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# Can you please give me just few lines on each question? 1. How would you classify your experience with software
###### Question:
Can you please give me just few lines on each question?
1. How would you classify your experience with software development? Beginner, Intermediate, or Advanced and why?
2. Have you built a web application that includes HTML, JavaScript, SQL, and a server-side technology, and is used in a production environment? If so, please describe.
3. Have you built a mobile or responsive application with HTML, CSS, and JavaScript? Please describe.
4. What JavaScript technologies do you have experience with? Have you worked with modern JavaScript frameworks such as Angular? If so, please describe.
5. How would you classify your experience with business process analysis and software design? Beginner, Intermediate, or Advanced and why? If you have analyzed a business process and designed a system from that analysis, please give a short summary.
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Oct 9 awarded Critic Aug 7 awarded Supporter Aug 1 comment Solving geometric problems using Linear Programming I would recommend the approach taken at CS.SE since that is $O(n)$ time versus $O(n \log n)$. I also think it lends itself to LP similar to constructing the "feasible region" (of convex shape). Aug 1 comment Solving geometric problems using Linear Programming I believe this could point you in the right direction. Jul 30 awarded Teacher Jul 30 answered How many rectangles can fit in a polygon with n-sides? Jul 30 awarded Autobiographer Jul 30 comment How many rectangles can fit in a polygon with n-sides? This question has been seen over at SO. Also, when you mention "must be consistent throughout", do you mean that all rectangles must be portrait or that their edges can only be perpendicular/parallel with each other (no weird angles)?
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# The compony had a profit gain of \$3,000,000,000 the first year and a profit gain of \$6,000,000 the second year. which number shows the change?
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# The unit of angular displacement in SI units is(a). Radians(b). Degrees(c). Degree Celsius(d). Radians/sec
Last updated date: 07th Sep 2024
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- Hint: To find the SI unit of angular displacement, understand its formula. With this formula, find the dimensional formula of angular displacement. Then find its units with help of its dimensional formula.
Complete step-by-step solution -
We all know about normal displacement, which tells you about the change in position of a body along a straight line. For example, if a body moves from point A to point B, then the displacement of that is the line joining the points A and B.
There is one more type of displacement called angular displacement. Sometimes, we tend to analyse certain cases with respect to a fixed point or fixed line (axis).
Then the moving body appears to be resolving around this fixed point/axis. This is where angular displacement comes into play. Now, we measure the change in the position of the body in terms of the distance of the body from the fixed point/axis and the angle the body makes with respect to the fixed point/axis while moving. This angle is called angular displacement. The SI unit of angular displacement is radians.
Now, let us consider a simple case. A body is revolving around a fixed point in a circular path with a fixed point as the centre. Suppose it moves from point A to point B, along the circular path, which makes arc AB. An angle is defined as the ratio of the length of the arc to the radius of the circle.
Therefore, $\theta =\dfrac{s}{r}$
The dimensional formula of angular displacement is $\dfrac{[L]}{[L]}=1$. Hence, it does not have any dimension and it is true that it will not have any units, but for some reason it has been given a SI unit i.e. radians. Other unit is degree.
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# Multiplying With 2 Digits Worksheets
Multiplying With 2 Digits Worksheets. Remember to put the extra zero in the ones place on the second line of multiplication. One where the questions are just written and children need to set them out and one where they are already set out ready.
One where the questions are just written and children need to set them out and one where they are already set out ready. 3 digit by 2 digit long multiplication practice worksheets; Click on the image to view or download the.
### While You’re Here, Make Sure You Check Out These Resources Too:
Both worksheets are set out in two ways: Practice two digit multiplication with this math worksheet (ws2). To the answer from 4 x 1 this 1 got carried during the.
### Teachers Can Use These 2 Digits By 2 Digit Multiplication Worksheets With Grids As Tests, Practice Assignments, Or Teaching Tools.
It is a faster technique than the grid method which is usually used until children reach year five and six. 597 46 27,462 × 7. It may be printed, downloaded or saved and used in your classroom, home school, or other educational.
### Remember To Put The Extra Zero In The Ones Place On The Second Line Of Multiplication.
One where the questions are just written and children need to set them out and one where they are already set out ready. Answer sheets are also included for. Worksheet #1 worksheet #2 worksheet #3 worksheet #4 worksheet #5 worksheet #6.
### Multiplication 2 Digits By 2 Digits Sheet 1.
Learning different methods of multiplying double digit numbers can help a lot. Once a student grasps the concept of it, this makes it easier to obtain the product of any two numbers irrespective of the number of digits it possesses. These free multiplying 3 digit by 2 digit worksheets exercises will have your kids engaged and entertained while they improve their skills.
### Multiplying 2 Digits With Grids Worksheet.
Multiplying 2 digit and 3 digit numbers by a 1 digit number by jonholliday: You can encourage them to begin by. They multiply congruent shapes together.
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# A spaceship of speed Vo travelling along +y axis suddenly shots out one fourth of its part with speed 2Vo along +x axis. xy axes are fixed with respect to the ground. What is the velocity of the remaining part?
Asked by 11th December 2011, 8:39 PM
say, vel of remaining part: x at an angle ? with horizontal.
now equate, the x comp of momentum and y comp of momentum before and after explosion.
mom coservation:
mv0=3/4 mx sin? + 0 y comp of mom
0=3/4 mx cos?+m/4 2v0
x sin ?=4v0/3
x cos?=-2v0/3
x=2?5 v0/3
Answered by Expert 12th December 2011, 11:32 AM
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CFD Online Discussion Forums (http://www.cfd-online.com/Forums/)
- Main CFD Forum (http://www.cfd-online.com/Forums/main/)
- - Lake flow (http://www.cfd-online.com/Forums/main/2511-lake-flow.html)
Robert Kunz August 22, 2000 09:51
Lake flow
Has anyone experience with computing the flow of heat and/or contaminants in large bodies of water? In our application, there are significnat thermal gradients, but velocites are very small (<1 inch/s) and relevant time scales are on the order of days. What are the important issues? Can commercial packages handle such flows?
John C. Chien August 22, 2000 10:10
Re: Lake flow
(1). The problem is you need to model it first. (2). The geometry, and the boundary conditions need to be specified first by you. (3). Some commercial codes should be able to handle the natural convection problem as well as the diffusion problem.
Mehdi BEN ELHADJ August 22, 2000 10:32
Re: Lake flow
I think that you must model the problem first and the geometry, and the boundary conditions need to be specified by you like says John.
For this kind of problem you can contact David Glynn at cfd@flowsolve.com in order to discuss a bout some difficulties of the model.
Jared August 22, 2000 11:13
Re: Lake flow
I totally agree with Mr. Chien's recommendation. Once you model your problem properly, you will have some ideas if you need to include the time term or the convective term in N-S equation. You, then, can narrow your choices of commercial codes.
Jared
Adrin Gharakhani August 22, 2000 15:19
Re: Lake flow
Start from the original equations. Non-dimensionalize them and look at order of magnitude of the various terms to decide which terms can be (safely) eliminated. Time scale in the order of days does not mean much; it all depends on what you are looking for - whether you want to see what happens during the first hour, what happens in days, or in a year ... This will determine whether you have a quasi-steady problem at hand or not. That may also determine whether you have a purely diffusive problem or whether natural convection should also be included. That's all one can say given the amount of info you've provided
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# MASS RELATIONS IN CHEMISTRY; STOICHIOMETRY
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1 MASS RELATIONS IN CHEMISTRY; STOICHIOMETRY [MH5; Ch. 3] Each element has its own unique mass. The mass of each element is found on the Periodic Table under the chemical symbol for the element (it is usually given to several decimal places); this may be referred to as the ATOMIC MASS (or ATOMIC WEIGHT). Atomic masses are relative measurements; based on how heavy an element is compared to another element. The mass of 1 atom of an element may be expressed in amu, short for Atomic Mass Units. The modern scale of atomic masses is based on the most common isotope of carbon; 12 C... This isotope is assigned a mass of 12 amu: Mass of one 12 C atom = 12 amu (exactly) So... 1 amu = 1 x mass of one C atom 12 6 When determining atomic weight, we take a large number of atoms with isotopes present in their natural abundances. Atomic weights of all elements except 12 C are subject to experimental error; ( 12 C has atomic mass of exactly 12 amu). For elements that are isotopically pure (consist of one isotope only; monoisotopic), we may talk of atomic mass (mass of one atom) 12 17
2 EXAMPLE: Sodium, Na (only natural isotope), At. Wt = So the mass of 1 atom of Na is amu. Notice that the atomic mass of the element is very close to its mass number... For elements that naturally contain more than one isotope, we can determine an average atomic mass; an average of the isotopic masses weighted according to their fractional relative natural abundances. This average is numerically the same as atomic weight... EXAMPLE 1: Chlorine 35 CR isotopic mass: amu, 75.77% 37 CR isotopic mass: amu, 24.23% (Note: these are atom % s, not mass % s) What is the average atomic mass? 18
3 EXAMPLE 2: Antimony, Sb, consists of two isotopes; 121 Sb (atomic mass ) and 123 Sb (atomic mass ) Calculate the relative abundances (in atom %) of these two isotopes. The Average Atomic Mass for any element is found on the Periodic Table. 19
4 How do we calculate the mass of a molecule? This is called Molecular Weight (or Mass): it is the sum of the atomic weights of all the atoms in the molecule. EXAMPLE 1: O 2 EXAMPLE 2: H 2 O EXAMPLE 3: Mg(NO 3 ) 2 We never actually measure the mass of one atom or one molecule of anything...an amu is an extremely small measurement. Chemists like to measure masses in grams......so what is the relationship between amu and grams? Before we can answer this question, we must get a very clear idea of what a mole is... 20
5 Demystifying The Mole A mole is a number; a very large number. It is a word we use to represent a certain number of things. We use it because it is easier to say the word than to say the number itself... We have commonly used names for many groups of items... Quantity Number of Items pair trio quartet dozen gross mole So, the mole (which we abbreviate as mol) contains Avogadro s Number (N A ) of items: N A = 6.02 x This is a really strange number, but we use it because of the official definition of the mole; which is the number of atoms in exactly 12 g of the isotope 12 C. It turns out that this number of atoms is Avogadro s Number, or 6.02 x Therefore, 1 mole of anything contains 6.02 x of those things. 12 g of 12 C is one mole of 12 C and contains N A (Avogadro s No.) atoms (by definition). 21
6 Atoms and molecules are really small, and in any sample of measurable size (usually in g or mg) we have a lot of them; so we use the term moles as a sort of shorthand way to express how many molecules or atoms we are actually talking about. Back to the question: Relating the amu to grams g of 12 C is one mole of 12 C and contains N A (Avogadro s Number) of atoms... This is another very small number...and remember that we normally do not measure the mass of just one atom or molecule. It is much more practical to measure the mass of a larger amount of a substance; so we use the mass of one mole of the substance... The MOLAR MASS is the mass in grams of 1 mole of compound. EXAMPLES: O 2 : H 2 O: Mg(NO 3 ) 2 : Notice that the number (from the Periodic Table) is the same as the Molecular Weight; it is the units that are different!! 22
7 One atom (or molecule) of substance will have a mass given in amu. One mole of atoms (or molecules) of a substance will have a mass given in grams. (Molar Mass is usually given units of g mol 1 ) Some compounds exist as "hydrated" molecules; for each molecule of compound, a certain number of water molecules are associated with that molecule. Think of it as a molecule of compound being surrounded by a certain number of water molecules. To determine the Molar Mass of one of these compounds, the mass of the water molecules associated with the molecule must be added to the mass of the molecule. EXAMPLE: The Molar Mass of CuSO 4 C 5 H 2 O We use Molar Mass to convert from moles to g or vice-versa. 23
8 EXAMPLE 1: a) How many moles of CO 2 molecules are there in a 55.0 g sample of CO 2? b) How many moles of CO 2 molecules are there in a 25.0 g sample of CO 2? 24
9 EXAMPLE 2: What is the mass of moles of C 2 H 5 OH? EXAMPLE 3: Suppose we have a sample of a compound with a mass of 2.68 g. An experiment has shown that in this sample there are 1.26 x 10 2 moles of the compound in this sample. What is the molar mass of the compound? 25
10 CHEMICAL FORMULAE Recall that a chemical formula is the ratio (in whole numbers) of the atoms in a compound... EXAMPLE: CH 4 (Methane) 1 molecule of CH 4 contains: 1 mole of CH 4 molecules contains: How many atoms would this be?? 26
11 PERCENTAGE COMPOSITION The composition of a compound is often expressed by mass as % s, because that is the way elemental analyses are reported. % composition by mass is found by using atomic masses and the formula of the compound. EXAMPLE: Using C 8 H 18, find the % composition; i) by mass ii) by moles i) C 8 H 18 has a Molar Mass of C 8 H 18 contains 8 C (at g mol -1 each) and 18 H (at g mol -1 ) % C (by mass) = % H (by mass) = We could use this method to determine the MASS FRACTION of one of the elements in a compound. In reporting the composition of C 8 H 18 by mass fractions, we would say that the fraction of C 8 H 18 that is Carbon is: and the fraction that is Hydrogen is: 27
12 ii) Mole percent uses the same principle, but is reporting the composition using moles instead of mass. In one mole of C 8 H 18 molecules, there are 8 moles of Carbon atoms and 18 moles of Hydrogen atoms. The total number of moles of atoms is twenty-six (26). The mole % of Carbon = The mole % of Hydrogen = The term MOLE FRACTION (denoted by X) is used frequently... The mole fraction is the mole percent reported as a fraction. We could say that: The mole fraction of Carbon in C 8 H 18 is: The mole fraction of Hydrogen is: 28
13 Can we deduce a formula from % composition? Yes!! EXAMPLE: A compound contains 80.1 % C and 19.9 % H by mass. Find the SIMPLEST FORMULA of the compound. (The simplest, or empirical formula shows the simplest ratio of atoms in the compound.) Given % by mass, we use a 100 g sample of compound; therefore, we have 80.1 g of C and 19.9 g of H. Find the number of moles of each element first... To find the simplest formula, we want a whole number ratio, so divide each # moles by the smallest # moles... 29
14 The MOLECULAR FORMULA of a compound is a multiple of the simplest formula; for CH 3, we could write the molecular formula as (CH 3 ) n. To find "n", we need the Molar Mass of the compound's molecular formula... EXAMPLE: If we know that the Molar Mass of (CH 3 ) n is 30 g mol -1, we can say: 30
15 Sometimes you will not be given the elemental analysis, only some experimental data. From the experimental data you will need to determine the amounts of each element present in the compound. EXAMPLE: A compound contains only C, H and O. When burned in excess O 2, a g sample of compound yields g CO 2 and g H 2 O. If the Molar Mass of the compound is 90 ± 5 g mol -1, find the molecular formula. To find molecular formula, we must first find simplest formula. We need either % composition, or the masses of each element in the compound. This will enable us to find the # moles of each element, and therefore the mole ratio ( the simplest formula). For C : 31
16 For H : For O: 32
17 Now find the formula... 33
18 Sometimes we do not know the identity of all the elements in a compound, but we do know the percentage of one element in said compound... We can use this percentage to determine the molar mass, and then the identity of the unknown element. EXAMPLE: An oxide of an unknown metal has the formula X 3 O 4. Analysis shows that this compound is 27.64% oxygen by mass. Determine the molar mass and identity of X. 34
19 MASS RELATIONS IN REACTIONS: [MH5; 3.4] Stoichiometry ; a term used to describe the relationship between amounts of reactants and products (in moles and grams) using balanced chemical equations. In a chemical equation, symbols of elements or compounds represent MOLES... 2 S + 3 O 2! 2 SO 3 two moles three moles two moles of of Sulfur of Oxygen Sulfur Trioxide ----REACTANTS PRODUCTS---- What is this equation telling us? 1) 2 moles of Sulfur will react with 3 moles of Oxygen. 2) 2 moles of Sulfur will produce 2 moles of Sulfur Trioxide. 3) 3 moles of Oxygen will produce 2 moles of Sulfur Trioxide. Notes: a) Equations must be balanced in mass of each element, and in charge. b) By convention, coefficients in an equation are usually integers, the smallest values possible to balance. S + O 2! SO 3 S + O 2! SO 3 S + O 2! SO 3 35
20 c) Use a one-way arrow (!) for a reaction going mainly forwards : C + O 2! CO 2 Use a two-way arrow ( º ) for an equilibrium: 2NO + O 2 º 2NO 2 (In an equilibrium situation, the reaction is proceeding in both directions.) d) Indicate the phase of reactants and products, if required, by the symbols: (s) solid (R) liquid (g) gas (aq) aqueous sol n 2 Na (s) + 2 H 2 O (R)! 2 Na + (aq) + 2 OH - (aq) + H 2 (g) e) Remember the difference between atoms and molecules. Don t write a symbol for an atom when an element occurs as a diatomic molecule: H 2 (g) N 2 (g) O 2 (g) F 2 (g) CR 2 (g) Br 2 (R) I 2 (s) Balancing Equations - this is essential! Some are easy: C(s) + O 2 (g)! CO 2 (g) Others need work: N 2 + H 2! NH 3 36
21 f) Combustion Reactions are important - what happens when octane (a component of gasoline) burns in oxygen? C 8 H 18 + O 2! CO 2 + H 2 O C 8 H 18 + O 2! CO 2 + H 2 O Always check your equation AFTER balancing it!! g) By knowing the ratios by which species react to form products, we can determine the number of moles of species formed or reacted, given the number of moles of another species present in the reaction. EXAMPLE 1: If 2.00 g of C 2 H 4 are burnt in excess O 2, how many grams of CO 2 are formed? First, balance the equation: C 2 H 4 + O 2! CO 2 + H 2 O The steps to solving this problem are: 1) Find number of moles of C 2 H 4. 2) Find number of moles of CO 2 by relating moles reactant (which we know) to moles product (which is what we want). 3) Find mass CO 2. 37
22 The solution is: 1) moles C 2 H 4 2) Relate # moles of reactant to # moles product. (from equation) 3) Mass CO 2 = EXAMPLE 2: White phosphorous, P 4, may be burned in oxygen, O 2, to produce P 4 O 10. What mass of oxygen is needed to completely burn (or react with) 1.20 g of white phosphorus? First, write the balanced equation: P 4 + O 2! P 4 O 10 Then find the number of moles of P 4 : (MM = 4 x gmol 1 ) 38
23 Write what is happening in words: Set up a ratio and solve it. Finish the calculation... Note: The Molar Mass you use must match the species in the equation! LIMITING REAGENT problems involve reactions where the amount of product formed is limited by the amount of reagent available to react. EXAMPLE 1: You are assembling bicycles. You have 6 frames and 10 wheels. How many bicycles can you build? 39
24 EXAMPLE 2: What mass of AR 2 (SO 4 ) 3 can be prepared by reacting 25.0 g of AR with 100 g of H 2 SO 4? The reaction is: AR + H 2 SO 4! AR 2 (SO 4 ) 3 + H 2 First things first: Make sure that the equation is balanced!!! The clue that this is a limiting reagent problem is that the amounts of both starting materials are given... These amounts may be given in various ways; they will all enable you to find the # moles of each reactant. Find # moles of each reactant: TO FIND THE LIMITING REAGENT: Divide the actual number of moles of each reactant by its coefficient (the required number of moles) from the balanced equation; then compare the resultant numbers. The smallest of these two numbers tells you the limiting reagent! 40
25 It is important to realize that you only use this step to determine the limiting reagent; you do not necessarily use these numbers again (although you often see them again!) You can only form as much product as you have limiting reagent available to react...so all calculations must relate the amount of limiting reagent to another species. Relate # moles of limiting reagent to # moles product... Set up a ratio: Finish the calculation: 41
26 YIELD CALCULATIONS Yield is usually expressed on a % basis, showing the amount of product obtained as a % of what would be expected from the equation: moles obtained x 100 % OR mass obtained x 100 % moles expected mass expected EXAMPLE 1: When 12.0 g iron, Fe, were burned in air, 13.8 g iron oxide, Fe 3 O 4, were produced. Calculate the % yield of Fe 3 O 4 in this reaction. For % Yield, remember... moles obtained x 100 % = mass obtained x 100% moles expected mass expected Write the balanced equation: Fe + O 2! Fe 3 O 4 Find number of moles of Fe: 42
27 From the equation, we would expect to obtain: We actually obtained 13.8 g; less than we would expect... The PERCENTAGE YIELD would be: EXAMPLE 2: A 5.00 g sample of S was chemically treated and several steps later resulted in the retrieval of 41.6 g of Na 2 SO 4 C10 H 2 O. Calculate the % yield. All the intermediate steps may be ignored! Since the final product contains 1 S atom; one mole S gives one mole of Na 2 SO 4 C10H 2 O (this statement serves as the balanced equation!!) 43
28 As usual, start by determining the number of moles... The molar mass of Na 2 SO 4 C10H 2 O is: We would expect a yield of: Mass of Na 2 SO 4 C10H 2 O expected: 44
29 EXAMPLE 3: We have seen that in most chemical reactions, one of the reactants is present in a lesser amount; this is called the Limiting Reagent. Some reactions take more than one step to occur; we saw that the moles of the various species carry all the way through the entire process. Here s an example which incorporates almost all the intricacies involved with stoichiometry. EXAMPLE: Copper metal is usually obtained from sulfide ores, such as Cu 2 S. The refining process takes two steps; the first is roasting the Cu 2 S in oxygen, O 2, to obtain copper (I) oxide, Cu 2 O. In the second step, the Cu 2 O is reacted with powdered carbon, a process which yields carbon monoxide gas and copper metal. 1) 2 Cu 2 S (s) + 3 O 2 (g)! 2 Cu 2 O (s) + 2 SO 2 (g) 2) Cu 2 O (s) + C (s)! 2 Cu (s) + CO (g) A) What is the overall equation? -45-
30 B) Suppose we react 10.0 grams of Cu 2 S with 4.50 grams of O 2. What yield of Cu 2 O would be expected in step 1)? C) If 5.16 g of Cu (s) results from the second step, what is the percentage yield from the entire process? -46-
31 % PURITY Some problems involve the use of reactants that are not 100 % pure. EXAMPLE 1: A solution of CS 2 (R) has a purity of 76.5 % (by mass) and has a density of 1.20 gml 1. What mass of SO 2 (g) could be produced if 15.0 ml of CS 2 were burned in excess oxygen? 47
32 EXAMPLE 2: An ore contains 35.0 % by mass iron, Fe. What mass of this ore is needed to produce 500 g of Fe 2 O 3? Write a balanced equation: Fe + O 2! Fe 2 O 3 Find the number of moles of whatever you can... Moles Fe 2 O 3 = From the equation, we know: 48
33 Mass of Fe required: Set up a ratio to relate the % purity of the ore to the amount of Fe that is needed: 49
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p70 Chapter Three: STOICHIOMETRY Contents p76 Stoichiometry - The study of quantities of materials consumed and produced in chemical reactions. p70 3-1 Counting by Weighing 3-2 Atomic Masses p78 Mass Mass
### Balance the following equation: KClO 3 + C 12 H 22 O 11 KCl + CO 2 + H 2 O
Balance the following equation: KClO 3 + C 12 H 22 O 11 KCl + CO 2 + H 2 O Ans: 8 KClO 3 + C 12 H 22 O 11 8 KCl + 12 CO 2 + 11 H 2 O 3.2 Chemical Symbols at Different levels Chemical symbols represent
### Element of same atomic number, but different atomic mass o Example: Hydrogen
Atomic mass: p + = protons; e - = electrons; n 0 = neutrons p + + n 0 = atomic mass o For carbon-12, 6p + + 6n 0 = atomic mass of 12.0 o For chlorine-35, 17p + + 18n 0 = atomic mass of 35.0 atomic mass
### Unit 6 The Mole Concept
Chemistry Form 3 Page 62 Ms. R. Buttigieg Unit 6 The Mole Concept See Chemistry for You Chapter 28 pg. 352-363 See GCSE Chemistry Chapter 5 pg. 70-79 6.1 Relative atomic mass. The relative atomic mass
### MOLE CONVERSION PROBLEMS. 2. How many moles are present in 34 grams of Cu(OH) 2? [0.35 moles]
MOLE CONVERSION PROBLEMS 1. What is the molar mass of MgO? [40.31 g/mol] 2. How many moles are present in 34 grams of Cu(OH) 2? [0.35 moles] 3. How many moles are present in 2.5 x 10 23 molecules of CH
### SYMBOLS, FORMULAS AND MOLAR MASSES
SYMBOLS, FORMULAS AND MOLAR MASSES OBJECTIVES 1. To correctly write and interpret chemical formulas 2. To calculate molecular weights from chemical formulas 3. To calculate moles from grams using chemical
### Moles. Balanced chemical equations Molar ratios Mass Composition Empirical and Molecular Mass Predicting Quantities Equations
Moles Balanced chemical equations Molar ratios Mass Composition Empirical and Molecular Mass Predicting Quantities Equations Micro World atoms & molecules Macro World grams Atomic mass is the mass of an
### Concept 1. The meaning and usefulness of the mole. The mole (or mol) represents a certain number of objects.
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### Chapter 3. Mass Relationships in Chemical Reactions
Chapter 3 Mass Relationships in Chemical Reactions This chapter uses the concepts of conservation of mass to assist the student in gaining an understanding of chemical changes. Upon completion of Chapter
### Moles and Chemical Reactions. Moles and Chemical Reactions. Molar mass = 2 x 12.011 + 6 x 1.008 + 1 x15.999 = 46.069 g/mol
We have used the mole concept to calculate mass relationships in chemical formulas Molar mass of ethanol (C 2 H 5 OH)? Molar mass = 2 x 12.011 + 6 x 1.008 + 1 x15.999 = 46.069 g/mol Mass percentage of
### Useful only for measuring the mass of very small objects atoms and molecules!
Chapter 9 Chemical Composition (Moles) Which weighs more, 1 atom of He or 1 atom of O? Units of mass: Pound Kilogram Atomic mass unit (AMU) There are others! 1 amu = 1.66 x 10-24 grams = mass of a proton
### Chemistry I: Using Chemical Formulas. Formula Mass The sum of the average atomic masses of all elements in the compound. Units are amu.
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### Chapter 3! Stoichiometry: Calculations with Chemical Formulas and Equations. Stoichiometry
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### IB Chemistry. DP Chemistry Review
DP Chemistry Review Topic 1: Quantitative chemistry 1.1 The mole concept and Avogadro s constant Assessment statement Apply the mole concept to substances. Determine the number of particles and the amount
### Lecture Notes Chemistry E-1. Chapter 3
Lecture Notes Chemistry E-1 Chapter 3 http://inserbia.info/news/wp-content/uploads/2013/05/tamiflu.jpg http://nutsforhealthcare.files.wordpress.com/2013/01/tamiflu-moa.jpg The Mole A mole is a certain
### Matter. Atomic weight, Molecular weight and Mole
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### Calculations and Chemical Equations. Example: Hydrogen atomic weight = 1.008 amu Carbon atomic weight = 12.001 amu
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### MOLAR MASS AND MOLECULAR WEIGHT Themolar mass of a molecule is the sum of the atomic weights of all atoms in the molecule. Molar Mass.
Counting Atoms Mg burns in air (O 2 ) to produce white magnesium oxide, MgO. How can we figure out how much oxide is produced from a given mass of Mg? PROBLEM: If If 0.200 g of Mg is is burned, how much
### Chapter 5, Calculations and the Chemical Equation
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### The Mole Notes. There are many ways to or measure things. In Chemistry we also have special ways to count and measure things, one of which is the.
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### Chemical Reactions. Chemistry 100. Bettelheim, Brown, Campbell & Farrell. Introduction to General, Organic and Biochemistry Chapter 4
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### Lecture 5, The Mole. What is a mole?
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### CH3 Stoichiometry. The violent chemical reaction of bromine and phosphorus. P.76
CH3 Stoichiometry The violent chemical reaction of bromine and phosphorus. P.76 Contents 3.1 Counting by Weighing 3.2 Atomic Masses 3.3 The Mole 3.4 Molar Mass 3.5 Percent Composition of Compounds 3.6
### KEY for Unit 1 Your Chemical Toolbox: Scientific Concepts, Fundamentals of Typical Calculations, the Atom and Much More
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### EXPERIMENT 12: Empirical Formula of a Compound
EXPERIMENT 12: Empirical Formula of a Compound INTRODUCTION Chemical formulas indicate the composition of compounds. A formula that gives only the simplest ratio of the relative number of atoms in a compound
### How much does a single atom weigh? Different elements weigh different amounts related to what makes them unique.
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### We know from the information given that we have an equal mass of each compound, but no real numbers to plug in and find moles. So what can we do?
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### TOPIC 7. CHEMICAL CALCULATIONS I - atomic and formula weights.
TOPIC 7. CHEMICAL CALCULATIONS I - atomic and formula weights. Atomic structure revisited. In Topic 2, atoms were described as ranging from the simplest atom, H, containing a single proton and usually
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1 Angstrom = 1.0E-10 Meter
## How to convert from Angstrom to Meter?
### Angstrom to Meter?
To convert Angstrom to Meter: Every 1 Angstrom equals 1.0E-10 Meter. For example, 100 Angstrom equal 100 * 1.0E-10 = 1.0E-8 Meter and so on..
Angstrom to Foot | Angstrom to Centimeter | Angstrom to Inch | Angstrom to Kilometer | Angstrom to Mile | Angstrom to Millimeter | Angstrom to Yard | Angstrom to Micrometer |
## Angstrom
What is angstrom?
Angstrom, named after the Swedish physicist Anders Jonas Angstrom, is a vert tiny length unit. It equals 10−10 meter.
Angstrom is not considered as a part of the SI system of units but it is defined in the metric system.
Now, let's mention some applications of Angstrom:
• measurement of wavelengths of light.
• express sizes of atoms, molecules and microscopic biological structures
• dimensions of integrated circuit parts.
## Meter
What is meter?
In 1793, The distance from the equator to the North Pole was measured and divided by 10 million. The result of the previous process was called as a meter.
Nowadays, meter or metre is the base unit of length in the International System of Units (SI).
Meter or metre is a metric measurement slightly longer than a yard. It is the base unit of length in the International System of Units (SI). Distance measures length. For example, the distance of a road is how long the road is.
Meter is used as a measure of distance across the globe. Distance expresses the length. For example, the distance of a road is how long the road is.
However most of countries use meter to measure length, US is the primary exception which is using imperial system.
Here we will show you how to convert meters to feet:
1 meter is equal to 0.3.28084 feet:
1 m = (1/0.3048) ft = 3.28084 ft
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```Introduction to Topographic Maps
Purpose:
The purpose of this exercise is to become familiar with the information presented on a
topographic map and how to interpret that information.
Essential Learning Outcomes:
Use representative and graphic scales to calculate distance
Determine elevation
Interpolate contour lines
Construct a topographic profile
Equipment supplied:
Baraboo, WI topographic map
Ruler
Calculator
Exercises:
Part 1 – Reading and interpreting a topographic map
Use the topographic map of Baraboo, WI to answer the following questions:
1. What is the scale of the map? One inch on the map equals how many feet on the ground?
1:24,000; 1” = 24,000” / 12”/foot = 2,000 feet
2. What is the contour interval of the map? So, if you cross three contour lines how much has
20 feet; 60 feet
3. What is the projection of the map? Why use a map projection?
Wisconsin coordinate system, south zone
Lambert conformal conic
4. What series does this map belong to? What does that mean? (Tip: Compare latitude and
longitude values in the four corners of the map)
7.5’ series, the map covers 7.5’ latitude and 7.5’ of longitude
Page | 1
5. Fill in the information below:
A. Highest point on map (name and elevation): Sauk Point; 1593 ft above sea level
B. Lowest point on map (name and elevation): Gallus Slough; 774 ft above sea level
C. Total relief of map (A – B): 1593 ft – 774 ft = 819 ft
6. Which gradient is steeper, the west bluff or the east bluff of Devils Lake? How can you tell?
West bluff; the contour lines are closer together
7. Is the Baraboo River flowing from west to east or from east to west? How can you tell?
West to east; the contour lines indicate a higher elevation to the west and water flows
from higher elevations to lower elevations
8. Using the graphic scale, determine the straight-line distance in miles between Brewster
Cemetery and Koerth Cemetery.
~1.95 miles
9. Using the representative scale, determine the straight-line distance in miles between Brewster
Cemetery and Koerth Cemetery. How does your answer compare to question 8? Which do
you think is more accurate?
5.125” x 24,000” = 123,000” / 12”/foot = 10,250 ft / 5,280 ft/mile = 1.94; my answer is
slightly less than answer 8; I think using the representative scale is more accurate
because you only estimate distance once and then use mathematics
10. Which ski lift for Devils Head Lodge (found on the right side of the map) has the steepest
Ski Lift
Top elev.
