
def evaluateCubicDPS(policy, current_value):
value = 0
for i in range(policy["length"]):
rbf = policy["rbfs"][i]
value += rbf["weight"] * abs((current_value - rbf["center"]) / rbf["radius"])**3
value = min(max(value, 0.01), 0.1)
return value
# Construct the lake problem
def lake_problem(policy, # the DPS policy
b = 0.42, # decay rate for P in lake (0.42 = irreversible)
q = 2.0, # recycling exponent
mean = 0.02, # mean of natural inflows
stdev = 0.001, # standard deviation of natural inflows
alpha = 0.4, # utility from pollution
delta = 0.98, # future utility discount rate
nsamples = 100, # monte carlo sampling of natural inflows
steps = 100): # the number of time steps (e.g., days)
Pcrit = root(lambda x: x**q/(1+x**q) - b*x, 0.01, 1.5)
X = np.zeros((steps,))
decisions = np.zeros((steps,))
average_daily_P = np.zeros((steps,))
reliability = 0.0
utility = 0.0
inertia = 0.0
for _ in range(nsamples):
X[0] = 0.0
natural_inflows = np.random.lognormal(
math.log(mean**2 / math.sqrt(stdev**2 + mean**2)),
math.sqrt(math.log(1.0 + stdev**2 / mean**2)),
size=steps)
for t in range(1,steps):
decisions[t-1] = evaluateCubicDPS(policy, X[t-1])
X[t] = (1-b)*X[t-1] + X[t-1]**q/(1+X[t-1]**q) + decisions[t-1] + natural_inflows[t-1]
average_daily_P[t] += X[t]/float(nsamples)
reliability += np.sum(X < Pcrit)/float(steps)
utility += np.sum(alpha*decisions*np.power(delta,np.arange(steps)))
inertia += np.sum(np.diff(decisions) > -0.01)/float(steps-1)
max_P = np.max(average_daily_P)
reliability /= float(nsamples)
utility /= float(nsamples)
inertia /= float(nsamples)
return (max_P, utility, inertia, reliability)
The formulation of the decision rule assumes that policy is a dict, which is composed of a set of variables generated either through sampling or through optimization. This is relatively straightforward to do in Rhodium, but not so easy to do in the workbench. In the workbench, a policy is a composition of policy levers, where each policy lever is either a range of real values, a range of integers, or an unordered set of categories. To adapt the DPS version of the lake problem to work with the workbench, we have to first replace the policy dict with the different variables explicitly.
def get_antropogenic_release(xt, c1, c2, r1, r2, w1):
'''
Parameters
----------
xt : float
polution in lake at time t
c1 : float
center rbf 1
c2 : float
center rbf 2
r1 : float
r2 : float
w1 : float
weight of rbf 1
note:: w2 = 1 - w1
'''
rule = w1*(abs(xt-c1/r1))**3+(1-w1)*(abs(xt-c2/r2))**3
at = min(max(rule, 0.01), 0.1)
return at
Next, we need to adapt the lake_problem function itself to use this adapted version of the decision rule. This requires 2 changes: replace policy in the function signature of the lake_model function with the actual underlying parameters c1, c2, r1, r2, and w1, and use these when calculating the anthropological pollution rate.
def lake_model(b=0.42, q=2.0, mean=0.02, stdev=0.001, alpha=0.4, delta=0.98,
c1=0.25, c2=0.25, r1=0.5, r2=0.5, w1=0.5, nsamples=100,
steps=100):
Pcrit = root(lambda x: x**q/(1+x**q) - b*x, 0.01, 1.5)
X = np.zeros((steps,))
decisions = np.zeros((steps,))
average_daily_P = np.zeros((steps,))
reliability = 0.0
utility = 0.0
inertia = 0.0
for _ in range(nsamples):
X[0] = 0.0
natural_inflows = np.random.lognormal(
math.log(mean**2 / math.sqrt(stdev**2 + mean**2)),
math.sqrt(math.log(1.0 + stdev**2 / mean**2)),
size=steps)
for t in range(1,steps):
decisions[t-1] = get_antropogenic_release(X[t-1], c1, c2, r1, r2, w1)
X[t] = (1-b)*X[t-1] + X[t-1]**q/(1+X[t-1]**q) + decisions[t-1] + natural_inflows[t-1]
average_daily_P[t] += X[t]/float(nsamples)
reliability += np.sum(X < Pcrit)/float(steps)
utility += np.sum(alpha*decisions*np.power(delta,np.arange(steps)))
inertia += np.sum(np.diff(decisions) > -0.01)/float(steps-1)
max_P = np.max(average_daily_P)
reliability /= float(nsamples)
utility /= float(nsamples)
inertia /= float(nsamples)
return (max_P, utility, inertia, reliability)
This version of the code can be combined with the workbench already. However, we can clean it up a bit more if we want to. Note how there are 2 for loops in the lake model. The outer loop generates stochastic realizations of the natural inflow, while the inner loop calculates the the dynamics of the system given a stochastic realization. The workbench can be made responsible for this outer loop.
A quick note on terminology is in order here. I have a background in transport modeling. Here we often use discrete event simulation models. These are intrinsically stochastic models. It is standard practice to run these models several times and take descriptive statistics over the set of runs. In discrete event simulation, and also in the context of agent based modeling, this is known as running replications. The workbench adopts this terminology and draws a sharp distinction between designing experiments over a set of deeply uncertain factors, and performing replications of each experiment to cope with stochastic uncertainty.
Some other notes on the code:
* To aid in debugging functions, it is good practice to make a function deterministic. In this case we can quite easily achieve this by including an optional argument for setting the seed of the random number generation.
* I have slightly changed the formulation of inertia, which is closer to the mathematical formulation used in the various papers.
* I have changes the for loop over t to get rid of virtually all the t-1 formulations
from __future__ import division # python2
import math
import numpy as np
from scipy.optimize import brentq
def lake_model(b=0.42, q=2.0, mean=0.02, stdev=0.001, alpha=0.4,
delta=0.98, c1=0.25, c2=0.25, r1=0.5, r2=0.5,
w1=0.5, nsamples=100, steps=100, seed=None):
'''runs the lake model for 1 stochastic realisation using specified
random seed.
Parameters
----------
b : float
decay rate for P in lake (0.42 = irreversible)
q : float
recycling exponent
mean : float
mean of natural inflows
stdev : float
standard deviation of natural inflows
alpha : float
utility from pollution
delta : float
future utility discount rate
c1 : float
c2 : float
r1 : float
r2 : float
w1 : float
steps : int
the number of time steps (e.g., days)
seed : int, optional
seed for the random number generator
'''
np.random.seed(seed)
Pcrit = brentq(lambda x: x**q/(1+x**q) - b*x, 0.01, 1.5)
X = np.zeros((steps,))
decisions = np.zeros((steps,))
X[0] = 0.0
natural_inflows = np.random.lognormal(
math.log(mean**2 / math.sqrt(stdev**2 + mean**2)),
math.sqrt(math.log(1.0 + stdev**2 / mean**2)),
size=steps)
for t in range(steps-1):
decisions[t] = get_antropogenic_release(X[t], c1, c2, r1, r2, w1)
X[t+1] = (1-b)*X[t] + X[t]**q/(1+X[t]**q) + decisions[t] + natural_inflows[t]
reliability = np.sum(X < Pcrit)/steps
utility = np.sum(alpha*decisions*np.power(delta,np.arange(steps)))
# note that I have slightly changed this formulation to retain
# consistency with the equations in the papers
inertia = np.sum(np.abs(np.diff(decisions)) < 0.01)/(steps-1)
return X, utility, inertia, reliability
Now we are ready to connect this model to the workbench. This is fairly similar to how you would do it with Rhodium. We have to specify the uncertainties, the outcomes, and the policy levers. For the uncertainties and the levers, we can use real valued parameters, integer valued parameters, and categorical parameters. For outcomes, we can use either scalar, single valued outcomes or time series outcomes. For convenience, we can also explicitly control constants in case we want to have them set to a value different from their default value.
In this particular case, we are running the replications with the workbench. We still have to specify the descriptive statistics we would like to gather over the set of replications. For this, we can pass a function to an outcome. This function will be called with the results over the set of replications.
import numpy as np
from ema_workbench import (RealParameter, ScalarOutcome, Constant,
ReplicatorModel)
model = ReplicatorModel('lakeproblem', function=lake_model)
model.replications = 150
#specify uncertainties
model.uncertainties = [RealParameter('b', 0.1, 0.45),
RealParameter('q', 2.0, 4.5),
RealParameter('mean', 0.01, 0.05),
RealParameter('stdev', 0.001, 0.005),
RealParameter('delta', 0.93, 0.99)]
# set levers
model.levers = [RealParameter("c1", -2, 2),
RealParameter("c2", -2, 2),
RealParameter("r1", 0, 2),
RealParameter("r2", 0, 2),
RealParameter("w1", 0, 1)]
def process_p(values):
values = np.asarray(values)
values = np.mean(values, axis=0)
return np.max(values)
#specify outcomes
model.outcomes = [ScalarOutcome('max_P', kind=ScalarOutcome.MINIMIZE,
function=process_p),
ScalarOutcome('utility', kind=ScalarOutcome.MAXIMIZE,
function=np.mean),
ScalarOutcome('inertia', kind=ScalarOutcome.MINIMIZE,
function=np.mean),
ScalarOutcome('reliability', kind=ScalarOutcome.MAXIMIZE,
function=np.mean)]
# override some of the defaults of the model
model.constants = [Constant('alpha', 0.41),
Constant('steps', 100)]
## Open exploration
Now that we have specified the model with the workbench, we are ready to perform experiments on it. We can use evaluators to distribute these experiments either over multiple cores on a single machine, or over a cluster using ipyparallel. Using any parallelization is an advanced topic, in particular if you are on a windows machine. The code as presented here will run fine in parallel on a mac or Linux machine. If you are trying to run this in parallel using multiprocessing on a windows machine, from within a jupyter notebook, it won’t work. The solution is to move the lake_model and get_antropogenic_release to a separate python module and import the lake model function into the notebook.
Another common practice when working with the exploratory modeling workbench is to turn on the logging functionality that it provides. This will report on the progress of the experiments, as well as provide more insight into what is happening in particular in case of errors.
If we want to perform experiments on the model we have just defined, we can use the perform_experiments method on the evaluator, or the stand alone perform_experiments function. We can perform experiments over the uncertainties and/or over the levers. Any policy is evaluated over each of the scenarios. So if we want to use 100 scenarios and 10 policies, this means that we will end up performing 100 * 10 = 1000 experiments. By default, the workbench uses Latin hypercube sampling for both sampling over levers and sampling over uncertainties. However, the workbench also offers support for full factorial, partial factorial, and Monte Carlo sampling, as well as wrappers for the various sampling schemes provided by SALib.
from ema_workbench import (MultiprocessingEvaluator, ema_logging,
perform_experiments)
ema_logging.log_to_stderr(ema_logging.INFO)
with MultiprocessingEvaluator(model) as evaluator:
results = evaluator.perform_experiments(scenarios=10, policies=10)
## Directed Search
Similarly, we can easily use the workbench to search for a good candidate strategy. This requires that platypus is installed. If platypus is installed, we can simply use the optimize method. By default, the workbench will use $\epsilon$-NSGAII. The workbench can be used to search over the levers in order to find a good candidate strategy as is common in Many-Objective Robust Decision Making. The workbench can also be used to search over the uncertainties in order to find for example the worst possible outcomes and the conditions under which they appear. This is a form of worst case discovery. The optimize method takes an optional reference argument. This can be used to set the scenario for which you want to find good policies, or for setting the policy for which you want to find the worst possible outcomes. This makes implementing the approach suggested in Watson & Kasprzyk (2017) very easy.
with MultiprocessingEvaluator(model) as evaluator:
results = evaluator.optimize(nfe=1000, searchover='levers',
epsilons=[0.1,]*len(model.outcomes))
## Robust optimization
A third possibility is to perform robust optimization. In this case, the search will take place over the levers, but a given policy is than evaluated for a set of scenarios and the performance is defined over this set. To do this, we need to explicitly define robustness. For this, we can use the outcome object we have used before. In the example below we are defining robustness as the worst 10th percentile over the set of scenarios. We need to pass a variable_name argument to explicitly link outcomes of the model to the robustness metrics.
import functools
percentile10 = functools.partial(np.percentile, q=10)
percentile90 = functools.partial(np.percentile, q=90)
MAXIMIZE = ScalarOutcome.MAXIMIZE
MINIMIZE = ScalarOutcome.MINIMIZE
robustnes_functions = [ScalarOutcome('90th percentile max_p', kind=MINIMIZE,
variable_name='max_P', function=percentile90),
ScalarOutcome('10th percentile reliability', kind=MAXIMIZE,
variable_name='reliability', function=percentile10),
ScalarOutcome('10th percentile inertia', kind=MAXIMIZE,
variable_name='inertia', function=percentile10),
ScalarOutcome('10th percentile utility', kind=MAXIMIZE,
variable_name='utility', function=percentile10)]
Given the specification of the robustness function, the remainder is straightforward and analogous to normal optimization.
<br />n_scenarios = 200
scenarios = sample_uncertainties(lake_model, n_scenarios)
nfe = 100000
with MultiprocessingEvaluator(lake_model) as evaluator:
robust_results = evaluator.robust_optimize(robustnes_functions, scenarios,
nfe=nfe, epsilons=[0.05,]*len(robustnes_functions))
This blog has introduced the exploratory modeling workbench and has shown its basic functionality for sampling or searching over uncertainties and levers. In subsequent blogs, I will take a more in depth look at this funcitonality, as well as demonstrate how the workbench facilitates the entire Many-Objective Robust Decision Making process.
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0
# How do you explain 6 in the number 364021 in the thousand place?
Updated: 9/20/2023
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Home Metamath Proof ExplorerTheorem List (p. 344 of 429) < Previous Next > Bad symbols? Try the GIF version. Mirrors > Metamath Home Page > MPE Home Page > Theorem List Contents > Recent Proofs This page: Page List
Color key: Metamath Proof Explorer (1-27903) Hilbert Space Explorer (27904-29428) Users' Mathboxes (29429-42879)
Theorem List for Metamath Proof Explorer - 34301-34400 *Has distinct variable group(s)
TypeLabelDescription
Statement
Theoremxrnres2 34301 Two ways to express restriction of range Cartesian product, cf. xrnres 34300, xrnres3 34302. (Contributed by Peter Mazsa, 6-Sep-2021.)
((𝑅𝑆) ↾ 𝐴) = (𝑅 ⋉ (𝑆𝐴))
Theoremxrnres3 34302 Two ways to express restriction of range Cartesian product, cf. xrnres 34300, xrnres2 34301. (Contributed by Peter Mazsa, 28-Mar-2020.)
((𝑅𝑆) ↾ 𝐴) = ((𝑅𝐴) ⋉ (𝑆𝐴))
Theoremxrnres4 34303 Two ways to express restriction of range Cartesian product. (Contributed by Peter Mazsa, 29-Dec-2020.)
((𝑅𝑆) ↾ 𝐴) = ((𝑅𝑆) ∩ (𝐴 × (ran (𝑅𝐴) × ran (𝑆𝐴))))
Theoremxrnresex 34304 Sufficient condition for a restricted range Cartesian product to be a set. (Contributed by Peter Mazsa, 16-Dec-2020.) (Revised by Peter Mazsa, 7-Sep-2021.)
((𝐴𝑉𝑅𝑊 ∧ (𝑆𝐴) ∈ 𝑋) → (𝑅 ⋉ (𝑆𝐴)) ∈ V)
Theoremxrnidresex 34305 Sufficient condition for a range Cartesian product with restricted identity to be a set. (Contributed by Peter Mazsa, 31-Dec-2021.)
((𝐴𝑉𝑅𝑊) → (𝑅 ⋉ ( I ↾ 𝐴)) ∈ V)
Theoremxrncnvepresex 34306 Sufficient condition for a range Cartesian product with restricted converse epsilon to be a set. (Contributed by Peter Mazsa, 16-Dec-2020.) (Revised by Peter Mazsa, 23-Sep-2021.)
((𝐴𝑉𝑅𝑊) → (𝑅 ⋉ ( E ↾ 𝐴)) ∈ V)
Theorembrin2 34307 Binary relation on an intersection is a special case of binary relation on range Cartesian product. (Contributed by Peter Mazsa, 21-Aug-2021.)
((𝐴𝑉𝐵𝑊) → (𝐴(𝑅𝑆)𝐵𝐴(𝑅𝑆)⟨𝐵, 𝐵⟩))
Theorembrin3 34308 Binary relation on an intersection is a special case of binary relation on range Cartesian product. (Contributed by Peter Mazsa, 21-Aug-2021.) (Avoid depending on this detail.)
((𝐴𝑉𝐵𝑊) → (𝐴(𝑅𝑆)𝐵𝐴(𝑅𝑆){{𝐵}}))
20.21.4 Cosets by ` R `
Definitiondf-coss 34309* Define the class of cosets by 𝑅: 𝑥 and 𝑦 are cosets by 𝑅 iff there exists a set 𝑢 such that both 𝑢𝑅𝑥 and 𝑢𝑅𝑦 hold, i.e., both 𝑥 and 𝑦 are are elements of the 𝑅 -coset of 𝑢 (cf. dfcoss2 34311 and the comment of dfec2 7790). 𝑅 is usually a relation.
This concept simplifies theorems relating partition and equivalence: the left side of these theorems relate to 𝑅, the right side relate to 𝑅 (cf. e.g. ~? pet ). Without the definition of 𝑅 we should have to relate the right side of these theorems to a composition of a converse (cf. dfcoss3 34312) or to the range of a range Cartesian product of classes (cf. dfcoss4 34313), which would make the theorems complicated and confusing. Alternate definition is dfcoss2 34311. Technically, we can define it via composition (dfcoss3 34312) or as the range of a range Cartesian product (dfcoss4 34313), but neither of these definitions reveal directly how the cosets by 𝑅 relate to each other. We define functions ( ~? df-funsALTV , ~? df-funALTV ) and disjoints ( ~? dfdisjs , ~? dfdisjs2 , ~? df-disjALTV , ~? dfdisjALTV2 ) with the help of it as well. (Contributed by Peter Mazsa, 9-Jan-2018.)
𝑅 = {⟨𝑥, 𝑦⟩ ∣ ∃𝑢(𝑢𝑅𝑥𝑢𝑅𝑦)}
Definitiondf-coels 34310 Define the class of coelements on the class 𝐴, cf. the alternate definition dfcoels 34325. Possible definitions are the special cases of dfcoss3 34312 and dfcoss4 34313. (Contributed by Peter Mazsa, 20-Nov-2019.)
𝐴 = ≀ ( E ↾ 𝐴)
Theoremdfcoss2 34311* Alternate definition of the class of cosets by 𝑅: 𝑥 and 𝑦 are cosets by 𝑅 iff there exists a set 𝑢 such that both 𝑥 and 𝑦 are are elements of the 𝑅-coset of 𝑢 (cf. the comment of dfec2 7790). 𝑅 is usually a relation. (Contributed by Peter Mazsa, 16-Jan-2018.)