Bottom
“Rise”
Distance on
“Run”
elev.
map
1 (west)
1,280 ft
945 ft
335 ft
0.875”
1,750 ft
19%
2
1,370 ft
940 ft
430 ft
1.375”
2.,750 ft
15.6%
3 (center) 1,380 ft
940 ft
440 ft
1.375”
2,750 ft
16%
4
1,410 ft
930 ft
480 ft
1.3125”
2,625 ft
18.1%
5 (east)
1,410 ft
950 ft
460 ft
1.1875”
2,375 ft
18.9%
Page | 2
Part 2 – Constructing a topographic profile
A topographic profile is a cross-section view of the topography of a landscape along a transect.
Use the Baraboo, WI topographic map and the profile grid below to construct a topographic
profile along the thin black line that passes west to east through Devils Lake, originating at 4811
on the west side of the map and ending at the 1300 ft index contour immediately west of the red
line in section 19.
To construct the profile, fold this paper so that the top of the profile grid forms the top edge of
this sheet of paper. Determine the interval and label the lines of the Y-axis; the base line should
be slightly less than the minimum elevation along the transect, the top line should be slightly
greater than the maximum elevation along the transect, and each interval line in between should
be equally-spaced.
For example, if the minimum elevation is 424 ft and the maximum elevation is 697 ft, you should
use base and top line values of 420 ft and 700 ft. This is a difference of 280 ft; since there are 14
interval lines, the interval between lines is 20 ft (280 ft / 14 = 20 ft).
Next, place the left side of the profile grid on the western edge of the map at point 4811 and align
the top edge of the profile grid with the thin black line. Project points along the thin black line
straight down to the interval line of equal value on the profile grid and place a dot at that spot on
the grid. Plot beginning and end points and points at each location that the thin black line crosses
an index contour (the dark brown lines). Connect all of the dots, interpolating between high and
low points, to complete the topographic profile.
1500’
1380’
1300’
1180’
1100’
1020’
940’
Page | 3
Part 3 – Drawing contour lines
In the diagram below, dashed lines represent stream channels and black dots with numbers
represent elevation points in feet above sea level. The 950 ft contour line has been provided.
Draw all other contour lines for the elevation data using a 10-ft contour interval. Label each end
of the contour line appropriately. To plot contour lines between two points you must interpolate
its position; that is, you must decide where best to draw it based on nearby data points (i.e., the
980 ft line should be drawn closer to the 982 ft data point than the 971 ft data point).
980’
971
970’
981
977
982
984
980’
977
970
975
967
970’
963
971
961
968
962
964
960
960
956
960’
960’
960
959
958
956
953
954
950
951
948
950
949
946
950’
950’
Page | 4
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Kids Worksheet: Worksheet Builder Printable First Grade Math Worksheets Expressions Kids Division Easy Games For Addition Kindergarten Geometry Text Book Free Fun Yakut Software Mathematics Questions And. integers worksheets grade 8 introduction to kids worksheet presentation Second Grade Subtraction Worksheets | Pembeforum
# Worksheet Builder Printable First Grade Math Worksheets Expressions Kids Division Easy Games For Addition Kindergarten Geometry Text Book Free Fun Yakut Software Mathematics Questions And
Published at Sunday, 01 September 2019. Kids Worksheet. By .
All my toys were one way or the other math related. I had puzzles, and tons of things Mom had me do as games on daily basis at home to get me ready for kindergarten! In fact, she continued guiding me towards being math friendly throughout kindergarten and first grade during which time 1st grade math worksheets was my constant companion. Practice surely makes perfect and I am very gratefully to Mom for taking her time to familiarize me with math even as a child.
I recommend getting one of these books when you first begin homeschooling and use it as a reference throughout your home-school journey. Regardless of how long you home-schooling, you will always have doubts and questions about how your child is performing.A scope and sequence book can put your mind at ease. Once you have a scope and sequence book, make a list of each area in math that he needs to work on for the school year.For example for grades three and four, by the end of the year in subtraction, your child should be able to: Solve vertical and horizontal computation problems,review subtraction of 2 numbers whose sums would be 18 or less,subtract 1- or 2-digit number from a 2-digit number with/without renaming, subtract 1-, 2-, or 3-digit numbers from 3- and 4-digit number with/without renaming,subtract 1-, 2-, 3-, 4-, or 5-digit number from a 5-digit number. When you have this list, begin searching online for free math worksheets that fit your child has scope and sequence for the year and the goals you have set for your child.
Worksheets are slowly becoming an important tool of learning for little children. Nowadays, worksheets are planned and created by many companies, publishers and schools. Some sell these worksheets both online and off line and others let people download them from the internet. Since there are so many worksheets available in the market, it may be difficult for parents to know which the appropriate and right worksheet is for their child. This article will take you through the basic elements of a good worksheet for children. Creating a worksheet requires a lot of planning and research. Things like the purpose of the worksheet, the age group for which it is being created and the resources available to solve the worksheets should be considered.
File name:
### Worksheet Builder Printable First Grade Math Worksheets Expressions Kids Division Easy Games For Addition Kindergarten Geometry Text Book Free Fun Yakut Software Mathematics Questions And
Image Size: 1811 x 2321 Pixels
File Type: Image/jpg
Total Gallery: 33 Pictures
File Size: 167 kb
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## OpenFOAM 3D Steady Turbulent Simulation (OF-1.4.1)
In this example we shall investigate how to make a 3D simulation with OpenFOAM. The case we want to study is a steady turbulent flow circulating inside a room. This example introduces almost all the basic steps which should be done for studying a CFD problem and explains how these things can be carried out in a quite simple way in OpenFOAM.
## 1 Problem specification
We want to study the motion of air circulating inside a room. The outline of the geometry is shown in Fig. 1. The room is crossed by a conduct. The air enters perpendicularly to one of the extremities of the conduct (inlet) at a velocity U = 0.5 m/s and flows out the other (outlet). There are two holes in the conduct and four small windows on one of the room walls through which the air can also flow. Two pillars are placed between the conduct and the wall with the windows.
### 1.1 Governing equations
As the case is steady-state and the fluid is supposed viscous and incompressible, the governing equations to be solved are:
Mass conservation for incompressible flow: ∇ ⋅ U = 0 Steady-flow momentum conservation: (U ⋅ ∇) U + (∇P/ρ) = ν ∇2U + g
where P, ρ, ν and g are, respectively, presuure, density, kinematic viscosity and gravitational field. We have already talked about the initial velocity above. The initial pressure has been set equal to 1 atm = 101325 Pa at all the boundaries except at the windows where its value is 0.995 atm to allow a pressure gradient.
### 1.2 Turbulence model
We choosed the standard k - ε turbulence model with coefficients Cμ = 0.09, C1 = 1.44, C2 = 1.92, α k = 1, α ε = 7.69. In this model the velocity of the turbulent flow is estimated as a fluctuation around an average value: U = ‹ U › + U'; k = (1/2) |U'|2 represents the turbulent kinetic energy per unit of mass and ε the energy dissipation rate. We assume the fluctuations at the inlet to be isotropic and 5% of U, thus Ux = Uy = Uz = 0.5 0.05 = 25 10-3 m/s and k = 9.375 10-4 m2/s2. The energy dissipation is related to the turbulence scale l which we estimated to be 10% of 0.5 m, which is the width of the inlet. The value of ε is thus set to ε = (C μ 0.75 k 3/2 )/l = 9.53 10 -5 m2/s3.
## 2 Mesh generation
The mesh was generated by converting a `.msh` GAMBIT file according to the procedure described in the section Mesh conversion. The domain consists of about 5000000 cells. In Fig. 2 are shown some of the meshed patches of the domain.
## 3 Boundary conditions and initial fields
The case requires initial and boundary conditions settings for all the involved fields: velocity U, pressure P, turbulent kinetic energy k and energy dissipation rate ε. The initial values for the fields are specified in the `casename/0/fieldname` files. The reader may consult the boundary conditions documentation section (in Italian) for more details. We report the rilevant sections of the `fieldname` files.
The initial velocity U has been set equal to 0.5 m/s in the y direction at the inlet: ``` dimensions [0 1 -1 0 0 0 0]; // dimension in m s -1 internalField uniform (0 0 0); boundaryField { inlet // inlet patch { type fixedValue; value uniform (0 0.5 0); // U x = U z = 0, U y = 0.5 m/s } ``` The initial turbulent kinetic energy has been set equal to k = 9.375 ⋅ 10-4 m2/s2 at the inlet: ``` dimensions [0 2 -2 0 0 0 0]; // dimension in m 2 s -2 internalField uniform 0.0009375; boundaryField { inlet // inlet patch { type fixedValue; value uniform 0.0009375; } ``` The initial dissipation energy rate has been set equal to ε = 9.53 ⋅ 10 -5 m2/s3 at the inlet: ``` dimensions [0 2 -3 0 0 0 0]; // dimension in m 2 s -3 internalField uniform 9.53e-05; boundaryField { inlet { type fixedValue; value uniform 0.0016875; } ``` The initial pressure field P has been set equal to 1 atm at the inlet and at all the outlets except at the windows where its value is 0.995 atm to allow internal flow circulation. The gradient of the pressure in the direction perpendicular to the the patches of type `wall` has been set equal to 0: ``` dimensions [0 2 -2 0 0 0 0]; dimension in m 2 s -2 internalField uniform 101325; boundaryField { wall_ceiling // wall type patch (the ceiling) { type zeroGradient; } inlet // inlet type patch (extremity of the conduct) { type fixedValue; value uniform 101325; } outlet_conduct // outlet type patch (extremity of the conduct) { type fixedValue; value uniform 1; } outlets_windows // outlet type patch (the windows) { type fixedValue; value uniform 108800; } `````` ```
## 4 Solver and case control
We used the `simpleFoam` solver, implemented for steady incompressible flow. The numerical schemes for terms such as derivatives, interpolation procedures etc. are set in the `casename/system/fvSchemes` file. In this file we specify that the case is steady-state by giving the `SteadyState` value to the keyword `timeScheme`. The terms ∇ and ∇ 2 controlled respectively by the keywords `gradSchemes` and `laplacianSchemes` have been set to the value `Gauss` because we adopted the standard finite volume discretization of Gaussian integration which requires the interpolation of values from cell centres to face centres. The divergence term ∇ , controlled by the keyword `divSchemes` has been set to `UD` to ensure boundedness.
## 5 Results
In Fig. 3 are shown the residuals of the involved fields after 1000 iterations. The case has been run in parallel using 64 cores.
Even if the maximum pressure gradient is only 0.5%, it is sufficient to cause a back flow with the air entering the room from the outlet as shown in Fig. 4.
The same simulation has been run with Fluent. In order to make a comparison we shall analyse the shape of the fields on two cutting planes, as shown in Fig. 5.
We observe that the magnitude of the velocity fields is of the same order. In the regions far from the two small holes placed on the conduct the velocity magnitude approaches to zero in both the two simulations. In the regions next to the conduct holes the velocity contour has a more diffusive shape in the OpenFOAM case. In Fig. 7 are shown the contour plots for the velocity components. An overall agreement is observed.
In Fig. 8 are shown the contour plots for the magnitude velocity field on the vertical plane. On average, the strength of the velocity fields at the conduct section are comparable. There is a nice agreement even for the size of the region where the velocity field doest not dump to zero.
In Fig. 8 are shown the pressure contour plots on the horizontal plane. The Fluent picture reports (in Pa) the pressure gradient with respect to 1 atm = 101325 Pa as reference value, thus the negative values mean a value of pressure lower tham 1 atm. On the other hand the OpenFOAM picture reports (in Pa) the absolute pressure values as shown in the dispayed range. Hence in this case too there is an overall agreement.
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# Trigonometric Functions! (Ferris Wheel)
5,054 results
Trigonometric Functions
A person is riding a ferris wheel that turns at a constant speeed. The lowest point of the ferris wheel is at ground level. Another person is standing at the side of the wheel on a platform 4m above the ground. She notes the times that the person on the wheel is at the same ...
Trigonometric Functions! (Ferris Wheel)
The Ferris wheel at a carnival has a diameter of 18m and descends to 2m above the ground at its lowest point. Assume that a rider enters a car at this point and rides the wheel for two revolutions My main problem is how to get the phase shift. (The answer says its 270 degrees)
Trigonometric Function
A Ferris wheel has a radius of 10 meters and is revolving 6 times each minute (wheel's frequency.) The wheel's center is 12 meters from the ground. If time starts on the ground, then find a sine function that shows the height from the ground for a car on the Ferris wheel at ...
Physics
A particular Ferris wheel takes you through one complete revolution every 25.3 seconds. If the radius of the Ferris wheel is 10.4 m, and your mass is 63.3 kg, calculate: (a) Your apparent weight, in N, when you are at the top of the Ferris wheel. (b) Your apparent weight, in N...
Physics
You and a friend are going to ride on a Ferris wheel. For a little extra fun, when your friend is at the very top of the Ferris wheel she's going to drop a tennis ball. How far around the Ferris wheel in radians from your friend should you sit so you can catch the ball? ...
Physics :(( help please
You and a friend are going to ride on a Ferris wheel. For a little extra fun, when your friend is at the very top of the Ferris wheel she's going to drop a tennis ball. How far around the Ferris wheel in radians from your friend should you sit so you can catch the ball? ...
trig question
A Ferris wheel has a deameter of 50m. The platform at the bottom, where you load the ferris wheel, is 3 m above the ground. The Ferris wheel rotates three times every two minutes. A stopwatch is started and you notice you are even with the center of the ferris wheel, going ...
physics
A ferris wheel bucket travels around at 3.40 m/s. The radius of the ferris wheel is 39.8 m. What is the ferris wheel's angular velocity?
calculus
Eric is standing on the ground, a distance of 70 ft from the bottom of Ferris wheel that has a 20 ft radius. His arm is at the same height as the bottom of the Ferris wheel. Janice is on Ferris wheel which makes one revolution counter clock wise every 16 secs. At the instant ...
pre-cal
A small ferris wheel: A student is riding a 20-m-diameter ferris wheel that is making three revolutions per minute. If there are eight equally spaced seats on the Ferris wheel, then what is the length of the arc between two adjacent seats?
math trig
A Ferris wheel with a diameter of 37 meters rotates at a rate of 4 minutes per revolution. Riders board the Ferris wheel 4 meters above the ground at the bottom of the wheel. A couple boards the Ferris wheel and the ride starts. Write a formula for the height of the couple t ...
Pre-Calc Trig
Create both a sine and cosine model for height of a passenger off of the ground for each of the following Ferris wheels. 1) customers must climb up 12 foot steps to get into the Ferris wheel (I.e) bottom of Ferris wheel is at the top of the steps Diameter 67ft Rotional speed 1...
Math
A Ferris wheel has a diameter of 60 feet. When you start at the bottom of the Ferris wheel, you are 2 feet from the ground.The Ferris wheel completes one rotation in 2 minutes. Create a graph that represents your height relative to the center of the Ferris wheel as a function ...
Math
Tony visits the local fair and sees one if therides, the ferris wheel. The ferris wheel has a diameter of 50 feet and is on a platform of 4 feet. if it takes 12 seconds to make one full revolution, what is the equation of the height of a person on the ferris wheel at any time...
Trig
A ferris wheel has a 14 meter diameter and turns counterclockwise at 6 rpm. a) Assuming that the center of the wheel is the origin of an x-y coordinate plane, write functions to find the position (x,y) of a rider that starts at the bottom of the wheel. b) Where is the rider 2 ...
Honors Pre Calculus
Tony visits the local fair and sees one of the rides, the ferris wheel. The ferris wheel has a diameter of 50 feet and is on a platform of 4 feet. If it takes 12 seconds to make one full revolution, what is the equation of the height of a person on the ferris wheel at any time t?
Advanced Functions
The vertical position, h, in metres, of a gondola on a Ferris wheel is modelled using the function h = 10 sin ((pi/15) (t-d)) + 12 where t is the time, in seconds. The gondola must assume a vertical position of 7 m at t=0 s , and be on its way upward. Determine the phase shift...
pre-cal
A small ferris wheel, a student is riding a 20-m-diameter ferris wheel that is making 3 revolutions per minute. What is the area of the sector between 3 of the seats (3/8) of the wheel?
pre cal
A small ferris wheel, a student is riding a 20-m-diameter ferris wheel that is making 3 revolutions per minute. What is the area of the sector between 3 of the seats (3/8) of the wheel?
Physics
Fairgoers ride a Ferris wheel with a radius of 5.00 {\rm m} . The wheel completes one revolution every 31.5 s What is the average speed of a rider on this Ferris wheel? If a rider accidentally drops a stuffed animal at the top of the wheel, where does it land relative to the ...
math
The height of a person above the ground on a Ferris wheel is given by the function h(t)=9 sin {pi/20 (t- pi/2))+ 12 where h(t) is the height in meters of person t seconds after getting on a) how long does it take the Ferris wheel to make one complete revolution b)what is the ...
Trig application
I'm having trouble with this trig application. The scenario goes: A ferris wheel has a diameter of approximately 65 meters. Assume it takes 110 seconds for the ferris wheel to make one complete rotation. find the angular speed of the ferris wheel in radians per minute and ...
Physics
A 2000kg ferris wheel accelerates from rest to an angular speed of 2.0 rad/s in 12 secs. Approximate the ferris wheel as a circular disk with a radius of 30m. what iss the net torque on the wheel
Integers
Suppose Frontier City Amusement Park bought a ferris wheel for $92,000. 42,512 people paid$2.00 each to ride the ferris wheel the first year. Ignoring the cost of operating how much money did the park lose on the ferris wheel it first year.
Math
A ferris wheel has a diameter of 10 m and takes 24 sec to make one revolution. The lowest point on the wheel is 1 m above the ground. (4 marks) a. Sketch a graph to show how a rider's height above the ground varies with time as the ferris wheel makes a rotation. Assume the ...
5th grade math (word problems)
I'm not sure how to solve this: A group of friends went to an amusement park. 10 of them rode the ferris wheel, 15 rode carousel, and 11 rode the roller coaster. 7 of them rode both the ferris wheel & roller coaster. 4 rode both the ferris wheel & carousel. 5 rode both the ...
College Algebra
A ferris wheel has a diameter of 320 feet and the bottom of the Ferris wheel is 9 feet above the ground. Find the equation of the wheel if the origin is placed on the ground directly below the center of the wheel.
math
a ferris wheel has the diameter of 240 feet and the bottom of the ferris wheel is 9 feet above the ground. find the equation of the wheel if the origin is placed on the ground directly below the center of the wheel
7th grade math
A group of 32 students went to the amusement park.All of them either rode the Ferris wheel or Roller coaster.Seventeen rode the Ferris wheel,23 rode the roller coaster,and 8 did both.How many rode only the Ferris wheel
Math (Trig)
a ferris wheel has a radius of 10m and is one meter above the ground. If the ferris wheel makes 1 revolution every 20 seconds, write an equation that gives the height above the ground of a person on the ferris wheel as a function of time if that person starts (t=0) 1/8th of a ...
Physics
At a county fair, a boy takes his teddy bear on the giant Ferris wheel. Unfortunately, at the top of the ride, he accidentally drops his stuffed buddy. The wheel has a diameter of 14.6 m, the bottom of the wheel is 1.1 m above the ground and its rim is moving at a speed of 1.0...
Physics
At a county fair, a boy takes his teddy bear on the giant Ferris wheel. Unfortunately, at the top of the ride, he accidentally drops his stuffed buddy. The wheel has a diameter of 14.6 m, the bottom of the wheel is 2.4 m above the ground and its rim is moving at a speed of 1.0...
Science
At a county fair, a boy takes his teddy bear on the giant Ferris wheel. Unfortunately, at the top of the ride, he accidentally drops his stuffed buddy. The wheel has a diameter of 15.3 m, the bottom of the wheel is 3.37 m above the ground and its rim is moving at a speed of 1....
Physic
Me and my friend are playing on a Ferris wheel which has Radius of 20 Meter and had an constant angular velocity of 0.2 rad/s . Then, When I'm at the very top of the Ferris wheel i'm going to drop a tennis ball( dropping without initial velocity). How far around the Ferris ...
Physics
In a ferris wheel, one person is at the top and another at the bttom. The ferris wheel rotates such that the magnitude of the tangetial velocity of each person is equal |v|top=|v|bottom. Determine for the person at the bottom, the force the seat exerts on the person. Use a ...
pre-calculus
A.) A Ferris wheel has a radius of 32 feet and completes one revolution every 5.4 minutes. What is the speed of the Ferris wheel (in radians per minute)? _______radians per minute B.) A Ferris wheel has a radius of 39 feet and travels 5.6 feet every second. What is the speed ...
maths
diameter of a ferris wheel is 15 metres. how far does a seat on the ferris wheel travel in 5 revolutions
math
the maximun height of a ferris wheel is 35m. the wheel takes 2 mins to make one revolution. passengers board the ferris wheel 2m above the ground at the bottom of the rotation. a. write an equation to model this b. how high is the passengers after 45s c. draw a graph
trig
A ferris wheel with a diameter of 100 feet rotates at a constant rate of 4 revolutions per minute. Let the center of the ferris wheel be at the origin. 1. Each of the ferris wheel's cars travels around a cirlce. a) Write an equation of the circle where x and y are measured in ...
CHHS
A a ferris wheel has a diameter of 100ft. The passengers ride in cars that make 5 revolutions in 7.5 minutes. Approximately how far does a car on the Ferris wheel travel in 1 minute
geometry
A person riding the original Ferris Wheel is traveling at a speed of 0.495 mph. If one revolution takes 9 minutes, what is the diameter of the Ferris Wheel? Show your work.
Trigonometry
A carnival Ferris wheel with a radius of 7 m makes one complete revolution every 16 seconds. The bottom of the wheel is 1.5 m above the ground. The ride starts at the bottom. Find the sinusoidal function that models this Ferris wheel in terms of time (t) in seconds.
Math
A carnival ferris wheel that has a diameter of 18m, goes 4 seconds every revolution, there are 5 revolutions, also the ferris wheel is 2m above the ground Cosine and Sine equation needed
physics
A Ferris wheel with radius 8.8 m rotates at a constant rate, completing one revolution every 34.6 s. Suppose the Ferris wheel begins to decelerate at the rate of 0.212 rad/s2 when a passenger is at the top of the wheel. Find the magnitude and direction of the passenger's ...
Advanced Functions
the ferris wheel at an amusement park measures 16m in diameter. the wheel does 3 rotations every minute. the bottom of the wheel is 1m above the ground... a) determine the simplest equation that models Megan's height above that ground(h) over time (t). give 2 more equations ...
math
The Ferris wheel at the county fair is 30 feet tall. If a person rides the Ferris wheel around 4 times, how far has he traveled? 60 ft 120 ft 90.8 ft 376.8 ft please answer as soon as possible
Physics
While riding a Ferris wheel, the rider determines that the Ferris wheel makes 1.5 revolutions per minute. a) knowing that the diameter of the Ferris wheel is 100 ft, determine the angular speed (in rad/s) of the Ferris wheel. b) Determine the linear speed (in ft/s) of the ...
A physics question
I was researching amusement park rides for a recent project (from this site I have noticed that many others are in the same boat as me). Anywho, I am doing the Ferris wheel or the Giant Wheel, and I was wondering if this was right, with respect to potential energy: this is due...
Advanced Functions
the ferris wheel at an amusement park measures 16m in diameter. the wheel does 3 rotations every minute. the bottom of the wheel is 1m above the ground... a) determine the simplest equation that models Megan's height above that ground(h) over time (t). give 2 more equations ...
Math (IB Calc)
A ferris wheel has a diameter of 50 meters. If point A is at the lowest point of the ferris wheel, 0 meters. What is the height of point A if the wheel moves 2pi/3 radians clockwise? What is the equation?
college trig word problem
A Ferris wheel has a radius of 25 feet.The wheel is rotating at two revolutions per minute.Find the linear speed, in feet per minute, of a seat on this ferris wheel.
Math
The first ferris wheel was built in1893 in Chicago. It's diameter was 250 feet.how many feet did the ferris wheel rotate with each complete turn?
Math
A Ferris wheel has a radius of 30 feet and travels 3.6 feet every second. How long does it take for the Ferris wheel to complete one revolution?
Math ( trig )
If a person on the Ferris wheel is 45 feet above the ground. The Gris wheel has a radius of 47.f and is 5 feet above the ground. At what degrees had the Ferris wheel rotated counted clockwise?
Physics
In order to stop a Ferris wheel, a force is applied that accelerates at a rate of -0.010rad/s^2. How long will it take the Ferris wheel to stop?
physics
Please help with these problems. Thanks! Suppose the circumference of a bicycle wheel is 2 meters. If it rotates at 1 revolution per second when you are riding the bicycle, then your speed will be? Suppose you're on a Ferris wheel at a carnival, seated 10 m from the Ferris ...
math
A ferris wheel is 35 meters in diameter,and can be boarded at ground level. The wheel turns in a counterclokcwise direction, completing one full revolution every 5 minutes. Suppose that a t=0 you are in the three o'clock position. Write a formula, using the sine function for ...
trig
The ferris wheel in an amusement park. It has a diameter of 24m, and it takes 40s to make one complete revolution. If Peter gets on a gondola which is vertically below the centre of the ferris wheel, find his rise in the height after 5s.
Physics
Please help with these problems, I don't understand them. Thanks! Suppose the circumference of a bicycle wheel is 2 meters. If it rotates at 1 revolution per second when you are riding the bicycle, then your speed will be? Suppose you're on a Ferris wheel at a carnival, seated...
physics
A Ferris wheel at an amusement park having a diameter of 14 m completes one rotation in 45 s. (a) What is the rotational speed of this Ferris wheel in revolution per minute and in degrees per second?
Physics
A Ferris wheel with radius 11.0 m rotates at a constant rate, completing one revolution every 33.9 s. Suppose the Ferris wheel begins to decelerate at the rate of 0.227 rad/s2 when a passenger is at the top of the wheel. Find the magnitude and direction (in degrees below ...
n
A ferris wheel is 20 meters in diameter and boarded from a platform that is 1 meters above the ground. The six o'clock position on the ferris wheel is level with the loading platform. The wheel completes 1 full revolution in 6 minutes. The function h = f(t) gives your height ...
precalculus
A ferris wheel is 20 meters in diameter and boarded from a platform that is 3 meters above the ground. The six o'clock position on the ferris wheel is level with the loading platform. The wheel completes 1 full revolution in 10 minutes. The function h = f(t) gives your height ...
Math Help Please I Don't Understand
A Ferris wheel has a diameter of 60 feet. When you start at the bottom of the Ferris wheel, you are 2 feet from the ground. The Ferris wheel completes one rotation in 2 minutes. Create a function that represents your height relative to the center of the Ferris wheel as a ...
Physics
A Ferris wheel with radius 9.8 m rotates at a constant rate, completing one revolution every 36.6 s. Suppose the Ferris wheel begins to decelerate at the rate of 0.227 rad/s2 when a passenger is at the top of the wheel. Find the magnitude and direction of the passenger's ...
Physics
A Ferris wheel with radius 9.8 m rotates at a constant rate, completing one revolution every 36.6 s. Suppose the Ferris wheel begins to decelerate at the rate of 0.227 rad/s2 when a passenger is at the top of the wheel. Find the magnitude and direction of the passenger's ...
Science
The rider to a Ferris wheel moves at a rate of 0.18m/s2. If the rider is 5.0m from the center of the Ferris wheel, what is his/her velocity?
Pre Algebra
Tickets on the ferris wheel cost .50 cents. At least how many tickets were sold? There are 36 cars on the ferris wheel that holds 60 people per car. Suppose every ticket holder rode for two turns of the wheel. At least how many turns did the wheel have to make? I think it's ...
Math
A carnival Ferris wheel with a radius of 7 m makes one complete revolution every 16s. The bottom of the wheel is 1.5 m above the ground. Find the amplitude, vertical shift, phase shift and k value and period. Do all calculations and then make your final equation. Predict how ...
trig
1. A Ferris wheel with a radius of 7m makes one complete revolution every 16 s. The bottom of the wheel is 1.5m above ground. a)Find the equation of the graph b)predict how the graph and the equation will change if the Ferris wheel turns more slowly c) test your predictions ...
math
A ferris wheel has radius of 25 m and its centre is 27 m above the ground. It rotates once every 40 seconds. Sandy gets on the ferris wheel at its lowest point and the wheel starts to rotate. Determine a sinusodial equation that gives her height, h, above the ground as a ...
Physics
A Ferris wheel has a circumference of 93 m and it completes one rotation in 2.3 minutes without stopping. What is the percentage change in apparent weight (=weight difference/weight = W/W) of a passenger between the highest and the lowest positions on this Ferris wheel?
Math
At the amusement park, you decide to ride the Ferris wheel, which has a maximum height of 50 meters and a diameter of 35 meters. It takes the wheel 5 minutes to make one revolution. If you start your ride at the midline and the ferris wheel rotates counter clockwise, how many ...
Math - need help ASAP!! Please!
The question relates to application of sine functions. -------------- A carnival Ferris wheel with a radius of 7m makes one complete revolution every 16 seconds. The bottom of the wheel is 1.5m above the ground. a) Draw a sketch to show a person's height (h) above the ground ...
Math
At the amusement park, you decide to ride the Ferris wheel, which has a maximum height of 100 meters and a diameter of 50 meters. It takes the wheel seven minutes to make one revolution. If you start your ride at the midline and the ferris wheel rotates counter clockwise, how ...
Math, Trigonometry
The height of a person on a ferris wheel is measured by function h(t)=38 sin 3.5 (t-4.0)+41.? h= height of person in feet t=represents the time in minutes. a) At what height, to the nearest tenth, are passengers loaded onto the Ferris wheel ? what characteristics of the graph ...
Trig - Ferris Wheel
A ferris wheel is 250 ft in diameter and revolves every 40 second when in motion. You step up to the seat on the wheel at the bottom 2 feet about the fround so you are sitting 4 feet about the ground at start. Derive the formula for the height of your seat at time (t). Then ...
Math
Billy is riding on a 20 meter diameter Ferris wheel that is making 3 revolutions per minute. What are your linear velocity and your angular velocity? If there are 8 equally spaced seats on the ferris wheel then what is the length of the arc between two adjacent seats?
Trig
A Ferris wheel is 40 meters in diameter and boarded from a platform that is 4 meters above the ground. The six o'clock position on the Ferris wheel is level with the loading platform. The wheel completes 1 full revolution in 2 minutes. How many minutes of the ride are spent ...
precalculus
A Ferris wheel is 10 meters in diameter and boarded from a platform that is 3 meters above the ground. The six o'clock position on the Ferris wheel is level with the loading platform. The wheel completes 1 full revolution in 4 minutes. How many minutes of the ride are spent ...
physics
a 60kg man is on a steadily rotating Ferris wheel. the man is standing on a bathroom scale and the scale reads 45kg at the top of the wheel. take g=10m/s^2 (a) what does the scale read when the man is at the bottom of the wheel? (b)what does the scale read when the man is ...
math
At the amusement park, you decide to ride the Ferris wheel, which has a maximum height of 50 meters and a diameter of 35 meters. It takes the wheel three minutes to make one revolution. If you start your ride at the midline and the ferris wheel rotates counter clockwise, how ...
Pre-Cal
A Ferris wheel is 40 meters in diameter and boarded from a platform that is 5 meters above the ground. The six o'clock position on the Ferris wheel is level with the loading platform. The wheel completes one full revolution every 7 minutes. You make two complete revolutions on...
math
a ferris wheel has a radius of 8m and rotates every 12 hrs. THe bottom of the ferris wheel is 1m above the ground. draw a graph describing how a person's height above the ground varies with time. Find an qeation for your graph.
Trig
As you ride a ferris wheel, your distance from the ground varies sinusoidally with time. When the last passenger boards the ferris wheel and the ride starts moving, let your position be modeled by the diagram provided. Let t be the number of seconds that have elapsed since you...
Physics
A Ferris wheel, rotating initially at an angular speed of 0.500 rad/s, accelerates over a 7.00-s interval at a rate of 0.04 rad/s^2. What angular displacement does the Ferris wheel undergo in this 7-s interval?
math
A Ferris wheel with a diameter of 100 feet rotates at a constant rate of 4 revolutions per minute. Let the center of the Ferris wheel be at the origin. The height h (in feet) of a Ferris wheel car located at the point (x, y) is given by h = 50 + y where y is related to the ...
Math
A ferris wheel is 30 meters in diameter and boarded in the six o'clock position from a platform that is 10 meters above the ground. The wheel completes one full revolution every 10 minutes. At the initial time t=0 you are in the twelve o'clock position.Find a formula, using ...
Math-trig
A Ferris wheel with a diameter of 100 feet rotates at a constant rate of 4 revolutions per minute. Let the center of the Ferris wheel be at the origin. The height h (in feet) of a Ferris wheel car located at the point (x, y) is given by h = 50 + y where y is related to the ...
Math(Correction)
Use the given function values and trigonometric identities (including the relationship between a trigonometric function and its cofunction of a complementary angle) to find the indicated trigonometric functions. Sorry for the mistake. They are two separate functions and i need...
precalculus
A Ferris wheel has a radius of 35 feet, and is rotating at 5 revolutions per minute. Find the linear speed, in feet per minute, of a seat on the Ferris wheel.
math
At the amusement park, you decide to ride the Ferris wheel, which has a maxmum height of 80 meters and a diameter of 40 meters. It takes the wheel seven minutes to make one revolution. If you start your ride at the midline and the ferris wheel rotates counter clockwise, how ...
Math
1. What percent of people chose the Ferris wheel or the carousel as their favorite ride? Other 19% Ferris wheel 18% Carousel 25% Roller coaster 38% A. 25% B. 37% C. 43% < -- C? D. 56% 2. Two hundred people were surveyed about their favorite ride at an amusement park. The ...
Honor Physics
Fairgoers ride a Ferris wheel with a radius of 5.00 . The wheel completes one revolution every 33.0 . If a rider accidentally drops a stuffed animal at the top of the wheel, where does it land relative to the base of the ride? (Note: The bottom of the wheel is 1.75 above the ...
Math (please help!)
The largest Ferris Wheel in the world is the London Eye in England. The height (in metres) of a rider on the London Eye after t minutes can be described by the function h(t) = 67 sin[0.2094(t-30)] + 70 Using this equation i'm supposed to find the diameter, the maximum (top of ...
trig
Brian is riding a Ferris wheel. The wheel has a radius of 25 feet, and at his lowest point, Brian is 8 feet off the ground. Brian times how long it takes to travel from the lowest point to the highest point and finds that it takes 8 seconds. Write a sinusoidal equation to ...
math
Matthew is riding a ferris wheel at a constant speed of 10km/h. The boarding height for the wheel is 1m, and the wheel has a radius of 7m. What is the equation for the function that describes Matthews height in the terms of time, assuming Matthew starts at the highest point on...
Math
The rim of the London Eye (a 135m diameter ferris wheel) moves 26 cm/sec, slow enough for passengers to safely get on the wheel from the platform (2 meters above ground level) without stopping the wheel at the bottom of its rotation. What's the height at the bottom of the ...
Science
The rider of a ferris wheel moves at a rate of 0.18 m/s2. If the rider is 5.0 m from the center of the ferris wheel, What is his/her Velocity? Can't figure out if the RADIUS and RATE are the same and Which one is the radius/rate?
Trig Question Help plz
A ferris wheel has a radius of 25 feet. A person takes a seat, and the wheel turns 5pie/6 radians. How far is the person above the ground? please explain to me how to solve this We are to suppose the person got on the wheel at the very bottom. You might want to convert this to...
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# IMPACT & COLLISION
1. Two particles of masses 10 kg and 20 kg are moving along a straight line towards each other
at velocities of 4m/s and 1 m/s respectively. If e = 0.6, determine the velocities of the particles
immediately after their collision and loss of K.E.
Ans: v = 4/3 m/s & 5/3 m/s in opposite direction & 53.33J
2. Three balls A, B and C of masses 2kg, 4kg and 8kg respectively move along the same straight
line and in the same direction, with velocities 4m/sec, 1m/sec and 0.75 m/s. If ball A collides
with ball B and ball B collides with C show that the balls A and B will be brought to rest by the
collision which will take place. Take coefficient of restitution as unity.
Ans: Vc = 2.25 m/s
A B C
2 Kg 4 Kg 8 Kg
## 3. Two identical balls each of mass m collide with A B
velocities as shown in figure. Find the final
velocities of the balls after impact. 1.6 m/s
450
Assume e=0.9
3.5 m/s
Ans. : - 2.27 m/sec, 2.84 m/sec 60.450with
Horizontal.
## 4. Two smooth balls A (mass 3 kg) and B (mass 4 600
kg) are moving with the velocities 25 m/sec and 40 m/s
40 m/sec respectively. Before impact, the
direction of velocity of two balls are 300 and 600
with the line joining the centers as shown in
figure. If e = 0.8, find the magnitude and direction
of velocities of these balls after impact.
Ans : 24.58m/s = 59.44, 36.7m/s = 19.30
25m/s
300
## 5. Two billiard balls of equal masses collide with
velocities v1 = 1.5 m/s and v2 = 2 m/s as shown in V1
figure. Find velocities of balls after impact and
percentage loss in K.E. Coefficient of restitution
is 0.9. 600
Ans.: For ball no. 1 v = 0.875 m/s and for ball no. 2
V2
v = 2.21 m/s, = 51.550, 9. 6 %
2 m/s A
6. Two identical balls of 120 gm collide when they
are moving with velocity perpendicular to each
other as shown in figure. Assuming that the line
of impact is in the direction of motion of ball B,
determine the velocity of ball A and ball B B
completely after the impact. Take e = 0.8.
Ans. : VA = 6.96m/s, VB = 0.6m/s
6 m/s
7. A ball is dropped from a height of 9m upon a horizontal slab. If it rebounds to a height of
6meters, show that the coefficient of restitution is 0.82
8. A boy throws a ball vertically downwards. He wants the ball to rebound from floor and just
touch the ceiling of room which is at a height of 4m from ground. If coefficient of restitution e
is 0.8, the velocity with which the ball should strike the floor taking g = 10m/s 2 is
Ans. : 11.18m/s
## 9. A ball is thrown against a wall with a velocity v forming an angle
of 300 with the horizontal. Assuming friction less conditions
and e = 0.50 determine the magnitude and direction of the
velocity of the ball as it rebounds from the wall.
Ans. : 0.661v, 49.10
## 10. A billiard ball moving 4m/sec strikes a smooth
horizontal plane at an angle of 350 as shown in fig. If the
coefficient of restitution is 0.6, what is the velocity with 4 m/s
which the ball rebounds?
Ans. : 3.55m/sec, 22.80
350
1.2 m
1. 2 m
11. A sphere of a mass 2 kg is released from rest and
strikes a block of mass 2.5 kg resting on a horizontal
surface. How far the block will move after the impact?
Assume e = 0.75 and the coefficient of friction between
the block and the floor is 0.25.
Ans. : Vb=3.77m/s , 2.9m
## 12. Figure shows spheres A and B suspended by cords 2.5
m and 2.0m long respectively. Masses of spheres A and
B are 1.5 kg and 2 kg respectively. Sphere A is pulled to
a position ‘A’, 500 mm above A and released from rest.
Sphere B is at rest when struck by A with direct impact.
After impact the sphere B rises to a height of 300 mm
above its lowest point. Determine the coefficient of 500 mm
restitution.
Ans. : 0.808 300
mm
A O
13. A simple pendulum OA when released from rest in the
horizontal position falls under gravity and strikes a
vertical wall at B as shown in figure. If the coefficient of
restitution between the wall and the ball is 0.5, find the Ɵ
angle defining the total rebound of the ball.
Ans.: = 41.420
A B
14. A ballistic pendulum consisting of a block of mass 30 kg
suspended from two wires each of length 1.8 m is used
to measure the muzzle velocity of the gun. If the
pendulum swings through a horizontal distance s = 25 1.8m
cm when a 40 gm bullet is fired into it, determine the
muzzle velocity v of the gun. u
Ans. : 439 m/sec
## 15. A 2kg mass falls 150mm onto a 1 kg platform mounted
on springs whose combined k = 730N/m. If the impact is 2 kg
fully plastic, determine the maximum distance the
platform moves down from its position before impact? 150 mm
Case 1: Considering pre-compression of spring.
Case 2: Neglecting pre-compression of spring.
Ans. : 121 mm
K= 730 N/m
750 kg
1.2 m
16. A 750 kg hammer of a drop hammer pile driver falls from
a height of 1.2m onto the top of a pile of mass 2250 kg.
The pile is driven 100mm into the ground. Assuming
perfectly plastic impact, determine the average
resistance of the ground to penetration. 2250 kg
Ans. : RestAvg= 51.5 KN
100 mm
17. A ball is thrown downwards with a velocity of 12m/s and at a 30 0 with the horizontal from the
top of a building 12m high. Find where the ball will hit the ground second time if coefficient of
restitution between ball and ground is e = 0.75.
Ans. : x = 37.3m
## 18. Determine the horizontal velocity at which vA
the girl must throw the ball so that it
bounces once on the smooth surface and
then lands into the cup at C. Take e=0.6. 3m
Neglect the size of the cup.
Ans: V = 12.375 m/s
8m
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# Math
posted by on .
I just need a description of the following:
Associative property of Multiplication
Commutative properties
Distributive Property
• Math - ,
associative a+(b+c) = (a+b) + c
commutative a+b = b+a
identity : there is a number 0 with
a+0=0+a = a
inverse a + -a = -a + a = 0
MULTIPLICATION:
associative a(bc) = (ab)c
commutative ab=ba
identity there is a number 1 with
a*1 = 1*a = a
inverse If a not zero then there is an 1/a such that a(1/a) = (1/a)a = 1
BOTH (the biggie!!)
distributive a(b+c) = ab + ac
• Math - ,
thank u soooooooooooooooo much, i really needed to understand this.
• Math - ,
Associative property of Multiplication:
(x*y)*z = x*(y*z)
(x + y) + z = x + (y + z)
Commutative properties:
x*y = y*x
x + y = y + x
Distributive Property :
x*(y + z) = x*y + x*z
Identity property: There exists a number 1 such that:
1*x = x for all x.
Zero property: There exists a number 0 such that:
0 + x = x for all x
Inverse properties:
For every x there exists a number -x, such that:
x + (-x) = 0
For every x not equal to zero there exists a number x^(-1), such that:
x*x^(-1) = 1
Examples:
There can only be one 0. Proof suppose there were two numbers 0 and 0' that both satisfy the property that the zero element has to satsify, then:
0 + 0' = 0'
because 0 is a zero element.
But because 0' is also a zero element, you also have:
0 + 0' = 0
This means that 0 = 0'
Example:
(-1)*x = -x
Proof:
Let's check of (-1)*x satisfies the criterium of being the inverse (relative to addition) of x:
x + (-1)*x =
1*x + (-1)*x =
(1 + (-1))*x = (use that -1 is the inverse relative to addition of 1)
0*x = 0
So, we can conclude that (-1)*x is the inverse of x relative to addition, which means that (-1)*x = -x
• Math - ,
thank u soooooo much!
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# Search by Topic
#### Resources tagged with Area similar to Symmetric Angles:
Filter by: Content type:
Stage:
Challenge level:
##### Other tags that relate to Symmetric Angles
Radius (radii) & diameters. Area. Investigations. Scale factors. Perimeters. Fractal. Inscribed circle. Logo. Circles. Angles. Similarity.
### There are 94 results
Broad Topics > Measures and Mensuration > Area
### Overlapping Squares
##### Stage: 2 Challenge Level:
Have a good look at these images. Can you describe what is happening? There are plenty more images like this on NRICH's Exploring Squares CD.
### A Day with Grandpa
##### Stage: 2 Challenge Level:
Grandpa was measuring a rug using yards, feet and inches. Can you help William to work out its area?
### Triangle Island
##### Stage: 2 Challenge Level:
You have pitched your tent (the red triangle) on an island. Can you move it to the position shown by the purple triangle making sure you obey the rules?
### Approaches to Area
##### Stage: 1 and 2
This article for teachers gives some food for thought when teaching ideas about area.
### Circle Panes
##### Stage: 2 Challenge Level:
Look at the mathematics that is all around us - this circular window is a wonderful example.
### Numerically Equal
##### Stage: 2 Challenge Level:
Can you draw a square in which the perimeter is numerically equal to the area?
### Being Resourceful - Primary Measures
##### Stage: 1 and 2 Challenge Level:
Measure problems at primary level that require careful consideration.
### Being Collaborative - Primary Measures
##### Stage: 1 and 2 Challenge Level:
Measure problems for primary learners to work on with others.
### Being Curious - Primary Measures
##### Stage: 1 and 2 Challenge Level:
Measure problems for inquiring primary learners.
### Being Resilient - Primary Measures
##### Stage: 1 and 2 Challenge Level:
Measure problems at primary level that may require resilience.
### Lawn Border
##### Stage: 1 and 2 Challenge Level:
If I use 12 green tiles to represent my lawn, how many different ways could I arrange them? How many border tiles would I need each time?
### Transformations on a Pegboard
##### Stage: 2 Challenge Level:
How would you move the bands on the pegboard to alter these shapes?
### It Must Be 2000
##### Stage: 2 Challenge Level:
Here are many ideas for you to investigate - all linked with the number 2000.
### Cylinder Cutting
##### Stage: 2 and 3 Challenge Level:
An activity for high-attaining learners which involves making a new cylinder from a cardboard tube.
### From One Shape to Another
##### Stage: 2
Read about David Hilbert who proved that any polygon could be cut up into a certain number of pieces that could be put back together to form any other polygon of equal area.
### Cutting it Out
##### Stage: 1 and 2 Challenge Level:
I cut this square into two different shapes. What can you say about the relationship between them?
### Shape Draw
##### Stage: 2 Challenge Level:
Use the information on these cards to draw the shape that is being described.
### Different Sizes
##### Stage: 1 and 2 Challenge Level:
A simple visual exploration into halving and doubling.
### Dicey Perimeter, Dicey Area
##### Stage: 2 Challenge Level:
In this game for two players, you throw two dice and find the product. How many shapes can you draw on the grid which have that area or perimeter?
### Through the Window
##### Stage: 2 Challenge Level:
My local DIY shop calculates the price of its windows according to the area of glass and the length of frame used. Can you work out how they arrived at these prices?
### Paper Halving
##### Stage: 1 and 2 Challenge Level:
In how many ways can you halve a piece of A4 paper? How do you know they are halves?
### Perimeter Possibilities
##### Stage: 3 Challenge Level:
I'm thinking of a rectangle with an area of 24. What could its perimeter be?
### Tiling Into Slanted Rectangles
##### Stage: 2 and 3 Challenge Level:
A follow-up activity to Tiles in the Garden.
### Area and Perimeter
##### Stage: 2 Challenge Level:
What can you say about these shapes? This problem challenges you to create shapes with different areas and perimeters.
### Growing Rectangles
##### Stage: 3 Challenge Level:
What happens to the area and volume of 2D and 3D shapes when you enlarge them?
### Changing Areas, Changing Perimeters
##### Stage: 3 Challenge Level:
How can you change the area of a shape but keep its perimeter the same? How can you change the perimeter but keep the area the same?
### Extending Great Squares
##### Stage: 2 and 3 Challenge Level:
Explore one of these five pictures.
### Tiles in the Garden
##### Stage: 2 Challenge Level:
How many tiles do we need to tile these patios?
### Great Squares
##### Stage: 2 and 3 Challenge Level:
Investigate how this pattern of squares continues. You could measure lengths, areas and angles.
### Rope Mat
##### Stage: 2 Challenge Level:
How many centimetres of rope will I need to make another mat just like the one I have here?
### Warmsnug Double Glazing
##### Stage: 3 Challenge Level:
How have "Warmsnug" arrived at the prices shown on their windows? Which window has been given an incorrect price?
### Pebbles
##### Stage: 2 and 3 Challenge Level:
Place four pebbles on the sand in the form of a square. Keep adding as few pebbles as necessary to double the area. How many extra pebbles are added each time?
### Can They Be Equal?
##### Stage: 3 Challenge Level:
Can you find rectangles where the value of the area is the same as the value of the perimeter?
### Fencing Lambs
##### Stage: 2 Challenge Level:
A thoughtful shepherd used bales of straw to protect the area around his lambs. Explore how you can arrange the bales.
### Square Areas
##### Stage: 3 Challenge Level:
Can you work out the area of the inner square and give an explanation of how you did it?
### The Big Cheese
##### Stage: 2 Challenge Level:
Investigate the area of 'slices' cut off this cube of cheese. What would happen if you had different-sized block of cheese to start with?
### Pie Cuts
##### Stage: 3 Challenge Level:
Investigate the different ways of cutting a perfectly circular pie into equal pieces using exactly 3 cuts. The cuts have to be along chords of the circle (which might be diameters).
### The Pillar of Chios
##### Stage: 3 Challenge Level:
Semicircles are drawn on the sides of a rectangle ABCD. A circle passing through points ABCD carves out four crescent-shaped regions. Prove that the sum of the areas of the four crescents is equal to. . . .
### Tilted Squares
##### Stage: 3 Challenge Level:
It's easy to work out the areas of most squares that we meet, but what if they were tilted?
### Shear Magic
##### Stage: 3 Challenge Level:
What are the areas of these triangles? What do you notice? Can you generalise to other "families" of triangles?
### Torn Shapes
##### Stage: 2 Challenge Level:
These rectangles have been torn. How many squares did each one have inside it before it was ripped?
### F'arc'tion
##### Stage: 3 Challenge Level:
At the corner of the cube circular arcs are drawn and the area enclosed shaded. What fraction of the surface area of the cube is shaded? Try working out the answer without recourse to pencil and. . . .
### Carpet Cuts
##### Stage: 3 Challenge Level:
You have a 12 by 9 foot carpet with an 8 by 1 foot hole exactly in the middle. Cut the carpet into two pieces to make a 10 by 10 foot square carpet.
### Covering Cups
##### Stage: 3 Challenge Level:
What is the shape and dimensions of a box that will contain six cups and have as small a surface area as possible.
### More Transformations on a Pegboard
##### Stage: 2 Challenge Level:
Use the interactivity to find all the different right-angled triangles you can make by just moving one corner of the starting triangle.
### Triangle Relations
##### Stage: 2 Challenge Level:
What do these two triangles have in common? How are they related?
### Tiling
##### Stage: 2 Challenge Level:
An investigation that gives you the opportunity to make and justify predictions.
### Kite
##### Stage: 3 Challenge Level:
Derive a formula for finding the area of any kite.
### Shaping It
##### Stage: 1 and 2 Challenge Level:
These pictures were made by starting with a square, finding the half-way point on each side and joining those points up. You could investigate your own starting shape.
### Place Your Orders
##### Stage: 3 Challenge Level:
Can you rank these sets of quantities in order, from smallest to largest? Can you provide convincing evidence for your rankings?
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## 12.3 – Logarithmic Functions
### Learning Objectives
• (12.3.1) – Define logarithmic functions
• Convert between logarithmic and exponential forms
• (12.3.2) – Evaluate Logarithms
• Evaluate logarithms without a calculator
• Define and evaluate natural logarithm
• Define and evaluate common logarithm
• (12.3.3) – Graph logarithmic functions
• Identify the domain of a logarithmic function
Figure 1. Devastation of March 11, 2011 earthquake in Honshu, Japan. (credit: Daniel Pierce)
In 2010, a major earthquake struck Haiti, destroying or damaging over 285,000 homes.[1] One year later, another, stronger earthquake devastated Honshu, Japan, destroying or damaging over 332,000 buildings,[2] like those shown in the picture above. Even though both caused substantial damage, the earthquake in 2011 was 100 times stronger than the earthquake in Haiti. How do we know? The magnitudes of earthquakes are measured on a scale known as the Richter Scale. The Haitian earthquake registered a 7.0 on the Richter Scale[3] whereas the Japanese earthquake registered a 9.0.[4]
The Richter Scale is a base-ten logarithmic scale. In other words, an earthquake of magnitude 8 is not twice as great as an earthquake of magnitude 4. It is ${10}^{8 - 4}={10}^{4}=10,000$ times as great! In this lesson, we will investigate the nature of the Richter Scale and the base-ten function upon which it depends.
In the last section, we defined inverse functions, logarithmic functions are the inverse of an exponential functions, and sometimes understanding this helps us make sense of what a logarithm is.
# (12.3.1) – Define Logarithmic Functions
In order to analyze the magnitude of earthquakes or compare the magnitudes of two different earthquakes, we need to be able to convert between logarithmic and exponential form. For example, suppose the amount of energy released from one earthquake were 500 times greater than the amount of energy released from another. We want to calculate the difference in magnitude. The equation that represents this problem is ${10}^{x}=500$, where $x$ represents the difference in magnitudes on the Richter Scale. How would we solve for $x$?
We have not yet learned a method for solving exponential equations. None of the algebraic tools discussed so far is sufficient to solve ${10}^{x}=500$. We know that ${10}^{2}=100$ and ${10}^{3}=1000$, so it is clear that $x$ must be some value between 2 and 3, since $y={10}^{x}$ is increasing. We can examine a graph to better estimate the solution.
Figure 2
Estimating from a graph, however, is imprecise. To find an algebraic solution, we must introduce a new function. Observe that the graph above passes the horizontal line test. The exponential function $y={b}^{x}$ is one-to-one, so its inverse, $x={b}^{y}$ is also a function. As is the case with all inverse functions, we simply interchange $x$ and $y$ and solve for $y$ to find the inverse function. To represent $y$ as a function of $x$, we use a logarithmic function of the form $y={\mathrm{log}}_{b}\left(x\right)$. The base $b$ logarithm of a number is the exponent by which we must raise $b$ to get that number.
We read a logarithmic expression as, “The logarithm with base $b$ of $x$ is equal to $y$,” or, simplified, “log base $b$ of $x$ is $y$.” We can also say, “$b$ raised to the power of $y$ is $x$,” because logs are exponents. For example, the base 2 logarithm of 32 is 5, because 5 is the exponent we must apply to 2 to get 32. Since ${2}^{5}=32$, we can write ${\mathrm{log}}_{2}32=5$. We read this as “log base 2 of 32 is 5.”
We can express the relationship between logarithmic form and its corresponding exponential form as follows:
${\mathrm{log}}_{b}\left(x\right)=y\Leftrightarrow {b}^{y}=x,\text{}b>0,b\ne 1$
Note that the base $b$ is always positive.
Because logarithm is a function, it is most correctly written as ${\mathrm{log}}_{b}\left(x\right)$, using parentheses to denote function evaluation, just as we would with $f\left(x\right)$. However, when the input is a single variable or number, it is common to see the parentheses dropped and the expression written without parentheses, as ${\mathrm{log}}_{b}x$. Note that many calculators require parentheses around the $x$.
We can illustrate the notation of logarithms as follows:
Notice that, comparing the logarithm function and the exponential function, the input and the output are switched. This means $y={\mathrm{log}}_{b}\left(x\right)$ and $y={b}^{x}$ are inverse functions.
### Definition of the Logarithmic Function
A logarithm base $b$ of a positive number $x$ satisfies the following definition.
For $x>0,b>0,b\ne 1$,
$y={\mathrm{log}}_{b}\left(x\right)\text{ is equivalent to }{b}^{y}=x$
where,
• we read ${\mathrm{log}}_{b}\left(x\right)$ as, “the logarithm with base $b$ of $x$” or the “log base $b$ of $x$.”
• the logarithm $y$ is the exponent to which $b$ must be raised to get $x$.
Also, since the logarithmic and exponential functions switch the $x$ and $y$ values, the domain and range of the exponential function are interchanged for the logarithmic function. Therefore,
• the domain of the logarithm function with base $b \text{ is} \left(0,\infty \right)$.
• the range of the logarithm function with base $b \text{ is} \left(-\infty ,\infty \right)$.
### Convert between logarithmic and exponential forms
In our first example we will convert logarithmic equations into exponential equations.
### Example
Write the following logarithmic equations in exponential form.
1. ${\mathrm{log}}_{6}\left(\sqrt{6}\right)=\frac{1}{2}$
2. ${\mathrm{log}}_{3}\left(9\right)=2$
In the following video we present more examples of rewriting logarithmic equations as exponential equations.
#### How To: Given an equation in logarithmic form ${\mathrm{log}}_{b}\left(x\right)=y$, convert it to exponential form.
1. Examine the equation $y={\mathrm{log}}_{b}x$ and identify $b$, $y$, and $x$.
2. Rewrite ${\mathrm{log}}_{b}x=y$ as ${b}^{y}=x$.
Can we take the logarithm of a negative number? Re-read the definition of a logarithm and formulate an answer. Think about the behavior of exponents. You can use the textbox below to formulate your ideas before you look at an answer.
#### Convert from exponential to logarithmic form
To convert from exponents to logarithms, we follow the same steps in reverse. We identify the base $b$, exponent $x$, and output $y$. Then we write $x={\mathrm{log}}_{b}\left(y\right)$.
### Example
Write the following exponential equations in logarithmic form.
1. ${2}^{3}=8$
2. ${5}^{2}=25$
3. $\displaystyle {10}^{-4}=\frac{1}{10,000}$
In the next video we show more examples of writing logarithmic equations as exponential equations.
# (12.3.2) – Evaluate Logarithms
Knowing the squares, cubes, and roots of numbers allows us to evaluate many logarithms mentally. For example, consider ${\mathrm{log}}_{2}8$. We ask, “To what exponent must 2 be raised in order to get 8?” Because we already know ${2}^{3}=8$, it follows that ${\mathrm{log}}_{2}8=3$.
Now consider solving ${\mathrm{log}}_{7}49$ and ${\mathrm{log}}_{3}27$ mentally.
• We ask, “To what exponent must 7 be raised in order to get 49?” We know ${7}^{2}=49$. Therefore, ${\mathrm{log}}_{7}49=2$
• We ask, “To what exponent must 3 be raised in order to get 27?” We know ${3}^{3}=27$. Therefore, ${\mathrm{log}}_{3}27=3$
Even some seemingly more complicated logarithms can be evaluated without a calculator. For example, let’s evaluate $\displaystyle {\mathrm{log}}_{\frac{2}{3}}\frac{4}{9}$ mentally.
• We ask, “To what exponent must $\frac{2}{3}$ be raised in order to get $\displaystyle \frac{4}{9}$? ” We know ${2}^{2}=4$ and ${3}^{2}=9$, so $\displaystyle {\left(\frac{2}{3}\right)}^{2}=\frac{4}{9}$. Therefore, $\displaystyle {\mathrm{log}}_{\frac{2}{3}}\left(\frac{4}{9}\right)=2$.
In our first example we will evaluate logarithms mentally (without a calculator).
### Example
Solve $y={\mathrm{log}}_{4}\left(64\right)$ without using a calculator.
In our first video we will show more examples of evaluating logarithms mentally, this helps you get familiar with what a logarithm represents.
In our next example we will evaluate the logarithm of a reciprocal.
### Example
Evaluate $y={\mathrm{log}}_{3}\left(\frac{1}{27}\right)$ without using a calculator.
#### How To: Given a logarithm of the form $y={\mathrm{log}}_{b}\left(x\right)$, evaluate it without a calculator.
1. Rewrite the argument $x$ as a power of $b$: ${b}^{y}=x$.
2. Use previous knowledge of powers of $b$ identify $y$ by asking, “To what exponent should $b$ be raised in order to get $x$?”
### Define and evaluate natural logarithm
The most frequently used base for logarithms is $e$. Base $e$ logarithms are important in calculus and some scientific applications; they are called natural logarithms. The base e logarithm, ${\mathrm{log}}_{e}\left(x\right)$, has its own notation, $\mathrm{ln}\left(x\right)$.
Most values of $\mathrm{ln}\left(x\right)$ can be found only using a calculator. The major exception is that, because the logarithm of 1 is always 0 in any base, $\mathrm{ln}1=0$. For other natural logarithms, we can use the $\mathrm{ln}$ key that can be found on most scientific calculators. We can also find the natural logarithm of any power of e using the inverse property of logarithms.
### A General Note: Definition of the Natural Logarithm
A natural logarithm is a logarithm with base e. We write ${\mathrm{log}}_{e}\left(x\right)$ simply as $\mathrm{ln}\left(x\right)$. The natural logarithm of a positive number $x$ satisfies the following definition.
For $x>0$,
$y=\mathrm{ln}\left(x\right)\text{ is equivalent to }{e}^{y}=x$
We read $\mathrm{ln}\left(x\right)$ as, “the logarithm with base $e$ of $x$” or “the natural logarithm of $x$.”
The logarithm $y$ is the exponent to which $e$ must be raised to get $x$.
Since the functions $y=e{}^{x}$ and $y=\mathrm{ln}\left(x\right)$ are inverse functions, $\mathrm{ln}\left({e}^{x}\right)=x$ for all $x$ and $e{}^{\mathrm{ln}\left(x\right)}=x$ for $x>0$.
In the next example, we will evaluate a natural logarithm using a calculator.
### Example
Evaluate $y=\mathrm{ln}\left(500\right)$ to four decimal places using a calculator.
In our next video, we show more examples of how to evaluate natural logarithms using a calculator.
### Define and evaluate common logarithm
Sometimes we may see a logarithm written without a base. In this case, we assume that the base is 10. In other words, the expression ${\mathrm{log}}_{}$ means ${\mathrm{log}}_{10}$ We call a base-10 logarithm a common logarithm. Common logarithms are used to measure the Richter Scale mentioned at the beginning of the section. Scales for measuring the brightness of stars and the pH of acids and bases also use common logarithms.
### Definition of Common Logarithm: Log is an exponent
A common logarithm is a logarithm with base 10. We write ${\mathrm{log}}_{10}(x)$ simplify as ${\mathrm{log}}_{}(x)$. The common logarithm of a positive number, $x$, satisfies the following definition:
For $x\gt0$
$y={\mathrm{log}}_{}(x)$ is equivalent to $10^y=x$
We read ${\mathrm{log}}_{}(x)$ as ” the logarithm with base 10 of $x$” or “log base 10 of $x$“.
The logarithm y is the exponent to which 10 must be raised to get $x$.
### Example
Evaluate ${\mathrm{log}}_{}(1000)$ without using a calculator.
### Example
Evaluate $y={\mathrm{log}}_{}(321)$ to four decimal places using a calculator.
In our last example we will use a logarithm to find the difference in magnitude of two different earthquakes.
### Example
The amount of energy released from one earthquake was 500 times greater than the amount of energy released from another. The equation $10^x=500$ represents this situation, where $x$ is the difference in magnitudes on the Richter Scale. To the nearest thousandth, what was the difference in magnitudes?
# (12.3.3) – Graph Logarithmic Functions
Before working with graphs, we will take a look at the domain (the set of input values) for which the logarithmic function is defined.
Recall that the exponential function is defined as $y={b}^{x}$ for any real number x and constant $b>0$, $b\ne 1$, where
• The domain of $x$ is $\left(-\infty ,\infty \right)$.
• The range of $y$ is $\left(0,\infty \right)$.
In the last section we learned that the logarithmic function $y={\mathrm{log}}_{b}\left(x\right)$ is the inverse of the exponential function $y={b}^{x}$. So, as inverse functions:
• The domain of $y={\mathrm{log}}_{b}\left(x\right)$ is the range of $y={b}^{x}$:$\left(0,\infty \right)$.
• The range of $y={\mathrm{log}}_{b}\left(x\right)$ is the domain of $y={b}^{x}$: $\left(-\infty ,\infty \right)$.
#### How To: Given a logarithmic function, identify the domain.
1. Set up an inequality showing the argument greater than zero.
2. Solve for $x$.
3. Write the domain in interval notation.
In our first example we will show how to identify the domain of a logarithmic function.
### Example
What is the domain of $f(x)={\mathrm{log}}_{2}\left(x+3\right)$ ?
Here is another example of how to identify the domain of a logarithmic function.
### Example
What is the domain of $f\left(x\right)=\mathrm{log}\left(5 - 2x\right)$?
Creating a graphical representation of most functions gives us another layer of insight for predicting future events. How do logarithmic graphs give us insight into situations? Because every logarithmic function is the inverse function of an exponential function, we can think of every output on a logarithmic graph as the input for the corresponding inverse exponential equation. In other words, logarithms give the cause for an effect.
To illustrate, suppose we invest \$2500 in an account that offers an annual interest rate of 5%, compounded continuously. We already know that the balance in our account for any year t can be found with the equation $A=2500{e}^{0.05t}$.
But what if we wanted to know the year for any balance? We would need to create a corresponding new function by interchanging the input and the output; thus we would need to create a logarithmic model for this situation. By graphing the model, we can see the output (year) for any input (account balance). For instance, what if we wanted to know how many years it would take for our initial investment to double? Figure 1 shows this point on the logarithmic graph.
Figure 1
Now that we have a feel for the set of values for which a logarithmic function is defined, we move on to graphing logarithmic functions. The family of logarithmic functions includes the parent function $y={\mathrm{log}}_{b}\left(x\right)$ along with all its transformations: shifts, stretches, compressions, and reflections.
We begin with the parent function $y={\mathrm{log}}_{b}\left(x\right)$. Because every logarithmic function of this form is the inverse of an exponential function with the form $y={b}^{x}$, their graphs will be reflections of each other across the line $y=x$. To illustrate this, we can observe the relationship between the input and output values of $y={2}^{x}$ and its equivalent $x={\mathrm{log}}_{2}\left(y\right)$ in the table below.
x –3 –2 –1 0 1 2 3 ${2}^{x}=y$ $\displaystyle \frac{1}{8}$ $\displaystyle \frac{1}{4}$ $\displaystyle \frac{1}{2}$ 1 2 4 8 ${\mathrm{log}}_{2}\left(y\right)=x$ –3 –2 –1 0 1 2 3
Using the inputs and outputs from the table above, we can build another table to observe the relationship between points on the graphs of the inverse functions $f\left(x\right)={2}^{x}$ and $g\left(x\right)={\mathrm{log}}_{2}\left(x\right)$.