𝑅 = {⟨𝑥, 𝑦⟩ ∣ ∃𝑢(𝑥 ∈ [𝑢]𝑅𝑦 ∈ [𝑢]𝑅)}
Theoremdfcoss3 34312 Alternate definition of the class of cosets by 𝑅 (cf. the comment of df-coss 34309). (Contributed by Peter Mazsa, 27-Dec-2018.)
𝑅 = (𝑅𝑅)
Theoremdfcoss4 34313 Alternate definition of the class of cosets by 𝑅 (cf. the comment of df-coss 34309). (Contributed by Peter Mazsa, 12-Jul-2021.)
𝑅 = ran (𝑅𝑅)
Theoremcossex 34314 If 𝐴 is a set then the class of cosets by 𝐴 is a set. (Contributed by Peter Mazsa, 4-Jan-2019.)
(𝐴𝑉 → ≀ 𝐴 ∈ V)
Theoremcosscnvex 34315 If 𝐴 is a set then the class of cosets by the converse of 𝐴 is a set. (Contributed by Peter Mazsa, 18-Oct-2019.)
(𝐴𝑉 → ≀ 𝐴 ∈ V)
Theorem1cosscnvepresex 34316 Sufficient condition for a restricted converse epsilon coset to be a set. (Contributed by Peter Mazsa, 24-Sep-2021.)
(𝐴𝑉 → ≀ ( E ↾ 𝐴) ∈ V)
Theorem1cossxrncnvepresex 34317 Sufficient condition for a restricted converse epsilon range Cartesian product to be a set. (Contributed by Peter Mazsa, 23-Sep-2021.)
((𝐴𝑉𝑅𝑊) → ≀ (𝑅 ⋉ ( E ↾ 𝐴)) ∈ V)
Theoremrelcoss 34318 Cosets by 𝑅 is a relation. (Contributed by Peter Mazsa, 27-Dec-2018.)
Rel ≀ 𝑅
Theoremrelcoels 34319 Coelements on 𝐴 is a relation. (Contributed by Peter Mazsa, 5-Oct-2021.)
Rel ∼ 𝐴
Theoremcossss 34320 Subclass theorem for the classes of cosets by 𝐴 and 𝐵. (Contributed by Peter Mazsa, 11-Nov-2019.)
(𝐴𝐵 → ≀ 𝐴 ⊆ ≀ 𝐵)
Theoremcosseq 34321 Equality theorem for the classes of cosets by 𝐴 and 𝐵. (Contributed by Peter Mazsa, 9-Jan-2018.)
(𝐴 = 𝐵 → ≀ 𝐴 = ≀ 𝐵)
Theoremcosseqi 34322 Equality theorem for the classes of cosets by 𝐴 and 𝐵, inference form. (Contributed by Peter Mazsa, 9-Jan-2018.)
𝐴 = 𝐵 𝐴 = ≀ 𝐵
Theoremcosseqd 34323 Equality theorem for the classes of cosets by 𝐴 and 𝐵, deduction form. (Contributed by Peter Mazsa, 4-Nov-2019.)
(𝜑𝐴 = 𝐵) (𝜑 → ≀ 𝐴 = ≀ 𝐵)
Theorem1cossres 34324* The class of cosets by a restriction. (Contributed by Peter Mazsa, 20-Apr-2019.)
≀ (𝑅𝐴) = {⟨𝑥, 𝑦⟩ ∣ ∃𝑢𝐴 (𝑢𝑅𝑥𝑢𝑅𝑦)}
Theoremdfcoels 34325* Alternate definition of the class of coelements on the class 𝐴. (Contributed by Peter Mazsa, 20-Apr-2019.)
𝐴 = {⟨𝑥, 𝑦⟩ ∣ ∃𝑢𝐴 (𝑥𝑢𝑦𝑢)}
Theorembrcoss 34326* 𝐴 and 𝐵 are cosets by 𝑅: a binary relation. (Contributed by Peter Mazsa, 27-Dec-2018.)
((𝐴𝑉𝐵𝑊) → (𝐴𝑅𝐵 ↔ ∃𝑢(𝑢𝑅𝐴𝑢𝑅𝐵)))
Theorembrcoss2 34327* Alternate form of the 𝐴 and 𝐵 are cosets by 𝑅 binary relation. (Contributed by Peter Mazsa, 26-Mar-2019.)
((𝐴𝑉𝐵𝑊) → (𝐴𝑅𝐵 ↔ ∃𝑢(𝐴 ∈ [𝑢]𝑅𝐵 ∈ [𝑢]𝑅)))
Theorembrcoss3 34328 Alternate form of the 𝐴 and 𝐵 are cosets by 𝑅 binary relation. (Contributed by Peter Mazsa, 26-Mar-2019.)
((𝐴𝑉𝐵𝑊) → (𝐴𝑅𝐵 ↔ ([𝐴]𝑅 ∩ [𝐵]𝑅) ≠ ∅))
Theorembrcosscnvcoss 34329 For sets, the 𝐴 and 𝐵 cosets by 𝑅 binary relation and the 𝐵 and 𝐴 cosets by 𝑅 binary relation are the same. (Contributed by Peter Mazsa, 27-Dec-2018.)
((𝐴𝑉𝐵𝑊) → (𝐴𝑅𝐵𝐵𝑅𝐴))
Theorembrcoels 34330* 𝐵 and 𝐶 are coelements : a binary relation. (Contributed by Peter Mazsa, 14-Jan-2020.) (Revised by Peter Mazsa, 5-Oct-2021.)
((𝐵𝑉𝐶𝑊) → (𝐵𝐴𝐶 ↔ ∃𝑢𝐴 (𝐵𝑢𝐶𝑢)))
Theoremcocossss 34331* Two ways of saying that cosets by cosets by 𝑅 is a subclass. (Contributed by Peter Mazsa, 17-Sep-2021.)
( ≀ ≀ 𝑅𝑆 ↔ ∀𝑥𝑦𝑧((𝑥𝑅𝑦𝑦𝑅𝑧) → 𝑥𝑆𝑧))
Theoremcnvcosseq 34332 The converse of cosets by 𝑅 are cosets by 𝑅. (Contributed by Peter Mazsa, 3-May-2019.)
𝑅 = ≀ 𝑅
Theorembr2coss 34333 Cosets by 𝑅 binary relation. (Contributed by Peter Mazsa, 25-Aug-2019.)
((𝐴𝑉𝐵𝑊) → (𝐴 ≀ ≀ 𝑅𝐵 ↔ ([𝐴] ≀ 𝑅 ∩ [𝐵] ≀ 𝑅) ≠ ∅))
Theorembr1cossres 34334* 𝐵 and 𝐶 are cosets by a restriction: a binary relation. (Contributed by Peter Mazsa, 30-Dec-2018.)
((𝐵𝑉𝐶𝑊) → (𝐵 ≀ (𝑅𝐴)𝐶 ↔ ∃𝑢𝐴 (𝑢𝑅𝐵𝑢𝑅𝐶)))
Theorembr1cossres2 34335* 𝐵 and 𝐶 are cosets by a restriction: a binary relation. (Contributed by Peter Mazsa, 3-Jan-2018.)
((𝐵𝑉𝐶𝑊) → (𝐵 ≀ (𝑅𝐴)𝐶 ↔ ∃𝑥𝐴 (𝐵 ∈ [𝑥]𝑅𝐶 ∈ [𝑥]𝑅)))
Theoremrelbrcoss 34336* 𝐴 and 𝐵 are cosets by relation 𝑅: a binary relation. (Contributed by Peter Mazsa, 22-Apr-2021.)
((𝐴𝑉𝐵𝑊) → (Rel 𝑅 → (𝐴𝑅𝐵 ↔ ∃𝑥 ∈ dom 𝑅(𝐴 ∈ [𝑥]𝑅𝐵 ∈ [𝑥]𝑅))))
Theorembr1cossinres 34337* 𝐵 and 𝐶 are cosets by an intersection with a restriction: a binary relation. (Contributed by Peter Mazsa, 31-Dec-2021.)
((𝐵𝑉𝐶𝑊) → (𝐵 ≀ (𝑅 ∩ (𝑆𝐴))𝐶 ↔ ∃𝑢𝐴 ((𝑢𝑆𝐵𝑢𝑅𝐵) ∧ (𝑢𝑆𝐶𝑢𝑅𝐶))))
Theorembr1cossxrnres 34338* 𝐵, 𝐶 and 𝐷, 𝐸 are cosets by an intersection with a restriction: a binary relation. (Contributed by Peter Mazsa, 8-Jun-2021.)
(((𝐵𝑉𝐶𝑊) ∧ (𝐷𝑋𝐸𝑌)) → (⟨𝐵, 𝐶⟩ ≀ (𝑅 ⋉ (𝑆𝐴))⟨𝐷, 𝐸⟩ ↔ ∃𝑢𝐴 ((𝑢𝑆𝐶𝑢𝑅𝐵) ∧ (𝑢𝑆𝐸𝑢𝑅𝐷))))
Theorembr1cossinidres 34339* 𝐵 and 𝐶 are cosets by an intersection with the restricted identity class: a binary relation. (Contributed by Peter Mazsa, 31-Dec-2021.)
((𝐵𝑉𝐶𝑊) → (𝐵 ≀ (𝑅 ∩ ( I ↾ 𝐴))𝐶 ↔ ∃𝑢𝐴 ((𝑢 = 𝐵𝑢𝑅𝐵) ∧ (𝑢 = 𝐶𝑢𝑅𝐶))))
Theorembr1cossincnvepres 34340* 𝐵 and 𝐶 are cosets by an intersection with the restricted converse epsilon class: a binary relation. (Contributed by Peter Mazsa, 31-Dec-2021.)
((𝐵𝑉𝐶𝑊) → (𝐵 ≀ (𝑅 ∩ ( E ↾ 𝐴))𝐶 ↔ ∃𝑢𝐴 ((𝐵𝑢𝑢𝑅𝐵) ∧ (𝐶𝑢𝑢𝑅𝐶))))
Theorembr1cossxrnidres 34341* 𝐵, 𝐶 and 𝐷, 𝐸 are cosets by a range Cartesian product with the restricted identity class: a binary relation. (Contributed by Peter Mazsa, 8-Jun-2021.)
(((𝐵𝑉𝐶𝑊) ∧ (𝐷𝑋𝐸𝑌)) → (⟨𝐵, 𝐶⟩ ≀ (𝑅 ⋉ ( I ↾ 𝐴))⟨𝐷, 𝐸⟩ ↔ ∃𝑢𝐴 ((𝑢 = 𝐶𝑢𝑅𝐵) ∧ (𝑢 = 𝐸𝑢𝑅𝐷))))
Theorembr1cossxrncnvepres 34342* 𝐵, 𝐶 and 𝐷, 𝐸 are cosets by a range Cartesian product with the restricted converse epsilon class: a binary relation. (Contributed by Peter Mazsa, 12-May-2021.)
(((𝐵𝑉𝐶𝑊) ∧ (𝐷𝑋𝐸𝑌)) → (⟨𝐵, 𝐶⟩ ≀ (𝑅 ⋉ ( E ↾ 𝐴))⟨𝐷, 𝐸⟩ ↔ ∃𝑢𝐴 ((𝐶𝑢𝑢𝑅𝐵) ∧ (𝐸𝑢𝑢𝑅𝐷))))
Theoremdmcoss3 34343 The domain of cosets is the domain of converse. (Contributed by Peter Mazsa, 4-Jan-2019.)
dom ≀ 𝑅 = dom 𝑅
Theoremdmcoss2 34344 The domain of cosets is the range. (Contributed by Peter Mazsa, 27-Dec-2018.)
dom ≀ 𝑅 = ran 𝑅
Theoremrncossdmcoss 34345 The range of cosets is the domain of them (this should be rncoss 5418 but there exists a theorem with this name already). (Contributed by Peter Mazsa, 12-Dec-2019.)
ran ≀ 𝑅 = dom ≀ 𝑅
Theoremdm1cosscnvepres 34346 The domain of cosets of the restricted converse epsilon relation is the union of the restriction. (Contributed by Peter Mazsa, 18-May-2019.) (Revised by Peter Mazsa, 26-Sep-2021.)
dom ≀ ( E ↾ 𝐴) = 𝐴
Theoremdmcoels 34347 The domain of coelements in 𝐴 is the union of 𝐴. (Contributed by Rodolfo Medina, 14-Oct-2010.) (Revised by Peter Mazsa, 5-Apr-2018.) (Revised by Peter Mazsa, 26-Sep-2021.)
dom ∼ 𝐴 = 𝐴
Theoremeldmcoss 34348* Elementhood in the domain of cosets. (Contributed by Peter Mazsa, 29-Mar-2019.)
(𝐴𝑉 → (𝐴 ∈ dom ≀ 𝑅 ↔ ∃𝑢 𝑢𝑅𝐴))
Theoremeldmcoss2 34349 Elementhood in the domain of cosets. (Contributed by Peter Mazsa, 28-Dec-2018.)
(𝐴𝑉 → (𝐴 ∈ dom ≀ 𝑅𝐴𝑅𝐴))
Theoremeldm1cossres 34350* Elementhood in the domain of restricted cosets. (Contributed by Peter Mazsa, 30-Dec-2018.)
(𝐵𝑉 → (𝐵 ∈ dom ≀ (𝑅𝐴) ↔ ∃𝑢𝐴 𝑢𝑅𝐵))
Theoremeldm1cossres2 34351* Elementhood in the domain of restricted cosets. (Contributed by Peter Mazsa, 30-Dec-2018.)
(𝐵𝑉 → (𝐵 ∈ dom ≀ (𝑅𝐴) ↔ ∃𝑥𝐴 𝐵 ∈ [𝑥]𝑅))
Theoremrefrelcosslem 34352 Lemma for the left side of the refrelcoss3 34353 reflexivity theorem. (Contributed by Peter Mazsa, 1-Apr-2019.)
𝑥 ∈ dom ≀ 𝑅𝑥𝑅𝑥
Theoremrefrelcoss3 34353* The class of cosets by 𝑅 is reflexive, cf. dfrefrel3 34406. (Contributed by Peter Mazsa, 30-Jul-2019.)
(∀𝑥 ∈ dom ≀ 𝑅𝑦 ∈ ran ≀ 𝑅(𝑥 = 𝑦𝑥𝑅𝑦) ∧ Rel ≀ 𝑅)
Theoremrefrelcoss2 34354 The class of cosets by 𝑅 is reflexive, cf. dfrefrel2 34405. (Contributed by Peter Mazsa, 30-Jul-2019.)
(( I ∩ (dom ≀ 𝑅 × ran ≀ 𝑅)) ⊆ ≀ 𝑅 ∧ Rel ≀ 𝑅)
Theoremsymrelcoss3 34355 The class of cosets by 𝑅 is symmetric, cf. dfsymrel3 34436. (Contributed by Peter Mazsa, 28-Mar-2019.) (Revised by Peter Mazsa, 17-Sep-2021.)
(∀𝑥𝑦(𝑥𝑅𝑦𝑦𝑅𝑥) ∧ Rel ≀ 𝑅)
Theoremsymrelcoss2 34356 The class of cosets by 𝑅 is symmetric, cf. dfsymrel2 34435. (Contributed by Peter Mazsa, 27-Dec-2018.)
(𝑅 ⊆ ≀ 𝑅 ∧ Rel ≀ 𝑅)
Theoremcossssid 34357 Equivalent expressions for the class of cosets by 𝑅 to be a subset of the identity class. (Contributed by Peter Mazsa, 27-Jul-2021.)
( ≀ 𝑅 ⊆ I ↔ ≀ 𝑅 ⊆ ( I ∩ (dom ≀ 𝑅 × ran ≀ 𝑅)))
Theoremcossssid2 34358* Equivalent expressions for the class of cosets by 𝑅 to be a subset of the identity class. (Contributed by Peter Mazsa, 10-Mar-2019.)
( ≀ 𝑅 ⊆ I ↔ ∀𝑥𝑦(∃𝑢(𝑢𝑅𝑥𝑢𝑅𝑦) → 𝑥 = 𝑦))
Theoremcossssid3 34359* Equivalent expressions for the class of cosets by 𝑅 to be a subset of the identity class. (Contributed by Peter Mazsa, 10-Mar-2019.)
( ≀ 𝑅 ⊆ I ↔ ∀𝑢𝑥𝑦((𝑢𝑅𝑥𝑢𝑅𝑦) → 𝑥 = 𝑦))
Theoremcossssid4 34360* Equivalent expressions for the class of cosets by 𝑅 to be a subset of the identity class. (Contributed by Peter Mazsa, 31-Aug-2021.)
( ≀ 𝑅 ⊆ I ↔ ∀𝑢∃*𝑥 𝑢𝑅𝑥)
Theoremcossssid5 34361* Equivalent expressions for the class of cosets by 𝑅 to be a subset of the identity class. (Contributed by Peter Mazsa, 5-Sep-2021.)
( ≀ 𝑅 ⊆ I ↔ ∀𝑥 ∈ ran 𝑅𝑦 ∈ ran 𝑅(𝑥 = 𝑦 ∨ ([𝑥]𝑅 ∩ [𝑦]𝑅) = ∅))
Theorembrcosscnv 34362* 𝐴 and 𝐵 are cosets by converse 𝑅: a binary relation. (Contributed by Peter Mazsa, 23-Jan-2019.)
((𝐴𝑉𝐵𝑊) → (𝐴𝑅𝐵 ↔ ∃𝑥(𝐴𝑅𝑥𝐵𝑅𝑥)))
Theorembrcosscnv2 34363 𝐴 and 𝐵 are cosets by converse 𝑅: a binary relation. (Contributed by Peter Mazsa, 12-Mar-2019.)
((𝐴𝑉𝐵𝑊) → (𝐴𝑅𝐵 ↔ ([𝐴]𝑅 ∩ [𝐵]𝑅) ≠ ∅))
Theorembr1cosscnvxrn 34364 𝐴 and 𝐵 are cosets by the converse range Cartesian product: a binary relation. (Contributed by Peter Mazsa, 19-Apr-2020.) (Revised by Peter Mazsa, 21-Sep-2021.)
((𝐴𝑉𝐵𝑊) → (𝐴(𝑅𝑆)𝐵 ↔ (𝐴𝑅𝐵𝐴𝑆𝐵)))
Theorem1cosscnvxrn 34365 Cosets by the converse range Cartesian product. (Contributed by Peter Mazsa, 19-Apr-2020.) (Revised by Peter Mazsa, 21-Sep-2021.)
(𝐴𝐵) = ( ≀ 𝐴 ∩ ≀ 𝐵)
Theoremcosscnvssid3 34366* Equivalent expressions for the class of cosets by the converse of 𝑅 to be a subset of the identity class. (Contributed by Peter Mazsa, 28-Jul-2021.)
( ≀ 𝑅 ⊆ I ↔ ∀𝑢𝑣𝑥((𝑢𝑅𝑥𝑣𝑅𝑥) → 𝑢 = 𝑣))
Theoremcosscnvssid4 34367* Equivalent expressions for the class of cosets by the converse of 𝑅 to be a subset of the identity class. (Contributed by Peter Mazsa, 31-Aug-2021.)
( ≀ 𝑅 ⊆ I ↔ ∀𝑥∃*𝑢 𝑢𝑅𝑥)
Theoremcosscnvssid5 34368* Equivalent expressions for the class of cosets by the converse of the relation 𝑅 to be a subset of the identity class. (Contributed by Peter Mazsa, 5-Sep-2021.)