$f\left(x\right)={2}^{x}$ $\displaystyle \left(-3,\frac{1}{8}\right)$ $\displaystyle \left(-2,\frac{1}{4}\right)$ $\displaystyle \left(-1,\frac{1}{2}\right)$ $\left(0,1\right)$ $\left(1,2\right)$ $\left(2,4\right)$ $\left(3,8\right)$ $g\left(x\right)={\mathrm{log}}_{2}\left(x\right)$ $\displaystyle \left(\frac{1}{8},-3\right)$ $\displaystyle\left(\frac{1}{4},-2\right)$ $\displaystyle \left(\frac{1}{2},-1\right)$ $\left(1,0\right)$ $\left(2,1\right)$ $\left(4,2\right)$ $\left(8,3\right)$
As we’d expect, the $x$– and $y$-coordinates are reversed for the inverse functions. The figure below shows the graph of $f$ and $g$.
Notice that the graphs of $f\left(x\right)={2}^{x}$ and $g\left(x\right)={\mathrm{log}}_{2}\left(x\right)$ are reflections about the line $y=x$.
Observe the following from the graph:
• $f\left(x\right)={2}^{x}$ has a $y$-intercept at $\left(0,1\right)$ and $g\left(x\right)={\mathrm{log}}_{2}\left(x\right)$ has an $x$-intercept at $\left(1,0\right)$.
• The domain of $f\left(x\right)={2}^{x}$, $\left(-\infty ,\infty \right)$, is the same as the range of $g\left(x\right)={\mathrm{log}}_{2}\left(x\right)$.
• The range of $f\left(x\right)={2}^{x}$, $\left(0,\infty \right)$, is the same as the domain of $g\left(x\right)={\mathrm{log}}_{2}\left(x\right)$.
### A General Note: Characteristics of the Graph $f(x)=\log_b{(x)}$
For any real number $x$ and constant $b>0$, $b\ne 1$, we can see the following characteristics in the graph of $f\left(x\right)={\mathrm{log}}_{b}\left(x\right)$:
• domain: $\left(0,\infty \right)$
• range: $\left(-\infty ,\infty \right)$
• $x$-intercept: $\left(1,0\right)$ and key point $\left(b,1\right)$
• $y$-intercept: none
• increasing if $b>1$
• decreasing if $0<b<1$
Figure 3 shows how changing the base $b$ in $f\left(x\right)={\mathrm{log}}_{b}\left(x\right)$ can affect the graphs. Observe that the graphs compress vertically as the value of the base increases. (Note: recall that the function $\mathrm{ln}\left(x\right)$ has base $e\approx \text{2}.\text{718.)}$
Figure 4. The graphs of three logarithmic functions with different bases, all greater than 1.
In our first example we will graph a logarithmic function of the form $f\left(x\right)={\mathrm{log}}_{b}\left(x\right)$.
### Example
Graph $f\left(x\right)={\mathrm{log}}_{5}\left(x\right)$. State the domain, range.
### How To: Given a logarithmic function with the form $f\left(x\right)={\mathrm{log}}_{b}\left(x\right)$, graph the function.
1. Plot the $x$-intercept, $\left(1,0\right)$.
2. Plot the key point $\left(b,1\right)$.
3. Draw a smooth curve through the points.
4. State the domain, $\left(0,\infty \right)$, the range, $\left(-\infty ,\infty \right)$.
### Summary
The base $b$ logarithm of a number is the exponent by which we must raise $b$ to get that number. Logarithmic functions are the inverse of Exponential functions, and it is often easier to understand them through this lens. We can never take the logarithm of a negative number, therefore ${\mathrm{log}}_{b}\left(x\right)=y$ is defined for $b>0$.
Knowing the squares, cubes, and roots of numbers allows us to evaluate many logarithms mentally because the logarithm is an exponent. Logarithms most commonly sue base 10, and often use base e. Logarithms can also be evaluated with most kinds of calculator.
To define the domain of a logarithmic function algebraically, set the argument greater than zero and solve. To plot a logarithmic function, it is easiest to find and plot the $x$-intercept, and the key point$\left(b,1\right)$.
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# Electron tube design by Radio Corporation of America. Electron Tube Division
By Radio Corporation of America. Electron Tube Division
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Extra resources for Electron tube design
Sample text
We’ll start our study, this chapter, with a brief recap of the ideas we covered in the last chapter. We saw then that resistors in series may be considered as a single equivalent resistor, whose resistance is found by adding together the resistance of each resistor. This is called the law of series resistors, which is given mathematically by: Similarly, there is a law of parallel resistors, by which the single equivalent resistance of a number of resistors connected in parallel is given by: Using these two laws many involved circuits may be broken down, step by step, into an equivalent circuit consisting of only one equivalent resistor.
Not end-to-end as series joined resistors are, but joined at both 44 On the boards ends. We say resistors joined together at both ends are in parallel. 8 shows a breadboard layout. Both these resistors are, again, 10 k resistors. What do you think the overall resistance will be? It’s certainly not 20 k! Measure it yourself using your multi-meter and breadboard. You should find that the overall resistance is 5 k. Odd, eh? Replace the two 10 k resistors with resistors of different value say, two 150 Ω resistors (brown, green, brown).
What is the overall resistance? You should find it’s about 1k3 — neither one thing nor the other! So, what’s the relationship? Well, a clue to the relationship between parallel resistors comes from the fact that, in a funny sort of way, parallel is the inverse of series. So if we inverted the formula for series resistors we saw earlier: 46 On the boards we would get: and this is the formula for parallel resistors. Let’s try it out on the resistors of this last experiment. Putting in the values, 10 k and 1k5 we get: which is about 1k3, the measured value.
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http://www.johndcook.com/blog/2015/05/17/fibonacci-number-system/
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Fibonacci number system
Every positive integer can be written as the sum of distinct Fibonacci numbers. For example, 10 = 8 + 2, the sum of the fifth Fibonacci number and the second.
This decomposition is unique if you impose the extra requirement that consecutive Fibonacci numbers are not allowed. [1] It’s easy to see that the rule against consecutive Fibonacci numbers is necessary for uniqueness. It’s not as easy to see that the rule is sufficient.
Every Fibonacci number is itself the sum of two consecutive Fibonacci numbers—that’s how they’re defined—so clearly there are at least two ways to write a Fibonacci number as the sum of Fibonacci numbers, either just itself or its two predecessors. In the example above, 8 = 5 + 3 and so you could write 10 as 5 + 3 + 2.
The nth Fibonacci number is approximately φn/√5 where φ = 1.618… is the golden ratio. So you could think of a Fibonacci sum representation for x as roughly a base φ representation for √5x.
You can find the Fibonacci representation of a number x using a greedy algorithm: Subtract the largest Fibonacci number from x that you can, then subtract the largest Fibonacci number you can from the remainder, etc.
Programming exercise: How would you implement a function that finds the largest Fibonacci number less than or equal to its input? Once you have this it’s easy to write a program to find Fibonacci representations.
* * *
[1] This is known as Zeckendorf’s theorem, published by E. Zeckendorf in 1972. However, C. G. Lekkerkerker had published the same result 20 years earlier.
9 thoughts on “Fibonacci number system”
1. Kevin Wheatley
Hi John,
looks like you have a typo in the 3rd paragraph, I think you meant 10 = 5 + 3 + 2.
Kevin
2. Thanks. Fixed.
3. I can’t find Lekkerkerker’s 1951 paper, nor Zeckendorf 1972 but my understanding is that Zeckendorf discovered his theorem in 1939. David Daykin’s 1960 JLMS paper on the subject says “Zeckendorf’s proof is given by C. G. Lekkerkerka (sic)”. Lekkerkerker’s paper, I understand, went on to present a theorem on the average number of summands in a Zeckendorf representation.
4. I could use the approximation for the n-th Fibonacci number you gave for the exercise and calculate:
x = log(input * sqrt(5)) / log(phi)
Now take the precise formula by Binet/Moivre:
fib(n) = (phi^n – (1 – phi)^n) / sqrt(5)
If fib(ceil(x)) equals the input, than that’s the solution, otherwise it’s fib(floor(x)).
5. Pseudonym
By the way, this numbering system makes a great alternative to Elias codes for compressing integers. Base-Fibonacci numbers greater than zero have a “canonical” form where there is a 1 in the leftmost place, and there are no two consecutive “1” digits. The first few are 1, 10, 100, 101, 1000, 1001, 1010, 10000.
If you reverse the digits and add a trailing 1, you have a canonical prefix-free encoding of the positive integers: 11, 011, 0011, 1011, 00011, 10011, 01011 etc.
One advantage of this system is that unlike the other logarithmic codes, you can detect the code boundaries without having to decode them; the code boundaries are exactly the double-1’s.
6. The previous binary observation could be used to find the largest Fibonacci term. Observing the binary representation of the input, go from LSB to MSB, and remove appropriate 1 bits whenever there are repeated sequences of 1 bits. It’s important to turn ones into zero bits such that you keep the largest number afterwards.
7. Nice Post !
8. Fredrik Meyer (@FredrikMeyer)
Interesting! After reading this, I made some plots. Since the representation is unique, you could ask about the length of the presentation, and then you get this really nice fractal-looking graph. http://i1.wp.com/cube.fredrikmeyer.net/wp-content/uploads/10000f.png (Fibonacci-numbers versus the length of their Zeckendorf-representation)
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A spectral refinement of the Bergelson-Host-Kra decomposition and new multiple ergodic theorems
# A spectral refinement of the Bergelson-Host-Kra decomposition and new multiple ergodic theorems
Joel Moreira
joel.moreira@northwestern.edu
Department of Mathematics
Northwestern University
Evanston, Illinois
Florian Karl Richter
richter.109@osu.edu
Department of Mathematics
The Ohio State University
Columbus, Ohio
###### Abstract
We investigate how spectral properties of a measure preserving system are reflected in the multiple ergodic averages arising from that system. For certain sequences we provide natural conditions on the spectrum such that for all ,
limN→∞1NN∑n=1k∏j=1Tja(n)fj=limN→∞1NN∑n=1k∏j=1Tjnfj
in -norm. In particular, our results apply to infinite arithmetic progressions , Beatty sequences , the sequence of squarefree numbers , and the sequence of prime numbers .
We also obtain a new refinement of Szemerédi’s theorem via Furstenberg’s correspondence principle.
## 1 Introduction
Let be a probability space and let be a measure preserving transformation. The discrete spectrum of the measure preserving system is the set of eigenvalues for which there exists a non-zero eigenfunction satisfying , where . It follows from the spectral theorem that given any functions , there exists a complex measure on the torus such that
α(n):=∫Xf⋅Tng\leavevmode\nobreak dμ=∫Te(nx)\leavevmode\nobreak dν(x).
Decomposing the measure into its discrete and continuous components , one can then represent the single-correlation sequence as
α(n)=ϕ(n)+ω(n), (1.1)
where is a Bohr almost periodic sequence111A sequence is Bohr almost periodic if there exists a compact abelian group (written multiplicatively), elements and a continuous function such that . and is a null-sequence222A sequence is a null-sequence if .. It is a further consequence of the spectral theorem that for any frequency which does not belong to the discrete spectrum , one has and hence
limN→∞1NN∑n=1α(n)e(−θn)=0.
This observation can be reformulated as
σ(α)⊂σ(T), (1.2)
where denotes the spectrum of the sequence in the sense introduced by Rauzy in [35], which we recall now.
###### Definition 1.1.
The spectrum of an arbitrary bounded sequence is the set of frequencies for which
limsupN→∞∣∣ ∣∣1NN∑n=1η(n)e(−θn)∣∣ ∣∣>0.
Throughout this paper we are only concerned with sequences for which the limit supremum in the above expression is an actual limit.
Formula (1.2) serves as the premise of our paper. Its importance lies in the many variations of the mean ergodic theorem that one can derive from it. For instance, for any , if the discrete spectrum of an ergodic system is disjoint from then it follows from (1.2) that for any and any ,
limN→∞1NN∑n=1Tqn+rf=∫Xf\leavevmode\nobreak dμin L2(X). (1.3)
More generally, for any real number , if the discrete spectrum of an ergodic system is disjoint from then for any and any ,
limN→∞1NN∑n=1T⌊θn+γ⌋f=∫Xf\leavevmode\nobreak dμin L2(X). (1.4)
Also, invoking classical equidistribution results of Vinogradov [38], one can derive from (1.2) the following ergodic theorem for totally ergodic systems: for all ,
limN→∞1π(N)∑p⩽Np\leavevmode\nobreak primeTpf=∫Xf\leavevmode\nobreak dμin L2(X), (1.5)
where denotes the prime-counting function.
In this paper we seek to extend (1.2) from single-correlation sequences to multi-correlation sequences, i.e., sequences of the form
α(n)=∫Xf0⋅Tnf1⋅…⋅Tknfk\leavevmode\nobreak dμ
where . Among other things, this will allow us to derive generalizations of (1.3), (1.4) and (1.5).
The theory of multi-correlation sequences was pioneered by Furstenberg in connection with his ergodic-theoretic proof of Szemerédi’s theorem [14]. A result of Bergelson, Host and Kra [3] offers a decomposition of multi-correlation sequences in analogy with (1.1):
α(n)=ϕ(n)+ω(n) (1.6)
where is a null-sequence and is a nilsequence333A sequence is a nilsequence if it can be approximated in by sequences of the form , where is a continuous function on the compact homogenous space of a nilpotent Lie group , and .. By examining the spectrum of the nilsystem from which the nilsequence in (1.6) arises, we show that the spectrum of the multicorrelation sequence is contained in the discrete spectrum of its originating system (cf. creftype 2.1 below). From this we derive several multiple ergodic theorems (see Theorems 2.5, 2.7, 2.9 and 2.10 in Section 2). As corollaries of Theorems 2.5 and 2.9, we obtain the following generalizations of (1.4) and (1.5): For any real number , if the discrete spectrum of an ergodic system is disjoint from then for any , and any
limN→∞1NN∑n=1k∏j=1Tj⌊nθ+γ⌋fj=limN→∞1NN∑n=1k∏j=1Tjnfjin L2(X).
For any totally ergodic system , any and any ,
limN→∞1π(N)∑p⩽Np\leavevmode\nobreak primek∏j=1Tjpfj=limN→∞1NN∑n=1k∏j=1Tjnfjin L2(X).
#### Structure of the paper:
In Section 2 we formulate the main results of this paper and present relevant examples as well as applications to combinatorics. In Section 3 we provide the necessary background on the theory of nilmanifolds and nilsystems, which is used in the rest of the paper.
Our main technical result, creftype 2.1, is proven in three steps: in Section 5, we reduce it to the special case of nilsystems. In Section 4 we derive a proof of this special case conditionally on a result involving the spectrum of nilsystems. The proof of the latter is provided in Section 7. In the remainder of Section 5 and also in Section 6 we deduce the other results stated in Section 2 from creftype 2.1.
#### Acknowledgements:
We thank V. Bergelson for many inspiring conversations, N. Frantzikinakis, B. Kra and A. Leibman for providing helpful references, and D. Glasscock for helpful comments regarding an earlier draft of this paper. The authors also thank the referee for several pertinent suggestions which greatly improved the final version of this paper.
## 2 Statement of results
In this section we state the main results of the paper; the proofs are presented in Sections 5 and 6. The following theorem is our main technical result. For a definition of nilsystems, see Section 3.
###### Theorem 2.1.
Let , let be an ergodic measure preserving system and let . For every there exists a decomposition of the form
α(n):=∫f0⋅Tnf1⋅…⋅Tknfk\leavevmode\nobreak dμ=ϕ(n)+ω(n)+γ(n),
where is a null-sequence, satisfies and for some and , where is a -step nilsystem whose discrete spectrum is contained in the discrete spectrum of .
###### Remark 2.2.
A natural question is whether analogues of creftype 2.1 hold for commuting transformations or for polynomial iterates. However these extensions seem to be out of reach by the methods used in the current paper.
The following is an immediate corollary of creftype 2.1 and generalizes (1.2).
###### Corollary 2.3.
Under the same assumptions as creftype 2.1, the spectrum of the multi-correlation sequence (see creftype 1.1) is contained in the discrete spectrum of the system .
From creftype 2.1 we derive various multiple ergodic theorems. The first theorem we derive this way is an extension of (1.3). An equivalent result was proven by Frantzikinakis in [8]. In the following we will use to denote the subgroup of generated by a subset . Subsets of are tacitly identified with their projections onto .
###### Theorem 2.4 (cf. [8, Theorem 6.4]).
Let , and let be an ergodic measure preserving system whose discrete spectrum satisfies . For any ,
limN−M→∞1N−MN∑n=Mk∏j=1Tj(qn+r)fj=limN→∞1NN∑n=1k∏j=1Tjnfj, (2.1)
where convergence takes place in . In particular, if is totally ergodic, then equation (2.1) holds for all .
The case of creftype 2.4 was proven by Host and Kra in [21]. In the same paper creftype 2.4 for was posed as a question ([21, Question 2]).
creftype 2.4 features multi-correlation sequences along infinite arithmetic progressions . The next theorem is a generalization in which infinite arithmetic progressions are replaced by more general Beatty sequences .
###### Theorem 2.5.
Let with , and let be an ergodic measure preserving system whose discrete spectrum satisfies . For any ,
limN−M→∞1N−MN∑n=Mk∏j=1Tj⌊θn+γ⌋fj=limN→∞1NN∑n=1k∏j=1Tjnfj, (2.2)
where convergence takes place in . In particular, since discrete spectra are always countable, we have that for any fixed system and for almost all equation (2.2) holds for all .
creftype 2.5, together with a standard application of Furstenberg’s correspondence principle (cf. [3, Proposition 3.1]), implies the following combinatorial result. Recall that the upper density of a set is defined by
¯d(A):=limsupN→∞∣∣A∩{1,…,N}∣∣N.
###### Corollary 2.6.
Let , and let have positive upper density. Then for almost every and every there exists a -term arithmetic progression in with common difference in the Beatty sequence .
In fact, under the assumptions of creftype 2.6, there are many arithmetic progressions contained in with common difference in a Beatty sequence. More precisely, there is a syndetic444A subset is called syndetic if it has bounded gaps, more precisely if there exists such that any interval of length contains at least one element of . set such that for every , there exist a set with positive upper density and with the property that for every , the set is contained in .
Recall that a bounded sequence is Besicovitch almost periodic if for every , there exists a trigonometric polynomial , where , and , such that
limsupN→∞1NN∑n=1|ϕ(n)−ρ(n)|<ε. (2.3)
The indicator function of a Beatty sequence is a Besicovitch almost periodic sequence with spectrum contained in the subgroup . Thus, sacrificing the uniformity in the Cesàro averages on the left hand side of (2.1) and (2.2), one can extend Theorems 2.4 and 2.5 as follows.
###### Theorem 2.7.
Let be a Besicovitch almost periodic sequence, and let be a measure preserving system whose discrete spectrum satisfies . For any ,
limN→∞1NN∑n=1ϕ(n)k∏j=1Tjnfj=(limN→∞1NN∑n=1ϕ(n))(limN→∞1NN∑n=1k∏j=1Tjnfj)
in .
###### Remark 2.8.
creftype 2.7 is not true for uniform Cesàro averages. One way of obtaining a version of creftype 2.7 with uniform Cesàro averages is by replacing Besicovitch almost periodic sequences with Weyl almost periodic sequences555A bounded sequence is Weyl almost periodic if for every , there exists a trigonometric polynomial such that . In fact, one can easily modify the proof of creftype 2.7 given below to obtain a proof of this variation.
An interesting application of creftype 2.7 concerns the sequence of squarefree numbers. Since the indicator function of the set of squarefree numbers is Besicovitch almost periodic with rational spectrum (cf. Section 3.4), it follows that for any totally ergodic ,
limN→∞1NN∑n=1k∏j=1Tjqnfj=limN→∞1NN∑n=1k∏j=1Tjnfj.
By combining creftype 2.1 with results of Green, Tao and Ziegler [17, 18, 19] on the asymptotic Gowers uniformity of the von Mangoldt function, we obtain the following multiple ergodic theorem along primes for totally ergodic systems.
###### Theorem 2.9.
Let and let be a totally ergodic system. For every ,
limN→∞1π(N)∑p⩽N,p\leavevmode\nobreak prime\leavevmode\nobreak k∏j=1Tjpfj=limN→∞1NN∑n=1\leavevmode\nobreak k∏j=1Tjnfjin L2(X).
The case of creftype 2.9 was obtained by Frantzikinakis, Host and Kra in [11, Theorem 5]. In the same paper they outline the proof of creftype 2.9 in full generality, conditional on the then unknown creftype 6.3.
Using creftype 2.7, we obtain a strengthening of the above result involving primes in Beatty sequences. Let and let .
###### Theorem 2.10.
Let with irrational and let be a measure preserving system whose discrete spectrum satisfies . For every ,
limN→∞1πθ,γ(N)∑p⩽N,p∈P(θ,γ)\leavevmode\nobreak k∏j=1Tjpfj=limN→∞1NN∑n=1\leavevmode\nobreak k∏j=1Tjnfjin L2(X). (2.4)
In particular, if is totally ergodic then for almost all equation (2.4) holds for all .
## 3 Preliminaries
In this section we give an overview of the theory of nilspaces and nilmanifolds.
### 3.1 Nilmanifolds and sub-nilmanifolds
Let be a Lie group with identity . The lower central series of is the sequence
G=G1⊵G2⊵G3⊵…⊵{1G}
where is, as usual, the subgroup of generated by all the commutators with and . If for some finite we say that is (-step) nilpotent. Each is a closed normal subgroup of (cf. [27, Section 2.11]).
Given a nilpotent Lie group and a uniform666A closed subgroup of is called uniform if is compact or, equivalently, if there exists a compact set such that . and discrete subgroup of , the quotient space is called a nilmanifold. Naturally, acts continuously and transitively on via left-multiplication.
Any element with the property that for some is called rational (or rational with respect to ). A closed subgroup of is then called rational (or rational with respect to ) if rational elements are dense in . For example, the subgroups in the lower central series of are rational with respect to any uniform and discrete subgroup of . (A proof of this fact can be found in [34, Corollary 1 of Theorem 2.1] for connected and in [27, Section 2.11] for the general case.)
###### Remark 3.1.
It is shown in [28] that a closed subgroup is rational if and only if is a uniform discrete subgroup of if and only if is closed in .
If is a nilmanifold, then a sub-nilmanifold of is any closed set of the form , where and where is a closed subgroup of . It is not true that for every closed subgroup of and for every element in the set is a sub-nilmanifold of ; as a matter of fact, from creftype 3.1 it follows that is closed in (and hence a sub-nilmanifold) if and only if the subgroup is rational with respect to .
### 3.2 Nilsystems and their dynamics
Let be a -step nilpotent Lie group and let be a nilmanifold. In the following we will use or ( if we want to emphasize the dependence on ) to denote the translation by a fixed element , i.e. . The map is called a nilrotation and the pair is called a (-step) nilsystem.
Every nilmanifold possesses a unique -invariant probability measure called the Haar measure on X (cf. [34, Lemma 1.4]). We will use to denote this measure.
Let us state some classical results regarding the dynamics of nilrotations.
###### Theorem 3.2 (see [1, 32] in the case of connected G and [27] in the general case).
Suppose is a nilsystem. Then the following are equivalent:
1. is transitive777A topological dynamical system is called transitive if there exists at least one point with dense orbit.;
2. is ergodic;
3. is strictly ergodic888A topological dynamical system is called strictly ergodic if there exists a unique -invariant probability measure on and additionally the orbit of every point in is dense.;
Moreover, the following are equivalent:
1. is connected and is ergodic.
2. is totally ergodic.
A theorem by Lesigne [31] asserts that for any nilmanifold with connected and any the closure of the set is a sub-nilmanifold of . (Actually, he shows that the sequence equidistributes with respect to the Haar measure on some sub-nilmanifold of , but in virtue of creftype 3.2 these two assertions are equivalent.) Leibman has extended this result as follows.
###### Theorem 3.3 ([26, Corollary 1.9]).
Let be a nilpotent Lie group and let be a uniform and discrete subgroup. Assume is a connected sub-nilmanifold of and . Then is a disjoint union of finitely many connected sub-nilmanifolds of .
### 3.3 The Kronecker factor of a nilsystem
Let be a nilmanifold and let be a normal, closed and rational subgroup of . Since is closed, the quotient topology on is Hausdorff and the map that sends elements to their right cosets is continuous and commutes with the action of . Therefore the nilsystem is a factor of with factor map .
An important tool in studying equidistribution of orbits on nilmanifolds is a theorem by Leon Green (see [1, 20, 33]). In [27] Leibman offers a refinement of this classical result of Green, a special case of which we state now. Here and throughout the text we denote by the connected component of containing the group identity .
###### Theorem 3.4 (cf. [27, Theorem 2.17]).
Let be a connected nilmanifold, let and let , where denotes the group generated by and . Then is ergodic on if and only if is ergodic on .
Note that in creftype 3.4 it is not explicitly stated but implied that is a normal, closed and rational subgroup of and hence the factor space is well defined.
Given a measure preserving dynamical system let denote the smallest sub--algebra of such that any eigenfunction of becomes measurable with respect to . The resulting factor system is called the (measure-theoretic) Kronecker factor of .
The following corollary of creftype 3.4 describes the Kronecker factor of a connected ergodic nilsystem.
###### Corollary 3.5.
Let be a connected nilmanifold, let and assume is ergodic. Define . Then the Kronecker factor of is .
For the proof of creftype 3.5 it will be convenient to recall the definition of vertical characters: Let be a connected nilmanifold and let be the lower central series of . The quotient is a connected compact abelian group and hence isomorphic to a torus . We call the vertical torus of . Since is contained in the center of , the vertical torus acts naturally on . A measurable function is called a vertical character if there exists a continuous group character of such that for all and almost every .
###### Proof of creftype 3.5.
Notice that the nilsystem is isomorphic to the nilsystem , where . We can therefore assume without loss of generality that . We proceed by induction on the nilpotency class of . Suppose is a -step nilpotent Lie group. If , is abelian and the result is trivial. Next, assume that and that creftype 3.5 has already been proven for all nilpotent Lie groups of step .
Observe that is a compact group and hence is contained in the Kronecker factor of . It thus suffices to show that for all eigenfunctions of the system one has
∀v∈Nf∘Rv=fin L2(X,μX). (3.1)
Let be an eigenvalue of the Koopman operator associated with , let be its (non-trivial) eigenspace and let .
Let denote the vertical torus of and note that the action of on commutes with the action of . In particular, leaves the eigenspace invariant. It thus follows from the Peter-Weyl theorem that decomposes into a direct sum of eigenspaces for the Koopman representation of . In other words, any -eigenfunction can be further decomposed into a sum of vertical characters that are also contained in . It therefore suffices to establish (3.1) in the special case where is a vertical character.
Now assume , is a group character of and for all and almost every . We distinguish two cases; the first case where is trivial and the second case where is non-trivial.
Let us first assume that is trivial, i.e. for all . This implies that is invariant. Let denote the nilpotent Lie group and let denote the natural quotient map. We define , which is a uniform and discrete subgroup of , and we define . Since is invariant it can be identified with a function on the nilmanifold and is then an eigenfunction for , where . Since is an -step nilpotent Lie group, we can invoke the induction hypothesis and conclude that
∀v′∈N′:=ξ(N)f′∘Rv′=f′in L2(X′,μX′). (3.2)
However, is invariant, and therefore (3.2) implies (3.1).
Now assume that is non-trivial. Since is connected, any non-trivial character has full range in the unit circle. In particular, there exists such that . Pick any element such that and define . Then from and from it follows that . Also, note that since the actions of and on the factor are identical (because ), it follows from the ergodicity of that acts ergodically on . Finally, the groups and are identical and hence it follows from creftype 3.4 that the ergodicity of lifts from to . We conclude that has to be a constant function, thereby satisfying (3.1). ∎
### 3.4 Spectrum of almost periodic sequences
In this section we collect a few facts about the spectrum of almost periodic sequences; we refer the reader to the book of Besicovitch [4] for a complete treatment on the theory of almost periodic sequences.
Almost periodic sequences were first introduced by Bohr in [5]. In his second paper on this subject [6] he proves that any Bohr almost periodic sequence can be approximated uniformly by trigonometric polynomials whose frequencies are all contained in the spectrum . This theorem is known as Bohr’s approximation theorem. An analogue of Bohr’s approximation theorem for Besicovitch almost periodic sequences was later obtained by Besicovitch. He showed that the spectrum of a Besicovitch almost periodic sequence is at most countable and then proved that any Besicovitch almost periodic sequence can be approximated in the Besicovitch seminorm by trigonometric polynomials whose frequencies are all contained in the spectrum . The precise statement of Besicovitch’s result is as follows999In his book, Besicovitch only deals with continuous almost periodic functions (i.e., almost periodic functions with domain ), but the proof of [4, Theorem II.8.2Â] also works for discrete almost periodic sequences (i.e., almost periodic functions with domain ); see also [2, Section 3]..
###### Theorem 3.6 (cf. [4, Theorem II.8.2Â∘(page 105)]).
Let be a Besicovitch almost periodic sequence with spectrum . Then for every there exists a trigonometric polynomial with and such that
limsupN→∞1NN∑n=1|ϕ(n)−ρ(n)|<ε. (3.3)
We will also make use of the following lemma regarding the spectrum of the product of two Besicovitch almost periodic sequences.
###### Lemma 3.7.
Let be bounded Besicovitch almost periodic sequences. The the product is also Besicovitch almost periodic with spectrum .
###### Proof.
In view of creftype 3.6, we can approximate each with a trigonometric polynomial whose spectrum is contained in . Observe that the product is a trigonometric polynomial with spectrum contained in . Finally, it is not hard to show that approximates , which finishes the proof. ∎
## 4 Proving creftype 2.1 for the special case of nilsystems
In this section we will prove creftype 2.1 for the special case of nilsystems. This will serve as an important intermediate step in obtaining creftype 2.1 in its full generality.
###### Theorem 4.1.
Let , let be an ergodic -step nilsystem and let . Then
∫f0⋅Rnf1R2nf2⋅…⋅Rknfk\leavevmode\nobreak dμX=ϕ(n)+ω(n), (4.1)
where is a null-sequence and for some and , where is a -step nilsystem whose discrete spectrum is contained in , the discrete spectrum of .
The main new ingredient used in the proof of creftype 4.1 is the following result.
###### Theorem 4.2.
Let , let be a connected nilmanifold and let be an ergodic nilrotation. Define and
YXΔ:=¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯{Sn(x,x,…,x):x∈X,\leavevmode\nobreak n∈Z}⊂Xk.
Then .
The proof of creftype 4.2 is postponed to Section 7.
Most of the ideas used in the rest of the proof of creftype 4.1 were already present in [3] and [29]. For completeness, we repeat the same arguments here, adapting them to our situation as needed.
Let be a nilmanifold and let denote the natural projection of onto . We will use to denote the point . Consider a closed subgroup of . As noticed in creftype 3.1 the set is a sub-nilmanifold of if and only if is rational. Let denote the normal closure of in , that is, let be the smallest normal subgroup of containing . One can show that if is closed and rational then so is (cf. [29]). In particular, the set is a sub-nilmanifold of containing . We call the normal closure of .
Note, every sub-nilmanifold of can be viewed as a nilmanifold on its own and in particular it has its own Haar measure . Moreover, for any , the Haar measure of the sub-nilmanifold coincides with the push forward of under .
###### Proposition 4.3 (cf. [29, Proposition 3.1]).
Assume is a connected nilmanifold, is the natural projection of onto and is a connected sub-nilmanifold of containing . Let and assume is dense in . If denotes the normal closure of , then for all we have
limN−M→∞1N−MN∑n=M∣∣∣∫bnWf\leavevmode\nobreak dμbnW−∫bnZf\leavevmode\nobreak dμbnZ∣∣∣=0. (4.2)
###### Proposition 4.4.
Let be a nilsystem, let be a connected sub-nilmanifold containing the origin and assume that is also a connected sub-nilmanifold of . Then there exists a factor of , and a point such that for any continuous function , there exists such that
limN−M→∞1N−MN∑n=M∣∣∣∫SnWf\leavevmode\nobreak dμSnW−F(Sny)∣∣∣=0. (4.3)
###### Proof.
Since is invariant under and , we can find a closed rational subgroup of such that . Therefore, is a uniform discrete subgroup of and the nilsystem is isomorphic to . In the following we will identify with and vice versa. Let be the normal closure of the sub-nilmanifold in and let denote the corresponding normal subgroup of such that .
Define and let denote the natural projection. As explained at the beginning of Section 3.3, is a well defined factor of with factor map .
Define and observe that . Note that for every , the set is a sub-nilmanifold of and therefore it possesses a Haar measure, which we denote by . Let and define the function as
F(z):=∫η−1(z)f\leavevmode\nobreak dμη−1(z).
Finally, observe that and so (4.3) follows immediately from Eq. 4.2 in creftype 4.3. ∎
To prove creftype 4.1 we will also require a technical lemma:
###### Lemma 4.5.
Let be an ergodic connected nilsystem of step and let . Then there exists an ergodic nilsystem of step with exactly connected components and such that the restriction of to each connected component of yields a system isomorphic to .
###### Proof.
First, we claim that one can embed the connected nilsystem into a nilflow , so that is a subnilmanifold of invariant under . Indeed, say . One can assume that the identity component of is simply connected, by passing to the universal cover if needed. Next one can use [34, Theorem 2.20] to find a connected simply connected nilpotent Lie group such that and is a uniform discrete subgroup of . In particular is a sub-nilmanifold of . Since is connected and simply connected, for any element the associated one-parameter subgroup is well defined (cf. [27, Subsection 2.4]). In particular, the nilrotation can be extended to a nilflow on by defining for all , .
Next, consider the product nilsystem , so that as a nilmanifold and the nilrotation is defied as . Finally, let be the orbit of under . Since preserves , we deduce that has precisely connected components. In fact,
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Features of the magic cube
Look at the most magic 4x4x4 cube of Walter Trump.
It consists of 4x4 cells in each of the 4 levels and the numbers 1 up to (4x4x4=) 64 are in it. I have numbered the cells as follows:
level I Level II Level III Level IV a1 a2 a3 a4 a1 a2 a3 a4 a1 a2 a3 a4 a1 a2 a3 a4 b1 b2 b3 b4 b1 b2 b3 b4 b1 b2 b3 b4 b1 b2 b3 b4 c1 c2 c3 c4 c1 c2 c3 c4 c1 c2 c3 c4 c1 c2 c3 c4 d1 d2 d3 d4 d1 d2 d3 d4 d1 d2 d3 d4 d1 d2 d3 d4
The possible magic features of a magic cube are:
[All rows in each level]
give the magic sum (in each of the levels I, II, III and IV: a1+a2+a3+a4 = b1+b2+b3+b4 = c1+c2+c3+c4 = d1+d2+d3+d4 = magic sum)
[All columns in each level]
give the magic sum (in each of the levels I, II, III and IV: a1+b1+c1+d1 = a2+b2+c2+d2 = a3+b3+c3+d3 = a4+b4+c4+d4 = magic sum)
[All main diagonals in each level] give the magic sum (in each of the levels I, II, III and IV: a1+b2+c3+d4 = a4+b3+c2+d1 = magic sum)
[All pandiagonals in each level] give the magic sum (in each of the levels I, II, III and IV: a2+b3+c4+d1 = a3+b4+c1+d2 = a4+b1+c2+d3 = b4+c3+d2+a1 = c4+d3+a2+b1 = d4+a3+b2+c1 = magic sum)
[All 2x2 subsquares in each level]
give the magic sum (in each of the levels I, II, III and IV: a1+a2+b1+b2 = a2+a3+b2+b3 = a3+a4+b3+b4 = b1+b2+c1+c2 = b2+b3+c2+c3 = b3+b4+c3+c4 = c1+c2+d1+d2 = c2+c3+d2+d3 = c3+c4+d3+d4 = magic sum)
[All pillars through the levels]
give the magic sum (I a1 + II a1 + III a1 + IV a1 = I a2 + II a2 + III a2 + IV a2 = ... = I d4 + II d4 + III d4 + IV d4 = magic sum {for all 16 pillars})
[All diagonals from left to right through the levels] give the magic sum (I a1 + II a2 + III a3 + IV a4 = I b1 + II b2 + III b3 + IV b4 = I c1 + II c2 + III c3 + IV c4 = I d1 + II d2 + III d3 + IV d4 = magic sum)
[All diagonals from right to left through the levels] give the magic sum (I a4 + II a3 + III a2 + IV a1 = I b4 + II b3 + III b2 + IV b1 = I c4 + II c3 + III c2 + IV c1 = I d4 + II d3 + III d2 + IV d1 = magic sum)
[All diagonals from up to down through the levels] give the magic sum (I a1 + II b1 + III c1 + IV d1 = I a2 + II b2 + III c2 + IV d2 = I a3 + II b3 + III c3 + IV d3 = I a4 + II b4 + III c4 + IV d4 = magic sum)
[All diagonals from down to up through the levels] give the magic sum (I d1 + II c1 + III b1 + IV a1 = I d2 + II c2 + III b2 + IV a2 = I d3 + II c3 + III b3 + IV a3 = I d4 + II c4 + III b4 + IV a4 = magic sum)
[All space diagonals through the levels]
give the magic sum (I a1 + II b2 + III c3 + IV d4 = I a4 + II b3 + III c2 + IV d1 = I d1 + II c2 + III b3 + IV a4 = I d4 + II c3 + III b2 + IV a1 = magic sum)
[All pandiagonals from left to right through the levels]
give the magic sum (I a2 + II a3 + III a4 + IV a1 = I a3 + II a4 + III a1 + IV a2 = I b2 + II b3 + III b4 + IV b1 = I b3 + II b4 + III b1 + IV b2 = I c2 + II c3 + III c4 + IV c1 = I c3 + II c4 + III c1 + IV c2 = I d2 + II d3 + III d4 + IV d1 = I d3 + II d4 + III d1 + IV d2 = magic sum)
[All pandiagonals from right to left through the levels]
give the magic sum (I a3 + II a2 + III a1 + IV a4 = I a2 + II a1 + III a4 + IV a3 = I b3 + II b2 + III b1 + IV b4 = I b2 + II b1 + III b4 + IV b3 = I c3 + II c2 + III c1 + IV c4 = I c2 + II c1 + III c4 + IV c3 = I d3 + II d2 + III d1 + IV d4 = I d2 + II d1 + III d4 + IV d3 = magic sum)
[All pandiagonals from up to down through the levels] give the magic sum (I b1 + II c1 + III d1 + IVa1 = I c1 + II d1 + III a1 + IV b1 = I b2 + II c2 + III d2 + IV a2 = I c2 + II d2 + III a2 + IV b2 = I b3 + II c3 + III d3 + IV a3 = I c3 + II d3 + III a3 + IV b3 = I b4 + II c4 + III d4 + IV a4 = I c4 + II d4 + III a4 + IV b4 = magic sum)
[All pandiagonals from down to up through the levels]
give the magic sum (I c1 + II b1 + III a1 + IV d1 = I b1 + II a1 + III d1 + IV c1 = I c2 + II b2 + III a2 + IV d2 = I b2 + II a2 + III d2 + IV c2 = I c3 + II b3 + III a3 + IV d3 = I b3 + II a3 + III d3 + IV c3 = I c4 + II b4 + III a4 + IV d4 = I b4 + II a4 + III d4 + IV c4 = magic sum)
[All pantriagonals, first direction, through the levels]
give the magic sum (I a2 + II b3 + III c4 + IV d1 = I a3 + II b4 + III c1 + IV d2 = I a4 + II b1 + III c2 + IV d3 = I b1 + II c2 + III d3 + IV a4 = I b2 + II c3 + III d4 + I a1 = I b3 + II c4 + III d1 + IV a2 = I b4 + II c1 + III d2 + IV a3 = I c1 + II d2 + III a3 + IV b4 = I c2 + II d3 + III a4 + IV b1 = I c3 + II d4 + III a1 + IV b2 = I c4 + II d1 + III a2 + IV b3 = I d1 + II a2 + III b3 + IV c4 = I d2 + II a3 + III b4 + IV c1 = I d3 + II a4 + III b1 + IV c2 = I d4 + II a1 + III b2 + IV c3 = magic sum)
[All pantriagonals, second direction, through the levels]
give the magic sum (I a1 + II b4 + III c3 + IV d2 = I a2 + II b1 + III c4 + IV d3 = I a3 + II b2 + II c1 + IV d4 = I b1 + II c4 + III d3 + IV a2 = I b2 + II c1 + III d4 + IV a3 = I b3 + II c2 + III d1 + IV a4 = I b4 + II c3 + III d2 + IV a1 = I c1 + II d4 + III a3 + IV b2 = I c2 + II d3 + III a4 + IV b1 = I c3 + II d2 + III a1 + IV b2 = I c4 + II d1 + III a2 + IV b3 = I d1 + II a4 + III b3 + IV c2 = I d2 + II a1 + III b4 + IV c3 = I d3 + II a2 + III b1 + IV c4 = I d4 + II a3 + III b2 + IV c1 = magic sum)
[All pantriagonals, third direction, through the levels]
give the magic sum (I a1 + II d2 + III c3 + IV b4 = I a2 + II d3 + III c4 + IV b1 = I a3 + II d4 + III c1 + IV b2 = I a4 + II d1 + III c2 + IV b3 = I b1 + II a2 + III d3 + IV c4 = I b2 + II a3 + III d4 + IV c1 = I b3 + II a4 + III d1 + IV c2 = I b4 + II a1 + III d2 + IV c3 = I c1 + II b2 + III a3 + IV d4 = I c2 + II b3 + III a4 + IV d1 = I c3 + II b4 + III a1 + IV d2 = I c4 + II b1 + III a2 + IV d3 = I d2 + II c3 + III b4 + IV a1 = I d3 + II c4 + III b1 + IV a2 = I d4 + II c1 + III b2 + IV a3 = magic sum)
[All pantriagonals, fourth direction, through the levels] give the magic sum (I a1 + II d4 + III c3 + IV b2 = I a2 + II d1 + III c4 + IV b3 = I a3 + II d2 + III c1 + IV b4 = I a4 + II d3 + III c2 + IV b1 = I b1 + II a4 + III d3 + IV c2 = I b2 + II a1 + III d4 + IV c3 = I b3 + II a2 + III d1 + IV c4 = I b4 + II a3 + III d2 + IV c1 = I c1 + II b4 + III a3 + IV d2 = I c2 + II b1 + III a4 + IV d3 = I c3 + II b2 + III a1 + IV d4 = I c4 + II b3 + III a2 + IV d1 = I d1 + II c4 + III b3 + IV a2 = I d2 + II c1 + III b4 + IV a3 = I d3 + II c2 + III b1 + IV a4 = magic sum)
[Symmetric]
In a symmetric magic cube each time addition of two digits, which can be connected with a straight line through the centre of the magic cube and which are at the same distance to the centre, gives the same sum. The centre of a(n even, for example the) 4x4x4 magic is the virtual crosspoint of the middle 2x2 cells of the second and third level. For example II b2 + III c3 = II b3 + III c2 = II c2 + III b3 = II c3 + III b2 = proportional part (= 1/2) of the magic sum ór I a1 + IV d4 = I a2 + IV d3 = I a3 + IV d2 = I a4 + IV d1 = proportional part (= 1/2) of the magic sum.
[half of]
the rows/columns/diagonals in each level or pillars/space diagonals through the levels give half of the magic sum.
• Simple magic cubes have the red marked (see above) magic features.
• Diagonal magic cubes have the red & orange marked (see above) magic features
• Pantriagonal cubes have the red & yellow marked (see above) magic futures
• Pandiagonal magic cubes have the red & orange & pink marked (see above) magic features.
• Perfect (Nasik) magic cubes have the red & orange & pink & yellow marked (see above) magic features.
• More than perfect magic cubes have red & orange & yellow & [part of the] green marked (see above) magic features.
Order 3 is simple magic. Diagonal magic cubes exist from order 5 and up. Pantriagonal magic cubes exist from order 4 and up. Pandiagonal magic cubes exist from order 7 and up. Nasik perfect (= pandiagonal & pantriagonal) magic cubes exist for odd orders from order 9 and up and for order 8, 16, 24, 32, …
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https://deepai.org/publication/functional-ghobber-jaming-uncertainty-principle
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# Functional Ghobber-Jaming Uncertainty Principle
Let ({f_j}_j=1^n, {τ_j}_j=1^n) and ({g_k}_k=1^n, {ω_k}_k=1^n) be two p-orthonormal bases for a finite dimensional Banach space 𝒳. Let M,N⊆{1, …, n} be such that o(M)^1/qo(N)^1/p< 1/max_1≤ j,k≤ n|g_k(τ_j) |, where q is the conjugate index of p. Then for all x ∈𝒳, we show that (1) x≤(1+1/1-o(M)^1/qo(N)^1/pmax_1≤ j,k≤ n|g_k(τ_j)|)[(∑_j∈ M^c|f_j(x)|^p)^1/p+(∑_k∈ N^c|g_k(x) |^p)^1/p]. We call Inequality (1) as Functional Ghobber-Jaming Uncertainty Principle. Inequality (1) improves the uncertainty principle obtained by Ghobber and Jaming [Linear Algebra Appl., 2011].
04/05/2023
Let ({f_j}_j=1^n, {τ_j}_j=1^n) and ({g_k}_k=1^m, {ω_k}_k=1^m) be p-Schau...
08/01/2023
### Functional Continuous Uncertainty Principle
Let (Ω, μ), (Δ, ν) be measure spaces. Let ({f_α}_α∈Ω, {τ_α}_α∈Ω) and ({g...
07/01/2023
### Functional Donoho-Stark Approximate Support Uncertainty Principle
Let ({f_j}_j=1^n, {τ_j}_j=1^n) and ({g_k}_k=1^n, {ω_k}_k=1^n) be two p-o...
10/17/2022
### Fourier theoretic inequalities for inclusion of simple C*-algebras
This paper originates from a naive attempt to establish various non-comm...
11/21/2022
### Sharpened Uncertainty Principle
For any finite group G, any finite G-set X and any field F, we consider ...
02/25/2022
### A Probabilistic Oracle Inequality and Quantification of Uncertainty of a modified Discrepancy Principle for Statistical Inverse Problems
In this note we consider spectral cut-off estimators to solve a statisti...
01/04/2022
### Dynamics of polynomial maps over finite fields
Let 𝔽_q be a finite field with q elements and let n be a positive intege...
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http://aimsciences.org/article/doi/10.3934/dcdsb.2016110
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# American Institute of Mathematical Sciences
2016, 21(10): 3551-3573. doi: 10.3934/dcdsb.2016110
## Global attracting set, exponential decay and stability in distribution of neutral SPDEs driven by additive $\alpha$-stable processes
1 School of Mathematical Sciences, Tianjin Normal University, Tianjin 300387, China 2 School of Information and Mathematics, Yangtze University, Jingzhou 434023, China
Received January 2016 Revised June 2016 Published November 2016
In this paper, we are concerned with a class of neutral stochastic partial differential equations driven by $\alpha$-stable processes. By combining some stochastic analysis techniques, tools from semigroup theory and delay integral inequalities, we identify the global attracting sets of the equations under investigation. Some sufficient conditions ensuring the exponential decay of mild solutions in the $p$-th moment to the stochastic systems are obtained. Subsequently, by employing a weak convergence approach, we try to establish some stability conditions in distribution of the segment processes of mild solutions to the stochastic systems under consideration. Last, an example is presented to illustrate our theory in the work.
Citation: Kai Liu, Zhi Li. Global attracting set, exponential decay and stability in distribution of neutral SPDEs driven by additive $\alpha$-stable processes. Discrete & Continuous Dynamical Systems - B, 2016, 21 (10) : 3551-3573. doi: 10.3934/dcdsb.2016110
##### References:
[1] D. Applebaum, Lévy Processes and Stochastic Calculus,, 2nd Edition. Cambridge University Press, (2009). doi: 10.1017/CBO9780511809781. [2] J. Bao, Z. Hou and C. Yuan, Stability in distribution of neutral stochastic differential delay equations with Markovian switching,, Statist. Probab. Lett., 79 (2009), 1663. doi: 10.1016/j.spl.2009.04.006. [3] J. Bao and C. Yuan, Numerical analysis for neutral SPDEs driven by $\alpha$-stable processes,, Infinite Dimen. Anal. Quant. Probab. Relat. Topics., 17 (2014). doi: 10.1142/S0219025714500313. [4] Z. Dong, L. Xu and X. C. Zhang, Invariant measures of stochastic 2D Navier-Stokes equation driven $\alpha$-stable processes,, Elec. Comm. Probab., 16 (2011), 678. doi: 10.1214/ECP.v16-1664. [5] U. Haagerup, The best constants in the Khintchine inequality,, Studia Math., 70 (1981), 231. [6] N. Ikeda and S. Watanable, Stochastic Differential Equations and Diffusion Processes,, North-Holland, (1981). [7] Y. Liu and J. L. Zhai, A note on time regularity of generalized Ornstein-Uhlenbeck processes with cylindrical stable noise,, C. R. Acad. Sci. Paris, 350 (2012), 97. doi: 10.1016/j.crma.2011.11.017. [8] S. Long, L. Teng and D. Xu, Global attracting set and stability of stochastic neutral partial functional differential equations with impulses,, Statist. Probab. Lett., 82 (2012), 1699. doi: 10.1016/j.spl.2012.05.018. [9] S. Mohammed, Stochastic Functional Differential Equation,, Pitman, (1984). [10] A. Pazy, Semigroup of Linear Operators and Applications to Partial Differential Equations,, Springer-Verlag, (1983). doi: 10.1007/978-1-4612-5561-1. [11] E. Priola and J. Zabczyk, Structural properties of semilinear SPDEs driven by cylindrical stable processes,, Probab. Theory Relat. Fields., 149 (2011), 97. doi: 10.1007/s00440-009-0243-5. [12] G. Samorodnitsky and M. S. Taqqu, Stable Non-Gaussian Random Processes: Stochastic Models with Infinite Variance,, Chapman & Hall, (1994). [13] K. Sato, Lévy Processes and Infinitely Divisible Distributions,, Cambridge University Press, (1999). [14] F. Y. Wang, Gradient estimate for Ornstein-Uhlenbeck jump processes,, Stoch. Proc. Appl., 121 (2011), 466. doi: 10.1016/j.spa.2010.12.002. [15] J. Y. Wang and Y. L. Rao, A note on stability of SPDEs driven by $\alpha$-stable noises,, Adv. Difference Equations., 2014 (2014). doi: 10.1186/1687-1847-2014-98. [16] L. L. Wang and X. C. Zhang, Harnack inequalities for SDEs driven by cylindrical $\alpha$-stable processes,, Potential Anal., 42 (2015), 657. doi: 10.1007/s11118-014-9451-4. [17] L. Xu, Ergodicity of the stochastic real Ginzburg-Landau equation driven by $\alpha$-stable noise,, Stoch. Proc. Appl., 123 (2013), 3710. doi: 10.1016/j.spa.2013.05.002. [18] D. Y. Xu and S. J. Long, Attracting and quasi-invariant sets of non-autonomous neural networks with delays,, Neurocomputing, 77 (2012), 222. doi: 10.1016/j.neucom.2011.09.004. [19] L. G. Xu and D. Y. Xu, $P$-attracting and $p$-invariant sets for a class of impulsive stochastic functional differential equations,, Comput. Math. Appl., 57 (2009), 54. doi: 10.1016/j.camwa.2008.09.027. [20] Y. C. Zang and J. P. Li, Stability in distribution of neutral stochastic partial differential delay equations driven by $\alpha$-stable process,, Adv. Difference Equations, 13 (2014). [21] X. C. Zhang, Derivative formulas and gradient estimates for SDEs driven by $\alpha$-stable processes,, Stoch. Proc. Appl., 123 (2013), 1213. doi: 10.1016/j.spa.2012.11.012. [22] Z. H. Zhao and J. G. Jian, Attracting and quasi-invariant sets for BAM neural networks of neutral-type with time-varying and infinite distributed delays,, Neurocomputing., 140 (2014), 265. doi: 10.1016/j.neucom.2014.03.015.
show all references
##### References:
[1] D. Applebaum, Lévy Processes and Stochastic Calculus,, 2nd Edition. Cambridge University Press, (2009). doi: 10.1017/CBO9780511809781. [2] J. Bao, Z. Hou and C. Yuan, Stability in distribution of neutral stochastic differential delay equations with Markovian switching,, Statist. Probab. Lett., 79 (2009), 1663. doi: 10.1016/j.spl.2009.04.006. [3] J. Bao and C. Yuan, Numerical analysis for neutral SPDEs driven by $\alpha$-stable processes,, Infinite Dimen. Anal. Quant. Probab. Relat. Topics., 17 (2014). doi: 10.1142/S0219025714500313. [4] Z. Dong, L. Xu and X. C. Zhang, Invariant measures of stochastic 2D Navier-Stokes equation driven $\alpha$-stable processes,, Elec. Comm. Probab., 16 (2011), 678. doi: 10.1214/ECP.v16-1664. [5] U. Haagerup, The best constants in the Khintchine inequality,, Studia Math., 70 (1981), 231. [6] N. Ikeda and S. Watanable, Stochastic Differential Equations and Diffusion Processes,, North-Holland, (1981). [7] Y. Liu and J. L. Zhai, A note on time regularity of generalized Ornstein-Uhlenbeck processes with cylindrical stable noise,, C. R. Acad. Sci. Paris, 350 (2012), 97. doi: 10.1016/j.crma.2011.11.017. [8] S. Long, L. Teng and D. Xu, Global attracting set and stability of stochastic neutral partial functional differential equations with impulses,, Statist. Probab. Lett., 82 (2012), 1699. doi: 10.1016/j.spl.2012.05.018. [9] S. Mohammed, Stochastic Functional Differential Equation,, Pitman, (1984). [10] A. Pazy, Semigroup of Linear Operators and Applications to Partial Differential Equations,, Springer-Verlag, (1983). doi: 10.1007/978-1-4612-5561-1. [11] E. Priola and J. Zabczyk, Structural properties of semilinear SPDEs driven by cylindrical stable processes,, Probab. Theory Relat. Fields., 149 (2011), 97. doi: 10.1007/s00440-009-0243-5. [12] G. Samorodnitsky and M. S. Taqqu, Stable Non-Gaussian Random Processes: Stochastic Models with Infinite Variance,, Chapman & Hall, (1994). [13] K. Sato, Lévy Processes and Infinitely Divisible Distributions,, Cambridge University Press, (1999). [14] F. Y. Wang, Gradient estimate for Ornstein-Uhlenbeck jump processes,, Stoch. Proc. Appl., 121 (2011), 466. doi: 10.1016/j.spa.2010.12.002. [15] J. Y. Wang and Y. L. Rao, A note on stability of SPDEs driven by $\alpha$-stable noises,, Adv. Difference Equations., 2014 (2014). doi: 10.1186/1687-1847-2014-98. [16] L. L. Wang and X. C. Zhang, Harnack inequalities for SDEs driven by cylindrical $\alpha$-stable processes,, Potential Anal., 42 (2015), 657. doi: 10.1007/s11118-014-9451-4. [17] L. Xu, Ergodicity of the stochastic real Ginzburg-Landau equation driven by $\alpha$-stable noise,, Stoch. Proc. Appl., 123 (2013), 3710. doi: 10.1016/j.spa.2013.05.002. [18] D. Y. Xu and S. J. Long, Attracting and quasi-invariant sets of non-autonomous neural networks with delays,, Neurocomputing, 77 (2012), 222. doi: 10.1016/j.neucom.2011.09.004. [19] L. G. Xu and D. Y. Xu, $P$-attracting and $p$-invariant sets for a class of impulsive stochastic functional differential equations,, Comput. Math. Appl., 57 (2009), 54. doi: 10.1016/j.camwa.2008.09.027. [20] Y. C. Zang and J. P. Li, Stability in distribution of neutral stochastic partial differential delay equations driven by $\alpha$-stable process,, Adv. Difference Equations, 13 (2014). [21] X. C. Zhang, Derivative formulas and gradient estimates for SDEs driven by $\alpha$-stable processes,, Stoch. Proc. Appl., 123 (2013), 1213. doi: 10.1016/j.spa.2012.11.012. [22] Z. H. Zhao and J. G. Jian, Attracting and quasi-invariant sets for BAM neural networks of neutral-type with time-varying and infinite distributed delays,, Neurocomputing., 140 (2014), 265. doi: 10.1016/j.neucom.2014.03.015.
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2016 Impact Factor: 0.994
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# Triple Exponential Smoothing
Triple Exponential Smoothing, commonly known as the Holt-Winters Method, extends upon Double Exponential Smoothing to address time series data that contains both a trend and a seasonal component. It incorporates three equations to capture the level, trend, and seasonality of a dataset, making it particularly useful for predicting values in more complex time series scenarios. Triple Exponential Smoothing is a cornerstone of time series forecasting, especially when dealing with datasets that exhibit both trend and seasonality.
Conceptual Framework
The essence of Triple Exponential Smoothing is to use three smoothing equations to update the level, trend, and seasonality:
1. α (alpha): Smoothing parameter for the level.
2. β (beta): Smoothing parameter for the trend.
3. γ (gamma): Smoothing parameter for the seasonality.
Key Equations
Given a time series yt, the forecasting and smoothing equations for Triple Exponential Smoothing are:
1. Level: Lt=αStmyt+(1−α)(Lt−1+Tt−1)
2. Trend: Tt=β(Lt−Lt−1)+(1−β)Tt−1
3. Seasonality: St=γLtyt+(1−γ)Stm
4. Forecast for t+p periods ahead: Ft+p=(Lt+pTt)S (tm+p mod m)
Where:
• Lt is the smoothed value at time t.
• Tt is the trend factor at time t.
• St is the seasonal factor at time t.
• m represents the seasonal period (e.g., 12 for monthly data with yearly seasonality).
• Ft+p is the forecast for p periods ahead.
• yt is the actual value at time t.
Example
Consider monthly sales data for a store that sells winter gear:
MonthSales
Jan200
Feb220
Dec210
Using Triple Exponential Smoothing, we would:
1. Initialize estimates for level, trend, and seasonality.
2. Use the smoothing equations to update these values through each month.
3. Predict sales for January of the next year, considering both the trend and the seasonal pattern observed in the past.
Applications
Triple Exponential Smoothing finds applications in:
• Retail for seasonal sales forecasting.
• The Tourism industry predicts tourist inflow during various seasons.
• Energy sector to project electricity consumption.
Limitations
• Assumes seasonality patterns remain consistent over time.
• Parameters α,β,γ need to be chosen appropriately, often requiring trial and error or optimization techniques.
• Does not handle multiple seasonalities well. For instance, a retail store that has both a daily and yearly seasonality might struggle with this method.
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Search All of the Math Forum:
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Topic: MUTUAL TIME DILATION : PARADOX OR ABSURDITY?
Replies: 2 Last Post: Jul 3, 2013 10:16 PM
Messages: [ Previous | Next ]
Pentcho Valev Posts: 6,212 Registered: 12/13/04
MUTUAL TIME DILATION : PARADOX OR ABSURDITY?
Posted: Jul 2, 2013 9:47 AM
http://www.einstein-online.info/dictionary/time-dilation
"Time dilation can be mutual: When two inertial observers speed past each other, each will find that the other's clocks go slower."
http://www.amazon.com/Relativity-Its-Roots-Banesh-Hoffmann/dp/0486406768
Relativity and Its Roots, Banesh Hoffmann, p. 105: "In one case your clock is checked against two of mine, while in the other case my clock is checked against two of yours, and this permits us each to find without contradiction that the other's clocks go more slowly than his own."
There is nothing in Banesh Hoffmann's text above that justifies the words "without contradiction". "Without contradiction" could be justified if Einsteinians had devised and discussed at least one scenario where EITHER observer's clock is checked against two of the other observer's clocks (at some distance from one another). Instead, discussions in Divine Albert's world are restricted to the misleading twin paradox scenario where the travelling twin is implicitly deprived of the setup "two clocks at some distance from one another against which the sedentary twin's clock could be checked" and so there is no way to demonstrate the slowness of the sedentary twin's clock. Only the slowness of the travelling twin's clock can be demonstrated, Divine Einstein, yes we all believe in relativity, relativity, relativity.
The following scenario allows EITHER observer's clock to be checked against two of the other observer's clocks. Two long inertial systems each carrying synchronous clocks pass one another:
..........Inertial system A moving to the right..........
..........Inertial system B moving to the left..........
The systems are so designed that, the moment they stop moving relative to one another, all clocks on both systems stop ticking. In this final static configuration clock A2 faces clock B1 and clock A1 faces clock B2:
..........A2..........A1..........
..........B1..........B2..........
Before reaching clock A2, clock B1 passed clock A1 and the difference in their readings, (A1then - B1then), was then registered. *Now*, in the final static configuration, clock B1 faces clock A2 and the difference in their readings is (A2now - B1now). Clearly clock B1 has been checked against two of Inertial system A's clocks so, according to special relativity, the following inequality holds:
(A2now - B1now) > (A1then - B1then) /1/
Before reaching clock B2, clock A1 passed clock B1 and the difference in their readings, (B1then - A1then), was then registered. *Now*, in the final static configuration, clock A1 faces clock B2 and the difference in their readings is (B2now - A1now). Clearly clock A1 has been checked against two of Inertial system B's clocks so, according to special relativity, the following inequality holds:
(B2now - A1now) > (B1then - A1then)
This inequality easily becomes:
(A1then - B1then) > (A1now - B2now)
Since clocks on Inertial system A were synchronous and stopped ticking simultaneously, A1now = A2now. For the same reason B2now = B1now. So the last inequality becomes:
(A1then - B1then) > (A2now - B1now) /2/
Inequalities /1/ and /2/ are contradictory and both are consequences of Einstein's 1905 light postulate. Reductio ad absurdum par excellence. The light postulate is false.
Pentcho Valev
Date Subject Author
7/2/13 Pentcho Valev
7/3/13 Pentcho Valev
7/3/13 grei
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# Thread: Finding a line that splits the integral into two equal regions.
1. ## Finding a line that splits the integral into two equal regions.
Ok so the problem is to find a line ex y=kx + m. That splits f(x)=4x-x^2 into two equal areas and the area is restricted by the x axis and the y axis. so its basically the area from x=0 to x=4 and splitting it. You are not allowed to split it parallel to any of the axises. so u cant make a line that is X=2.
My attempt to the solution was using y=kx So i integrated (4x-x^2)-kx from 0 to b. and i set it equal to half of area of 4x-x^2. The integral of f(x) is equal to F(x)=(2x^2)-(x^3)/3. and the area is 32/3 Area units. So half the area is 16/3 area units. The other region is integral of kx from 0 to b, + integral of 4x-x^2 from b to 4. So i end up with the following equations.
16/3 = (2x^2)-(x^3)/3 -(kx^2)/2 = (kx^2)/2 + 32/3 - 2b^2 + (b^3)/3.
And i get stuck there and dont know how to continue.
Im thankful for any help that i can get.
2. All good - just express k in terms of b, by dividing the-height-of-the-curve-at-b by b.
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Work output
Encyclopedia
In physics
Physics
Physics is a natural science that involves the study of matter and its motion through spacetime, along with related concepts such as energy and force. More broadly, it is the general analysis of nature, conducted in order to understand how the universe behaves.Physics is one of the oldest academic...
, work output is the work
Mechanical work
In physics, work is a scalar quantity that can be described as the product of a force times the distance through which it acts, and it is called the work of the force. Only the component of a force in the direction of the movement of its point of application does work...
done by a simple machine
Simple machine
A simple machine is a mechanical device that changes the direction or magnitude of a force.In general, they can be defined as the simplest mechanisms that use mechanical advantage to multiply force. A simple machine uses a single applied force to do work against a single load force...
, compound machine, or any type of engine model. In common terms, it is the energy output, which for simple machines is always less than the energy input, even though the force
Force
In physics, a force is any influence that causes an object to undergo a change in speed, a change in direction, or a change in shape. In other words, a force is that which can cause an object with mass to change its velocity , i.e., to accelerate, or which can cause a flexible object to deform...
s may be drastically different.
In thermodynamics
Thermodynamics
Thermodynamics is a physical science that studies the effects on material bodies, and on radiation in regions of space, of transfer of heat and of work done on or by the bodies or radiation...
, work output can refer to the thermodynamic work
Work (thermodynamics)
In thermodynamics, work performed by a system is the energy transferred to another system that is measured by the external generalized mechanical constraints on the system. As such, thermodynamic work is a generalization of the concept of mechanical work in mechanics. Thermodynamic work encompasses...
done by a heat engine
Heat engine
In thermodynamics, a heat engine is a system that performs the conversion of heat or thermal energy to mechanical work. It does this by bringing a working substance from a high temperature state to a lower temperature state. A heat "source" generates thermal energy that brings the working substance...
, in which case the amount of work output must be less than the input as energy is lost to heat, as determined by the engine's efficiency
Energy conversion efficiency
Energy conversion efficiency is the ratio between the useful output of an energy conversion machine and the input, in energy terms. The useful output may be electric power, mechanical work, or heat.-Overview:...
.
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# Search results
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Listed under: Parabolas
### Fountain parabolas - mathematics activities
The trajectory of the water from the Captain Cook Memorial Fountain in Lake Burley Griffin, Canberra, provides a good approximation to a parabola. By having students examine other fountains, different 'shapes' can be detected. The mathematical demands increase accordingly. Teachers are encouraged to scan all the ideas suggested ...
### EagleCat: parabola
Explore the graphs of quadratic equations in two forms: (a) the turning point form, y = a(x – h)² + k, and (b) the intercept form, y = (x – a)(x – b). Observe changes to the turning point and the shape of parabolic graphs through various transformations. Alternately, change the equation and observe changes in the x-intercepts ...
### TIMES Module 34: Number and Algebra: quadratic equations - teacher guide
This is a 19-page guide for teachers. It introduces quadratic equations and methods for solving them.
### TIMES Module 35: Number and Algebra: the quadratic function - teacher guide
This is a 29-page guide for teachers. It introduces graphing of quadratic functions.