(( ≀ 𝑅 ⊆ I ∧ Rel 𝑅) ↔ (∀𝑢 ∈ dom 𝑅𝑣 ∈ dom 𝑅(𝑢 = 𝑣 ∨ ([𝑢]𝑅 ∩ [𝑣]𝑅) = ∅) ∧ Rel 𝑅))
Theoremcoss0 34369 Cosets by the empty set are the empty set. (Contributed by Peter Mazsa, 22-Oct-2019.)
≀ ∅ = ∅
Theoremcossid 34370 Cosets by the identity relation are the identity relation. (Contributed by Peter Mazsa, 16-Jan-2019.)
≀ I = I
Theoremcosscnvid 34371 Cosets by the converse identity relation are the identity relation. (Contributed by Peter Mazsa, 27-Sep-2021.)
I = I
Theoremtrcoss 34372* Sufficient condition for the transitivity of cosets by 𝑅. (Contributed by Peter Mazsa, 26-Dec-2018.)
(∀𝑦∃*𝑢 𝑢𝑅𝑦 → ∀𝑥𝑦𝑧((𝑥𝑅𝑦𝑦𝑅𝑧) → 𝑥𝑅𝑧))
Theoremeleccossin 34373 Two ways of saying that the coset of 𝐴 and the coset of 𝐶 have the common element 𝐵. (Contributed by Peter Mazsa, 15-Oct-2021.)
((𝐵𝑉𝐶𝑊) → (𝐵 ∈ ([𝐴] ≀ 𝑅 ∩ [𝐶] ≀ 𝑅) ↔ (𝐴𝑅𝐵𝐵𝑅𝐶)))
Theoremtrcoss2 34374* Equivalent expressions for the transitivity of cosets by 𝑅. (Contributed by Peter Mazsa, 4-Jul-2020.) (Revised by Peter Mazsa, 16-Oct-2021.)
(∀𝑥𝑦𝑧((𝑥𝑅𝑦𝑦𝑅𝑧) → 𝑥𝑅𝑧) ↔ ∀𝑥𝑧(([𝑥] ≀ 𝑅 ∩ [𝑧] ≀ 𝑅) ≠ ∅ → ([𝑥]𝑅 ∩ [𝑧]𝑅) ≠ ∅))
20.21.5 Relations
Definitiondf-rels 34375 Define the relations class. Proper class relations (like I, cf. reli 5282) are not elements of it. The element of this class and the relation predicate are the same when 𝑅 is a set (cf. elrelsrel 34377).
The class of relations is a great tool we can use when we define classes of different relations as nullary class constants as required by the 2. point in our Guidelines http://us.metamath.org/mpeuni/mathbox.html. When we want to define a specific class of relations as a nullary class constant, the appropriate method is the following:
1. We define the specific nullary class constant for general sets (cf. e.g. df-refs 34400), then
2. we get the required class of relations by the intersection of the class of general sets above with the class of relations df-rels 34375 (cf. df-refrels 34401 and the resulting dfrefrels2 34403 and dfrefrels3 34404).
3. Finally, in order to be able to work with proper classes (like iprc 7143) as well, we define the predicate of the relation (cf. df-refrel 34402) so that it is true for the relevant proper classes (cf. refrelid 34411), and that the element of the class of the required relations (e.g. elrefrels3 34408) and this predicate are the same in case of sets (cf. elrefrelsrel 34409). (Contributed by Peter Mazsa, 13-Jun-2018.)
Rels = 𝒫 (V × V)
Theoremelrels2 34376 The element of the relations class (df-rels 34375) and the relation predicate (df-rel 5150) are the same when 𝑅 is a set. (Contributed by Peter Mazsa, 14-Jun-2018.)
(𝑅𝑉 → (𝑅 ∈ Rels ↔ 𝑅 ⊆ (V × V)))
Theoremelrelsrel 34377 The element of the relations class (df-rels 34375) and the relation predicate are the same when 𝑅 is a set. (Contributed by Peter Mazsa, 24-Nov-2018.)
(𝑅𝑉 → (𝑅 ∈ Rels ↔ Rel 𝑅))
Theoremelrelsrelim 34378 The element of the relations class is a relation. (Contributed by Peter Mazsa, 20-Jul-2019.)
(𝑅 ∈ Rels → Rel 𝑅)
Theoremelrels5 34379 Equivalent expressions for an element of the relations class. (Contributed by Peter Mazsa, 21-Jul-2021.)
(𝑅𝑉 → (𝑅 ∈ Rels ↔ (𝑅 ↾ dom 𝑅) = 𝑅))
Theoremelrels6 34380 Equivalent expressions for an element of the relations class. (Contributed by Peter Mazsa, 21-Jul-2021.)
(𝑅𝑉 → (𝑅 ∈ Rels ↔ (𝑅 ∩ (dom 𝑅 × ran 𝑅)) = 𝑅))
Theoremelrelscnveq3 34381* Two ways of saying a relation is symmetric. (Contributed by Peter Mazsa, 22-Aug-2021.)
(𝑅 ∈ Rels → (𝑅 = 𝑅 ↔ ∀𝑥𝑦(𝑥𝑅𝑦𝑦𝑅𝑥)))
Theoremelrelscnveq 34382 Two ways of saying a relation is symmetric. (Contributed by Peter Mazsa, 22-Aug-2021.)
(𝑅 ∈ Rels → (𝑅𝑅𝑅 = 𝑅))
Theoremelrelscnveq2 34383* Two ways of saying a relation is symmetric. (Contributed by Peter Mazsa, 22-Aug-2021.)
(𝑅 ∈ Rels → (𝑅 = 𝑅 ↔ ∀𝑥𝑦(𝑥𝑅𝑦𝑦𝑅𝑥)))
Theoremelrelscnveq4 34384* Two ways of saying a relation is symmetric. (Contributed by Peter Mazsa, 22-Aug-2021.)
(𝑅 ∈ Rels → (𝑅𝑅 ↔ ∀𝑥𝑦(𝑥𝑅𝑦𝑦𝑅𝑥)))
Theoremcnvelrels 34385 The converse of a set is an element of the class of relations. (Contributed by Peter Mazsa, 18-Aug-2019.)
(𝐴𝑉𝐴 ∈ Rels )
Theoremcosselrels 34386 Cosets of sets are elements of the relations class. Implies (𝑅 ∈ Rels → ≀ 𝑅 ∈ Rels ). (Contributed by Peter Mazsa, 25-Aug-2021.)
(𝐴𝑉 → ≀ 𝐴 ∈ Rels )
Theoremcosscnvelrels 34387 Cosets of converse sets are elements of the relations class. (Contributed by Peter Mazsa, 31-Aug-2021.)
(𝐴𝑉 → ≀ 𝐴 ∈ Rels )
20.21.6 Subset relations
Definitiondf-ssr 34388* Define the subsets class or the class of all subset relations. Similar to definitions of epsilon relation (df-eprel 5058) and identity relation (df-id 5053) classes. Subset relation class and Scott Fenton's subset class df-sset 32088 are the same: S = SSet (compare dfssr2 34389 with df-sset 32088, cf. comment of df-xrn 34273), the only reason we do not use dfssr2 34389 as the base definition of the subsets class is the way we defined the epsilon relation and the identity relation classes.
The binary relation on the class of all subsets and the subclass relationship (df-ss 3621) are the same, that is, (𝐴 S 𝐵𝐴𝐵) when 𝐵 is a set, cf. brssr 34391. Yet in general we use the subclass relation 𝐴𝐵 both for classes and for sets, cf. the comment of df-ss 3621. The only exception (aside from directly investigating the class S e.g. in relssr 34390 or in extssr 34399) is when we have a specific purpose with its usage, like in case of df-refs 34400 vs. df-cnvrefs 34413, where we need S to define the class of reflexive sets in order to be able to define the class of converse reflexive sets with the help of the converse of S.
The subsets class S has another place in set.mm as well: if we define extensional relation based on the common property in extid 34222, extep 34189 and extssr 34399, then "extrelssr" " |- ExtRel _S " is a theorem along with "extrelep" " |- ExtRel _E " and "extrelid" " |- ExtRel _I ". (Contributed by Peter Mazsa, 25-Jul-2019.)
S = {⟨𝑥, 𝑦⟩ ∣ 𝑥𝑦}
Theoremdfssr2 34389 Alternate definition of the subset relation. (Contributed by Peter Mazsa, 9-Aug-2021.)
S = ((V × V) ∖ ran ( E ⋉ (V ∖ E )))
Theoremrelssr 34390 The subset relation is a relation. (Contributed by Peter Mazsa, 1-Aug-2019.)
Rel S
Theorembrssr 34391 The subset relation and subclass relationship (df-ss 3621) are the same, that is, (𝐴 S 𝐵𝐴𝐵) when 𝐵 is a set. (Contributed by Peter Mazsa, 31-Jul-2019.)
(𝐵𝑉 → (𝐴 S 𝐵𝐴𝐵))
Theorembrssrid 34392 Any set is a subset of itself. (Contributed by Peter Mazsa, 1-Aug-2019.)
(𝐴𝑉𝐴 S 𝐴)
Theoremissetssr 34393 Two ways of expressing set existence. (Contributed by Peter Mazsa, 1-Aug-2019.)
(𝐴 ∈ V ↔ 𝐴 S 𝐴)
Theorembrssrres 34394 Restricted subset binary relation. (Contributed by Peter Mazsa, 25-Nov-2019.)
(𝐶𝑉 → (𝐵( S ↾ 𝐴)𝐶 ↔ (𝐵𝐴𝐵𝐶)))
Theorembr1cnvssrres 34395 Restricted converse subset binary relation. (Contributed by Peter Mazsa, 25-Nov-2019.)
(𝐵𝑉 → (𝐵( S ↾ 𝐴)𝐶 ↔ (𝐶𝐴𝐶𝐵)))
Theorembrcnvssr 34396 The converse of a subset relation swaps arguments. (Contributed by Peter Mazsa, 1-Aug-2019.)
(𝐴𝑉 → (𝐴 S 𝐵𝐵𝐴))
Theorembrcnvssrid 34397 Any set is a converse subset of itself. (Contributed by Peter Mazsa, 9-Jun-2021.)
(𝐴𝑉𝐴 S 𝐴)
Theorembr1cossxrncnvssrres 34398* 𝐵, 𝐶 and 𝐷, 𝐸 are cosets by tail Cartesian product with restricted converse subsets class: a binary relation. (Contributed by Peter Mazsa, 9-Jun-2021.)
(((𝐵𝑉𝐶𝑊) ∧ (𝐷𝑋𝐸𝑌)) → (⟨𝐵, 𝐶⟩ ≀ (𝑅 ⋉ ( S ↾ 𝐴))⟨𝐷, 𝐸⟩ ↔ ∃𝑢𝐴 ((𝐶𝑢𝑢𝑅𝐵) ∧ (𝐸𝑢𝑢𝑅𝐷))))
Theoremextssr 34399 Property of subset relation, cf. extid 34222, extep 34189 and the comment of df-ssr 34388. (Contributed by Peter Mazsa, 10-Jul-2019.)
((𝐴𝑉𝐵𝑊) → ([𝐴] S = [𝐵] S ↔ 𝐴 = 𝐵))
20.21.7 Reflexivity
Definitiondf-refs 34400 Define the class of all reflexive sets. It is used only by df-refrels 34401. We use subset relation S (df-ssr 34388) here to be able to define converse reflexivity (df-cnvrefs 34413), cf. the comment of df-ssr 34388. The elements of this class are not necessarily relations (vs. df-refrels 34401).
Note the similarity of the definitions df-refs 34400, df-syms 34428 and ~? df-trs , cf. the comments of dfrefrels2 34403. (Contributed by Peter Mazsa, 19-Jul-2019.)
Refs = {𝑥 ∣ ( I ∩ (dom 𝑥 × ran 𝑥)) S (𝑥 ∩ (dom 𝑥 × ran 𝑥))}
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14301-14400 145 14401-14500 146 14501-14600 147 14601-14700 148 14701-14800 149 14801-14900 150 14901-15000 151 15001-15100 152 15101-15200 153 15201-15300 154 15301-15400 155 15401-15500 156 15501-15600 157 15601-15700 158 15701-15800 159 15801-15900 160 15901-16000 161 16001-16100 162 16101-16200 163 16201-16300 164 16301-16400 165 16401-16500 166 16501-16600 167 16601-16700 168 16701-16800 169 16801-16900 170 16901-17000 171 17001-17100 172 17101-17200 173 17201-17300 174 17301-17400 175 17401-17500 176 17501-17600 177 17601-17700 178 17701-17800 179 17801-17900 180 17901-18000 181 18001-18100 182 18101-18200 183 18201-18300 184 18301-18400 185 18401-18500 186 18501-18600 187 18601-18700 188 18701-18800 189 18801-18900 190 18901-19000 191 19001-19100 192 19101-19200 193 19201-19300 194 19301-19400 195 19401-19500 196 19501-19600 197 19601-19700 198 19701-19800 199 19801-19900 200 19901-20000 201 20001-20100 202 20101-20200 203 20201-20300 204 20301-20400 205 20401-20500 206 20501-20600 207 20601-20700 208 20701-20800 209 20801-20900 210 20901-21000 211 21001-21100 212 21101-21200 213 21201-21300 214 21301-21400 215 21401-21500 216 21501-21600 217 21601-21700 218 21701-21800 219 21801-21900 220 21901-22000 221 22001-22100 222 22101-22200 223 22201-22300 224 22301-22400 225 22401-22500 226 22501-22600 227 22601-22700 228 22701-22800 229 22801-22900 230 22901-23000 231 23001-23100 232 23101-23200 233 23201-23300 234 23301-23400 235 23401-23500 236 23501-23600 237 23601-23700 238 23701-23800 239 23801-23900 240 23901-24000 241 24001-24100 242 24101-24200 243 24201-24300 244 24301-24400 245 24401-24500 246 24501-24600 247 24601-24700 248 24701-24800 249 24801-24900 250 24901-25000 251 25001-25100 252 25101-25200 253 25201-25300 254 25301-25400 255 25401-25500 256 25501-25600 257 25601-25700 258 25701-25800 259 25801-25900 260 25901-26000 261 26001-26100 262 26101-26200 263 26201-26300 264 26301-26400 265 26401-26500 266 26501-26600 267 26601-26700 268 26701-26800 269 26801-26900 270 26901-27000 271 27001-27100 272 27101-27200 273 27201-27300 274 27301-27400 275 27401-27500 276 27501-27600 277 27601-27700 278 27701-27800 279 27801-27900 280 27901-28000 281 28001-28100 282 28101-28200 283 28201-28300 284 28301-28400 285 28401-28500 286 28501-28600 287 28601-28700 288 28701-28800 289 28801-28900 290 28901-29000 291 29001-29100 292 29101-29200 293 29201-29300 294 29301-29400 295 29401-29500 296 29501-29600 297 29601-29700 298 29701-29800 299 29801-29900 300 29901-30000 301 30001-30100 302 30101-30200 303 30201-30300 304 30301-30400 305 30401-30500 306 30501-30600 307 30601-30700 308 30701-30800 309 30801-30900 310 30901-31000 311 31001-31100 312 31101-31200 313 31201-31300 314 31301-31400 315 31401-31500 316 31501-31600 317 31601-31700 318 31701-31800 319 31801-31900 320 31901-32000 321 32001-32100 322 32101-32200 323 32201-32300 324 32301-32400 325 32401-32500 326 32501-32600 327 32601-32700 328 32701-32800 329 32801-32900 330 32901-33000 331 33001-33100 332 33101-33200 333 33201-33300 334 33301-33400 335 33401-33500 336 33501-33600 337 33601-33700 338 33701-33800 339 33801-33900 340 33901-34000 341 34001-34100 342 34101-34200 343 34201-34300 344 34301-34400 345 34401-34500 346 34501-34600 347 34601-34700 348 34701-34800 349 34801-34900 350 34901-35000 351 35001-35100 352 35101-35200 353 35201-35300 354 35301-35400 355 35401-35500 356 35501-35600 357 35601-35700 358 35701-35800 359 35801-35900 360 35901-36000 361 36001-36100 362 36101-36200 363 36201-36300 364 36301-36400 365 36401-36500 366 36501-36600 367 36601-36700 368 36701-36800 369 36801-36900 370 36901-37000 371 37001-37100 372 37101-37200 373 37201-37300 374 37301-37400 375 37401-37500 376 37501-37600 377 37601-37700 378 37701-37800 379 37801-37900 380 37901-38000 381 38001-38100 382 38101-38200 383 38201-38300 384 38301-38400 385 38401-38500 386 38501-38600 387 38601-38700 388 38701-38800 389 38801-38900 390 38901-39000 391 39001-39100 392 39101-39200 393 39201-39300 394 39301-39400 395 39401-39500 396 39501-39600 397 39601-39700 398 39701-39800 399 39801-39900 400 39901-40000 401 40001-40100 402 40101-40200 403 40201-40300 404 40301-40400 405 40401-40500 406 40501-40600 407 40601-40700 408 40701-40800 409 40801-40900 410 40901-41000 411 41001-41100 412 41101-41200 413 41201-41300 414 41301-41400 415 41401-41500 416 41501-41600 417 41601-41700 418 41701-41800 419 41801-41900 420 41901-42000 421 42001-42100 422 42101-42200 423 42201-42300 424 42301-42400 425 42401-42500 426 42501-42600 427 42601-42700 428 42701-42800 429 42801-42879
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## Solved! Leetcode 139. Word Break
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source: https://leetcode.com/problems/number-of-music-playlists/ Table of ContentsNumber of Music PlaylistsDescriptionExample 1:Example 2:Example 3:Constraints:Solution Number of Music Playlists Description Your music player contains n different songs. You want to listen to goal songs (not necessarily different) during your trip. To avoid boredom, you will create a playlist so that: Every song is played at least once. A song ...
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source: https://leetcode.com/problems/single-number-ii/description/ Table of ContentsSingle Number IIDescriptionExample 1:Example 2:Constraints:Solution Single Number II Description Given an integer array nums where every element appears three times except for one, which appears exactly once. Find the single element and return it. You must implement a solution with a linear runtime complexity and use only constant extra space. Example ...
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source: https://leetcode.com/problems/path-with-maximum-probability/description/ Table of ContentsPath with Maximum ProbabilityDescriptionExample 1:Example 2:Example 3:Constraints:Solution Path with Maximum Probability Description You are given an undirected weighted graph of n nodes (0-indexed), represented by an edge list where edges[i] = [a, b] is an undirected edge connecting the nodes a and b with a probability of success of traversing that ...
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source: https://leetcode.com/problems/maximize-the-confusion-of-an-exam/description/ Table of ContentsMaximize the Confusion of an ExamDescriptionExample 1:Example 2:Example 3:Constraints:Solution Maximize the Confusion of an Exam Description A teacher is writing a test with n true/false questions, with ‘T’ denoting true and ‘F’ denoting false. He wants to confuse the students by maximizing the number of consecutive questions with the same answer ...
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ETO Results | Functions: Floor and Ceil
BO 4.3 Platform
Updated over a week ago
When working in ETO Results, you may want to round a number down or up in order to make your data more succinct. In order to do this, you can use the floor or ceil functions.