### Non-linear systems of equations
This digital resource presents a video demonstration, with commentary, of the pen-and-paper technique of constructing the graphs of linear and quadratic functions to find the solutions to a non-linear system of equations. In a single example, a step-by-step illustration of the written method is presented, complemented by ...
### Identifying a possible non-linear rule for a given table of values
This is a five-page HTML resource about solving problems involving the identification of possible non-linear rules matching given tables of values. It contains five questions, two of which are interactive, and one video. The resource discusses and explains identifying a possible non-linear rule that matches a given table ...
This is a teacher resource for quadratics consisting of a website and a PDF with identical content. The website contains a number of screencasts discussing the solutions of exercises, a number of interactives demonstrating properties of quadratics and accompanying screencasts describing the interactives. The resource contains ...
### Coordinate geometry
This is a website designed for both teachers and students that addresses coordinate geometry from the Australian Curriculum for year 9 students. It contains material that shows the connection between algebra and geometry through graphs of lines and curves. There are pages for both teachers and students. The student pages ...
### Graphing calculator by Desmos - iTunes app
Plot functions, create tables, add sliders and animate your graphs. Touch points of interest on the graph to show maximums, minimums and points of intersection. Type in an equation and watch the calculator solve the problem. Free when reviewed on 12/5/2015.
### Desmos Graphing Calculator - Google Play app
Plot functions, create tables, add sliders and animate your graphs. Touch points of interest on the graph to show maximums, minimums and points of intersection. Type in an equation and watch the calculator solve the problem. Free when reviewed on 12/5/2015.
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open import Cat.Diagram.Limit.Finite
open import Cat.Functor.Properties
open import Cat.Diagram.Terminal
open import Cat.Functor.Pullback
open import Cat.Instances.Slice
open import Cat.Functor.Base
open import Cat.Prelude
import Cat.Reasoning
module Cat.Functor.Slice where
# Slicing functorsπ
Let $F : \mathcal{C} \to \mathcal{D}$ be a functor and $X : \mathcal{C}$ an object. By a standard tool in category theory (namely βwhacking an entire commutative diagram with a functorβ), $F$ restricts to a functor $F/X : \mathcal{C}/X \to \mathcal{D}/F(X)$. We call this βslicingβ the functor $F$. This module investigates some of the properties of sliced functors.
open Functor
open /-Obj
open /-Hom
open _=>_
open _β£_
Sliced : β {o β o' β'} {C : Precategory o β} {D : Precategory o' β'}
β (F : Functor C D)
β (X : Precategory.Ob C)
β Functor (Slice C X) (Slice D (F .Fβ X))
Sliced F X .Fβ ob = cut (F .Fβ (ob .map))
Sliced F X .Fβ sh = sh' where
sh' : /-Hom _ _
sh' .map = F .Fβ (sh .map)
sh' .commutes = sym (F .F-β _ _) β ap (F .Fβ) (sh .commutes)
Sliced F X .F-id = ext (F .F-id)
Sliced F X .F-β f g = ext (F .F-β _ _)
# Faithful, fully faithfulπ
Slicing preserves faithfulness and fully-faithfulness. It does not preserve fullness: Even if, by fullness, we get a map $f : x \to y \in \mathcal{C}$ from a map $h : F(x) \to F(y) \in \mathcal{D}/F(X)$, it does not necessarily restrict to a map in $\mathcal{C}/X$. Weβd have to show $F(y)h=F(x)$ and $h=F(f)$ implies $yf=x$, which is possible only if $F$ is faithful.
module _ {o β o' β'} {C : Precategory o β} {D : Precategory o' β'}
{F : Functor C D} {X : Precategory.Ob C} where
private
module D = Cat.Reasoning D
However, if $F$ is faithful, then so are any of its slices $F/X$, and additionally, if $F$ is full, then the sliced functors are also fully faithful.
Sliced-faithful : is-faithful F β is-faithful (Sliced F X)
Sliced-faithful faith p = ext (faith (ap map p))
Sliced-ff : is-fully-faithful F β is-fully-faithful (Sliced F X)
Sliced-ff eqv = is-isoβis-equiv isom where
isom : is-iso _
isom .is-iso.inv sh = record
{ map = equivβinverse eqv (sh .map)
; commutes = ap fst \$ is-contrβis-prop (eqv .is-eqv _)
(_ , F .F-β _ _ β apβ D._β_ refl (equivβcounit eqv _) β sh .commutes) (_ , refl)
}
isom .is-iso.rinv x = ext (equivβcounit eqv _)
isom .is-iso.linv x = ext (equivβunit eqv _)
# Left exactnessπ
If $F$ is left exact (meaning it preserves pullbacks and the terminal object), then $F/X$ is lex as well. We note that it (by definition) preserves products, since products in $\mathcal{C}/X$ are equivalently pullbacks in $\mathcal{C}/X$. Pullbacks are also immediately shown to be preserved, since a square in $\mathcal{C}/X$ is a pullback iff it is a pullback in $\mathcal{C}$.
Sliced-lex
: β {o β o' β'} {C : Precategory o β} {D : Precategory o' β'}
β {F : Functor C D} {X : Precategory.Ob C}
β is-lex F
β is-lex (Sliced F X)
Sliced-lex {C = C} {D = D} {F = F} {X = X} flex = lex where
module D = Cat.Reasoning D
module Dx = Cat.Reasoning (Slice D (F .Fβ X))
module C = Cat.Reasoning C
open is-lex
lex : is-lex (Sliced F X)
lex .pres-pullback = pullback-aboveβpullback-below
β flex .pres-pullback
β pullback-belowβpullback-above
That the slice of lex functor preserves the terminal object is slightly more involved, but not by a lot. The gist of the argument is that we know what the terminal object is: Itβs the identity map! So we can cheat: Since we know that $T$ is terminal, we know that $T \cong \operatorname{id}_{}$ β but $F$ preserves this isomorphism, so we have $F(T) \cong F(\operatorname{id}_{})$. But $F(\operatorname{id}_{})$ is $\operatorname{id}_{}$ again, now in $\mathcal{D}$, so $F(T)$, being isomorphic to the terminal object, is itself terminal!
lex .pres-β€ {T = T} term =
is-terminal-iso (Slice D (F .Fβ X))
(subst (Dx._β
cut (F .Fβ (T .map))) (ap cut (F .F-id))
(F-map-iso (Sliced F X)
(β€-unique (Slice C X) Slice-terminal-object (record { hasβ€ = term }))))
Slice-terminal-object'
A very important property of sliced functors is that, if $L \dashv R$, then $R/X$ is also a right adjoint. The left adjoint isnβt quite $L/X$, because the types there donβt match, nor is it $L/R(x)$ β but itβs quite close. We can adjust that functor by postcomposition with the counit $\varepsilon : LR(x) \to x$A to get a functor left adjoint to $R/X$.
Sliced-adjoints
: β {o β o' β'} {C : Precategory o β} {D : Precategory o' β'}
β {L : Functor C D} {R : Functor D C} (adj : L β£ R) {X : Precategory.Ob D}
β (Ξ£f (adj .counit .Ξ· _) Fβ Sliced L (R .Fβ X)) β£ Sliced R X
module L = Functor L
module R = Functor R
module C = Cat.Reasoning C
module D = Cat.Reasoning D
adj' : (Ξ£f (adj .counit .Ξ· _) Fβ Sliced L (R .Fβ X)) β£ Sliced R X
adj' .unit .is-natural x y f = ext (adj.unit.is-natural _ _ _)
adj' .counit .is-natural x y f = ext (adj.counit.is-natural _ _ _)
adj' .unit .Ξ· x .commutes =
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# kendall
1. ## Math
Rational numbers are fractions and their opposites. All of these numbers are rational numbers. Show that they are rational by writing them in the form a/b or -a/b. 0.2-1/5 -√4 0.333-333/1000 √100 -1.000001 √1/9 All rational numbers have decimal
2. ## math
Shannon can paint a room in 16 hours. Emily can paint the same room in 20 hours. How long does it take for both Shannon and Emily to paint the room it they are working together?
3. ## Math
I am three years younger than my brother, and i am 2 years older than my sister. My mom's age is one less than three times my brother's age. When you add all our ages, you get 87. What are our ages?
4. ## mental arithmetic
if p=3 and q=8 find the value of 3p+2-q how do I find out the value of 3p? can you please explain?
5. ## Algebra
Karrie is moving from Washington D.C. to Denver, CO. She has plotted her 1,670 mile trip on the following graph. (x-min = 0, x-max = 30, x-scale: 3; y-min = 0, y-max = 1800, y-scale = 200) How far is Karrie’s destination after she has traveled 6 hours?
6. ## math
how do I find the possibilities here: x^2 blank 24 x^2 +2x blank It is confusing
7. ## Math
If (a) varies directly as (b) and b= 20 when a= 4, find (b) when a=60 I don't understand this question at all
8. ## Math
Cathy wakes up at 7:00 am each morning. She spends 1/10 hour making her bed, 1/5 hour eating breakfast, and 1/2 hour getting ready for school. How long does Cathy spend doing these things each morning?
9. ## Physics
A squash ball typically rebounds from a surface with 25% of the speed with which it initially struck the surface. Suppose a squash ball is served in a shallow trajectory, from a height above the floor of 39 cm, at a launch angle of 6.0° above the
10. ## Physics
The roof of a two-story house makes an angle of 29° with the horizontal. A ball rolling down the roof rolls off the edge at a speed of 4.5 m/s. The distance to the ground from that point is 8.2 m. (a) How long is the ball in the air? (b) How far from the
11. ## Physics
Part A of this question asks for the sum of vector A = 3x+5y and vector B = 1x-3y. I did this and got the vector A + vector B = 4x+2y. Part B then asks what the magnitude and direction of vector A + vector B? The answer in the back of the textbook says
12. ## Chemistry
Determine the mass of PbSO4 produced when 10mL of 0.2M CuSO4 reacts with 10mL of 0.2M Pb(NO3)2? Balance the equation CuSO4 + Pb(NO3)2 → PbSO4 + Cu(NO3)2 Find the number of moles for CuSO4 c=n/v n=c × v n=0.2 × 0.01 n=0.002 mol Since it is a 1:1 ratio,
13. ## Chemistry
What are some errors that can be made in colorimetry?
14. ## chemistry
is gravimetric analysis or colourimetry more accurate when it comes to precipitation reactions where only around 6g and 100ml of the products are used?
15. ## chemistry
Ammonia is often produced by reacting nitrogen gas with hydrogen gas. How many liters of ammonia gas can be formed from 12.9 L of hydrogen gas at 93.0°C and a pressure of 43.5 kPa? (Assume an excess amount of nitrogen gas.) I have absolutely no idea how
16. ## science
sound travels in theses from its source
17. ## mixture problems
A chemist has two solutions of sulfuric acid. One is a 25% solution, and the other is a 50% solution. How many liters of each does the chemist mix to get 10 liters of a 40% solution? 25% solution L: 50% solution L:
18. ## mixture problems
How many gallons of 45% butterfat cream must be mixed with 500 gallons of 3% butterfat milk to obtain a 5% butterfat milk?
19. ## math
A manufacturer of lapel buttons test marketed a new item at a university. It was found that 1,000 items could be sold if they were priced at \$4, but only 600 items could be sold if the price were raised to \$8. On the other hand, the manufacturer finds that
20. ## Physics
Bats are extremely adept at catching insects in midair. If a 62-g bat flying in one direction at 9 m/s catches a 6-g insect flying in the opposite direction at 4 m/s, what is the speed of the bat immediately after catching the insect?
21. ## Physics
A skateboarder of mass 38.0 kg is riding her 3.90-kg skateboard at a speed of 5.40 m/s. She jumps backward off her skateboard, sending the skateboard forward at a speed of 8.5 m/s. At what speed is the skateboarder moving when her feet hit the ground?
22. ## Algebra
Fiona invested \$1000 at 8% compounded continuously. At the same time Maria invested \$1100 at 8% compounded daily. How long will it take for their investments to be equal in value? Assume there are 365 days in every year. Please help I have tried everything
23. ## science
On average, froghopper insects have a mass of 12.3 mg and jump to a height of 428 mm. The takeoff velocity is achieved as the little critter flexes its leg over a distance of approximately 2.0 mm. Assume a vertical jump with constant acceleration. How long
24. ## science
On average, froghopper insects have a mass of 12.3 mg and jump to a height of 428 mm. The takeoff velocity is achieved as the little critter flexes its leg over a distance of approximately 2.0 mm. Assume a vertical jump with constant acceleration. How long
25. ## Math
16. Alison uses a total of 94.5 cups of dog food each week to feed her three dogs. Each dog eats the same amount. How much does each dog eat in one day? 19. | x - 2 | = 2 Choose from -4, -3, -2, -1, 0, 1, 2, 3, 4. Find all of the numbers that are solutions
26. ## Algebra
When writing an equation in slope-intercept form, does be have to be a positive number or can it be negative? For example, y=-3x-5.
27. ## Algebra
Use properties of logarithms to find the exact value of each expression. Do not use a calculator. 8^log 8^5
28. ## math
What do you get when you double 1/8?
29. ## math
Erica had 7/8 of a pizza left. Her friend ate 1/6 f the pizza that was left. What fraction of the whole pizza did her friend eat? PLEASE HELP!
30. ## Pre-Cal Trig
Two lookout towers are situated on mountain tops A and B, 4 mi from each other. A helicopter firefighting team is located in a valley at point C, 3 mi from A and 2 mi from B. Using the line between A and B as a reference, a lookout spots a fire at an angle
31. ## Pre-Cal
Angles of elevation to an airplane are measured from the top and the base of a building that is 40 m tall. The angle from the top of the building is 37°, and the angle from the base of the building is 40°. Find the altitude of the airplane. (Round your
32. ## Physics
A 21.3-g bullet is fired from a rifle. It takes 3.77 × 10-3 s for the bullet to travel the length of the barrel, and it exits the barrel with a speed of 779 m/s. Assuming that the acceleration of the bullet is constant, find the average net force exerted
33. ## physics
Over a time interval of 1.72 years, the velocity of a planet orbiting a distant star reverses direction, changing from +18.2 km/s to -18.8 km/s. Find (a) the total change in the planet's velocity (in m/s) and (b) its average acceleration (in m/s2) during
34. ## physics
A motorcycle has a constant acceleration of 4.33 m/s2. Both the velocity and acceleration of the motorcycle point in the same direction. How much time is required for the motorcycle to change its speed from (a) 40.0 to 50.0 m/s, and (b) 70.0 to 80.0 m/s?
35. ## physics
Two motorcycles are traveling due east with different velocities. However, 5.81 seconds later, they have the same velocity. During this 5.81 -second interval, motorcycle A has an average acceleration of 4.07 m/s2 due east, while motorcycle B has an average
36. ## physic
The initial velocity and acceleration of four moving objects at a given instant in time are given in the following table. Determine the final speed of each of the objects, assuming that the time elapsed since t = 0 s is 2.2 s.
37. ## physics
A sprinter explodes out of the starting block with an acceleration of + 2.29 m/s2, which she sustains for 1.13 s. Then, her acceleration drops to zero for the rest of the race.
38. ## finance
Templeton Extended Care Facilities, INC. is considering the acquisition of a chain of cemeteries for \$370 million. Since the primary asset of this business is real estate, Templeton’s management has determined that they will be able to borrow the
39. ## Math
This is the error message I see, how do I fix this: listarray:argument must be an array;found:b--anerror.2 debug this try: debugmode(true)
40. ## chemistry
Aluminum reacts with hydrochloric acid to produce aluminum chloride and hydrogen gas. What mass of hydrochloric acid reacts when 87.7 grams of aluminum dissolves? The equation is 2Al+6HCl=2AlCl3+3H2
41. ## Physics
A deorbit burn has been performed. During this deorbit burn a pre-calculated Delta V (change in velocity) of 290 ft/s (or 88.4 m/s) will be used to decrease the Shuttle’s altitude from 205 miles to 60 miles at perigee. The Shuttle’s Orbital Maneuvering
42. ## Chemistry
The heads of matches are made with P4S3. this material is made by heating a mixture of red phosphorus and sulfur: 8P4 + 3S8----> 8P4S3 If a match company uses 5435 grams of phosphorous (p4) with an excess of sulfure, how many grams of P4S3 can be made? Can
43. ## Chemistry
Please help. Calculate the pH at the following points during the titration of 100.0 mL of 0.20 M acetic acid (Ka for acetic acid = 1.8 x 10-5) with 0.10M sodium hydroxide 1. before addition of any base 2. after addition of 30.0mL of base 3. after addition
44. ## Chemistry
Calculate [OH-] and pH for the following strong base solution: 10.0 mL of 0.0105 M Ca(OH)2 diluted to 500.0 mL
45. ## physics
There is a box attached to a spring. The spring has a spring constant of 39 N/m. The ì between the box and the surface is 0.3. The box has a mass of 1 kg. What is the maximum distance that the spring can be stretched from equilibrium before the box begins
46. ## math
-18 > x/6-10 let "x" be the unknown number solve the inequality.
47. ## Math
I am having a hard time with this word problem: A road crew has 3/4 ton of stone to divide evenly among 4 sidewalks. How much stone does the crew need for each sidewalk. a. 3/8t b. 3/16t c. 5/16t d. 5/24t This is what I did: 4 divided by 3/4 is 4/1 x 4/3
What kind of inheritance governs foot length?
Write a pedigree in which there are parents and 4 children, 1 girl and 1 boy are color blind and the other girl and boy are not. I HAVE A FEW QUESTIONS- ?- For a girl to be colorblind does her mother have to be colorblind too or just a carrier?her father?
50. ## THANK YOU
THANK YOU SO MUCH BOBPURSLY.you are a great help to me, i kind of already knew that but it reassured me that i was doing the right thing, i was confused on what side to put my variable on but now i see that i can put it on either side
"x" is the variable 11x=8x-6
"x" is the variable 11x=8x-6
Light travels approximately 5.87 X 10^12 miles in one year. This distance is called a light year. Suppose a star is 2 X 10^4 light years away. How many miles away is that star? Please help with detailed steps to solve problem. Thanks.
Light travels approximately 5.87 X 10^12 miles in one year. This distance is called a light year. Suppose a star is 2 X 10^4 light years away. How many miles away is that star? Please help with detailed steps to solve problem. Thanks.
How do you solve this equations? (xy^-3)^-5 (3x^4)(4x^12) ab^-6 ______ c^0d^-4 x^3y^0(z^-4) Can you please show it step by step? Thanks.
56. ## social studies - geography
How did texas geography affect the native american in the gulf coast?
What was the Geography and Climate in 1600-1700 Colonial America?
How do I find a unit rate?
59. ## Math
Find the indicated outputs for f(x)= 5x^2-2x The 2 after the ^ is an exponent. f(0)=0 f(-1)= f(2)=
60. ## Math
Find the indicated outputs for f(x)=5x^2-2x. f(0)= f(-1)= f(2)=
61. ## Math Geometry
the area of a triangle is 1/2 bh find the area of the given triangle: when b=50 inches h= 15 inches
9 - -9 =
63. ## math
40 35 - 2+i 2-i How do I solve this? there is a division sign between the 40 and 2=i and another one between 35 and 2-i
64. ## math
Help! I am gettting a C in math and I am so cofussed! How do I study better but have fun while I do it? I love math, I take it as a game, I take the #'s and use their equations. I know you might not like problem solving q?, but try to get a tutor & I know
65. ## help
hi i just need help with what the greatest common factor of 51 and 34 is 4+0
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# How do you find the axis of symmetry, graph and find the maximum or minimum value of the function y=x^2+6x+2?
$x = - 3 \text{ }$ is the axis of symmetry
#### Explanation:
the vertex of the parabola is at $\left(h , k\right) = \left(- 3 , - 7\right)$
by the formula
$h = - \frac{b}{2 a}$ and $k = c - {b}^{2} / \left(4 a\right)$
from the given $y = {x}^{2} + 6 x + 2$
$a = 1$, $b = 6$, and $c = 2$
$h = - \frac{b}{2 a} = - \frac{6}{2 \cdot 1} = - 3$
$k = c - {b}^{2} / \left(4 a\right) = 2 - {6}^{2} / \left(4 \cdot 1\right) = 2 - \frac{36}{4} = 2 - 9 = - 7$
The minimum point is the vertex $\left(- 3 , - 7\right)$
graph{y=x^2+6x+2 [-13.58, 6.42, -8.6, 1.4]}
and clearly $x = - 3$ a vertical line which passes thru $\left(- 3 , - 7\right)$ is the line of symmetry.
God bless....I hope the explanation is useful.
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Main Question:
What is the readability level of your hometown newspaper? Does national newspaper, such as USA Today, have higher readability level than your hometown newspaper?
Objective
The primary objectives of this project are to learn how to use random numbers to collect data, organize data, display data and write a statistics project report.
Activities
According to Dr. Fry’s Readability formula below, the readability of an article depends on the average number of words and the average number of syllables in a sentence. For an article if every 100 words have 8 sentences and 120 syllables in average, then according to the Fry Readability Formula, the readability of that article is at second grade level.
It was told that the readability of newspapers is lower than 5th grade level. This project is to test the readability of your hometown newspaper. In order to complete this project, you need to collect data from your hometown newspaper to know how many syllables and how many sentences in average per 100 words of your hometown newspaper. Notice that you need to use random numbers to get unbiased data to make a convincing conclusion.
1. Select a copy of newspaper. (Which copy of the newspaper is representing?)
2. Select articles of the newspaper. (Which articles of the newspaper are representing?)
3. Select 30 sentences from the articles you selected. (Which sentences are representing?)
4. Count number of words and syllables of each sentence. (How to count words and syllables of each sentence? (How many words and syllables for the phrase ’16 GB thumb drive for \$24.95?’)
Your data should be a table as following:
Word count Syllable count
Sentence 1 Xxx Xxx
… Sentence 30 Xxx Xxx
Total Xxxxxxxx5. Create frequency distribution and histogram for your word count and syllable count.
6. Calculate average number of sentence and syllables per hundred words.
The total word count /30 = average word count per sentence
N = 100/average word count per sentence = number of sentence per 100 words
The total syllables count/30 = average syllables count per sentence
M = N*average syllables count per sentence = number of syllables per hundred words
Apply N as vertical coordinate and M as horizontal coordinate to the Fry Graph Readability Formula, you will find out the readability level of the newspaper.
Remark:
2. There is no right or wrong conclusion. As long as your research designs are logical and you can convince your readers, you have a good project.
Edward Fry developed one of the more popular Reading Formulas – the Fry Graph Readability Formula. Fry, who worked as a Fulbright Scholar in Uganda, also helped teachers to teach English as a Second Language (ESL) for a few years, from 1963 and onwards. During his early days, Fry developed readability tests based on graph [A Readability Formula That Saves Time, Journal of Reading (1968)]. This graph-based test determined readability through high school; it was validated with materials from primary and secondary schools and with results of other readability formulas.In 1969, Fry extended the graph to primary levels. In his book Elementary Reading Instruction in 1977, Fry extended the graph to test through the college years. Fry advised that an individual’s vocabulary continues to grow during college years, yet the reading ability varies depending on the individual and the subjects taught.
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The coordinates below represent two linear equations. How many solutions does this system of equations have? Line 1 x y –3 6 3 4 Line 2 x y –6 7 9 2 A. 0 B. exactly 1 C. exactly 2 D. infinitely many I really need help
1
by WendellSproul728
2015-01-05T21:54:23-05:00
Do point slope for each line since you have two points for each
slope: 4-6/3+3=-2/6
This simplifies to -1/3
The equation for the first line is:
y-6=-1/3(x+3)
y-6=-1/3x-1
y=-1/3x+5
The second line's slope :
2-7/9+6= -5/15
This simplifies to -1/3
The equation for this line is:
y-6=-1/3(x+3)
y-6=-1/3x-1
y=-1/3x+5
Since these equations are the same the solution is only one becasue the line only has one x-intercept
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This is a classic example used in many textbooks, papers, and talks on forecast verification to illustrate the characteristics of the various categorical verification scores (e.g., Murphy, 1996; Stephenson, 2000).
In March 1884 Sergeant John Finley initiated twice daily tornado forecasts for eighteen regions in the United States, east of the Rocky Mountains. Finley claimed 95.6% to 98.6% overall accuracy for the first 3-month period, depending on the time and district, with some districts achieving 100% accuracy for all 3 months. A critic of the results pointed out that 98.2% accuracy could be had by merely forecasting "no tornado"! This clearly illustrates the need for more meaningful verification scores.
The contingency table for Finley's (1884) forecasts is:
The standard categorical verification scores are computed and interpreted below:
Accuracy (fraction correct) Overall, 96.6% of the forecasts were correct.
Bias score (frequency bias) Tornados were predicted roughly twice as often as they occurred.
Probability of detection (hit rate) Slightly more than half of the tornados that occurred were correctly predicted to occur.
False alarm ratio 72% of the forecasts for tornados turned out to be false alarms (no tornado occurred).
Threat score (critical success index) Of the tornado events that were either forecast or observed, 23% of those were correctly forecast.
Equitable threat score (Gilbert skill score) -
where Since one could expect to get very few positive tornado forecasts correct due to random chance (only about 2 out of a total of 2803 forecasts), the ETS is almost the same as the TS.
Hanssen and Kuipers discriminant (true skill statistic, Pierce's skill score)
The forecasts were 52% able to separate the "yes" cases from the "no" cases. Because the correct negatives term dominates the others in the contingency table, the HK tends toward the POD when "yes" events are rare.
Heidke skill score
There was a 36% improvement in forecast accuracy when compared to random chance.
Odds ratio The odds were roughly 45 to 1 that a forecast for a tornado would be a hit as opposed to a false alarm.
References:
Finley, J.P., 1884: Tornado predictions. Amer. Meteor. J., 1, 85-88.
Murphy, A.H., 1996: The Finley affair: A signal event in the history of forecast verification. Wea. Forecasting, 11, 3-20.
Stephenson, D.B., 2000: Use of the "odds ratio" for diagnosing forecast skill. Wea. Forecasting, 15, 221-232.
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https://chemistry.stackexchange.com/questions/68922/how-does-density-of-alkali-metals-increase-down-the-group
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# How does density of alkali metals increase down the group?
Size of alkali metals increases down the group, so volume also shows increment, and since volume is inversely proportional to density, how does density also increase down the group?
• Because mass is also increasing, presumably faster than volume. Commented Feb 20, 2017 at 15:49
• Do you mean "why"? "How" suggests you may want to know whether the increase is linear or some other mathematical function. Commented Feb 20, 2017 at 16:10
• The molar volume increases from 13.10 cm$^{3}$/mol for Li to 71.07 cm$^{3}$/mol for Cs. Mass increases from 6.9 gm/mol for Li to 132.9 gm/mol for Cs. So, the mass per atom increases faster than the atomic density decreases. Commented Feb 20, 2017 at 17:05
## 2 Answers
$Density=mass/volume$
1) If mass is increasing and volume is decreasing, then density (mass/volume) will increase.
2)If mass is decreasing and volume is increasing simultaneously, then the density (mass/volume) will decrease.
*3)If both mass and volume are increasing, then we need to check which one of them is increasing at a faster rate(since both are contradictory factors)
a) If mass is increasing at a faster rate than volume, then density will increase.
b) If volume is increasing at a faster rate than mass (i.e denominator in mass/volume is increasing making the overall fraction smaller), then density decreases.
You can apply same logic when both volume and mass are decreasing simultaneously.
Generally, we see that in alkali metals the rate of increase of mass is greater than rate of increase of volume, therefore the density increases down the group.
Also, note that there is no definite reason for why mass is increasing at a faster rate than volume. So asking "why" does it happen isn't a great question.
Its just because density is directly proportional to mass. And the mass increases faster than that of volume expect for the case of potassium which is lighter than sodium.
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Actuarial Outpost Pseudorandom number generation in Excel
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#1
07-16-2009, 02:09 PM
campbell Mary Pat Campbell SOA AAA Join Date: Nov 2003 Location: NY Studying for duolingo and coursera Favorite beer: Murphy's Irish Stout Posts: 82,421 Blog Entries: 6
Pseudorandom number generation in Excel
I was asked as to why RAND() in Excel sucks in a different thread.
I believe this is an important enough topic I'm spinning it off to here.
I'm going to give two practical reasons up front - and this holds for all versions of Excel that have a RAND() function. I can talk about tests of randomness at a later time. The first issue is relatively minor, if annoying, but the second issue is a big bold X over using the function for serious Monte Carlo modeling.
1. RAND() is volatile
That sucker reevaluates every time the spreadsheet recalculates, which could be every time you change a cell if you have it on automatic calculation. If you have too many RAND() calls, your spreadsheet will be slower than molasses even if you just type a short string in a cell.
A bit of info on volatile functions:
http://www.decisionmodels.com/calcsecretsi.htm
You could keep your spreadsheet in manual calculation mode, but that has dangers of its own.
2. RAND() is an unseeded PRNG
This is important for serious Monte Carlo applications. You need to be able to reproduce a particular sequence of numbers. The way this is accomplished in most PRNGs is by having a seed for the PRNG, which is an argument passed to the generator. This will set the initial "random" number for a sequence, and then you will get a specific "random" sequence out for however long it takes until it repeats.
One of the real reasons you need this seed is so you can put the PRNG through a bunch of tests - the Diehard suite of tests is common, but there are other tests you can get from the National Institute of Standards and Tech [NIST].
When you combine 1 & 2, you basically get an untestable algorithm, and I really can't rely on Microsoft statements such as this one:
http://support.microsoft.com/kb/828795
They claim that the current [Excel 2003 & 2007] RAND() has a period of >10^13 [so it will take over 10 trillion calls before it repeats] and that the underlying PRNG satisfies the Diehard tests.
But how can I test that? Sure, I can get around the volatility issue by setting calculation to manual, but then I have the issue of generating over 10 trillion numbers to try to catch the repeat... but because of the structure of Excel, I'm not quite sure what the "order" of these numbers is supposed to be. With seeded PRNGs, you can say "I want the 1000th number generated in the sequences seeded with 2362357" [or whatever]. I can't do that with Excel's RAND().
Of course, I can get around the reproducibility problem by saving all the numbers generated in my pseudorandom sequence... but why should I have to when there are perfectly decent seeded PRNGs out there?
More later.
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#2
07-16-2009, 02:19 PM
tommie frazier Member Join Date: Aug 2003 Favorite beer: The kind with 2 e's Posts: 22,678
thanks campbell. I have always heard that it was lousy for serious simulations, but hadn't cared to dig into the why of it.
#3
07-16-2009, 02:30 PM
ADoubleDot Member SOA Join Date: Nov 2007 Location: Slightly Dusty South Studying for the rest of my life Posts: 37,127
Thanks, that makes sense. The volatility thing is truly annoying, but I guess excel will work for what I need. I just wanted to make sure it wasn't wrong.
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#4
07-16-2009, 03:03 PM
1695814 Member SOA AAA Join Date: Aug 2002 Studying for GED Favorite beer: Root Posts: 34,496
Quote:
Originally Posted by campbell This is important for serious Monte Carlo applications. You need to be able to reproduce a particular sequence of numbers.
why is that?
because of this?
Quote:
Originally Posted by campbell One of the real reasons you need this seed is so you can put the PRNG through a bunch of tests.
and why is that? to show that it's random-enough?
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#5
07-16-2009, 03:05 PM
JMO Carol Marler Non-Actuary Join Date: Sep 2001 Location: Back home again in Indiana Studying for Nothing actuarial. Posts: 37,188
Quote:
Originally Posted by 1695814 why is that?
For your auditors who want you to document your MonteCarlo process. For doing reruns of a stochastic model with corrected data. For coming up with a winning strategy at roulette.
Quote:
because of this?and why is that? to show that it's random-enough?
yes, exactly.
And, when I first started doing actuarial work, PCs had not been invented.
So we had a statistics book that included a table of random numbers. It was used to sample our activities at various times so that we wouldn't have to keep a detailed time log. This data was then used in expense allocations.
It was also useful for selecting samples, such as for selecting data for the auditors. Really.
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#6
07-16-2009, 03:06 PM
Bubba Colbert Member Join Date: Mar 2008 Posts: 1,918
Pseudolus is one of a kind, I don't think you can randomly generate more of him. Although Mrs. Pseudolus has the best shot at coming close.
#7
07-16-2009, 03:43 PM
DeepPurple Member Join Date: Jun 2004 Posts: 4,022
I think that you can seed the Excel PRNG in VB.
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#8
07-16-2009, 03:47 PM
E Eddie Smith SOA AAA Join Date: May 2003 College: UGA Posts: 8,999
Quote:
Originally Posted by Bubba Colbert Pseudolus is one of a kind, I don't think you can randomly generate more of him. Although Mrs. Pseudolus has the best shot at coming close.
Now I think you're alluding to the RANDY() function.
Spoiler:
sarcasm <> hijacking
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#9
07-16-2009, 04:09 PM
echo Member Join Date: Nov 2002 Posts: 848
Quote:
Originally Posted by campbell 1. RAND() is volatile That sucker reevaluates every time the spreadsheet recalculates, which could be every time you change a cell if you have it on automatic calculation.