The floor function rounds the number down to the nearest whole number. Formulas using this function will be set up as follows:
=floor([Variable with Number])
For example, let us say you have a variable [Age in Years] that is equal to 1.8. Since it is years, you would want to round down to 1. To do this, you would use the formula =floor([Age in Years]).
Alternatively, the ceil function will round up. Formulas using this function will be set up as follows:
=ceil([Variable with Number])
For example, let us say you have a variable [Average Days in Program] that is equal to 30.15. You would want to round up to 31. To do this, you would use the formula =ceil([Average Days in Program]).
One helpful thing to remember is that in floors are the lowest point in a room, and the floor function lowers a number, while ceilings are the highest point in a room, and the ceil function raises a number up.
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https://writernursing.com/what-similarity-does-a-z-test-have-to-a-simple-z-or-standard-score/
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5/5
# Browse Over 10 Million Questions
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# What similarity does a z test have to a simple z or standard score?
## Writernursing.com
Some questions in Part A require that you access data from Statistics for People Who (Think They) Hate Statistics. This data is available on the student website under the Student Test Resources link.
For the following research questions, create one null hypothesis, one directional research hypothesis, and one nondirectional research hypothesis.
What are the effects of attention on out-of-seat classroom behavior?
Research Hypothesis: There will be a relationship between the effects of attention on out-of-seat classroom behavior versus in-seat-classroom behavior.
What is the relationship between the quality of a marriage and the quality of the spouses’ relationships with their siblings?
Null Hypothesis: There will be no relationship in the relationship between the quality of a marriage and the quality of the spouses’ relationship with their siblings.
What is the best way to treat an eating disorder?
One Directional Research Hypothesis:
Provide one research hypothesis and an equation for each of the following topics:
The amount of money spent on food among undergraduate students and undergraduate student-athletes
The average amount of time taken by white and brown rats to get out of a maze
The effects of Drug A and Drug B on a disease
The time to complete a task in Method 1 and Method 2
Why does the null hypothesis presume no relationship between variables?
Create a research hypothesis tested using a one-tailed test and a research hypothesis tested using a two-tailed test.
What does the critical value represent?
Given the following information, would your decision be to reject or fail to reject the null hypothesis? Setting the level of significance at .05 for decision making, provide an explanation for your conclusion.
The null hypothesis that there is no relationship between the type of music a person listens to and his crime rate (p < .05).
In Hypothesis Testing, we typically deem a research hypothesis to be significant, if the odds of two means actually being equal are no greater than 1 in 20 or .05 (5%) or less.
The null hypothesis that there is no relationship between the amount of coffee consumption and GPA (p = .62).
The null hypothesis that there is a negative relationship between the number of hours worked and level of job satisfaction (p = .51).
Why is it harder to find a significant outcome (all other things being equal) when the research hypothesis is being tested at the .01 rather than the .05 level of significance?
At the .01 level, there is less room for error because the test is more rigorous.
Why should we think in terms of “failing to reject” the null rather than just accepting it?
When is it appropriate to use the one-sample z test?
What similarity does a z test have to a simple z or standard score?
For the following situations, write out a research hypothesis:
Bob wants to know if the weight loss for his group on the chocolate-only diet is representative of weight loss in a large population of middle-aged men.
The health department is charged with finding out if the rate of flu per thousand citizens for this past flu season is comparable to the average rate of the past 50 seasons.
Blair is almost sure that his monthly costs for the past year are not representative of his average monthly costs over the past 20 years.
There were about 15 flu cases per week, this flu season, in the Oshkosh school system. The weekly average for the entire state is 16 and the standard deviation, is 2.35. Are the kids in Oshkosh as sick as the kids throughout the state?
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nit 1 – Advanced Nur Unit 1 Discussion – Advanced It is anticipated that the initial discussion response should be in the range of 250-300
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Course: Algebra (all content)>Unit 10
Lesson 8: Multiplying binomials by polynomials
Multiplying binomials by polynomials review
A review of how to multiply binomials like 1 + x by polynomials with more than two terms like x^2 - 5x - 6
We already know how to multiply binomials like $\left(x+2\right)\left(x-7\right)$. In this article, we review a slightly more complicated skill: multiplying binomials by polynomials with more than two terms.
Example
Expand and simplify.
$\left(1+x\right)\left({x}^{2}-5x-6\right)$
Apply the distributive property.
$\begin{array}{rl}& \left(1+x\right)\left({x}^{2}-5x-6\right)\\ \\ =& 1\left({x}^{2}-5x-6\right)+x\left({x}^{2}-5x-6\right)\end{array}$
Apply the distributive property again.
$=1\left({x}^{2}\right)+1\left(-5x\right)+1\left(-6\right)+x\left({x}^{2}\right)+x\left(-5x\right)+x\left(-6\right)$
Notice the pattern. We multiplied each term in the binomial by each term in the trinomial.
Simplify.
$\begin{array}{rl}=& {x}^{2}-5x-6+{x}^{3}-5{x}^{2}-6x\\ \\ =& {x}^{3}-4{x}^{2}-11x-6\end{array}$
Practice
Expand and simplify.
$\left({c}^{2}-6\right)\left(2{c}^{2}+3c-1\right)$
Want more practice? Check out this exercise.
Want to join the conversation?
• How do you simplify an equation
• You just need to combine like terms. Example- t to the fourth plus t to the fourth; 2w squared plus w squared. I hope this helps and God bless!
• What would be considered the fastest method of multiplying long polynomials?
• Great question! Yes there is a shortcut, which efficiently combines all the terms for each power of x.
If only one variable, say x, appears in the two polynomials to be multiplied together, then you can use a technique similar to the Vedic multiplication arithmetic technique called vertical and crosswise (which you can look up online to get the idea). Write the terms of each polynomial in order of descending powers of x. For any missing exponents on x, use a coefficient of zero.
Example: multiply (6x^4 + 3x^3 - 5x + 7) by (2x^3 - 4x^2 - x + 8).
Write
(6x^4 + 3x^3 + 0x^2 - 5x + 7)
(0x^4 + 2x^3 - 4x^2 - x + 8).
These two polynomials are now each written with five coefficients.
The idea is to multiply the first coefficient by the first coefficient, then cross multiply the first two coefficients by the first two coefficients, then the first three by the first three, then the first four by the first four, then the first five by the first five, then the last four by the last four, then the last three by the last three, then the last two by the last two, then finally the last one by the last one.
There are no terms of degree 9 or higher.
Coefficient of x^8 is 6(0) = 0.
Coefficient of x^7 is 6(2) + 3(0) = 12.
Coefficient of x^6 is 6(-4) + 3(2) + 0(0) = -18.
Coefficient of x^5 is 6(-1) + 3(-4) + 0(2) + (-5)(0) = -18.
Coefficient of x^4 is 6(8) + 3(-1) + 0(-4) + (-5)(2) + 7(0) = 35.
Coefficient of x^3 is 3(8) + 0(-1) + (-5)(-4) + 7(2) = 58.
Coefficient of x^2 is 0(8) + (-5)(-1) + 7(-4) = -23.
Coefficient of x is -5(8) + 7(-1) = -47.
Constant term (coefficient on x^0) is 7(8) = 56.
12x^7 - 18x^6 - 18x^5 + 35x^4 + 58x^3 - 23x^2 - 47x + 56.
Have a blessed, wonderful day!
• So, you can apply the distributive property to one of these equations, or you can use the graph/colored boxes thing? Which way is more efficient?
• There is FOIL, double distribution, and box method as 3 ways of doing the exact same thing, they all give the same 4 terms with the middle terms usually combined. FOIL is probably less efficient because it is limited to multiplying two binomials. The box method may require more writing, but might make more sense in exactly what is going on. Double distribution and box method can easily be expanded to larger polynomials in the future. Efficient may be in the eye of the beholder, so find which one works best for you. Many teachers will require you to learn all so that you can make an educated decision which is best, and if you do it enough and have a pretty good math brain, you will be able to do it in your head.
• in the vid they never clearly said which goes first they just said its the only one cubed and do I do y squared first or y alone first
• Generally, you should write your answer in standard form. This would have the term with y^2 first, then y, then the constant. However, it is usually not required unless the instructions specify your answer must be in standard form.
• is this apart of the algebra 2 course
• I would say this is something you learn in Algebra 1. This seems more like a review in Algebra 2.
• Am I ever going to need this type of math in the real world?
• I mean, you'll need it for the SAT or the ACT - the math in those is primarily Algebra 2, and stuff like this, but as far as the real world, it depends on your job... but we have computer software to do this stuff nowdays :)
• What do you think is the best way to multiply and simplify a problem?
• In my opinion, the area model is easier because I can see out all the terms and not get things mixed up. I tend to get numbers mixed up and write things wrong when I do the distributive property. However, it is your preference in the end!
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The links in this article can be very useful for that. fit_constrained (constraints[, start_params]) fit the model subject to linear equality constraints. You can regard polynomial regression as a generalized case of linear regression. Fits a generalized linear model for a given family. That’s one of the reasons why Python is among the main programming languages for machine learning. Stacking for Regression It’s a powerful Python package for the estimation of statistical models, performing tests, and more. It takes the input array as the argument and returns the modified array. Now if we have relaxed conditions on the coefficients, then the constrained regions can get bigger and eventually they will hit the centre of the ellipse. In the following example, we will use multiple linear regression to predict the stock index price (i.e., the dependent variable) of a fictitious economy by using 2 independent/input variables: 1. When I read explanation on how to do that stuff in Python, Logit Regression can handle multi class. Linear regression is implemented with the following: Both approaches are worth learning how to use and exploring further. In addition to numpy and sklearn.linear_model.LinearRegression, you should also import the class PolynomialFeatures from sklearn.preprocessing: The import is now done, and you have everything you need to work with. Enjoy free courses, on us →, by Mirko Stojiljković But to have a regression, Y must depend on X in some way. This is how the next statement looks: The variable model again corresponds to the new input array x_. They look very similar and are both linear functions of the unknowns ₀, ₁, and ₂. You can notice that .intercept_ is a scalar, while .coef_ is an array. I am trying to implement a linear regression model in Tensorflow, with additional constraints (coming from the domain) that the W and b terms must be non-negative. This is how you can obtain one: You should be careful here! Share Your goal is to calculate the optimal values of the predicted weights ₀ and ₁ that minimize SSR and determine the estimated regression function. It takes the input array x as an argument and returns a new array with the column of ones inserted at the beginning. Let’s create an instance of this class: The variable transformer refers to an instance of PolynomialFeatures which you can use to transform the input x. Tweet The team members who worked on this tutorial are: Master Real-World Python Skills With Unlimited Access to Real Python. This equation is the regression equation. In many cases, however, this is an overfitted model. Steps 1 and 2: Import packages and classes, and provide data. We introduce c-lasso, a Python package that enables sparse and robust linear regression and classification with linear equality constraints. c-lasso is a Python package that enables sparse and robust linear regression and classification with linear equality constraints on the model parameters. The inputs, however, can be continuous, discrete, or even categorical data such as gender, nationality, brand, and so on. It is the value of the estimated response () for = 0. Regression analysis is one of the most important fields in statistics and machine learning. You apply .transform() to do that: That’s the transformation of the input array with .transform(). Now, remember that you want to calculate ₀, ₁, and ₂, which minimize SSR. I do know I can constrain the coefficients with some python libraries but couldn't find one where I can constrain the intercept. Observations: 8 AIC: 54.63, Df Residuals: 5 BIC: 54.87, coef std err t P>|t| [0.025 0.975], ------------------------------------------------------------------------------, const 5.5226 4.431 1.246 0.268 -5.867 16.912, x1 0.4471 0.285 1.567 0.178 -0.286 1.180, x2 0.2550 0.453 0.563 0.598 -0.910 1.420, Omnibus: 0.561 Durbin-Watson: 3.268, Prob(Omnibus): 0.755 Jarque-Bera (JB): 0.534, Skew: 0.380 Prob(JB): 0.766, Kurtosis: 1.987 Cond. This approach yields the following results, which are similar to the previous case: You see that now .intercept_ is zero, but .coef_ actually contains ₀ as its first element. Its importance rises every day with the availability of large amounts of data and increased awareness of the practical value of data. In practice, regression models are often applied for forecasts. There are many regression methods available. It’s advisable to learn it first and then proceed towards more complex methods. The model has a value of ² that is satisfactory in many cases and shows trends nicely. Stacking for Classification 4. To check the performance of a model, you should test it with new data, that is with observations not used to fit (train) the model. site design / logo © 2020 Stack Exchange Inc; user contributions licensed under cc by-sa. Scipy's curve_fit will accept bounds. It returns self, which is the variable model itself. Disclaimer: This is a very lengthy blog post and involves mathematical proofs and python implementations for various optimization algorithms Optimization, one ⦠The goal of regression is to determine the values of the weights ₀, ₁, and ₂ such that this plane is as close as possible to the actual responses and yield the minimal SSR. However, in real-world situations, having a complex model and ² very close to 1 might also be a sign of overfitting. The procedure is similar to that of scikit-learn. He is a Pythonista who applies hybrid optimization and machine learning methods to support decision making in the energy sector. The matrix is a general constraint matrix. The intercept is already included with the leftmost column of ones, and you don’t need to include it again when creating the instance of LinearRegression. You can obtain the properties of the model the same way as in the case of simple linear regression: You obtain the value of ² using .score() and the values of the estimators of regression coefficients with .intercept_ and .coef_. The value of ₀, also called the intercept, shows the point where the estimated regression line crosses the axis. You can do this by replacing x with x.reshape(-1), x.flatten(), or x.ravel() when multiplying it with model.coef_. The case of more than two independent variables is similar, but more general. Following the assumption that (at least) one of the features depends on the others, you try to establish a relation among them. There are several more optional parameters. They are the distances between the green circles and red squares. The predicted response is now a two-dimensional array, while in the previous case, it had one dimension. This is why you can solve the polynomial regression problem as a linear problem with the term ² regarded as an input variable. You should, however, be aware of two problems that might follow the choice of the degree: underfitting and overfitting. Explaining them is far beyond the scope of this article, but you’ll learn here how to extract them. You can provide several optional parameters to LinearRegression: This example uses the default values of all parameters. Some of them are support vector machines, decision trees, random forest, and neural networks. These pairs are your observations. For example, the leftmost observation (green circle) has the input = 5 and the actual output (response) = 5. In this instance, this might be the optimal degree for modeling this data. brightness_4. It’s open source as well. That’s why .reshape() is used. You can provide the inputs and outputs the same way as you did when you were using scikit-learn: The input and output arrays are created, but the job is not done yet. That’s why you can replace the last two statements with this one: This statement does the same thing as the previous two. Before applying transformer, you need to fit it with .fit(): Once transformer is fitted, it’s ready to create a new, modified input. The forward model is assumed to be: By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy. As for enforcing the sum, the constraint equation reduces the number of degrees of freedom. Regression is also useful when you want to forecast a response using a new set of predictors. No. $\begingroup$ @Vic. Simple linear regression is an approach for predicting a response using a single feature.It is assumed that the two variables are linearly related. You need to add the column of ones to the inputs if you want statsmodels to calculate the intercept ₀. For that reason, you should transform the input array x to contain the additional column(s) with the values of ² (and eventually more features). The regression analysis page on Wikipedia, Wikipedia’s linear regression article, as well as Khan Academy’s linear regression article are good starting points. Of course, it’s open source. You can print x and y to see how they look now: In multiple linear regression, x is a two-dimensional array with at least two columns, while y is usually a one-dimensional array. Join us and get access to hundreds of tutorials, hands-on video courses, and a community of expert Pythonistas: Real Python Comment Policy: The most useful comments are those written with the goal of learning from or helping out other readers—after reading the whole article and all the earlier comments. By using our site, you acknowledge that you have read and understand our Cookie Policy, Privacy Policy, and our Terms of Service. Importing all the required libraries. Join us and get access to hundreds of tutorials, hands-on video courses, and a community of expert Pythonistas: Master Real-World Python SkillsWith Unlimited Access to Real Python. Multiple linear regression uses a linear function to predict the value of a target variable y, containing the function n independent variable x=[xâ,xâ,xâ,â¦,xâ]. It’s ready for application. When performing linear regression in Python, you can follow these steps: Import the packages and classes you need; Provide data to work with and eventually do appropriate transformations; Create a regression model and fit it with existing data; Check the results of model fitting to know whether the model is satisfactory; Apply the model for predictions How to force zero interception in linear regression? You can find many statistical values associated with linear regression including ², ₀, ₁, and ₂. lowerbound<=intercept<=upperbound. where XÌ is the mean of X values and Ȳ is the mean of Y values.. There are five basic steps when you’re implementing linear regression: These steps are more or less general for most of the regression approaches and implementations. constrained linear regression / quadratic programming python, How to carry out constrained regression in R, Multiple linear regression with fixed coefficient for a feature. You can obtain the predicted response on the input values used for creating the model using .fittedvalues or .predict() with the input array as the argument: This is the predicted response for known inputs. You'll want to get familiar with linear regression because you'll need to use it if you're trying to measure the relationship between two or more continuous values.A deep dive into the theory and implementation of linear regression will help you understand this valuable machine learning algorithm. Related Tutorial Categories: In other words, a model learns the existing data too well. For example, the case of flipping a coin (Head/Tail). import pandas as pd. How can a company reduce my number of shares? The independent features are called the independent variables, inputs, or predictors. This is how the new input array looks: The modified input array contains two columns: one with the original inputs and the other with their squares. What is the difference between "wire" and "bank" transfer? rev 2020.12.2.38106, Stack Overflow works best with JavaScript enabled, Where developers & technologists share private knowledge with coworkers, Programming & related technical career opportunities, Recruit tech talent & build your employer brand, Reach developers & technologists worldwide. You can implement multiple linear regression following the same steps as you would for simple regression. @seed the question was changed to ask about a range for the intercept, and no longer asks about a fixed value. As you can see, x has two dimensions, and x.shape is (6, 1), while y has a single dimension, and y.shape is (6,). Basically, all you should do is apply the proper packages and their functions and classes. The regression model based on ordinary least squares is an instance