This is Microsoft's solution. From the help manual under RAND():
"If you want to use RAND to generate a random number but don't want the numbers to change every time the cell is calculated, you can enter =RAND() in the formula bar, and then press F9 to change the formula to a random number. "
What a wonderful solution! What could be better? Interesting that they don't even mention copy/ paste special(values) or as you suggested setting the sheet to manual calculation.
#10
07-16-2009, 04:20 PM
spencerhs5 Member Join Date: Jan 2007 Posts: 1,475
If doing simulation wouldnt you be likely to be using VBAs rnd function? Which if I recall has limitations as well but can be initialized so that its truely random. I could be wrong, but I think this is what I remember.
Tags data science, excel, predictive analytics, prngs, pseudorandom numbers, rand, random
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Author: acroduster1 Time: 01/15/04 03:28 PST This is a reply to message no. 13004 by dfischer Reply | Original Message | New Topic | List Topics | List Messages on This Topic
Current Topic:Gearing formulas Care for a right answer? I'm running very tight on time, so I'll do the one I can do from my head, then I'll tackle the rest later: For #2: Say the worm's rotational velocity is 1rpm, the worm wheel's velocity is then 1/40rpm. At 1rpm, the edge of the worm wheel is moving at 3/4"/minute because pitch indicates how far you move with one revolution. If you do 1 rpm, the edge moves 3/4" in that minute. Armed with the rotational velocity of the worm wheel (1/40rpm) and the edge velocity (3/4"/min.) we can calculate the diameter of the wheel. The formula for radius using edge velocity and rotational velocity is: v=omega*r, where omega is your rotational velocity. Put in the numbers and do your units bookkeeping: 3/4"/min = 1/40 rpm * ((2*pi rad/min)/1rpm) * r Put everything on the left side: 4.775" = r D = 2r, so D = 9.55", your answer is a. A note to gerdb: stay away from the philosophy books, and pay more attention to your engineering texts...
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If the results of Bill Chen's performance at the 2006 World Series of Poker are any indication, poker is a numbers game.
Try these figures on for size: two gold bracelets, \$900,000 in tournament winnings, one poker room sponsorship and one published book, The Mathematics of Poker.
Chen - that's Dr. Chen, actually - has a Ph.D. in math which he's put to use as a software designer, quantitative analyst and (here's where poker comes in) game theorist.
Even so, after winning his second bracelet at the 2006 series, Chen had one explanation for his sudden success.
"Mostly, a lot of luck," he told PokerListings.com.
He doesn't really mean that. When push comes to shove, Chen is a math guy to the core, and quick to dispute any player's assertion that the system makes for boring poker. As a researcher of the game, Chen formed his own opinions about what it takes to dominate on the felt.
"In the tournaments that I've played in, it's been luck at the right times, you know, pocket jacks against ace-king, that sort of thing," he said.
"But I think that it shows that following the mathematical concepts that we have, balancing our play, calling with the right frequencies, and so forth, trying to figure out what your opponent's range of hands is, actually does work, somewhat."
Despite his success, Chen has not turned professional. He still works in the statistical arbitrage department at Susquehanna International Group, where he is allotted six weeks per year to compete in the World Series of Poker.
Chen's first WSOP cash came in 2000 when he placed 11th in the \$1,500 Ace-to-Five Lowball.
For the next five years he played frequently - mostly \$50 and \$100 live tournaments - but it wasn't until the 2006 World Series that Chen earned a reputation as a serious rounder for cashing in eight events.
Two of those cashes were first place wins, giving Chen two bracelets and throwing weight behind his mathematical approach. First he took down the \$3,000 Limit Hold'em event for \$343,618 and then went on to claim the \$2,500 Short-Handed No-Limit Hold'em event for \$442,511.
Chen wrapped up his 2006 WSOP performance by final tabling at the \$1,500 No-Limit Hold'em tournament, where he went out in second and went home as one of poker's players to watch.
Prior to his big wins, though, Chen was a Ph.D. candidate in math at the University of California - Berkeley. The career student discovered poker in 1993, when he started to subsidize his education with poker winnings from the Oaks Card Club in Emeryville, Calif.
By the time he graduated in 1999, Chen was armed with an interest in game theory, his Ph.D. and a job with financial trading company SIG, known for its poker culture.
The company is so poker-friendly, in fact, it lets Chen take time off to compete at the World Series of Poker and other tournaments each year. At these events, the good doctor plays for Team PokerStars.
In addition to expounding the mathematical theories of various poker games, Chen also practices yoga in his spare time at his home in Lafayette Hill, Pa. He is also known for frequenting the forums, notably the rec.gambling.poker news group, and attending the Big August Rec.Gambling Excursion in Las Vegas.
## Tournament Placing
Place Winnings Tournament
6 \$49,304.00 WSOP 2016 - Event 28 - \$10,000 Limit Hold'em Championship
136 \$1,554.00 WSOP 2016 - Event 19 - \$1000 Pot-Limit Omaha
7 \$30,466.00 WSOP 2016 - Event 3 - \$10,000 Seven Card Stud Championship
23 \$3,399.00 WSOP 2015 - Event 48 - \$1,500 Seven Card Stud
6 \$57,706.00 WSOP 2014 - Event 52 - \$10,000 Limit Hold'em
34 \$3,995.00 WSOP 2014 - Event 50 - \$1,500 8-Game Mix
4 \$38,735.00 WSOP 2014 - Event 41 - \$1,500 Dealer's Choice
7 \$65,273.00 WSOP 2014 - Event 22 - \$10,000 HORSE
23 \$4,082.00 WSOP 2014 - Event 16 - \$1,500 2-7 Triple Draw
15 \$4,447.00 WSOP 2014 - Event 7 - \$1,500 Razz
112 \$3,841.00 WSOP 2013 - Event 60 - \$1,500 No-Limit Hold'em
25 \$3,889.00 WSOP 2013 - Event 23 - \$2,500 Stud
63 \$4,366.00 WSOP 2013 - Event 20 - \$1,500 Omaha Hi-Low
7 \$205,856.00 WSOP 2012 - Event 45 - \$50,000 Players Championship
14 \$32,057.00 WSOP 2011 - Event 33 - \$10,000 Stud Hi-Lo Championship
4 \$100,200.00 WSOP 2011 - Event 12 - \$1,500 Triple Chance
169 \$2,954.00 WSOP 2011 - Event 10 - \$1,500 Six-Max No-Limit Hold'em
18 \$7,585.00 WSOP 2010 - Event 48 - \$2,500 Mixed Game
2 \$203,802.00 WSOP 2010 - Event 37 - \$3,000 HORSE
81 \$10,446.00 2010 Special - NAPT Venetian Las Vegas Main Event
42 \$17,271.00 2009 WSOP - Event 56 - \$5,000 Six-Handed No-Limit Hold'em
35 \$4,871.00 2009 WSOP - Event 26 - \$1,500 Limit Hold'em
70 \$4,062.00 2009 WSOP - Event 9 - \$1,500 Sih-Handed No-Limit Hold'em
24 \$5,474.00 2008 WSOP - Event 40, 2-7 Triple Draw Lowball (Limit)
19 \$15,594.00 2008 WSOP - Event 31, No-Limit Hold'em Six-Handed
36 \$3,238.00 2007 Special - APPT Macau Main Event
32 \$6,634.00 2007 WSOP - Event 53, Limit Hold'em Shootout
28 \$5,345.00 2007 WSOP - Event 42, Pot-Limit Omaha Hi-low Split-8 or Better
75 \$6,757.00 2007 WSOP - Event 21, No-Limit Hold'em Shootout
36 \$3,612.00 2007 WSOP - Event 14, Seven Card Stud
1 \$442,511.00 2006 WSOP - Event 21, No-Limit Hold'em- Short handed 6/table
19 \$12,295.00 2006 WSOP - Event 16, Pot-Limit Omaha
1 \$343,618.00 2006 WSOP - Event 7 Limit Hold'em
64 \$4,738.00 2006 WSOP - Event 5, No-limit Hold'em Short Handed, 6/table
61 \$7,578.00 2006 WSOP - Event 2, No-Limit Hold'em
35 \$3,305.00 2005 WSOP - Event 21, \$2,500 Omaha High-Low Split
25 \$4,970.00 2005 WSOP - Event 11, \$2,000 Pot-limit Hold'em
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# Assocative predicates
Peter David ROSS petdr at cs.mu.OZ.AU
Thu Apr 16 17:10:50 AEST 1998
```Hi,
Having implemented a portion of accumulator introduction (it works
for list__length), the solution needs to be made more general. For
that to happen we need to know for some predicates called by a
predicate if that predicate is associative (ie (A+B)+C = A+(B+C))
My preferred solution is that the user annotates the predicates the
user knows to be associative by some method. With the onus on the
user to ensure that the predicate is actually assocative (unless anyone
knows who to prove this in general).
The issues that arise are that
- these annotations need to be accessible by modules which import the
predicate.
- the assocativity of a predicate depends on the mode of the
predicate.
:- func int * int = int.
:- mode in * in = out. ( is assocative )
:- mode out * in = in. ( is not assocative since effectively division
and division not assocative )
Suggestions on the syntax for this?
Pete.
PS.
The predicate below can be transformed to use accumulator recursion
l__multiply([], 1).
l__multiply([H|T], R) :-
l__multiply(T, R0),
R = R0 * H.
If however R = R0 * H becomes R0 = R * H this is equivalent to
R = H / R0
which cannot use accumulator recursion since (A/B)/C != A/(B/C)
```
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https://metanumbers.com/636523
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# 636523 (number)
636,523 (six hundred thirty-six thousand five hundred twenty-three) is an odd six-digits composite number following 636522 and preceding 636524. In scientific notation, it is written as 6.36523 × 105. The sum of its digits is 25. It has a total of 2 prime factors and 4 positive divisors. There are 615,960 positive integers (up to 636523) that are relatively prime to 636523.
## Basic properties
• Is Prime? No
• Number parity Odd
• Number length 6
• Sum of Digits 25
• Digital Root 7
## Name
Short name 636 thousand 523 six hundred thirty-six thousand five hundred twenty-three
## Notation
Scientific notation 6.36523 × 105 636.523 × 103
## Prime Factorization of 636523
Prime Factorization 31 × 20533
Composite number
Distinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 2 Total number of prime factors rad(n) 636523 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) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 636,523 is 31 × 20533. Since it has a total of 2 prime factors, 636,523 is a composite number.
## Divisors of 636523
4 divisors
Even divisors 0 4 2 2
Total Divisors Sum of Divisors Aliquot Sum τ(n) 4 Total number of the positive divisors of n σ(n) 657088 Sum of all the positive divisors of n s(n) 20565 Sum of the proper positive divisors of n A(n) 164272 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 797.824 Returns the nth root of the product of n divisors H(n) 3.87481 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 636,523 can be divided by 4 positive divisors (out of which 0 are even, and 4 are odd). The sum of these divisors (counting 636,523) is 657,088, the average is 164,272.
## Other Arithmetic Functions (n = 636523)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 615960 Total number of positive integers not greater than n that are coprime to n λ(n) 102660 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 51703 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 615,960 positive integers (less than 636,523) that are coprime with 636,523. And there are approximately 51,703 prime numbers less than or equal to 636,523.
## Divisibility of 636523
m n mod m 2 3 4 5 6 7 8 9 1 1 3 3 1 6 3 7
636,523 is not divisible by any number less than or equal to 9.
## Classification of 636523
• Arithmetic
• Semiprime
• Deficient
• Polite
• Square Free
### Other numbers
• LucasCarmichael
## Base conversion (636523)
Base System Value
2 Binary 10011011011001101011
3 Ternary 1012100010221
4 Quaternary 2123121223
5 Quinary 130332043
6 Senary 21350511
8 Octal 2333153
10 Decimal 636523
12 Duodecimal 268437
20 Vigesimal 3jb63
36 Base36 dn57
## Basic calculations (n = 636523)
### Multiplication
n×y
n×2 1273046 1909569 2546092 3182615
### Division
n÷y
n÷2 318262 212174 159131 127305
### Exponentiation
ny
n2 405161529529 257894632260387667 164155865010278738961841 104488983663937653760207918843
### Nth Root
y√n
2√n 797.824 86.021 28.2458 14.4798
## 636523 as geometric shapes
### Circle
Diameter 1.27305e+06 3.99939e+06 1.27285e+12
### Sphere
Volume 1.08027e+18 5.09141e+12 3.99939e+06
### Square
Length = n
Perimeter 2.54609e+06 4.05162e+11 900179
### Cube
Length = n
Surface area 2.43097e+12 2.57895e+17 1.10249e+06
### Equilateral Triangle
Length = n
Perimeter 1.90957e+06 1.7544e+11 551245
### Triangular Pyramid
Length = n
Surface area 7.0176e+11 3.03932e+16 519719
## Cryptographic Hash Functions
md5 8a13d10adc14d39e5be5c21251965839 eaae23e2126bdac0d8da5ac7d5073df53e2c64c1 7f846ea55a69f464df87db752ed039cb5be1e2c89b082bc776c6be4391c6353a 9d56493f1970bd4078423d41d0964f9f6cf76cf2134d58407e1d79029e807c2f3e2aa9c88406c9b959f9e143513f283abae94c2eb5f0ca176c70f0af8c34d4e4 23e57924f221e4dd79853aef7f49c37987c4b560
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| 3.625
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CC-MAIN-2021-49
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https://www.convertunits.com/from/klafter+%5BSwitzerland%5D/to/Q
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## ››Convert klafter [Switzerland] to Q
klafter [Switzerland] Q
Did you mean to convert klafter [Austria] klafter [Switzerland] to Q
How many klafter [Switzerland] in 1 Q? The answer is 0.00013888888888889.
We assume you are converting between klafter [Switzerland] and Q.
You can view more details on each measurement unit:
klafter [Switzerland] or Q
The SI base unit for length is the metre.
1 metre is equal to 0.55555555555556 klafter [Switzerland], or 4000 Q.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between klafters and Q.
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of klafter [Switzerland] to Q
1 klafter [Switzerland] to Q = 7200 Q
2 klafter [Switzerland] to Q = 14400 Q
3 klafter [Switzerland] to Q = 21600 Q
4 klafter [Switzerland] to Q = 28800 Q
5 klafter [Switzerland] to Q = 36000 Q
6 klafter [Switzerland] to Q = 43200 Q
7 klafter [Switzerland] to Q = 50400 Q
8 klafter [Switzerland] to Q = 57600 Q
9 klafter [Switzerland] to Q = 64800 Q
10 klafter [Switzerland] to Q = 72000 Q
## ››Want other units?
You can do the reverse unit conversion from Q to klafter [Switzerland], or enter any two units below:
## Enter two units to convert
From: To:
## ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more!
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https://qa.auth.gr/en/class/1/600010715
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# Numerical Analysis
Title ΑΡΙΘΜΗΤΙΚΗ ΑΝΑΛΥΣΗ / Numerical Analysis Code ΓΕ1100 Faculty Engineering School Civil Engineering Cycle / Level 1st / Undergraduate Teaching Period Winter Common No Status Active Course ID 20000106
### Programme of Study: PPS TPM - EISAKTEOI APO 2022 KAI EXĪS
Registered students: 0
OrientationAttendance TypeSemesterYearECTS
### Programme of Study: PPS TPM (EISACΗTHENTES EŌS KAI 2021)
Registered students: 276
OrientationAttendance TypeSemesterYearECTS
Core program for all studentsCompulsory Course324
Academic Year 2015 – 2016 Class Period Winter Faculty Instructors Weekly Hours 4 Class ID 600010715
SectionInstructors
1. ΓΕ1100Athanasios Mygdalas
Course Type 2016-2020
• Background
Course Type 2011-2015
General Foundation
Mode of Delivery
• Face to face
Language of Instruction
• Greek (Instruction, Examination)
Prerequisites
General Prerequisites
This course is an introduction to "Numerical Analysis " oriented towards computational solution of problems in engineering and sciences. With relative mathematical rigor, it introduces the student to modeling certain typical problems and to the development of methods to solve them on the computer. Introduces the fundamentals of floating point computer calculations and their implications for the accuracy of the obtained solutions. It emphasizes the errors of the methods and their computational efficiency . Furthermore, the student is informed on the existence and use of Numerical Libraries (eg. LAPACK, EISPAC, MINPACK, ODEPACK, SLATEC, NAG, IMSL, MARTRAN), graphic libraries (eg. PGPLOT, MUESLI, DISLIN) and Algebraic Systems (eg. Octave, Scilab, Matlab) with the emphasis on Free and Open Source Software.
Learning Outcomes
(1) modeling typical problems in engineering (2) developing methods to solve these problems on a computer (3) knowledge of floating point operations in and their impact on computer calculations (4) knowledge of the errors committed by numerical methods (5) knowledge of computational efficiency and effectiveness of numerical methods (6) information on the existence and use of Numerical Libraries and other related tools
General Competences
• Apply knowledge in practice
• Make decisions
• Work autonomously
• Work in teams
• Work in an international context
• Work in an interdisciplinary team
• Generate new research ideas
• Advance free, creative and causative thinking
Course Content (Syllabus)
Errors, Root finding, Interpolation , Numerical integration, Numerical Linear Algebra and especially Solving Systems of Linear Equations with exact (direct) and iterative methods , Numerical Solution of Ordinary Differential Equations and Function Fitting by Least Squares and Householder Factorization.
Keywords
Errors, Root finding, Interpolation, Numerical integration, Numerical Linear Algebra, Numerical Solution of Ordinary Differential Equations, Least Squares, Householder Factorization
Educational Material Types
• Notes
• Book
Course Organization
Lectures
Tutorial
Total
Student Assessment
Description
questionnaire
Student Assessment methods
• Written Exam with Short Answer Questions (Formative, Summative)
• Written Exam with Problem Solving (Formative, Summative)
Bibliography
Course Bibliography (Eudoxus)
Βιβλίο [50657724] Εισαγωγή στην Αριθμητική Ανάλυση, 2η Έκδοση, Πιτσούλης Λεωνίδας Βιβλίο [239]: ΕΙΣΑΓΩΓΗ ΣΤΗΝ ΑΡΙΘΜΗΤΙΚΗ ΑΝΑΛΥΣΗ, ΑΚΡΙΒΗΣ Γ.Δ., ΔΟΥΓΑΛΗΣ Β.Α.
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# matplotlib.bezier¶
A module providing some utility functions regarding Bezier path manipulation.
class matplotlib.bezier.BezierSegment(control_points)[source]
Bases: object
A d-dimensional Bezier segment.
Parameters: control_points(N, d) arrayLocation of the N control points.
axis_aligned_extrema(self)[source]
Return the dimension and location of the curve's interior extrema.
The extrema are the points along the curve where one of its partial derivatives is zero.
Returns: dimsint, array_likeIndex $$i$$ of the partial derivative which is zero at each interior extrema. dzerosfloat, array_likeOf same size as dims. The $$t$$ such that $$d/dx_i B(t) = 0$$
property control_points
The control points of the curve.
property degree
Degree of the polynomial. One less the number of control points.
property dimension
The dimension of the curve.
point_at_t(self, t)[source]
Evaluate curve at a single point t. Returns a Tuple[float*d].
property polynomial_coefficients
The polynomial coefficients of the Bezier curve.
Warning
Follows opposite convention from numpy.polyval.
Returns: float, (n+1, d) array_likeCoefficients after expanding in polynomial basis, where $$n$$ is the degree of the bezier curve and $$d$$ its dimension. These are the numbers ($$C_j$$) such that the curve can be written $$\sum_{j=0}^n C_j t^j$$.
Notes
The coefficients are calculated as
${n \choose j} \sum_{i=0}^j (-1)^{i+j} {j \choose i} P_i$
where $$P_i$$ are the control points of the curve.
exception matplotlib.bezier.NonIntersectingPathException[source]
Bases: ValueError
matplotlib.bezier.check_if_parallel(dx1, dy1, dx2, dy2, tolerance=1e-05)[source]
Check if two lines are parallel.
Parameters: dx1, dy1, dx2, dy2floatThe gradients dy/dx of the two lines. tolerancefloatThe angular tolerance in radians up to which the lines are considered parallel. is_parallel 1 if two lines are parallel in same direction. -1 if two lines are parallel in opposite direction. False otherwise.
matplotlib.bezier.concatenate_paths(paths)[source]
[Deprecated] Concatenate a list of paths into a single path.
Notes
Deprecated since version 3.3.
matplotlib.bezier.find_bezier_t_intersecting_with_closedpath(bezier_point_at_t, inside_closedpath, t0=0.0, t1=1.0, tolerance=0.01)[source]
Find the intersection of the Bezier curve with a closed path.
The intersection point t is approximated by two parameters t0, t1 such that t0 <= t <= t1.
Search starts from t0 and t1 and uses a simple bisecting algorithm therefore one of the end points must be inside the path while the other doesn't. The search stops when the distance of the points parametrized by t0 and t1 gets smaller than the given tolerance.
Parameters: bezier_point_at_tcallableA function returning x, y coordinates of the Bezier at parameter t. It must have the signature: bezier_point_at_t(t: float) -> Tuple[float, float] inside_closedpathcallableA function returning True if a given point (x, y) is inside the closed path. It must have the signature: inside_closedpath(point: Tuple[float, float]) -> bool t0, t1floatStart parameters for the search. tolerancefloatMaximal allowed distance between the final points. t0, t1floatThe Bezier path parameters.
matplotlib.bezier.find_control_points(c1x, c1y, mmx, mmy, c2x, c2y)[source]
Find control points of the Bezier curve passing through (c1x, c1y), (mmx, mmy), and (c2x, c2y), at parametric values 0, 0.5, and 1.
matplotlib.bezier.get_cos_sin(x0, y0, x1, y1)[source]
matplotlib.bezier.get_intersection(cx1, cy1, cos_t1, sin_t1, cx2, cy2, cos_t2, sin_t2)[source]
Return the intersection between the line through (cx1, cy1) at angle t1 and the line through (cx2, cy2) at angle t2.
matplotlib.bezier.get_normal_points(cx, cy, cos_t, sin_t, length)[source]
For a line passing through (cx, cy) and having an angle t, return locations of the two points located along its perpendicular line at the distance of length.
matplotlib.bezier.get_parallels(bezier2, width)[source]
Given the quadratic Bezier control points bezier2, returns control points of quadratic Bezier lines roughly parallel to given one separated by width.
matplotlib.bezier.inside_circle(cx, cy, r)[source]
Return a function that checks whether a point is in a circle with center (cx, cy) and radius r.
The returned function has the signature:
f(xy: Tuple[float, float]) -> bool
matplotlib.bezier.make_path_regular(p)[source]
[Deprecated] If the codes attribute of Path p is None, return a copy of p with codes set to (MOVETO, LINETO, LINETO, ..., LINETO); otherwise return p itself.
Notes
Deprecated since version 3.3.
matplotlib.bezier.make_wedged_bezier2(bezier2, width, w1=1.0, wm=0.5, w2=0.0)[source]
Being similar to get_parallels, returns control points of two quadratic Bezier lines having a width roughly parallel to given one separated by width.
matplotlib.bezier.split_bezier_intersecting_with_closedpath(bezier, inside_closedpath, tolerance=0.01)[source]
Split a Bezier curve into two at the intersection with a closed path.
Parameters: bezierarray-like(N, 2)Control points of the Bezier segment. See BezierSegment. inside_closedpathcallableA function returning True if a given point (x, y) is inside the closed path. See also find_bezier_t_intersecting_with_closedpath. tolerancefloatThe tolerance for the intersection. See also find_bezier_t_intersecting_with_closedpath. left, rightLists of control points for the two Bezier segments.
matplotlib.bezier.split_de_casteljau(beta, t)[source]
Split a Bezier segment defined by its control points beta into two separate segments divided at t and return their control points.
matplotlib.bezier.split_path_inout(path, inside, tolerance=0.01, reorder_inout=False)[source]
Divide a path into two segments at the point where inside(x, y) becomes False.
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CC-MAIN-2022-05
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http://math.stackexchange.com/questions/124033/monotone-convergence-theorem-for-non-negative-decreasing-sequence-of-measurable
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# Monotone Convergence Theorem for non-negative decreasing sequence of measurable functions
Let $(X,\mathcal{M},\mu)$ be a measure space and suppose $\{f_n\}$ are non-negative measurable functions decreasing pointwise to $f$. Suppose also that $\int f_1 \lt \infty$. Then $$\int_X f~d\mu = \lim_{n\to\infty}\int_X f_n~d\mu.$$
Atempt:
Since $\{f_n\}$ are decreasing, and converges pointwise to $f$, then $\{-f_n\}$ is increasing pointwise to $f$. So by the monotone convergence theorem $$\int_X -f~d\mu = \lim_{n\to\infty}\int_X -f_n ~d\mu$$ and so $$\int_X f~d\mu = \lim_{n\to\infty}\int_X f_n~d\mu.$$
-
Your attempt is on the right track but is not quite right. In particular, you might think about the hypothesis $\int f_1 < \infty$ and whether you've used it. Hint: What do you know about $g_n = f_1 - f_n$? – cardinal Mar 24 '12 at 20:06
@cardinal: oh yes....$g_n \geq 0$...Thanks – Kuku Mar 24 '12 at 20:07
Yes, $g_n \geq 0$...and, what else? Davide's answer lays out the details. (+1 for showing your work.) – cardinal Mar 24 '12 at 20:08
@Cardinal...Is not homework. I saw it being used here: math.stackexchange.com/questions/86676 and thought I might try and prove it. – Kuku Mar 24 '12 at 20:16
Fair enough. Sorry, being a "standard" result, it sounded a bit like homework. Cheers. :) – cardinal Mar 24 '12 at 20:18
The problem is that $-f_n$ increases to $-f$ which is not non-negative, so we can't apply directly to $-f_n$ the monotone convergence theorem. But if we take $g_n:=f_1-f_n$, then $\{g_n\}$ is an increasing sequence of non-negative measurable functions, which converges pointwise to $f_1-f$. Monotone convergence theorem yields: $$\lim_{n\to +\infty}\int_X (f_1-f_n)d\mu=\int_X\lim_{n\to +\infty} (f_1-f_n)d\mu=\int_X f_1d\mu-\int_X fd\mu$$ so $\lim_{n\to +\infty}\int_X f_nd\mu=\int_X fd\mu$.
Note that the fact that there is an integrable function in the sequence is primordial, indeed, if you take $X$ the real line, $\mathcal M$ its Borel $\sigma$-algebra and $\mu$ the Lebesgue measure, and $f_n(x)=\begin{cases} 1&\mbox{ if }x\geq n\\\ 0&\mbox{ otherwise} \end{cases}$ the sequence $f_n$ decreases to $0$ but $\int_{\mathbb R}f_nd\lambda =+\infty$ for all $n$.
@cardinal Nice proof, yet I have a question: You say that $g_n:= f-f_n$ is an increasing sequence of function, I understand this, but why is she positive? I've thought that since $f_n$ is decreasing to f than $f_n \ge f \forall n$ but this would mean that $f-f_n$ is negative. Where is my error? I can't find it Thank you in advance if you answer :-) – Bman72 Jan 17 '14 at 7:53
I used it with $f_1$ not $f$. – Davide Giraudo Jan 17 '14 at 10:17
@Ale. $f_1\geq f_n$, since $f_n$ being decreasing and hence $f_1-f_n\geq 0$ for each $n$. – Alexander Jun 30 at 5:44
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The area of four walls of a room of height 5 m is 144 m2. If the total surface area of the room is 244 mthen its length is
Asked by arajeevshashank | 29 Nov, 2019, 06:35: PM
The area of four walls of a room of height 5 m is 144 m2. If the total surface area of the room is 244 m2
Length = 5.84 m
Answered by Arun | 30 Nov, 2019, 05:31: PM
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# how prove that $|f(x)|\ge2^n (n+1)$ under these conditions?
Assume $f:[0,1]\mapsto\mathbb{R}$ is continuous and satisfies
1. $\int_0^1x^kf(x) \, dx=0 \quad\forall k\in\{0,1,2,\ldots,n-1\}$,
2. $\int_0^1x^n f(x) \, dx=1$.
How do you prove that $\exists x\in[0 ,1]$ such that $|f(x)|\ge2^n (n+1)$?
-
This a puzzler. :) If no one solves it here, but you, would appreciate if you post a solution as i failed fir an hour (except for the tremendous insight it doesn't hold for $n=0$...) – gnometorule Jan 28 '13 at 19:40
It doesn't hold for $n=0$? Why not? – user7530 Jan 28 '13 at 19:43
@User7530: my bad. – gnometorule Jan 28 '13 at 19:57
## 1 Answer
Let $\tilde P_n$ be the $n$-th shifted Legendre polynomial. Then $\tilde P_n$ has degree $n$, leading coefficient equal to $\binom{2n}n$ and
$$\int_0^1\tilde P_n(x)\tilde P_m(x)\,dx=\frac{\delta_{mn}}{2n+1}\ (\delta\ \style{font-family:inherit;}{\text{denotes the Kronecker delta}})\,.$$
The hypotheses on $f$ imply that $\int_0^1\tilde P_n(x)f(x)\,dx=\binom{2n}n$. If $a_n=(2n+1)\binom{2n}n$, then by the ortogonality of the polynomials $\tilde P_n$ we have
$$0\leq\int_0^1\bigl(f(x)-a_n\tilde P_n(x)\bigr)^2\,dx=\int_0^1f(x)^2\,dx-2a_n\binom{2n}n+\frac{a_n^2}{2n+1}$$
$$=\int_0^1f(x)^2\,dx-(2n+1)\binom{2n}n^2\,.$$
Since $f$ is continuous, then for some $x_0\in[0,1]$ we have $f(x_0)^2\geq(2n+1)\binom{2n}n^2\geq4^n(n+1)^2$ (this inequality can be easily proved by induction on $\boldsymbol{n\geq2}$) , and so $|f(x_0)|\geq2^n(n+1)$, as desired.
-
You probably meant $f(x_0)^2\geq (2n+1)\binom{2n}{n}^2$. – 1015 Jan 28 '13 at 19:49
@julien I mean equality, by invoking mean value theorem for integrals, but yes, you can think it as an inequality. – Matemáticos Chibchas Jan 28 '13 at 19:53
Can you tell me how the mean value theorem for integrals gives you an equality here? – 1015 Jan 28 '13 at 19:58
@julien Apply mean value theorem to $F(x)=\int_0^x f(t)^2\,dt$, together with fundamental theorem of calculus. – Matemáticos Chibchas Jan 28 '13 at 19:59
If you mean $F(1)-F(0)=F'(x_0)(1-0)=f^2(x_0)$, I still only see $f(x_0)^2\geq (2n+1)\binom{2n}{n}^2$. But maybe you meant otherwise. – 1015 Jan 28 '13 at 20:06
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# What is the Convolutional Neural Network Architecture?
Phani Ratan 06 Nov, 2023 • 8 min read
This article was published as a part of the Data Science Blogathon.
## Introduction
Working on a Project on image recognition or Object Detection but didn’t have the basics to build an architecture?
 In this article, we will see what are convolutional neural network architectures right from basic and we will take a basic architecture as a case study to apply our learnings, The only pre-requisite is you just need to know how convolution works But don’t worry it is very simple !!
Let us take a simple Convolutional neural network,
First, there a few things to learn from layer 1 that is striding and padding, we will see each of them in brief with examples
Let us suppose this in the input matrix of 5×5 and a filter of matrix 3X3, for those who don’t know what a filter is a set of weights in a matrix applied on an image or a matrix to obtain the required features, please search on convolution if this is your first time!
Note: We always take the sum or average of all the values while doing a convolution.
A filter can be of any depth, if a filter is having a depth d it can go to a depth of d layers and convolute i.e sum all the (weights x inputs) of d layers
Here the input is of size 5×5 after applying a 3×3 kernel or filters you obtain a 3×3 output feature map so let us try to formulate this
So the output height is formulated and the same with o/p width also…
While applying convolutions we will not obtain the output dimensions the same as input we will lose data over borders so we append a border of zeros and recalculate the convolution covering all the input values.
We will try to formulate this,
Here 2 is for two columns of zeros along with height and width, and formulate the same for width also
#### Striding
Some times we do not want to capture all the data or information available so we skip some neighboring cells let us visualize it,
Here the input matrix or image is of dimensions 5×5 with a filter of 3×3 and a stride of 2 so every time we skip two columns and convolute, let us formulate this
If the dimensions are in float you can take ceil() on the output i.e (next close integer)
Here H refers to height, so the output height is formulated and the same with o/p width also and here 2 is the stride value so you can make it as S in the formulae.
## Pooling
In general terms pooling refers to a small portion, so here we take a small portion of the input and try to take the average value referred to as average pooling or take a maximum value termed as max pooling, so by doing pooling on an image we are not taking out all the values we are taking a summarized value over all the values present !!!
here this is an example of max pooling so here taking a stride of two we are taking the maximum value present in the matrix
## Activation function
The activation function is a node that is put at the end of or in between Neural Networks. They help to decide if the neuron would fire or not. We have different types of activation functions just as in the figure above, but for this post, my focus will be on Rectified Linear Unit (ReLU)
Don’t drop your jaws, this is not that complex this function simply returns 0 if your value is negative else it returns the same value you gave, nothing but eliminates negative outputs and maintains values between 0 to +infinity
Now, that we have learned all the basics needed let us study a basic neural net called LeNet.