of the class statsmodels.regression.linear_model.OLS. It might also be important that a straight line can’t take into account the fact that the actual response increases as moves away from 25 towards zero. The output here differs from the previous example only in dimensions. This kind of problem is well known as linear programming. Most notably, you have to make sure that a linear relationship exists between the depe⦠It doesn’t takes ₀ into account by default. Overfitting happens when a model learns both dependencies among data and random fluctuations. This is how the modified input array looks in this case: The first column of x_ contains ones, the second has the values of x, while the third holds the squares of x. First, you import numpy and sklearn.linear_model.LinearRegression and provide known inputs and output: That’s a simple way to define the input x and output y. It’s just shorter. Linear regression is sometimes not appropriate, especially for non-linear models of high complexity. This kind of problem is well known as linear programming. I do know I can constrain the coefficients with some python libraries but couldn't find one where I can constrain the intercept. You apply linear regression for five inputs: ₁, ₂, ₁², ₁₂, and ₂². This approach is called the method of ordinary least squares. Of course, there are more general problems, but this should be enough to illustrate the point. You can provide several optional parameters to PolynomialFeatures: This example uses the default values of all parameters, but you’ll sometimes want to experiment with the degree of the function, and it can be beneficial to provide this argument anyway. Therefore x_ should be passed as the first argument instead of x. You now know what linear regression is and how you can implement it with Python and three open-source packages: NumPy, scikit-learn, and statsmodels. Generally, in regression analysis, you usually consider some phenomenon of interest and have a number of observations. You can call .summary() to get the table with the results of linear regression: This table is very comprehensive. Multiple or multivariate linear regression is a case of linear regression with two or more independent variables. And the package used above for constrained regression is a custom library made for our Marketing Mix Model tool. The estimated or predicted response, (ᵢ), for each observation = 1, …, , should be as close as possible to the corresponding actual response ᵢ. For detailed info, one can check the documentation. First, you need to call .fit() on model: With .fit(), you calculate the optimal values of the weights ₀ and ₁, using the existing input and output (x and y) as the arguments. Provide data to work with and eventually do appropriate transformations, Create a regression model and fit it with existing data, Check the results of model fitting to know whether the model is satisfactory. linear regression. You can also use .fit_transform() to replace the three previous statements with only one: That’s fitting and transforming the input array in one statement with .fit_transform(). That’s exactly what the argument (-1, 1) of .reshape() specifies. In the case of two variables and the polynomial of degree 2, the regression function has this form: (₁, ₂) = ₀ + ₁₁ + ₂₂ + ₃₁² + ₄₁₂ + ₅₂². You should notice that you can provide y as a two-dimensional array as well. © 2012–2020 Real Python ⋅ Newsletter ⋅ Podcast ⋅ YouTube ⋅ Twitter ⋅ Facebook ⋅ Instagram ⋅ Python Tutorials ⋅ Search ⋅ Privacy Policy ⋅ Energy Policy ⋅ Advertise ⋅ Contact❤️ Happy Pythoning! The procedure for solving the problem is identical to the previous case. See the section marked UPDATE in my answer for the multivariate fitting example. When you implement linear regression, you are actually trying to minimize these distances and make the red squares as close to the predefined green circles as possible. Leave a comment below and let us know. In other words, you need to find a function that maps some features or variables to others sufficiently well. Ordinary least squares Linear Regression. Is it there a way for when several independent variables are required in the function?. By Nagesh Singh Chauhan , Data Science Enthusiast. I do want to make a constrained linear regression with the intercept value to be like: lowerbound<=intercept<=upperbound. Why does the FAA require special authorization to act as PIC in the North American T-28 Trojan? The predicted responses (red squares) are the points on the regression line that correspond to the input values. There are numerous Python libraries for regression using these techniques. Thus, you cannot fit a generalized linear model or multi-variate regression using this. The coefficient of determination, denoted as ², tells you which amount of variation in can be explained by the dependence on using the particular regression model. You can find more information on statsmodels on its official web site. The next step is to create a linear regression model and fit it using the existing data. Regression problems usually have one continuous and unbounded dependent variable. # Constrained Multiple Linear Regression import numpy as np nd = 100 # number of data sets nc = 5 # number of inputs x = np.random.rand(nd,nc) y = np.random.rand(nd) from gekko import GEKKO m = GEKKO(remote=False); m.options.IMODE=2 c = m.Array(m.FV,nc+1) for ci in c: ci.STATUS=1 ci.LOWER=0 xd = m.Array(m.Param,nc) for i in range(nc): xd[i].value = x[:,i] yd = m.Param(y); yp = ⦠machine-learning The following figure illustrates simple linear regression: When implementing simple linear regression, you typically start with a given set of input-output (-) pairs (green circles). Like NumPy, scikit-learn is also open source. The estimation creates a new model with transformed design matrix, exog, and converts the results back to the original parameterization. There are a lot of resources where you can find more information about regression in general and linear regression in particular. How to draw a seven point star with one path in Adobe Illustrator. If there are just two independent variables, the estimated regression function is (₁, ₂) = ₀ + ₁₁ + ₂₂. c-lasso: a Python package for constrained sparse regression and classification. Why not just make the substitution $\beta_i = \omega_i^2$? Each actual response equals its corresponding prediction. I ⦠The next figure illustrates the underfitted, well-fitted, and overfitted models: The top left plot shows a linear regression line that has a low ². If there are two or more independent variables, they can be represented as the vector = (₁, …, ᵣ), where is the number of inputs. The residuals (vertical dashed gray lines) can be calculated as ᵢ - (ᵢ) = ᵢ - ₀ - ₁ᵢ for = 1, …, . Simple or single-variate linear regression is the simplest case of linear regression with a single independent variable, = . The package scikit-learn provides the means for using other regression techniques in a very similar way to what you’ve seen. To find more information about the results of linear regression, please visit the official documentation page. You’ll have an input array with more than one column, but everything else is the same. In this example, the intercept is approximately 5.52, and this is the value of the predicted response when ₁ = ₂ = 0. Asking for help, clarification, or responding to other answers. The bottom left plot presents polynomial regression with the degree equal to 3. Provide data to work with and eventually do appropriate transformations. It depends on the case. Stacked Generalization 2. Keep in mind that you need the input to be a two-dimensional array. Curated by the Real Python team. This custom library coupled with Bayesian Optimization , fuels our Marketing Mix Platform â âSurgeâ as an ingenious and advanced AI tool for maximizing ROI and simulating Sales. We’re living in the era of large amounts of data, powerful computers, and artificial intelligence. The package NumPy is a fundamental Python scientific package that allows many high-performance operations on single- and multi-dimensional arrays. Are there any Pokemon that get smaller when they evolve? If you want to implement linear regression and need the functionality beyond the scope of scikit-learn, you should consider statsmodels. Finally, on the bottom right plot, you can see the perfect fit: six points and the polynomial line of the degree 5 (or higher) yield ² = 1. The elliptical contours are the cost function of linear regression (eq. It often yields a low ² with known data and bad generalization capabilities when applied with new data. Does your organization need a developer evangelist? The underlying statistical forward model is assumed to be of the following form: Here, is a given design matrix and the vector is a continuous or binary response vector. Linear Regression From Scratch. This column corresponds to the intercept. There is only one extra step: you need to transform the array of inputs to include non-linear terms such as ². You can find more information about LinearRegression on the official documentation page. The specific problem I'm trying to solve is this: I have an unknown X (Nx1), I have M (Nx1) u vectors and M (NxN) s matrices.. max [5th percentile of (ui_T*X), i in 1 to M] st 0<=X<=1 and [95th percentile of (X_T*si*X), i in 1 to M]<= constant Once you have your model fitted, you can get the results to check whether the model works satisfactorily and interpret it. Once your model is created, you can apply .fit() on it: By calling .fit(), you obtain the variable results, which is an instance of the class statsmodels.regression.linear_model.RegressionResultsWrapper. I do want to make a constrained linear regression with the intercept value to be like: If you want to get the predicted response, just use .predict(), but remember that the argument should be the modified input x_ instead of the old x: As you can see, the prediction works almost the same way as in the case of linear regression. Why does the Gemara use gamma to compare shapes and not reish or chaf sofit? It’s possible to transform the input array in several ways (like using insert() from numpy), but the class PolynomialFeatures is very convenient for this purpose. You should keep in mind that the first argument of .fit() is the modified input array x_ and not the original x. Panshin's "savage review" of World of Ptavvs. Check the results of model fitting to know whether the model is satisfactory. data-science This step is also the same as in the case of linear regression. Stack Overflow for Teams is a private, secure spot for you and [1] Standard Errors assume that the covariance matrix of the errors is correctly specified. Consider âlstatâ as independent and âmedvâ as dependent variables Step 1: Load the Boston dataset Step 2: Have a glance at the shape Step 3: Have a glance at the dependent and independent variables Step 4: Visualize the change in the variables Step 5: Divide the data into independent and dependent variables Step 6: Split the data into train and test sets Step 7: Shape of the train and test sets Step 8: Train the algorith⦠The estimated regression function (black line) has the equation () = ₀ + ₁. However, there is also an additional inherent variance of the output. It has many learning algorithms, for regression, classification, clustering and dimensionality reduction. Thus, you can provide fit_intercept=False. Once there is a satisfactory model, you can use it for predictions with either existing or new data. How are you going to put your newfound skills to use? intermediate coefficient of determination: 0.8615939258756777, adjusted coefficient of determination: 0.8062314962259488, regression coefficients: [5.52257928 0.44706965 0.25502548], Simple Linear Regression With scikit-learn, Multiple Linear Regression With scikit-learn, Advanced Linear Regression With statsmodels, Click here to get access to a free NumPy Resources Guide, Look Ma, No For-Loops: Array Programming With NumPy, Pure Python vs NumPy vs TensorFlow Performance Comparison, Split Your Dataset With scikit-learn’s train_test_split(), How to implement linear regression in Python, step by step.
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# Maths in IT #3: Algebra of sets
Hey everyone and welcome back! I’ve already announced this posts subject in my previous post, algebra of sets. So algebra is difficult and scary. At least that’s what people say (hmm hmm), so let’s shake that off. I’m sorry if you got the reference 🙂
If you haven’t read my previous posts I really suggest you do so as this post will continue where the previous one left off.
1. Maths in IT #1: Basic set theory
2. Maths in IT #2: Venn diagrams
3. Maths in IT #3: Algebra of sets
4. Coming soon…
Here’s a quick cheat sheet of symbols introduced in the previous post:
Intersection: $A \cap B = \{x | x \in A \land x \in B\}$
Union: $A \cup B = \{x | x \in A \lor x \in B\}$
Universe U: The collection of all elements we consider for our scenario.
Absolute complement (or simply complement): $A^c = \{x | x \in U \land x \notin A\}$
Relative complement of A in B: $B \cap A^c = A \setminus B$
## Algebra
What exactly is algebra? It’s about the study of symbols, or operations, and laws for manipulating these operations. Elementary algebra, for example, is a set of laws for operations (+, -, /, *) on numbers. Just like that $\cap, \cup$ and $A^c$ are operations on sets and the laws used to manipulating them is called the algebra of sets.
These laws can help us to simplify formulas. So let’s get started.
### Commutative property
Consider the following addition: 2 + 3. I’m pretty sure we all know the answer is 5 and the answer doesn’t change if we switch the numbers to 3 + 2. More abstract we can say that for every two numbers x and y goes that $x + y = y + x$.
When the order of operands (that is the objects on which an operation is performed) in an operation doesn’t matter for the outcome of that operation we call that operation commutative.
Multiplication (on real numbers) is commutative too, because $2 \cdot 3 = 3 \cdot 2$ (you may be used to seeing x or * as multiplication symbol, but you should remember $\cdot$ from high school). Minus isn’t commutative, because $3 - 2 \neq 2 - 3$ and neither is division, $2/3 \neq 3/2$.
When working with sets we can say that $A \cap B = B \cap A$ and $A \cup B = B \cup A$. That means both $\cap$ and $\cup$ are commutative.
### Associative property
When the order in which the same operation is performed in a single formula is unimportant for the outcome of that formula we say that the operation is associative. For example $(1 + 2) + 3 = 1 + (2 + 3)$. It doesn’t matter if we first evaluate 1 + 2 and then add 3 or first evaluate 2 + 3 and add it to 1, the outcome is always 6.
The same holds true for multiplication, because $(2 \cdot 3) \cdot 4 = 2 \cdot (3 \cdot 4)$ but not for subtraction, $(10 - 8) - 2 \neq 10 - (8 - 2)$ and division, $(16/4)/2 \neq 16/(4/2)$.
When looking at sets we find that both $\cap$ and $\cup$ are associative, because $(A \cap B) \cap C = A \cap (B \cap C)$ and $(A \cup B) \cup C = A \cup (B \cup C)$.
That means parenthesis can be omitted, removing any confusion on operator precedence.
And with the commutative and associative properties of both intersection and union we can now simplify
$C \cap (D \cap A) \cap B$
to $A \cap B \cap C \cap D$
or $C \cup (D \cup A) \cup B$
to $A \cup B \cup C \cup D$.
And when we combine intersection and union we can simplify as well. $D \cap (C \cup (B \cup A))$ can be simplified to $(A \cup B \cup C) \cap D$. Notice that we still need parenthesis to define the order of the $\cap$ and $\cup$ operations.
We can now also take the intersection or union of any number of sets because we don’t have to worry about the order of operands or operators. $A_1 \cap A_2 \cap A_3...$ gets tiresome really quick though. So suppose we have $A_1$ to $A_{100}$. The intersection or union of all those sets can now be expressed as follows:
$\displaystyle\bigcup_{i=1}^{100} A_i$ or $\displaystyle\bigcap_{i=1}^{100} A_i$
And inline this look like $\cup_{i=1}^{100} A_i$ and $\cap_{i=1}^{100} A_i$.
Sometimes we don’t know if a set has 100 sets. Especially when a set is an infinite set of sets the above notation is not feasible. We can now use index notation to intersect or union all sets within the set.
$\displaystyle\bigcap_{i \in I} A_i$ or $\displaystyle\bigcup_{i \in I} A_i$
$\cap_{i \in I} A_i$ and $\cup_{i \in I} A_i$ basically means “intersect/union for every set i in set I” (think of a foreach loop on a List of Lists in C#).
## Distributive property
The distributive property is difficult to describe in words. Just remember that multiplication is distributive over addition, meaning that $3 \cdot (4 + 5) = (3 \cdot 4) + (3 \cdot 5)$. More generally $x(y + z) = xy + xz$ (the multiplication symbol is often omitted between letters).
$\cap$ is distributive over $\cup$, meaning that
$A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$
Likewise $\cup$ is distributive over $\cap$, so
$A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$
We can now show that $(A \cap B) \cap (A \cup B) = A \cap B$. When we apply the distributive property on $(A \cap B) \cap (A \cup B)$ we get $((A \cap B) \cap A) \cup ((A \cap B) \cap B)$.
Here it is again, but with colors (and non-mathematical notation) so you can follow where everything went:
(A n B) n (A u B) = ((A n B) n A) u ((A n B) n B).
Applying commutativity and associativity we can now simplify further:
$((A \cap B) \cap A) \cup ((A \cap B) \cap B) = (A \cap A \cap B) \cup (A \cap B \cap B)$.
In the previous article, Maths in IT #2: Venn diagrams, we have seen that $A \cap A = A$ and $A \cup A = A$. This is called idempotence and can be applied here (twice) to simplify further:
$(A \cap A \cap B) \cup (A \cap B \cap B) = (A \cap B) \cup (A \cap B) = A \cap B$
Let’s repeat that:
$(A \cap B) \cap (A \cup B)$
Distributivity
$((A \cap B) \cap A) \cup ((A \cap B) \cap B)$
Associativity
$(A \cap B \cap A) \cup (A \cap B \cap B)$
Commutativity
$(A \cap A \cap B) \cup (A \cap B \cap B)$
Idempotence
$(A \cap B) \cup (A \cap B)$
Idempotence
$A \cap B$
So we could make use of distributivity, associativity, commutativity, and idempotence to make a rather difficult formula pretty easy! Unfortunately it was a lot less easy to do…
## Other properties
Unfortunately we’re not done yet. There are a few more properties, some of which we’ve already encountered, that I’d like to go over.
First we have the absorption law, which states that $A \cap (A \cup B) = A$ and $A \cup (A \cap B) = A$.
We’ve seen the so called null element, or empty set, $\emptyset$. Just like with numbers ($x + 0 = x$, $x - 0 = x$ and $x \cdot 0 = 0$) there are some rules for working with $\emptyset$. Luckily they’re not that hard.
$A \cap \emptyset = \emptyset$ and $A \cup \emptyset = A$.
When a universe U is defined we can say the following:
$A \cap U = A$ and $A \cup U = U$.
I’ve shown you that a double complement returns the original: $(A^c)^c = A$.
Also, $A \cap A^c = \emptyset$ and $A \cup A^c = U$.
## De Morgan’s Law
Finally we’ll look at the law of De Morgan, named after its inventor, the British mathematician Augustus De Morgan.
Suppose my blog can only be read by people who have not studied maths or IT. Let $A = \{x | \text{x has studied maths}\}$ and let $B = \{x | \text{x has studied IT}\}$. So now the people who may read my blog are people who haven’t studied maths or IT: $(A \cup B)^c$. I could also say people who haven’t studied maths and people who haven’t studied IT may read my blog: $A^c \cap B^c$. But that’s the same set of people!
In the following Venn diagram this is represented by the white part.
So in short $(A \cup B)^c = A^c \cap B^c$.
Now suppose my blog may be read only by people who haven’t studied both maths and IT. So that’s $(A \cap B)^c$. Or we could say people who haven’t studied maths and who haven’t studied IT, $A^c \cup B^c$. And again this is the same set of people. In the Venn diagram it’s the part where A and B don’t overlap (so almost everything).
And so that means $(A \cap B)^c = A^c \cup B^c$.
Now let’s take another example of how to simplify a formula. Suppose we have $B \cap (A \cap B)^c$. So let’s first apply De Morgan’s law: $B \cap (A \cap B)^c = B \cap (A^c \cup B^c)$. And now you may recognize a nice pattern for distributivity! So let’s apply distributivity, $B \cap (A^c \cup B^c) = (B \cap A^c) \cup (B \cap B^c)$. That’s nice, because we know that $B \cap B^c = \emptyset$. And now we know that $(B \cap A^c) \cup \emptyset = B \cap A^c$. And maybe you’ll remember this from my previous post, this is $B \setminus A$. And there you have it!
Again, let’s repeat:
$B \cap (A \cap B)^c$
De Morgan’s law
$B \cap (A^c \cup B^c)$
Distributivity
$(B \cap A^c) \cup (B \cap B^c)$
Complement rule
$(B \cap A^c) \cup \emptyset$
Null element
$B \cap A^c$
Relative complement
$B \setminus A$
And there you go! You algebra master, you!
So how will this help you in your day to day programming tasks? For most of us very little I’m afraid, except that you can work out some difficult set operations on paper and possibly simplify requirements. I’m no database developer, but I can imagine you’d need this when developing databases (and to a lesser extent when working with databases).
We did venture into the world of algebra though, and if you’re still with me I say congratulations! Algebra is a bit too abstract and difficult for many people.
I’m leaving you with a little cheat sheet of today’s lessons and I hope to see you again next time (no algebra, I promise)!
Commutativity:
$A \cap B = B \cap A$
$A \cup B = B \cup A$
Associativity:
$(A \cap B) \cap C = A \cap (B \cap C)$
$(A \cup B) \cup C = A \cup (B \cup C)$
Distributivity:
$A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$
$A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$
Idempotence:
$A \cup A = A$
$A \cap A = A$
Absorption:
$A \cup (A \cap B) = A$
$A \cap (A \cup B) = A$
Null element:
$A \cap \emptyset = \emptyset$
$A \cup \emptyset = A$
Universe:
$A \cap U = A$
$A \cup U = U$
Complement:
$A \cap A^c = \emptyset$
$A \cup A^c = U$
Double complement:
$(A^c)^c = A$
Relative complement:
$A \cap B^c = A \setminus B$
De Morgan’s Law:
$(A \cap B)^c = A^c \cup B^c$
$(A \cup B)^c = A^c \cap B^c$
# Maths in IT #2: Venn diagrams
Hey everyone, welcome back to part two of the Maths in IT series. I got a lot of positive response, so I guess I should just keep doing what I was already doing. This post will continue where part one left off, so if you haven’t read it I suggest you do so now before continuing.
1. Maths in IT #1: Basic set theory
2. Maths in IT #2: Venn diagrams
3. Maths in IT #3: Algebra of sets
4. Coming soon…
Here’s a quick cheat sheet with symbols I’ll use in this article:
Explicit definition: $A = \{a, b, c\}$
Implicit definition: $A = \{x | \text{ x is a letter in the alphabet}\}$
a is an element of A: $a \in A$
a is not an element of A: $a \notin A$
A is a subset of B: $A \subset B$
Empty set: $\emptyset$
## Venn diagrams
As promised we’ll use this post to combine sets. Before we do that let’s take a look at how to visually represent collection. We can do this using a Venn diagram. A Venn diagram is about as simple as it gets (although they can be pretty complex too). Each set is represented by a circle. The diagram can illustrate relationships between the represented sets.
Let’s look at an example. We have two collections, $A = \{a, b, c\}$ and $B = \{d, e, f\}$. Let’s show them in a Venn diagram.
When two sets have no shared elements, or for every $x \in A$ goes that $x \notin B$, we say the sets are disjoint.
Now suppose $A \subset B$ (A is a subset of B). We can show this in a Venn diagram and you’ll recognize it immediately.
When A is a subset of B then B overlaps A completely.
In the next sections we’ll combine sets and use Venn diagrams to visualize what elements we’re interested in.
## Intersection
So suppose we have two collections, $A = \{a, b, c, d\}$ and $B = \{c, d, e, f\}$. If I asked you which elements are in both A and B you’d answer c and d.
This is called the intersection of A and B.
We can write this as $A \cap B$.
In this example we can say $A \cap B = \{c, d\}$
More formally we say that $A \cap B = \{x | x \in A \text{ and } x \in B\}$.
And the symbol for “and” is actually $\land$, so to formalize it completely:
$A \cap B = \{x | x \in A \land x \in B\}$
Phew, that looks a whole lot like maths! You should read that as “the intersection of A and B is the collection of every x where x is an element of A and x is an element of B.”
And here is the Venn diagram, which visualizes this nicely. The intersection is the part in the thick black line where A and B overlap.
With an intersection we can give a formal definition of disjoint sets. When $A \cap B = \emptyset$ then A and B are disjoint.
Furthermore we can say that for any collection A goes that $A \cap \emptyset = \emptyset$.
Also $A \cap A = A$.
And when $A \subset B$ then $A \cap B = A$ (check the subset Venn diagram).
## Union
The intersection of two sets is the set of elements x where x is in A and B. Likewise, the union of two sets is the set of elements that are in set A or B. So when we have $A = \{a, b, c, d\}$ and $B = \{c, d, e, f\}$ then the union of A and B is $\{a, b, c, d, e, f\}$ (no doubles).
We can write a union of A and B as $A \cup B$.
Like $\land$ is the symbol for “and” $\lor$ is the symbol for “or”. So the formal definition of union is as follows:
$A \cup B = \{x | x \in A \lor x \in B\}$
Read that as “the union of A and B is the collection of every x where x is an element of A or x is an element of B.”
In a Venn diagram the union is basically just both sets (the part in the thick black line).
Unlike an intersection, a union of disjoint sets is not an empty set (notice that both sets have a thick black line).
Now for any collection A goes that $A \cup \emptyset = A$.
And, again, $A \cup A = A$.
Also when $A \subset B$ then $A \cup B = B$ (check the subset Venn diagram).
## Universe and Complement
When we’re talking about sets we’re usually not talking about that set in isolation. When I tell you that non-smokers live longer you understand that they live longer compared to people that do smoke. And you also understand that non-smokers and smokers combined make up for the worlds population. In this case we’re saying that we have a set of non-smokers in a universe of all people. We denote a universe with the capital letter U.
For every set $A_1, A_2, A_3... A_n$ goes that $A_n \subset U$.
That makes sense as U represents all elements we wish to consider for our case. No collection can ever contain an element that is not a part of all elements.
In a Venn diagram we draw a universe as a rectangle in which all our sets are drawn. In the following example N is the set of non-smokers and U is the universe containing all people.
Now with the notion of a universe we can say we want all elements that are not in any collection A. We call this the complement of A and we write it as $A^c$. In the following Venn diagram the white part represents $A^c$.
We can now formally define the complement:
$A^c = \{ x | x \in U \land x \notin A \}$
It goes without saying, but $\emptyset^c = U$ and $U^c = \emptyset$.
A little less obvious is that $(A^c)^c = A$. It makes sense though, as we first take everything that isn’t A (the complement of A) and then we take everything that isn’t in the resulting set, but that is A. Try drawing it in a Venn diagram and you’ll see what I mean.
A complement relative to a universe is called an absolute complement. A complement can also be relative to other sets. For example, when no universe is defined and we have sets A and B then the relative complement of A in B is the set of elements of B that are not in A. This takes the form of $B \cap A^c$ or $A \setminus B$.
$A \setminus B = \{ x | x \in B \land x \notin A \}$
In a Venn diagram $A \setminus B$ look as follows:
## Combining sets
We can now combine sets using intersection, union and complements. For example, let’s say our universe U is all living creatures on earth. Within U we have sets M, containing all mammals, B, containing all birds, and E, containing all animals that lay eggs. Formally:
$U = \{ x | \text{x is an animal} \}$
$M = \{ x | x \in U \land \text{x is a mammal} \}$
$B = \{ x | x \in U \land \text{x is a bird} \}$
$E = \{ x | x \in U \land \text{x lays eggs} \}$
Giving the following Venn diagram we can already draw some conclusions.
We can see that all birds lay eggs. Some mammals lay eggs too. No animal is both a bird and a mammal. As you see Venn diagrams can be really useful.
Now suppose we want the set of all mammals that lay eggs and also all birds. This is the collection $(M \cap E) \cup B$. That is, we take the intersection of M and E and union the result with B. In a Venn diagram we can see this collection (the red colored parts).
Now we can get every combination of sets using intersection, union and complement. It’s not always easy, but it’s possible.
## Some code
As promised I’m going to keep things practical. So what’s the practical use of all this? Well, most languages let you work with exactly these functions!
For example, take a look at this SQL expression.
(SELECT 1
UNION SELECT 2
UNION SELECT 3)
INTERSECT
(SELECT 3
UNION SELECT 4)
What will that return? It returns only 3 because 3 is an element of both collections. This example also shows the UNION operator. Basically this is the set $(\{1\} \cup \{2\} \cup \{3\}) \cap (\{3\} \cup \{4\})$.
And SQL knows complement too.
(SELECT 1
UNION SELECT 2
UNION SELECT 3)
EXCEPT
(SELECT 3
UNION SELECT 4)
What happens here is that we have (an implied) universe $U = \{1, 2, 3, 4\}$ and EXCEPT is the complement. So this query is basically the formula $\{3, 4\}^c$.
We could also say that U is not defined and this is the relative complement of $\{3, 4\}$ in $\{1, 2, 3\}$: $\{3, 4\} \setminus \{1, 2, 3\} = \{1, 2\}$.
But what about C#? In my previous post we’ve seen HashSet<T>, but I’m not going to use that now. Instead I’m just going with a List<T> as LINQ provides various extension methods for working with sets. Notice that HashSet<T> has its own methods in addition to the LINQ methods.
List<int> a = new List<int>();
a.Add(1);
a.Add(2);
a.Add(3);
List<int> b = new List<int>();
b.Add(3);
b.Add(4);
List<int> intersect = a.Intersect(b).ToList();
List<int> union = a.Union(b).ToList();
List<int> except = a.Except(b).ToList();
You may wonder what that looks like in Haskell (since I’ve shown you Haskell the last time too), but it’s not really that different.
Prelude> import Data.List
Prelude Data.List> [1, 2, 3] intersect [3, 4]
[3]
Prelude Data.List> [1, 2, 3] union [3, 4]
[1,2,3,4]
Prelude Data.List> [1, 2, 3] \\ [3, 4]
[1,2]
I’ve actually needed this stuff in my day to day work. It’s not that hard and sometimes it’s an explicit business requirement. For example I needed all sales orders from The Netherlands, Belgium and Luxembourg (the BeNeLux). That’s really just a union! And I’ve needed intersect too, give me all Dutch customers that are not in some list of customers. And how about all non-Dutch customers? That’s just the complement!
Next time I’d like to continue with algebra and algebra of sets in specific. Does that sound like it will give you nightmares? Don’t worry, I’ll be gentle!
See you next time!
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# Ockham algebra
A bounded distributive lattice together with a dual lattice endomorphism , i.e., a mapping such that the de Morgan laws and hold for all . The class of Ockham algebras is equational (i.e., is a variety; cf. also Algebraic systems, variety of). The Berman class is the subclass obtained by imposing on the dual endomorphism the restriction (, ). The Berman classes are related as follows:
The smallest Berman class is therefore the class described by the equation and is the class of de Morgan algebras. Perhaps the most important Berman class is , described by . This can be characterized as the class of Ockham algebras such that . It contains also the class of -algebras , and, in particular, the class of of Stone algebras (add the relation ).
An Ockham algebra congruence is an equivalence relation that has the substitution property for both the lattice operations and the unary operation . A basic congruence is , defined by
If , then, for , , where indicates an isomorphism when is even and a dual isomorphism when is odd.
An Ockham algebra is subdirectly irreducible if it has a smallest non-trivial congruence. Every Berman class contains only finitely many subdirectly irreducible algebras, each of which is finite.
The class of is given by
it is a locally finite generalized variety that contains all of the Berman classes. If , then is subdirectly irreducible if and only if the lattice of congruences of reduces to the chain
where . If , then .
Ockham algebras can also be obtained by topological duality. Recall that a set in a partially ordered set is called a down-set if , , implies . Dually, is called an up-set if , , implies . An ordered topological space (cf. also Order (on a set)) is said to be totally order-disconnected if, whenever , there exists a closed-and-open down-set such that and . A Priestley space is a compact totally order-disconnected space. An Ockham space is a Priestley space endowed with a continuous order-reversing mapping . The important connection with Ockham algebras was established by A. Urquhart and is as follows. If is an Ockham space and if denotes the family of closed-and-open down-sets of , then is an Ockham algebra, where is given by . Conversely, if is an Ockham algebra and if denotes the set of prime ideals of , then, if is equipped with the topology which has as base the sets and for every , is an Ockham space, where . Moreover, these constructions give a dual categorical equivalence. In the finite case the topology "evaporates" ; the dual space of a finite Ockham algebra consists of the ordered set of join-irreducible elements together with the order-reversing mapping .
Duality produces further classes of Ockham algebras. For , let be the subclass of formed by the algebras whose dual space satisfies . Then every Berman class is a ; more precisely, . If is the dual space of , let, for every , . If is finite, then is subdirectly irreducible if and only if there exists an such that . The dual space of a subdirectly irreducible Ockham algebra in can therefore be represented as follows (here the order is ignored and the arrows indicate the action of ):
Figure: o110030a
The subdirectly irreducible Ockham algebra that corresponds to this discretely ordered space is denoted by . In particular, is the algebra whose dual space is
Figure: o110030b
and is described as follows:
Figure: o110030c
The subdirectly irreducible algebras in are the nineteen subalgebras of . Using a standard theorem of B.A. Davey from universal algebra, it is possible to describe completely the lattice of subvarieties of .
#### References
[a1] T.S. Blyth, J.C. Varlet, "Ockham algebras" , Oxford Univ. Press (1994) [a2] J. Berman, "Distributive lattices with an additional unary operation" Aequationes Math. , 16 (1977) pp. 165–171 [a3] H.A. Priestley, "Ordered sets and duality for distributive lattices" Ann. Discrete Math. , 23 (1984) pp. 39–60 [a4] A. Urquhart, "Lattices with a dual homomorphic operation" Studia Logica , 38 (1979) pp. 201–209 [a5] B.A. Davey, "On the lattice of subvarieties" Houston J. Math. , 5 (1979) pp. 183–192
How to Cite This Entry:
Ockham algebra. Encyclopedia of Mathematics. URL: http://www.encyclopediaofmath.org/index.php?title=Ockham_algebra&oldid=39748
This article was adapted from an original article by T.S. Blyth (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article
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2 added 19 characters in body
Which finite subsets $S \subset \mathbb{N}$ have the following property : every countable group $G$ embeds into a finitely generated group $\Gamma$ such that $H_i(\Gamma;\mathbb{Z})=0$ for all $i \in S$.
The only positive answer I know here is that $S=\{1\}$ works since every countable group can be embedded into a simple group. I don't know any negative answers.
I'm especially interested in singleton sets $S$ (in particular, $S=\{2\}$ and $S=\{3\}$).
Also, is the question easier if I restrict myself to finitely generated or finitely presentable groups?
1
# Embedding groups into groups with some vanishing homology groups
Which finite subsets $S \subset \mathbb{N}$ have the following property : every countable group $G$ embeds into a group $\Gamma$ such that $H_i(\Gamma;\mathbb{Z})=0$ for all $i \in S$.
The only positive answer I know here is that $S=\{1\}$ works since every countable group can be embedded into a simple group. I don't know any negative answers.
I'm especially interested in singleton sets $S$ (in particular, $S=\{2\}$ and $S=\{3\}$).
Also, is the question easier if I restrict myself to finitely generated or finitely presentable groups?
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# Heinz, a manufacturer of ketchup, uses a particular machines to dispense 16 ounces of its ketchup... 1 answer below »
Heinz, a manufacturer of ketchup, uses a particular machines to dispense 16 ounces of its ketchup into containers. From many years experience with the particular dispensing machine, Heinz knows the amount of product in each container follows a normal contribution with a means of 16 ounces and a standard derivation of 0.15 ounce. A sample f 50 containers filled last hour revealed the mean amount per container was 16.017 ounces. Does this suggest that mean amount dispensed is different form 16 ounces? Use 0.05 significant level.
Swati V
Solution:
Null Hypothesis (Ho): mu is equal to 16
Alternative Hypothesis (Ha): mu is not equal to...
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## Explore printable Properties of Multiplication worksheets for 5th Class
Properties of Multiplication worksheets for Class 5 are essential tools for teachers who aim to strengthen their students' understanding of multiplication concepts. These worksheets are specifically designed to cater to the learning needs of Class 5 students, focusing on various aspects of multiplication, such as the commutative, associative, and distributive properties. By incorporating these worksheets into their lesson plans, teachers can effectively engage their students in practicing and mastering multiplication skills. Moreover, these worksheets can be utilized as supplementary materials for homework assignments, in-class activities, or even as assessment tools to evaluate students' progress in Math. With a wide range of exercises and problems, Properties of Multiplication worksheets for Class 5 provide teachers with a valuable resource to enhance their students' mathematical abilities.
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Community Profile
# Christine Tobler
### MathWorks
216 2015 이후 총 참여 횟수
Professional Interests: numerical linear algebra, graph algorithms
#### Christine Tobler's 배지
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Sort eigenvectors matrix.
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Eigenvectors of an SPD matrix being saved as complex doubles
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Given a matrix A^n. Comparing normal multiplication versus Diagonalization. I expect the former to be faster but its not in my case
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EigenValues of a Vibrating system
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issues with Cholesky decomposition
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how to do forward neighbor discovery in un-directed graph?
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Using eigs with singular matrix
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Change the alignment and font size of edgelabels
The edge labels provided with the plot of a graph can't be modified in terms of their alignment. However, you can add standard t...
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a question for defining Custom Deep Learning Layer
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Control edge alpha via edge weights to visualize a dynamic network
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how to create a symmetric Toeplitz matrix with bounds on eigenvalues?
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Interpolation for n-dimensional array data
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eigs gives wrong eigenvalues
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Which Right Eigenvector to report?
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How to add svd(singular value decomposition) in a custom layer
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Open just some Graph Edges in Variable Editor and let the user modify them?
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how get graph?
Since MATLAB R2015b, there are Graph and Network Algorithms in MATLAB that will allow you to construct and plot a graph of these...
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Stretching and Reflecting Transformations
## Transformations of parent functions produced by multiplying by a constant.
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Practice Stretching and Reflecting Transformations
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Stretching and Reflecting Transformations
Understanding how changes in the equation of a function result in stretching and/or reflecting the graph of the function is a great way to take some of the mystery out of graphing more complicated equations. By recognizing the family to which a more complex equation belongs, and then identifying what changes have been made to the parent of that family, the graph of even quite detailed functions can be made much more understandable.
See if you can identify what parts of the equation, y=15x2\begin{align*}y = \frac {-1}{5} x^{2}\end{align*} , represent either a stretch or a reflection of the parent function y=x2\begin{align*}y = x^{2}\end{align*} before the review at the end of this lesson.
Embedded Video:
### Guidance
Stretching and compressing graphs
If we multiply a function by a coefficient, the graph of the function will be stretched or compressed.
Given a function f(x), we can formalize compressing and stretching the graph of f(x) as follows:
• A function g(x) represents a vertical stretch of f(x) if g(x) = cf(x) and c > 1.
• A function g(x) represents a vertical compression of f(x) if g(x) = cf(x) and 0 < c < 1.
• A function h(x) represents a horizontal compression of f(x) if h(x) = f(cx) and c > 1.
• A function h(x) represents a horizontal stretch of f(x) if h(x) = f(cx) and 0 < c < 1.
Notice that a vertical compression or a horizontal stretch occurs when the coefficient is a number between 0 and 1.
Reflecting graphs over the y-axis and x-axis
Consider the graphs of the functions y = x2 and y = -x2, shown below.
The graph of y = -x2 represents a reflection of y = x2, over the x-axis. That is, every function value of y = -x2 is the negative of a function value of y = x2. In general, g(x) = -f(x) has a graph that is the graph of f(x), reflected over the x-axis.
#### Example 1
Identify the graph of the function y = (3x)2.
Solution
We have multiplied x by 3. This should affect the graph horizontally. However, if we simplify the equation, we get y = 9x2. Therefore the graph of this parabola will be taller/thinner than y = x2. Multiplying x by a number greater than 1 creates a vertical stretch.
#### Example 2
Identify the transformation described by y = (12x)2\begin{align*}\left (\frac {1}{2}x \right )^2\end{align*} .
Solution
If we simplify this equation, we get y = \begin{align*}\frac {1}{4}\end{align*} x2. Therefore multiplying x by a number between 0 and 1 creates a vertical compression. That is, the parabola will be shorter/wider.
#### Example 3
Sketch a graph of y = x3 and y = -x3 on the same axes.
Solution:
At first the two functions might look like two parabolas. If you graph by hand, or if you set your calculator to sequential mode (and not simultaneous), you can see that the graph of y = -x3 is in fact a reflection of y = x3 over the x-axis.
However, if you look at the graph, you can see that it is a reflection over the y-axis as well. This is the case because in order to obtain a reflection over the y-axis, we negate x. In other words, h(x) = f(-x) is a reflection of f(x) over the y-axis. For the function y = x3, h(x) = (-x)3 = (-x) (-x) (-x) = -x3. This is the same function as the one we have already graphed.
It is important to note that y = xis a special case. The graph of y = x2 is also a special case. If we want to reflect y = x2 over the y-axis, we will just get the same graph! This can be explained algebraically: y = (-x)2 = (-x) (-x) = x2.
Concept question wrap-up:
Are you able to identify the transformations described in the beginning of the lesson now?
The function: \begin{align*}y = \frac {-1}{5}x^{2}\end{align*} is the result of transforming \begin{align*}y = x^{2}\end{align*} by:
reflecting it over the x axis, because of the negative co-efficient on the x. and:
vertically compressing it (making it wider), because of the co-efficient being a fraction between 0 and 1.
### Vocabulary
Reflections are transformations which result in a "mirror image" of a parent function. They are a result of differing signs between parent and child functions.
Stretches are transformations which result in the width of a graph being increased or decreased. They are the result of the co-efficient of the x term being between 0 and 1.
Example 4
Graph the functions
\begin{align*}y = \sqrt{x}\end{align*}
and
\begin{align*}y = \sqrt{-x}\end{align*}
Solution
The equation \begin{align*}y = \sqrt{-x}\end{align*} might look confusing because of the -x under the square root. It is important to keep in mind that -x means the opposite of x. Therefore the domain of this function is restricted to values of x ≤ 0. For example, if x = - 4, \begin{align*}y = \sqrt{-(-4)}= \sqrt{4} = 2\end{align*}. It is this domain, which includes all real numbers not in the domain of \begin{align*}y = \sqrt{x}\end{align*} except zero, that gives us a graph that is a reflection over the y-axis.
In sum, a graph represents a reflection over the x-axis if the function has been negated (i.e. the y has been negated if we think of y = f(x)). The graph represents a reflection over the y-axis if the variable x has been negated.
Example 5
Sketch the graph of \begin{align*}y = 3x^2\end{align*} by appropriately stretching the parent graph \begin{align*}y = x^2\end{align*}.
Solution
The graph of \begin{align*}y = 3x^2\end{align*} is the graph of the parent, \begin{align*}y = x^2\end{align*}, with each y-coordinate multiplied by 3. The image below shows both the parent and the child function on the same axes. The parent function has been stretched vertically.
Example 6
Sketch the graph of \begin{align*}y = -3x^2\end{align*} by reflecting the graph of \begin{align*}y = 3x^2\end{align*}.
Solution
The graph of \begin{align*}y = -3x^2\end{align*} is the graph of \begin{align*}y = 3x^2\end{align*}reflected over the x-axis, the image below shows both functions.
Example 7
Sketch the graph of \begin{align*}y = \sqrt{3x}\end{align*} by appropriately stretching \begin{align*}y = \sqrt{x}\end{align*}.
Solution
The graph of \begin{align*}y = \sqrt{3x}\end{align*} is the graph of \begin{align*}y = \sqrt{x}\end{align*}with each coordinate multiplied by 3, the image below shows both graphs.
Example 8
Sketch the graph of \begin{align*}y = \sqrt{x}\end{align*} reflected over both axes and identify the functions.
Solution
To reflect the graph of \begin{align*}y = \sqrt{x}\end{align*} over both axes, the function must be negated both outside and inside the root: \begin{align*}y = -\sqrt{-x}\end{align*}. The negation (negative) outside of the root has the effect of reflecting the graph vertically (over the x-axis), and the negation inside of the root reflects the graph horizontally (over the y-axis). The image below shows three versions:
a) (BLUE) \begin{align*}y = \sqrt{x}\end{align*}
b) (GREEN) \begin{align*}y = -\sqrt{x}\end{align*}
c) (RED) \begin{align*}y = \sqrt{-x}\end{align*}
### Review Questions
1. If a function is multiplied by a coefficient, what will happen to the graph of the function?
2. What does multiplying x by a number greater than one create?
3. What happens when we multiply x by a number between 0 and 1
4. In order to obtain a reflection over the y axis what do we have to do to x?
5. How do we obtain a reflection over the x-axis?
6. Write a function that will create a horizontal compression of the following:\begin{align*}f(x) = x^2 + 3\end{align*}
7. Write a function that will horizontally stretch the following: \begin{align*}f(x) = x^2 - 6\end{align*}
8. Rewrite this function\begin{align*}f(x) = -\sqrt{x}\end{align*}to get a reflection over the x-axis.
9. Rewrite this function\begin{align*}f(x) = \sqrt{x}\end{align*}to get a reflection over the y-axis.
Graph each of the following using transformations. Identify the translations and reflections.
1. \begin{align*}f(x) = |x| -2\end{align*}
2. \begin{align*}h(x) = \sqrt{x + 3}\end{align*}
3. \begin{align*}g(x) = \frac{1}{x + 1}\end{align*}
4. \begin{align*}f(x) = -4x^3\end{align*}
5. \begin{align*}h(x) = (x + 3)^3 + 1\end{align*}
6. \begin{align*}f(x) = \frac{1}{3}(x - 3)^2 + 1\end{align*}
7. \begin{align*}f(x) = -4\sqrt{x + 1} - 2\end{align*}
8. \begin{align*}f(x) = \frac{2}{3(x - 2)} + \frac{1}{4}\end{align*}
Let \begin{align*}y = f(x)\end{align*} be the function defined by the line segment connecting the points (-1, 4) and (2, 5). Graph each of the following transformations of \begin{align*} y = f(x)\end{align*}.
1. \begin{align*}y = f(x) + 1\end{align*}
2. \begin{align*}y = f(x+ 2)\end{align*}
3. \begin{align*}y = f(-x)\end{align*}
4. \begin{align*}y = f(x + 3) - 2\end{align*}
The graph of y = x is shown below. Sketch the graph of each of the following transformations of y = x
1. \begin{align*}y = x + 3\end{align*}
2. \begin{align*}y = x - 2\end{align*}
3. \begin{align*}y = - x\end{align*}
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# 3.2: The Improved Euler Method and Related Methods
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In Section 3.1, we saw that the global truncation error of Euler’s method is $$O(h)$$, which would seem to imply that we can achieve arbitrarily accurate results with Euler’s method by simply choosing the step size sufficiently small. However, this is not a good idea, for two reasons. First, after a certain point decreasing the step size will increase roundoff errors to the point where the accuracy will deteriorate rather than improve. The second and more important reason is that in most applications of numerical methods to an initial value problem
$\label{eq:3.2.1} y'=f(x,y),\quad y(x_0)=y_0,$
the expensive part of the computation is the evaluation of $$f$$. Therefore we want methods that give good results for a given number of such evaluations. This is what motivates us to look for numerical methods better than Euler’s.
To clarify this point, suppose we want to approximate the value of $$e$$ by applying Euler’s method to the initial value problem
$y'=y,\quad y(0)=1 \nonumber$
(with solution $$y=e^x$$) on $$[0,1]$$, with $$h=1/12$$, $$1/24$$, and $$1/48$$, respectively. Since each step in Euler’s method requires one evaluation of $$f$$, the number of evaluations of $$f$$ in each of these attempts is $$n=12$$, $$24$$, and $$48$$, respectively. In each case we accept $$y_n$$ as an approximation to $$e$$. The second column of Table $$\PageIndex{1}$$ shows the results. The first column of the table indicates the number of evaluations of $$f$$ required to obtain the approximation, and the last column contains the value of $$e$$ rounded to ten significant figures.
In this section we will study the improved Euler method, which requires two evaluations of $$f$$ at each step. We’ve used this method with $$h=1/6$$, $$1/12$$, and $$1/24$$. The required number of evaluations of $$f$$ were 12, 24, and $$48$$, as in the three applications of Euler’s method; however, you can see from the third column of Table $$\PageIndex{1}$$ that the approximation to $$e$$ obtained by the improved Euler method with only 12 evaluations of $$f$$ is better than the approximation obtained by Euler’s method with 48 evaluations.
In Section 3.3, we will study the Runge- Kutta method, which requires four evaluations of $$f$$ at each step. We’ve used this method with $$h=1/3$$, $$1/6$$, and $$1/12$$. The required number of evaluations of $$f$$ were again 12, 24, and $$48$$, as in the three applications of Euler’s method and the improved Euler method; however, you can see from the fourth column of Table $$\PageIndex{1}$$ that the approximation to $$e$$ obtained by the Runge-Kutta method with only 12 evaluations of $$f$$ is better than the approximation obtained by the improved Euler method with 48 evaluations.
Table $$\PageIndex{1}$$: Approximations to $$e$$ obtained by three numerical methods.
$$n$$ Euler Improved Euler Runge-Kutta Exact
12 2.613035290 2.707188994 2.718069764 2.718281828
24 2.663731258 2.715327371 2.718266612 2.718281828
48 2.690496599 2.717519565 2.718280809 2.718281828
## The Improved Euler Method
The improved Euler method for solving the initial value problem Equation \ref{eq:3.2.1} is based on approximating the integral curve of Equation \ref{eq:3.2.1} at $$(x_i,y(x_i))$$ by the line through $$(x_i,y(x_i))$$ with slope
$m_i={f(x_i,y(x_i))+f(x_{i+1},y(x_{i+1}))\over2};$
that is, $$m_i$$ is the average of the slopes of the tangents to the integral curve at the endpoints of $$[x_i,x_{i+1}]$$. The equation of the approximating line is therefore
$\label{eq:3.2.2} y=y(x_i)+{f(x_i,y(x_i))+f(x_{i+1},y(x_{i+1}))\over2}(x-x_i).$
Setting $$x=x_{i+1}=x_i+h$$ in Equation \ref{eq:3.2.2} yields
$\label{eq:3.2.3} y_{i+1}=y(x_i)+{h\over2}\left(f(x_i,y(x_i))+f(x_{i+1},y(x_{i+1}))\right)$
as an approximation to $$y(x_{i+1})$$. As in our derivation of Euler’s method, we replace $$y(x_i)$$ (unknown if $$i>0$$) by its approximate value $$y_i$$; then Equation \ref{eq:3.2.3} becomes
$y_{i+1}=y_i+{h\over2}\left(f(x_i,y_i)+f(x_{i+1},y(x_{i+1})\right).$
However, this still will not work, because we do not know $$y(x_{i+1})$$, which appears on the right. We overcome this by replacing $$y(x_{i+1})$$ by $$y_i+hf(x_i,y_i)$$, the value that the Euler method would assign to $$y_{i+1}$$. Thus, the improved Euler method starts with the known value $$y(x_0)=y_0$$ and computes $$y_1$$, $$y_2$$, …, $$y_n$$ successively with the formula
$\label{eq:3.2.4} y_{i+1}=y_i+{h\over2}\left(f(x_i,y_i)+f(x_{i+1},y_i+hf(x_i,y_i))\right).$
The computation indicated here can be conveniently organized as follows: given $$y_i$$, compute
\begin{aligned} k_{1i}&=&f(x_i,y_i),\\ k_{2i}&=&f\left(x_i+h,y_i+hk_{1i}\right),\\ y_{i+1}&=&y_i+{h\over2}(k_{1i}+k_{2i}).\end{aligned}
The improved Euler method requires two evaluations of $$f(x,y)$$ per step, while Euler’s method requires only one. However, we will see at the end of this section that if $$f$$ satisfies appropriate assumptions, the local truncation error with the improved Euler method is $$O(h^3)$$, rather than $$O(h^2)$$ as with Euler’s method. Therefore the global truncation error with the improved Euler method is $$O(h^2)$$; however, we will not prove this.
We note that the magnitude of the local truncation error in the improved Euler method and other methods discussed in this section is determined by the third derivative $$y'''$$ of the solution of the initial value problem. Therefore the local truncation error will be larger where $$|y'''|$$ is large, or smaller where $$|y'''|$$ is small.
The next example, which deals with the initial value problem considered in Example $$\PageIndex{1}$$, illustrates the computational procedure indicated in the improved Euler method.
Example $$\PageIndex{1}$$
Use the improved Euler method with $$h=0.1$$ to find approximate values of the solution of the initial value problem
$\label{eq:3.2.5} y'+2y=x^3e^{-2x},\quad y(0)=1$
at $$x=0.1,0.2,0.3$$.
Solution
As in Example [example:3.1.1}, we rewrite Equation \ref{eq:3.2.5} as
$y'=-2y+x^3e^{-2x},\quad y(0)=1,$
which is of the form Equation \ref{eq:3.2.1}, with
$f(x,y)=-2y+x^3e^{-2x},\; x_0=0,\mbox{\; and}\; y_0=1.$
The improved Euler method yields
\begin{aligned} k_{10} & = & f(x_0,y_0) = f(0,1)=-2,\\ k_{20} & = & f(x_1,y_0+hk_{10})=f(0.1,1+(0.1)(-2))\\ &=& f(0.1,0.8)=-2(0.8)+(0.1)^3e^{-0.2}=-1.599181269,\\ y_1&=&y_0+{h\over2}(k_{10}+k_{20}),\\ &=&1+(0.05)(-2-1.599181269)=0.820040937,\\[4pt] k_{11} & = & f(x_1,y_1) = f(0.1,0.820040937)= -2(0.820040937)+(0.1)^3e^{-0.2}=-1.639263142,\\ k_{21} & = & f(x_2,y_1+hk_{11})=f(0.2,0.820040937+0.1(-1.639263142)),\\ &=& f(0.2,0.656114622)=-2(0.656114622)+(.2)^3e^{-0.4}=-1.306866684,\\ y_2&=&y_1+{h\over2}(k_{11}+k_{21}),\\ &=&.820040937+(.05)(-1.639263142-1.306866684)=0.672734445,\\[4pt] k_{12} & = & f(x_2,y_2) = f(.2,.672734445)= -2(.672734445)+(.2)^3e^{-.4}=-1.340106330,\\ k_{22} & = & f(x_3,y_2+hk_{12})=f(.3,.672734445+.1(-1.340106330)),\\ &=& f(.3,.538723812)=-2(.538723812)+(.3)^3e^{-.6}=-1.062629710,\\ y_3&=&y_2+{h\over2}(k_{12}+k_{22})\\ &=&.672734445+(.05)(-1.340106330-1.062629710)=0.552597643.\end{aligned}
Example $$\PageIndex{2}$$
Table $$\PageIndex{2}$$ shows results of using the improved Euler method with step sizes $$h=0.1$$ and $$h=0.05$$ to find approximate values of the solution of the initial value problem
$y'+2y=x^3e^{-2x},\quad y(0)=1$
at $$x=0$$, $$0.1$$, $$0.2$$, $$0.3$$, …, $$1.0$$. For comparison, it also shows the corresponding approximate values obtained with Euler’s method in [example:3.1.2}, and the values of the exact solution
$y={e^{-2x}\over4}(x^4+4). \nonumber$
The results obtained by the improved Euler method with $$h=0.1$$ are better than those obtained by Euler’s method with $$h=0.05$$.
Table $$\PageIndex{2}$$: Numerical solution of $$y'+2y=x^3e^{-2x},\ y(0)=1$$, by Euler’s method and the improved Euler method.
Euler Improved Euler Exact
$$x$$ $$h=0.1$$ $$h=0.05$$ $$h=0.1$$ $$h=0.05$$ Exact
0.0 1.000000000 1.000000000 1.000000000 1.000000000 1.000000000
0.1 0.800000000 0.810005655 0.820040937 0.819050572 0.818751221
0.2 0.640081873 0.656266437 0.672734445 0.671086455 0.670588174
0.3 0.512601754 0.532290981 0.552597643 0.550543878 0.549922980
0.4 0.411563195 0.432887056 0.455160637 0.452890616 0.452204669
0.5 0.332126261 0.353785015 0.376681251 0.374335747 0.373627557
0.6 0.270299502 0.291404256 0.313970920 0.311652239 0.310952904
0.7 0.222745397 0.242707257 0.264287611 0.262067624 0.261398947
0.8 0.186654593 0.205105754 0.225267702 0.223194281 0.222570721
0.9 0.159660776 0.176396883 0.194879501 0.192981757 0.192412038
1.0 0.139778910 0.154715925 0.171388070 0.169680673 0.169169104
Example $$\PageIndex{3}$$
Table $$\PageIndex{3}$$ shows analogous results for the nonlinear initial value problem
$y'=-2y^2+xy+x^2,\ y(0)=1. \nonumber$
We applied Euler’s method to this problem in Example $$\PageIndex{3}$$.
Table $$\PageIndex{3}$$: Numerical solution of $$y'=-2y^2+xy+x^2,\ y(0)=1$$, by Euler’s method and the improved Euler method.
Euler Improved Euler Exact
$$x$$ $$h=0.1$$ $$h=0.05$$ $$h=0.1$$ $$h=0.05$$ Exact
0.0 1.000000000 1.000000000 1.000000000 1.000000000 1.000000000
0.1 0.800000000 0.821375000 0.840500000 0.838288371 0.837584494
0.2 0.681000000 0.707795377 0.733430846 0.730556677 0.729641890
0.3 0.605867800 0.633776590 0.661600806 0.658552190 0.657580377
0.4 0.559628676 0.587454526 0.615961841 0.612884493 0.611901791
0.5 0.535376972 0.562906169 0.591634742 0.588558952 0.587575491
0.6 0.529820120 0.557143535 0.586006935 0.582927224 0.581942225
0.7 0.541467455 0.568716935 0.597712120 0.594618012 0.593629526
0.8 0.569732776 0.596951988 0.626008824 0.622898279 0.621907458
0.9 0.614392311 0.641457729 0.670351225 0.667237617 0.666250842
1.0 0.675192037 0.701764495 0.730069610 0.726985837 0.726015790
Example $$\PageIndex{3}$$
Use step sizes $$h=0.2$$, $$h=0.1$$, and $$h=0.05$$ to find approximate values of the solution of
$\label{eq:3.2.6} y'-2xy=1,\quad y(0)=3$
at $$x=0$$, $$0.2$$, $$0.4$$, $$0.6$$, …, $$2.0$$ by (a) the improved Euler method; (b) the improved Euler semilinear method. (We used Euler’s method and the Euler semilinear method on this problem in [example:3.1.4}.)
Solution
Rewriting Equation \ref{eq:3.2.6} as
$y'=1+2xy,\quad y(0)=3 \nonumber$
and applying the improved Euler method with $$f(x,y)=1+2xy$$ yields the results shown in Table $$\PageIndex{4}$$.
Table $$\PageIndex{4}$$: Numerical solution of $$y'-2xy=1,\ y(0)=3$$, by the improved Euler method.
$$x$$ $$h=0.2$$ $$h=0.1$$ $$h=0.05$$ Exact
0.0 3.000000000 3.000000000 3.000000000 3.000000000
0.2 3.328000000 3.328182400 3.327973600 3.327851973
0.4 3.964659200 3.966340117 3.966216690 3.966059348
0.6 5.057712497 5.065700515 5.066848381 5.067039535
0.8 6.900088156 6.928648973 6.934862367 6.936700945
1.0 10.065725534 10.154872547 10.177430736 10.184923955
1.2 15.708954420 15.970033261 16.041904862 16.067111677
1.4 26.244894192 26.991620960 27.210001715 27.289392347
1.6 46.958915746 49.096125524 49.754131060 50.000377775
1.8 89.982312641 96.200506218 98.210577385 98.982969504
2.0 184.563776288 203.151922739 209.464744495 211.954462214
Since $$y_1=e^{x^2}$$ is a solution of the complementary equation $$y'-2xy=0$$, we can apply the improved Euler semilinear method to Equation \ref{eq:3.2.6}, with
$y=ue^{x^2}\quad \text{and} \quad u'=e^{-x^2},\quad u(0)=3. \nonumber$
The results listed in Table $$\PageIndex{5}$$ are clearly better than those obtained by the improved Euler method.
Table $$\PageIndex{5}$$: Numerical solution of $$y'-2xy=1,\ y(0)=3$$, by the improved Euler semilinear method.
$$x$$ $$h=0.2$$ $$h=0.1$$ $$h=0.05$$ Exact
0.0 3.000000000 3.000000000 3.000000000 3.000000000
0.2 3.326513400 3.327518315 3.327768620 3.327851973
0.4 3.963383070 3.965392084 3.965892644 3.966059348
0.6 5.063027290 5.066038774 5.066789487 5.067039535
0.8 6.931355329 6.935366847 6.936367564 6.936700945
1.0 10.178248417 10.183256733 10.184507253 10.184923955
1.2 16.059110511 16.065111599 16.066611672 16.067111677
1.4 27.280070674 27.287059732 27.288809058 27.289392347
1.6 49.989741531 49.997712997 49.999711226 50.000377775
1.8 98.971025420 98.979972988 98.982219722 98.982969504
2.0 211.941217796 211.951134436 211.953629228 211.954462214
## A Family of Methods with O(h³) Local Truncation Error
We will now derive a class of methods with $$O(h^3)$$ local truncation error for solving Equation \ref{eq:3.2.1}. For simplicity, we assume that $$f$$, $$f_x$$, $$f_y$$, $$f_{xx}$$, $$f_{yy}$$, and $$f_{xy}$$ are continuous and bounded for all $$(x,y)$$. This implies that if $$y$$ is the solution of Equation \ref{eq:3.2.1} then $$y''$$ and $$y'''$$ are bounded (Exercise [exer:3.2.31}.
We begin by approximating the integral curve of Equation \ref{eq:3.2.1} at $$(x_i,y(x_i))$$ by the line through $$(x_i,y(x_i))$$ with slope
$m_i=\sigma y'(x_i)+\rho y'(x_i+\theta h),$
where $$\sigma$$, $$\rho$$, and $$\theta$$ are constants that we will soon specify; however, we insist at the outset that $$0<\theta\le 1$$, so that
$x_i<x_i+\theta h\le x_{i+1}.$
The equation of the approximating line is
$\label{eq:3.2.7} \begin{array}{rcl} y&=&y(x_i)+m_i(x-x_i)\\ &=&y(x_i)+\left[\sigma y'(x_i)+\rho y'(x_i+\theta h)\right](x-x_i). \end{array}$
Setting $$x=x_{i+1}=x_i+h$$ in Equation \ref{eq:3.2.7} yields
$\hat y_{i+1}=y(x_i)+h\left[\sigma y'(x_i)+\rho y'(x_i+\theta h)\right]$
as an approximation to $$y(x_{i+1})$$.
To determine $$\sigma$$, $$\rho$$, and $$\theta$$ so that the error
$\label{eq:3.2.8} \begin{array}{rcl} E_i&=&y(x_{i+1})-\hat y_{i+1}\\ &=&y(x_{i+1})-y(x_i)-h\left[\sigma y'(x_i)+\rho y'(x_i+\theta h)\right] \end{array}$
in this approximation is $$O(h^3)$$, we begin by recalling from Taylor’s theorem that
$y(x_{i+1})=y(x_i)+hy'(x_i)+{h^2\over2}y''(x_i)+{h^3\over6}y'''(\hat x_i),$
where $$\hat x_i$$ is in $$(x_i,x_{i+1})$$. Since $$y'''$$ is bounded this implies that
$y(x_{i+1})-y(x_i)-hy'(x_i)-{h^2\over2}y''(x_i)=O(h^3).$
Comparing this with Equation \ref{eq:3.2.8} shows that $$E_i=O(h^3)$$ if
$\label{eq:3.2.9} \sigma y'(x_i)+\rho y'(x_i+\theta h)=y'(x_i)+{h\over2}y''(x_i) +O(h^2).$
However, applying Taylor’s theorem to $$y'$$ shows that
$y'(x_i+\theta h)=y'(x_i)+\theta h y''(x_i)+{(\theta h)^2\over2}y'''(\overline x_i),$
where $$\overline x_i$$ is in $$(x_i,x_i+\theta h)$$. Since $$y'''$$ is bounded, this implies that
$y'(x_i+\theta h)=y'(x_i)+\theta h y''(x_i)+O(h^2).$
Substituting this into Equation \ref{eq:3.2.9} and noting that the sum of two $$O(h^2)$$ terms is again $$O(h^2)$$ shows that $$E_i=O(h^3)$$ if
$(\sigma+\rho)y'(x_i)+\rho\theta h y''(x_i)= y'(x_i)+{h\over2}y''(x_i),$
which is true if
$\label{eq:3.2.10} \sigma+\rho=1 \quad \text{and} \quad \rho\theta={1\over2}.$
Since $$y'=f(x,y)$$, we can now conclude from Equation \ref{eq:3.2.8} that
$\label{eq:3.2.11} y(x_{i+1})=y(x_i)+h\left[\sigma f(x_i,y_i)+\rho f(x_i+\theta h,y(x_i+\theta h))\right]+O(h^3)$
if $$\sigma$$, $$\rho$$, and $$\theta$$ satisfy Equation \ref{eq:3.2.10}. However, this formula would not be useful even if we knew $$y(x_i)$$ exactly (as we would for $$i=0$$), since we still wouldn’t know $$y(x_i+\theta h)$$ exactly. To overcome this difficulty, we again use Taylor’s theorem to write
$y(x_i+\theta h)=y(x_i)+\theta h y'(x_i)+{h^2\over2}y''(\tilde x_i),$
where $$\tilde x_i$$ is in $$(x_i,x_i+\theta h)$$. Since $$y'(x_i)=f(x_i,y(x_i))$$ and $$y''$$ is bounded, this implies that
$\label{eq:3.2.12} |y(x_i+\theta h)-y(x_i)-\theta h f(x_i,y(x_i))|\le Kh^2$
for some constant $$K$$. Since $$f_y$$ is bounded, the mean value theorem implies that
$|f(x_i+\theta h,u)-f(x_i+\theta h,v)|\le M|u-v|$
for some constant $$M$$. Letting
$u=y(x_i+\theta h)\quad \text{and} \quad v=y(x_i)+\theta h f(x_i,y(x_i))$
and recalling Equation \ref{eq:3.2.12} shows that
$f(x_i+\theta h,y(x_i+\theta h))=f(x_i+\theta h,y(x_i)+\theta h f(x_i,y(x_i)))+O(h^2).$
Substituting this into Equation \ref{eq:3.2.11} yields
\begin{aligned} y(x_{i+1})&=&y(x_i)+h\left[\sigma f(x_i,y(x_i))+\right.\\&&\left.\rho f(x_i+\theta h,y(x_i)+\theta hf(x_i,y(x_i)))\right]+O(h^3).\end{aligned}
This implies that the formula
$y_{i+1}=y_i+h\left[\sigma f(x_i,y_i)+\rho f(x_i+\theta h,y_i+\theta hf(x_i,y_i))\right]$
has $$O(h^3)$$ local truncation error if $$\sigma$$, $$\rho$$, and $$\theta$$ satisfy Equation \ref{eq:3.2.10}. Substituting $$\sigma=1-\rho$$ and $$\theta=1/2\rho$$ here yields
$\label{eq:3.2.13} y_{i+1}=y_i+h\left[(1-\rho)f(x_i,y_i)+\rho f\left(x_i+{h\over2\rho}, y_i+{h\over2\rho}f(x_i,y_i)\right)\right].$
The computation indicated here can be conveniently organized as follows: given $$y_i$$, compute
\begin{aligned} k_{1i}&=&f(x_i,y_i),\\ k_{2i}&=&f\left(x_i+{h\over2\rho}, y_i+{h\over2\rho}k_{1i}\right),\\ y_{i+1}&=&y_i+h[(1-\rho)k_{1i}+\rho k_{2i}].\end{aligned}
Consistent with our requirement that $$0<\theta<1$$, we require that $$\rho\ge1/2$$. Letting $$\rho=1/2$$ in Equation \ref{eq:3.2.13} yields the improved Euler method Equation \ref{eq:3.2.4}. Letting $$\rho=3/4$$ yields Heun’s method,
$y_{i+1}=y_i+h\left[{1\over4}f(x_i,y_i)+{3\over4}f\left(x_i+{2\over3}h,y_i+{2\over3}hf(x_i,y_i)\right)\right],$
which can be organized as
\begin{aligned} k_{1i}&=&f(x_i,y_i),\\ k_{2i}&=&f\left(x_i+{2h\over3}, y_i+{2h\over3}k_{1i}\right),\\ y_{i+1}&=&y_i+{h\over4}(k_{1i}+3k_{2i}).\end{aligned}
Letting $$\rho=1$$ yields the midpoint method,
$y_{i+1}=y_i+hf\left(x_i+{h\over2},y_i+{h\over2}f(x_i,y_i)\right),$
which can be organized as
\begin{aligned} k_{1i}&=&f(x_i,y_i),\\ k_{2i}&=&f\left(x_i+{h\over2}, y_i+{h\over2}k_{1i}\right),\\ y_{i+1}&=&y_i+hk_{2i}.\end{aligned}
Examples involving the midpoint method and Heun’s method are given in Exercises [exer:3.2.23}- [exer:3.2.30}.
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http://mathhelpforum.com/trigonometry/33560-proving-trig-identities.html
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1. ## proving trig identities
i need some help with this identity
(sinx+tanx)/1+secx=sinx
2. (sinx+tanx)/1+secx=sinx
First, the numerator:
(sinx+tanx) = (sinx + sinx/cosx) = (sinx) (1+(1/cosx)) = (sinx) (1+secx)
The denominator (of the left side of equation) is (1+secx)
The (1+secx) on the top and the (1+sec x) on the bottom cancel out.
You are left with sinx = sinx.
-Andy
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https://cascience.wordpress.com/2008/09/10/chemistry-law-of-combining-volumes-and-avogadros-hypothesis-sept-10-2008/
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# Chemistry- Law of Combining Volumes and Avogadro’s Hypothesis, Sept. 10, 2008
Today we discussed The Law of Combining Volumes (suggested by French chemist Joseph Louis Gay Lussac). The law states that when gases react, the volume of the reactants and products, when measure at equal temperature and pressure, are always in whole number ratios.
Example: 2 volumes of hydrogen combined with one volume of oxygen will give 2 volumes of water
This law eventually led to Avogadro’s Hypothesis. Avogadro postulated that equal volumes of all ideal gases, at the same temperature and pressure, contain the same number of molecules. Below, we have 3 containers with the same pressure (expressed in atms instead of kPa) and temperature; we can see that 1 mole of each of the 3 different molecules will occupy the same amount of space.
This law could be described mathematically as:
n1/V1 = n2/V2
n= kV
n α V
where n represents the number of moles, V represents volume and k is some constant.
Molar Volume (MV) describes the space occupied by one mole of gas. The unit for molar volume is L/mol.
MV = V/n
Standard Molar Volume: The molar volume of a gas at STP is 22.4 L/mol.
Note about Ideal Gases: ideal gases are a theoretically volumeless and without attraction. In real life, this does not stand true for many gases; however, we will apply the same formulae and laws to real gases for now.
For homework, refer to page 73 of the text and complete # 38 to 43.
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https://www.vedantu.com/question-answer/the-result-of-left-54327-times-3572-times-00057-class-8-maths-cbse-5f62d4e5eee2a36606a576f8
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Question
# The result of $\left( {{\text{54}}{\text{.327$\times$357}}{\text{.2$\times$0}}{\text{.0057}}} \right)$ is same asA. ${\text{5}}{\text{.4327$\times$3}}{\text{.572$\times$5}}{\text{.7}}$B. ${\text{5}}{\text{.4327$\times$3}}{\text{.572$\times$0}}{\text{.57}}$C. ${\text{54327$\times$3572$\times$0}}{\text{.0000057}}$D. ${\text{5432}}{\text{.7$\times$3}}{\text{.572$\times$0}}{\text{.000057}}$
Hint: The expressions given to us are product of decimals. The number of decimal places before and after multiplication will be equal. So, the option with an equal number of decimal places as in the question will be the correct answer.
We have the expression $\left( {{\text{54}}{\text{.327$\times$357}}{\text{.2$\times$0}}{\text{.0057}}} \right)$. This has a total of 8 decimal places.
The 1st option is ${\text{5}}{\text{.4327$\times$3}}{\text{.572$\times$5}}{\text{.7}}$. It has a total decimal place of ${\text{4 + 3 + 1 = 8}}$.
The second option is ${\text{5}}{\text{.4327$\times$3}}{\text{.572$\times$0}}{\text{.57}}$. The total number of decimal places of this expression is ${\text{4 + 3 + 2 = 9}}$
The 3rd option is ${\text{54327$\times$3572$\times$0}}{\text{.0000057}}$. It has a total number of decimal places of 7
The last option is ${\text{5432}}{\text{.7$\times$3}}{\text{.572$\times$0}}{\text{.000057}}$ and have a total number of decimal places equal to${\text{1 + 3 + 6 = 10}}$
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https://fr.slideshare.net/TELKOMNIKAJournal/the-flow-of-baseline-estimation-using-a-single-omnidirectional-camera
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# The flow of baseline estimation using a single omnidirectional camera
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## Consultez-les par la suite
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# The flow of baseline estimation using a single omnidirectional camera
Baseline is a distance between two cameras, but we cannot get information from a single camera. Baseline is one of the important parameters to find the depth of objects in stereo image triangulation. The flow of baseline is produced by moving the camera in horizontal axis from its original location. Using baseline estimation, we can determined the depth of an object by using only an omnidirectional camera. This research focus on determining the flow of baseline before calculating the disparity map. To estimate the flow and to tracking the object, we use three and four points in the surface of an object from two different data (panoramic image) that were already chosen. By moving the camera horizontally, we get the tracks of them. The obtained tracks are visually similar. Each track represent the coordinate of each tracking point. Two of four tracks have a graphical representation similar to second order polynomial.
Baseline is a distance between two cameras, but we cannot get information from a single camera. Baseline is one of the important parameters to find the depth of objects in stereo image triangulation. The flow of baseline is produced by moving the camera in horizontal axis from its original location. Using baseline estimation, we can determined the depth of an object by using only an omnidirectional camera. This research focus on determining the flow of baseline before calculating the disparity map. To estimate the flow and to tracking the object, we use three and four points in the surface of an object from two different data (panoramic image) that were already chosen. By moving the camera horizontally, we get the tracks of them. The obtained tracks are visually similar. Each track represent the coordinate of each tracking point. Two of four tracks have a graphical representation similar to second order polynomial.
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https://circlecoder.com/maximum-sum-of-3-non-overlapping-subarrays/
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# Maximum Sum Of 3 Non-Overlapping Subarrays Problem
## Description
LeetCode Problem 689.
Given an integer array nums and an integer k, find three non-overlapping subarrays of length k with maximum sum and return them.
Return the result as a list of indices representing the starting position of each interal (0-indexed). If there are multiple answers, return the lexicographically smallest one.
Example 1:
``````1
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Input: nums = [1,2,1,2,6,7,5,1], k = 2
Output: [0,3,5]
Explanation: Subarrays [1, 2], [2, 6], [7, 5] correspond to the starting indices [0, 3, 5].
We could have also taken [2, 1], but an answer of [1, 3, 5] would be lexicographically larger.
``````
Example 2:
``````1
2
Input: nums = [1,2,1,2,1,2,1,2,1], k = 2
Output: [0,2,4]
``````
Constraints:
• 1 <= nums.length <= 2 * 10^4
• 1 <= nums[i] <2^16
• 1 <= k <= floor(nums.length / 3)
## Sample C++ Code
``````1
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class Solution {
public:
vector<int> maxSumOfThreeSubarrays(vector<int>& nums, int k) {
int n = nums.size();
vector<int> prefixSum(n+1, 0);
for (int i = 1; i <= n; i++) {
prefixSum[i] = prefixSum[i-1] + nums[i-1];
}
vector<int> left(n+1, 0);
vector<int> leftInd(n+1, 0);
vector<int> right(n+1, 0);
vector<int> rightInd(n+1, 0);
for (int i = k; i <= n; i++) {
left[i] = max(left[i-1], prefixSum[i] - prefixSum[i-k]);
if (left[i] == left[i-1])
leftInd[i] = leftInd[i-1];
else
leftInd[i] = i;
}
for (int i = n-k; i >= 0; i--) {
right[i] = max(right[i+1], prefixSum[i+k] - prefixSum[i]);
if (right[i] == prefixSum[i+k] - prefixSum[i])
rightInd[i] = i;
else
rightInd[i] = rightInd[i+1];
}
int maxSum = 0, a, b, c;
for (int i = k; i <= n - 2 * k; i++) {
if (maxSum < left[i] + (prefixSum[i+k] - prefixSum[i]) + right[i+k]) {
maxSum = left[i] + (prefixSum[i+k] - prefixSum[i]) + right[i+k];
a = leftInd[i] - k;
b = i;
c = rightInd[i+k];
}
}
return {a, b, c};
}
};
``````
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http://www.jiskha.com/display.cgi?id=1359959637
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# Trigonometry
posted by .
Height of a mountain
While traveling across flat land, you notice a mountain directly in front of you. The angle of elevation to the peak is 2.5°. After you drive 18 miles closer to the mountain, the angle of elevation is 10°. Approximate the height of the mountain.
• Trigonometry -
draw a diagram. if the height is h,
h/tan2.5° = h/tan10° + 18
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https://www.esaral.com/q/evaluate-the-following-integrals-18128
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Deepak Scored 45->99%ile with Bounce Back Crack Course. You can do it too!
# Evaluate the following integrals:
Question:
Evaluate the following integrals:
$\int \frac{1}{\sqrt{(x-\alpha)(\beta-x)}} d x,(\beta>\alpha)$
Solution:
let $\mathrm{I}=\int \frac{1}{\sqrt{(\mathrm{x}-\mathrm{\alpha})(\beta-\mathrm{x})}} \mathrm{dx},(\operatorname{as} \beta>\alpha)$
$=\int \frac{1}{\sqrt{-x^{2}-x(\alpha+\beta)-\alpha \beta}} d x$
$=\int \frac{1}{\sqrt{-\left[\mathrm{x}^{2}-2 \mathrm{x}\left(\frac{\alpha+\beta}{2}\right)+\left(\frac{\alpha+\beta}{2}\right)^{2}-\left(\frac{\alpha+\beta}{2}\right)^{2}+\alpha \beta\right]}} d \mathrm{x}$
$=\int \frac{1}{\sqrt{-\left[\left(x-\frac{\alpha+\beta}{2}\right)^{2}-\left(\frac{\alpha+\beta}{2}\right)^{2}\right]}} d x$
$=\int \frac{1}{\sqrt{\left[\left(\frac{\beta-\alpha}{2}\right)^{2}-\left(x-\frac{\alpha+\beta}{2}\right)^{2}\right]}} d x \quad[\beta>\alpha]$
Let $(x-(\alpha+\beta) / 2)=t$
$d x=d t$
$I=\int \frac{1}{\sqrt{\left(\frac{\beta-\alpha}{2}\right)^{2}-t^{2}}} d t$
$=\sin ^{-1}\left(\frac{\mathrm{t}}{\frac{\beta-\alpha}{2}}\right)+\mathrm{c}$
$I=\sin ^{-1}\left(2 \frac{x-\frac{\alpha+\beta}{2}}{\beta-\alpha}\right)+c$
$I=\sin ^{-1}\left(\frac{2 x-\alpha-\beta}{\beta-\alpha}\right)$
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