## LeNet-5
Before starting we will see what are the architectures designed to date. These models were tested on ImageNet data where we have over a million images and 1000 classes to predict
What are the inputs and outputs (Layer 0 and Layer N) :
Here we are predicting digits based on the input image given, note that here the image is of dimensions height = 32 pixels, width = 32 pixels, and a depth of 1, so we can assume that it is a grayscale image or a black and white one, keeping that in mind the output is a softmax of all the 10 values, here softmax gives probabilities or ratios for all the 10 digits, we can take the number as output with highest probability or ratio.
Convolution 1 (Layer 1) :
Here we are taking the input and convoluting with filters of size 5 x 5 thereby producing an output of size 28 x 28 check the formula above to calculate the output dimensions, the thing here is we have taken 6 such filters and therefore the depth of conv1 is 6, hence its dimensions were, 28 x 28 x 6 now pass this to the pooling layer
Pooling 1 (Layer 2) :
Here we are taking the 28 x 28 x 6 as input and applying average pooling of a matrix of 2×2 and a stride of 2 i.e hovering a 2 x 2 matrix on the input and taking the average of all those four pixels and jumping with a skip of 2 columns every time thus giving 14 x 14 x 6 as output we are computing the pooling for every layer so here the output depth is 6
Convolution 2 (Layer 3) :
Here we are taking the 14 x 14 x 6 i.e the previous o/p and convoluting with a filter of size 5 x5, with a stride of 1 i.e (no skip), and with zero paddings so we get a 10 x 10 output, now here we are taking 16 such filters of depth 6 and convoluting thus obtaining an output of 10 x 10 x 16
Pooling 2 (Layer 4):
Here we are taking the output of the previous layer and performing average pooling with a stride of 2 i.e (skip two columns) and with a filter of size 2 x 2, here we superimpose this filter on the 10 x 10 x 16 layers therefore for each 10 x 10 we obtain 5 x 5 outputs, therefore, obtaining 5 x 5 x 16
Layer (N-2) and Layer (N-1) :
Finally, we flatten all the 5 x 5 x 16 to a single layer of size 400 values an inputting them to a feed-forward neural network of 120 neurons having a weight matrix of size [400,120] and a hidden layer of 84 neurons connected by the 120 neurons with a weight matrix of [120,84] and these 84 neurons indeed are connected to a 10 output neurons
These o/p neurons finalize the predicted number by softmaxing .
## How does a Convolutional Neural Network work actually?
It works through weight sharing and sparse connectivity,
So here as you can see the convolution has some weights these weights are shared by all the input neurons, not each input has a separate weight called weight sharing, and not all input neurons are connected to the output neuron a’o only some which are convoluted are fired known as sparse connectivity, CNN is no different from feed-forward neural networks these two properties make them special !!!
## How to build layers used to construct ConvNets
Here are the types of layers used to build ConvNets:
1. Input Layer: Think of it as the ground floor, where the raw image data enters the CNN.
2. Convolutional Layer: This is where the magic happens! Like skilled workers constructing walls, convolutional filters slide across the image, detecting patterns and extracting features.
3. Pooling Layer: This is like a foreman optimizing the construction process. Pooling reduces the image size, making computations faster and reducing memory usage.
4. Activation Layer: This is like adding color and vibrancy to the building. Activation functions introduce non-linearity, allowing the network to learn complex relationships between features.
5. Fully Connected Layer: This is like the top floor, where all the features come together for the final decision. The network takes in the extracted features and classifies the image.
6. Output Layer: This is like the roof, where the final classification result is displayed
## Points to look at
1. After every convolution the output is sent to an activation function so as to obtain better features and maintaining positivity eg: ReLu
2. Sparse connectivity and weight sharing are the main reason for a convolutional neural network to work
3. The concept of choosing a number of filters in between layers and padding and stride and filter dimensions are taken on doing a number of experimentations, don’t worry about that, focus on building foundation, someday you will do those experiments and build a more productive one!!!
## FAQs
Q1. Is CNN supervised or unsupervised?
CNNs are versatile machine learning algorithms capable of both supervised and unsupervised learning.. In supervised learning, the CNN is trained on labeled data, while in unsupervised learning, it is trained on unlabeled data.
Q2. Why CNN is better than Ann?
Differences between CNNs and ANNs:
2. capable of extracting local features from images.
3. Can share weights across different parts of the image.
4. Can learn hierarchical representations of images.
5. Tasks: Image classification, object detection, image segmentation, facial recognition. ANN
6. Suitable for a wider range of tasks, including image classification, but not as specialized as CNN.
7. capable of learning from non-image data.
8. Tasks: regression, classification of non-image data, time series forecasting, natural language processing
Q3. What is loss layer in CNN?
In CNNs, the loss layer measures how well the network’s predictions match the actual data. It helps the network improve its performance by adjusting its weights.
Phani Ratan 06 Nov 2023
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# Liberal arts math
posted by .
if this is the eaquation of a line y=2x+3 what is the parallel line and perpendicular line?
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please check my homework given P(7,2) and the line y + 3x=7 find the equation for the line parallel line: y+3x+7 y=-3(x-7) +7 y-2=-3x+21+7 y=-3x+28+2 y=-3x+30 perpendicular line: y-2=1/3(x-7) +7 y-2=1/3x-7/3 +21/3 y-2==1/3x + 14 y=1/3x …
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# Unraveling the Enigma: A Comprehensive Guide to the World of Logic Puzzles
Logic puzzles have been a part of human culture for centuries, providing entertainment and sharpening the mind. From ancient riddles to modern-day crosswords, these brain teasers come in many forms, each with its unique set of challenges. In this guide, we will explore the various types of logic puzzles, their history, and what makes them so captivating. Get ready to unravel the enigma and discover the thrill of solving these intellectual puzzles.
## The Thrill of Logic Puzzles: A Brief History
### The Roots of Logic Puzzles
Logic puzzles have been a part of human history for centuries, with their origins dating back to ancient Greece. The great philosopher Aristotle, in his works, explored the art of reasoning and logic, laying the foundation for what would later become the basis of modern logic puzzles.
In medieval Europe, puzzles and brain teasers were often used as a tool for teaching and learning. The famous book “The Art of Discovery” by Charles Lull, published in 1314, featured a collection of puzzles designed to improve problem-solving skills and critical thinking.
With the advent of modern printing, logic puzzles gained popularity in the form of mathematical problems and riddles. The “Enumerations of the Mathematician” by John Wallis, published in 1657, contained a series of brain teasers that required logical thinking and deductive reasoning.
In the 19th century, puzzles and games became more widespread, with the publication of books such as “The Book of Brain Problems” by Henry Ernest Dudeney and “The Golden Age of Puzzles” by Sam Loyd. These books featured a variety of puzzles, including logic puzzles, that challenged readers to think critically and creatively.
Today, logic puzzles continue to be a popular form of entertainment and a useful tool for improving cognitive skills. With the rise of the internet, a wealth of logic puzzles can be found online, making them accessible to people all over the world. Whether you’re a fan of crosswords, Sudoku, or other types of puzzles, the world of logic puzzles offers a wealth of challenges and opportunities for intellectual growth.
### The Evolution of Logic Puzzles
• Origins:
• Puzzles have been a part of human culture for centuries, serving as a source of entertainment and intellectual stimulation.
• Logic puzzles, in particular, have their roots in ancient Greece, where philosophers such as Aristotle and Plato explored the principles of reasoning and deductive reasoning.
• Medieval Period:
• During the medieval period, puzzles became more prevalent in the form of riddles and brain teasers, often used as a tool for education and moral instruction.
• One notable example is the “Western riddle” or “carousel puzzle,” which was used to teach young people about the Catholic Church’s sacraments.
• 19th Century:
• The 19th century saw the rise of more complex puzzles, such as the “15-puzzle” and “sliding puzzle,” which laid the foundation for modern logic puzzles.
• These puzzles required strategic thinking and problem-solving skills, capturing the imagination of people across all ages and social classes.
• 20th Century:
• The 20th century saw a surge in the popularity of logic puzzles, particularly with the publication of books like “The Code Book” by Simon Singh and “The Puzzle Palace” by David Kahn.
• The advent of computer technology allowed for the creation of new and innovative puzzles, such as “escape rooms” and “point-and-click” adventure games.
• Present Day:
• Today, logic puzzles continue to evolve and adapt to new technologies, with mobile apps and online platforms providing a wide range of puzzles for players to enjoy.
• The global community of puzzle enthusiasts continues to grow, fostering a shared passion for intellectual challenge and creative problem-solving.
### The Modern Age of Logic Puzzles
#### The Resurgence of Logic Puzzles in the Digital Age
• The proliferation of digital technology and the internet has led to a resurgence in the popularity of logic puzzles.
• Online platforms and mobile apps have made it easier for people to access and engage with logic puzzles, regardless of their location.
• The convenience and accessibility of digital puzzles have contributed to their growing popularity among people of all ages and backgrounds.
#### The Emergence of New Puzzle Formats and Genres
• The modern age of logic puzzles has seen the emergence of new puzzle formats and genres, such as escape rooms, hidden object games, and point-and-click adventures.
• These new formats often incorporate elements of storytelling, graphics, and interactivity, making them more engaging and immersive for players.
• The variety of puzzle formats and genres has expanded the appeal of logic puzzles to a wider audience, including those who may not have been interested in traditional puzzles.
#### The Influence of Logic Puzzles on Other Media
• The popularity of logic puzzles has also had an impact on other media, such as literature, film, and television.
• Many works of fiction now incorporate logic puzzles as plot elements, and some even use them as the central focus of the story.
• This has led to a greater appreciation of logic puzzles as a form of storytelling and a creative tool for writers and filmmakers.
#### The Growing Competitive Scene for Logic Puzzles
• The modern age of logic puzzles has also seen the emergence of a competitive scene, with tournaments and championships being held around the world.
• These competitions often involve solving a series of puzzles within a set time limit, with the fastest and most accurate solvers winning prizes and recognition.
• The competitive aspect of logic puzzles has added a new dimension to the hobby, and has inspired many people to improve their skills and push themselves to new limits.
## Types of Logic Puzzles: An Overview
Key takeaway: Logic puzzles have been a part of human history for centuries, with their origins dating back to ancient Greece. The modern age of logic puzzles has seen a resurgence in their popularity, due to the proliferation of digital technology and the internet. This has led to the emergence of new puzzle formats and genres, such as escape rooms, hidden object games, and point-and-click adventure games. Logic puzzles offer a wealth of challenges and opportunities for intellectual growth, and can be found online, in newspapers, and in a variety of book formats. Whether you’re a fan of crosswords, Sudoku, or other types of puzzles, the world of logic puzzles offers a wealth of challenges and opportunities for intellectual challenge and creative problem-solving.
### Crosswords and Word Searches
Crosswords and word searches are two of the most popular types of logic puzzles. Both of these puzzles involve finding words hidden in a grid or pattern, but they differ in their approach and rules.
#### Crosswords
Crosswords are puzzles where words are filled in a grid of white and black squares. The words are filled horizontally and vertically, and each word must be connected to another word by a black square. The words can be filled in any direction, but they must be connected to the words that are already filled.
To solve a crossword puzzle, the solver must use the clues provided to find the words that fit in the grid. The clues are usually given at the beginning of each row and column, and they provide hints to the words that can be filled in that particular row or column.
Crosswords can be classified into two types: themed and non-themed. Themed crosswords have a specific theme that ties all the clues together, while non-themed crosswords do not have a specific theme.
#### Word Searches
Word searches are puzzles where words are hidden in a grid of letters. The words can be hidden horizontally, vertically, or diagonally, and the solver must find them by searching the grid.
To solve a word search puzzle, the solver must use the clues provided to find the words that are hidden in the grid. The clues are usually given at the beginning of the puzzle, and they provide hints to the words that can be found in the grid.
Word searches can be classified into two types: themed and non-themed. Themed word searches have a specific theme that ties all the words together, while non-themed word searches do not have a specific theme.
Both crosswords and word searches are great ways to exercise the brain and improve vocabulary. They are also a fun way to pass the time and challenge the mind.
### Sudoku and Number Placement Puzzles
Sudoku, a puzzle game that originated in Japan, has become a global phenomenon, captivating millions of players worldwide. It is a number-placement puzzle that requires the player to fill a grid with numbers so that each row, column, and region (subgrid) contains every number from 1 to 9. The puzzle is created by filling in some of the grid cells with numbers, leaving the rest for the player to solve.
The goal of Sudoku is to fill in the empty cells with the correct numbers, following a set of rules. The rules are as follows:
1. Each row must contain the numbers 1-9 without repetition.
2. Each column must contain the numbers 1-9 without repetition.
3. Each region (subgrid) must contain the numbers 1-9 without repetition.
These rules may seem simple, but they can lead to challenging and complex puzzles that require critical thinking and logical reasoning. The beauty of Sudoku lies in its simplicity and the endless possibilities it offers for creating new and exciting puzzles.
Sudoku puzzles come in different difficulty levels, from beginner to expert, and can be found in newspapers, magazines, and online. The popularity of Sudoku has led to the creation of various variants, such as Samurai Sudoku, Killer Sudoku, and Conceptis Puzzles, each with its unique set of rules and challenges.
In addition to Sudoku, there are other number-placement puzzles, such as Battleships, Crosswords, and Wordsearches, that offer a fun and engaging way to challenge your brain and improve your cognitive skills.
Whether you are a seasoned puzzle solver or a newcomer to the world of logic puzzles, Sudoku and number-placement puzzles offer endless opportunities for fun and intellectual stimulation. So, grab a pen and paper, and get ready to unleash your inner detective as you unravel the mysteries of these fascinating puzzles.
### Mazes and Labyrinths
Mazes and labyrinths are two types of logic puzzles that have been popular for centuries. These puzzles are characterized by a series of paths that the solver must navigate in order to reach a specific destination.
#### Mazes
A maze is a puzzle that consists of a path with branches and dead ends. The objective of a maze is to find the shortest path from the starting point to the end point. Mazes can be found in many different forms, including physical mazes etched into paper or carved into walls, and virtual mazes found in video games and computer programs.
#### Labyrinths
A labyrinth is a puzzle that consists of a single path that winds its way through a series of twists and turns. Unlike a maze, a labyrinth has no dead ends, and the objective is simply to reach the center of the labyrinth. Labyrinths can be found in many different forms, including physical labyrinths etched into the ground or carved into walls, and virtual labyrinths found in video games and computer programs.
Both mazes and labyrinths require the solver to use logical thinking and problem-solving skills to navigate the path to the destination. They can be solved using a variety of techniques, including visualization, pattern recognition, and trial and error.
Some common variations of mazes and labyrinths include:
• Alice in Wonderland Mazes: These mazes often have unusual and unexpected twists and turns, and may feature imaginative themes and illustrations.
• 3D Mazes: These mazes are created in three dimensions, and may have physical obstacles that the solver must navigate around.
• Knotted Labyrinths: These labyrinths are created by knotting a rope or string into a specific pattern, and the solver must follow the twists and turns of the rope to reach the center.
Mazes and labyrinths are popular puzzles because they challenge the solver to think creatively and use logic to solve the puzzle. They can be enjoyed by people of all ages and skill levels, and are a fun way to challenge your mind and improve your problem-solving skills.
Sudoku, a puzzle game that originated in Japan, has become a worldwide phenomenon since its inception in the late 1970s. It is a number-placement puzzle that requires the player to fill a grid with numbers so that each row, column, and 3×3 subgrid contains every number from 1 to 9. The game is named after the Japanese word “sudoku,” which means “single number.”
The objective of Sudoku is to fill the grid with numbers from 1 to 9, adhering to the following rules:
1. Each row must contain the numbers 1 to 9 without repetition.
2. Each column must contain the numbers 1 to 9 without repetition.
3. Each 3×3 subgrid must contain the numbers 1 to 9 without repetition.
Sudoku puzzles come in varying degrees of difficulty, ranging from beginner to expert level. They can be found in newspapers, magazines, and online, and they have become a popular pastime for people of all ages.
Number placement puzzles, like Sudoku, are a type of logic puzzle that involves filling a grid with numbers so that each row, column, and defined section of the grid meets specific criteria. The most common form of number placement puzzle is Sudoku, but there are many other variations, such as Kakuro, Cross Sum, and Fill-ins.
The popularity of Sudoku and other number placement puzzles can be attributed to their simplicity and accessibility. They require no specialized knowledge or skills, and can be enjoyed by anyone with an interest in puzzles. Additionally, they offer a range of difficulty levels, making them suitable for both beginners and experienced puzzlers.
Overall, Sudoku and number placement puzzles are an enjoyable and challenging way to exercise the mind, and they are a great way to pass the time. Whether you are a seasoned puzzler or a newcomer to the world of logic puzzles, there is a Sudoku or number placement puzzle out there for you.
### Tic-Tac-Toe and Connect Four
Tic-Tac-Toe and Connect Four are two popular logic puzzles that are played by people of all ages. Both of these games are played on a grid, with Tic-Tac-Toe being played on a 3×3 grid and Connect Four being played on a 7×6 grid.
#### Tic-Tac-Toe
Tic-Tac-Toe is a simple game that is played by two players, X and O. The game is played on a 3×3 grid, with each player taking turns placing their mark, either X or O, on an empty square. The first player to get three of their marks in a row, either horizontally, vertically, or diagonally, wins the game. If all of the squares are filled and no player has three in a row, the game is a draw.
Tic-Tac-Toe is a game of strategy, and players must think several moves ahead to ensure that they win. It is a game that can be played by people of all ages, and it is a great way to improve one’s problem-solving skills.
#### Connect Four
Connect Four is a game that is played by two players, X and O. The game is played on a 7×6 grid, with each player taking turns placing their mark, either X or O, on an empty square. The goal of the game is to get four of your marks in a row, either horizontally, vertically, or diagonally.
Connect Four is a game of strategy, and players must think several moves ahead to ensure that they win. It is a game that can be played by people of all ages, and it is a great way to improve one’s problem-solving skills.
In conclusion, Tic-Tac-Toe and Connect Four are two popular logic puzzles that are played by people of all ages. Both of these games are played on a grid, with Tic-Tac-Toe being played on a 3×3 grid and Connect Four being played on a 7×6 grid. These games are a great way to improve one’s problem-solving skills and are fun for people of all ages.
### Cryptograms and Ciphers
Cryptograms and ciphers are two types of logic puzzles that involve the use of codes and secret messages.
#### Cryptograms
Cryptograms are puzzles that involve replacing letters in a message with other letters or symbols to create a new message. The goal is to decode the message by figuring out the substitutions used to create it. Cryptograms can be solved using a variety of techniques, including frequency analysis, letter-based substitutions, and word-based substitutions.
#### Ciphers
Ciphers are puzzles that involve encrypting a message using a specific algorithm or rule. The goal is to decrypt the message by figuring out the algorithm or rule used to encrypt it. Ciphers can be solved using techniques such as transposition, substitution, and frequency analysis.
Both cryptograms and ciphers have a long history, dating back to ancient times. Cryptograms were used by ancient cultures to communicate secret messages, while ciphers were used by military forces to protect sensitive information. Today, cryptograms and ciphers are still popular as puzzles and can be found in books, newspapers, and online.
While some cryptograms and ciphers can be solved using simple techniques, others can be quite challenging and require a great deal of analytical thinking and problem-solving skills. For those who enjoy a good challenge, these types of logic puzzles can be a lot of fun to solve.
## Decoding Logic Puzzles: Strategies and Techniques
### The Art of Observation
The ability to observe the details of a puzzle is a crucial skill in decoding logic puzzles. This section will explore the art of observation and the different techniques that can be used to help solve these types of puzzles.
#### The Importance of Observation
Observation is the act of carefully looking at something in order to notice and understand details. In the context of logic puzzles, observation is essential because it allows the solver to identify clues and patterns that may be hidden or disguised within the puzzle.
For example, in a classic logic puzzle like the “missing number” puzzle, observation skills can help the solver notice that one number is missing from a sequence of numbers. Similarly, in a more complex puzzle like a Sudoku, observation skills can help the solver identify groups of numbers that belong together.
#### Techniques for Improving Observation Skills
Improving observation skills is an important part of becoming a skilled puzzle solver. Here are some techniques that can help:
1. Slow Down: It’s easy to overlook important details when rushing through a puzzle. To improve observation skills, try taking a slow and deliberate approach to solving puzzles.
2. Look for Patterns: Patterns can be a powerful tool in solving logic puzzles. Look for repeated patterns or sequences in the puzzle, and try to identify any underlying rules or principles that might apply.
3. Use Mental Imagery: Mental imagery can help you visualize different possibilities and make connections between different elements of a puzzle. Try visualizing the puzzle in your mind, and see if this helps you notice any patterns or clues that you might have missed.
4. Take Breaks: Sometimes, taking a break from a puzzle can help you see things more clearly. If you’re feeling stuck, try stepping away from the puzzle for a few minutes and then come back to it with fresh eyes.
By developing your observation skills, you’ll be better equipped to solve a wide range of logic puzzles. So next time you encounter a puzzle, take a moment to carefully observe the details, and see if you can uncover any hidden clues or patterns.
### Analyzing Patterns and Trends
Puzzles are designed to challenge the human mind and provide a sense of accomplishment when solved. The ability to analyze patterns and trends is a crucial skill when it comes to solving logic puzzles. By identifying recurring patterns, you can uncover the underlying structure of the puzzle and use that information to make educated guesses about the solution.
One of the most common patterns in logic puzzles is the use of symmetry. Symmetry is when a puzzle has a mirror-image pattern that can be used to solve one half of the puzzle, which can then be used to solve the other half. For example, a Sudoku puzzle may have a diagonal line of numbers that are the same on both sides of the puzzle. By solving one side of the puzzle, you can use the symmetry to solve the other side as well.
Another pattern that is often used in logic puzzles is the use of deductions. Deductions are based on the principle that if a statement is true, then a related statement must also be true. For example, if it is known that a certain character cannot be in a certain location, then that character must be in another location. By making deductions based on the information given in the puzzle, you can eliminate possible solutions and arrive at the correct answer.
It is also important to pay attention to the language used in the puzzle. Puzzles are often designed to mislead the solver, and the language used can be a clue to the solution. For example, a clue in a crossword puzzle may be written in a way that suggests a different answer than the actual solution. By analyzing the language used in the puzzle, you can uncover hidden clues and arrive at the solution more quickly.
Finally, it is important to approach logic puzzles with an open mind. Many puzzles require a shift in perspective or a creative solution, so it is important to be willing to think outside the box. By analyzing patterns and trends, making deductions, paying attention to language, and approaching the puzzle with an open mind, you can unlock the secrets of logic puzzles and solve even the most challenging puzzles.
### Eliminating Possibilities
When it comes to solving logic puzzles, one of the most effective strategies is to eliminate possibilities. This technique involves narrowing down the possible solutions by systematically eliminating those that cannot be true based on the given clues. By doing so, you can reduce the number of potential solutions and make the puzzle easier to solve.
There are several ways to eliminate possibilities in logic puzzles. One common method is to use the process of elimination. This involves starting with all the possible solutions and then systematically eliminating those that cannot be true based on the clues provided. For example, if a puzzle has three suspects and two clues, you can start by eliminating one suspect based on the first clue, and then eliminate another suspect based on the second clue. This leaves you with one suspect who must be the culprit.
Another technique for eliminating possibilities is to use deduction. This involves using the clues to make logical deductions about the possible solutions. For example, if a puzzle has two clues that are contradictory, then none of the solutions can be true. By using deduction, you can eliminate one or more of the possible solutions based on the given clues.
In addition to elimination and deduction, there are other techniques for eliminating possibilities in logic puzzles. For example, you can use the process of induction to generate hypotheses about the possible solutions and then test those hypotheses against the clues. You can also use the process of reversal to start with the solution and work backward to find the clues that support it.
Overall, eliminating possibilities is a powerful strategy for solving logic puzzles. By systematically narrowing down the possible solutions, you can make the puzzle easier to solve and increase your chances of finding the correct answer.
### Backtracking and Trial and Error
Backtracking and trial and error are two common techniques used in solving logic puzzles. These techniques involve trying different solutions and evaluating them until the correct solution is found.
Backtracking is a systematic approach to solving a problem by trying all possible solutions and eliminating those that do not work. This technique is particularly useful in solving puzzles that have a large number of possible solutions.
Trial and error, on the other hand, involves trying different solutions until the correct one is found. This technique is more intuitive and can be used in situations where backtracking is not practical.
Both techniques require a certain level of creativity and problem-solving skills. It is important to approach the puzzle with an open mind and to be willing to try different solutions until the correct one is found.
It is also important to keep track of the progress made in solving the puzzle. This can be done by writing down the solutions that have been tried and eliminating those that do not work. This helps to avoid repeating the same solutions and saves time in the solving process.
In conclusion, backtracking and trial and error are two useful techniques in solving logic puzzles. They require creativity and problem-solving skills, and it is important to keep track of the progress made in solving the puzzle.
### Using Logic and Reasoning
• Employing deductive reasoning to narrow down possible solutions
• Identifying patterns and relationships between elements
• Applying mathematical concepts when appropriate
• Eliminating incorrect assumptions and conclusions
Deductive Reasoning
• Begin by reading the problem carefully to identify the given information
• Formulate a plan or strategy to approach the problem
• Identify the possible solutions based on the information provided
• Deduce the correct solution by logically eliminating incorrect ones
Identifying Patterns and Relationships
• Look for recurring themes or motifs in the puzzle
• Analyze the connections between different elements of the puzzle
• Identify any hidden rules or principles that may be governing the puzzle
Applying Mathematical Concepts
• Use algebra, geometry, or other mathematical tools to solve puzzles
• Identify patterns or relationships that can be represented mathematically
• Apply logical principles to solve problems that involve numerical data
Eliminating Incorrect Assumptions and Conclusions
• Avoid jumping to conclusions without evidence
• Be open to revising assumptions or conclusions based on new information
• Use logic to identify inconsistencies or contradictions in the puzzle
• Continuously refine and re-evaluate your reasoning throughout the problem-solving process
## Popular Logic Puzzles: A Deep Dive
### The Challenges of Crosswords
Crosswords are a popular type of logic puzzle that involve filling in a grid of words with clues provided for each entry. Despite their widespread appeal, crosswords present a number of challenges that make them a truly engaging and mentally stimulating activity.
One of the main challenges of crosswords is their requirement for both linguistic and analytical skills. In order to solve a crossword, a solver must not only have a strong vocabulary, but also be able to deduce the meaning of words based on their context clues. Additionally, crosswords often involve an element of logic, as clues may require the solver to use deductive reasoning to determine the correct answer.
Another challenge of crosswords is their inherent time pressure. Unlike other types of puzzles, crosswords have a time limit for completion, adding an extra layer of pressure to the activity. This pressure can be heightened further by the presence of deadlines, such as those found in newspaper crosswords, which can make the activity even more thrilling.
Finally, crosswords also present a challenge in terms of the sheer volume of information that must be processed in order to solve them. A single crossword can contain dozens or even hundreds of clues, each with its own set of variables and constraints. Solving a crossword requires the ability to process and remember large amounts of information, making it a true test of mental acuity.
Overall, the challenges of crosswords make them a highly engaging and rewarding activity for puzzle enthusiasts of all levels. Whether you are looking to improve your vocabulary, hone your analytical skills, or simply pass the time, crosswords offer a unique and stimulating challenge that is sure to keep you engaged from start to finish.
### Unraveling the Mystery of Sudoku
Sudoku, a puzzle game that originated in Japan, has become a worldwide phenomenon, captivating the minds of millions. The name “Sudoku” is derived from the Japanese words “Su” meaning “number” and “Doku” meaning “single”. The objective of the game is to fill a grid of squares with numbers so that each row, column, and 3×3 subgrid contains every number from 1 to 9.
The game is played on a 9×9 grid, with some squares already filled with numbers. The goal is to fill in the remaining squares, following the rules:
1. Each 3×3 subgrid must contain the numbers 1-9 without repetition.
To solve a Sudoku puzzle, one can use a variety of techniques such as the “hiding’ technique, “x-wing” technique, “candidates” technique, and “forcing chains” technique. These techniques involve identifying patterns and relationships between the numbers in the grid, and using them to eliminate possibilities and find the solution.
The beauty of Sudoku lies in its simplicity and the endless variety of puzzles that can be created. Whether you’re a beginner or an experienced player, Sudoku offers a fun and challenging way to exercise your mind and improve your problem-solving skills.
### The Challenge of Mazes and Labyrinths
Mazes and labyrinths are among the most popular types of logic puzzles. They have been enjoyed by people of all ages for centuries, and their appeal lies in the challenge of finding one’s way through a complex path.
#### Types of Mazes and Labyrinths
There are several types of mazes and labyrinths, each with its own unique characteristics. Some of the most common types include:
• Classic Mazes: These are the most traditional type of maze, featuring a winding path that leads from the entrance to the exit. The walls of the maze are typically impenetrable, and the goal is to reach the exit without retracing your steps.
• Single-Path Mazes: These mazes have only one path that leads from the entrance to the exit. The challenge lies in finding the correct path, which may be obscured or hidden in some way.
• Multi-Path Mazes: These mazes have multiple paths that lead from the entrance to the exit. The challenge lies in finding the correct path, or in finding all of the paths.
• Puzzle Mazes: These mazes are often based on a specific theme or concept, such as math or logic. They may have different rules or requirements, such as only moving in certain directions or using certain clues.
#### Solving Mazes and Labyrinths
Solving mazes and labyrinths requires a combination of logical thinking and problem-solving skills. Here are some tips for successfully navigating these types of puzzles:
• Observe the Layout: Take note of the layout of the maze, including the location of walls, dead ends, and other features. This can help you to develop a strategy for finding the exit.
• Use Clues: Many mazes and labyrinths include clues that can help you to navigate the path. These may be hidden in plain sight, or they may require you to use your imagination to decipher them.
• Think Logically: Mazes and labyrinths are all about logical thinking. Use your brain to figure out the best way to get from the entrance to the exit, and don’t be afraid to backtrack if you get lost.
• Stay Focused: It can be easy to get lost in a maze or labyrinth, especially if you are not paying attention to your surroundings. Stay focused on your goal, and don’t let distractions sidetrack you.
In conclusion, mazes and labyrinths are a challenging and rewarding type of logic puzzle. By observing the layout, using clues, thinking logically, and staying focused, you can successfully navigate these complex paths and find your way to the exit.
### Strategies for Tic-Tac-Toe and Connect Four
Tic-Tac-Toe and Connect Four are two of the most popular logic puzzles in the world. They are simple yet challenging games that require players to use their logical thinking skills to win. Here are some strategies for playing these games:
In Tic-Tac-Toe, players take turns placing their marks on a 3×3 grid. The player who places three of their marks in a row, column, or diagonal wins the game. Here are some strategies for playing Tic-Tac-Toe:
1. Take the center square: The center square is the most valuable square on the board because it gives the player the most opportunities to make a winning move.
2. Block your opponent: If your opponent has already placed a mark on the board, try to block them from making a winning move by placing your mark in a strategic location.
3. Force a draw: If you are unable to make a winning move, try to force a draw by placing your mark in a way that makes it impossible for your opponent to make a winning move.
In Connect Four, players take turns placing their marks on a vertical grid of 7×6. The player who places four of their marks in a row, column, or diagonal wins the game. Here are some strategies for playing Connect Four:
1. Aim for the center: The center of the grid is the most valuable location in Connect Four because it gives the player the most opportunities to make a winning move.
2. Look for potential winning moves: When it’s your turn, look for potential winning moves by looking for sequences of four marks in a row, column, or diagonal.
3. Force a draw: If you are unable to make a winning move, try to force a draw by placing your mark in a way that makes it impossible for your opponent to make a winning move.
Overall, these strategies can help you improve your chances of winning at Tic-Tac-Toe and Connect Four. However, it’s important to remember that the best strategy is to always think ahead and anticipate your opponent’s moves.
### Deciphering Cryptograms and Ciphers
Cryptograms and ciphers are types of logic puzzles that involve deciphering codes and ciphers to uncover hidden messages. These puzzles have been around for centuries and have been used for various purposes, including military communication, secret messages, and even entertainment.
Cryptograms are puzzles that involve deciphering encrypted messages. The puzzle consists of a clue or a set of clues that are encrypted using a simple substitution cipher. The objective of the puzzle is to decrypt the message by figuring out the correct substitution pattern.
There are different types of cryptograms, including:
• Monoalphabetic Ciphers: This type of cipher involves replacing each letter in the plaintext with another letter or symbol. For example, the letter ‘A’ might be replaced by the letter ‘B’, while the letter ‘E’ might be replaced by the letter ‘F’.
• Polyalphabetic Ciphers: This type of cipher involves using multiple substitutions to encrypt the message. The Vigenere Cipher is an example of a polyalphabetic cipher.
• Transposition Ciphers: This type of cipher involves rearranging the letters in the plaintext without changing them. For example, the letters might be written in a different order or repeated.
Ciphers are puzzles that involve deciphering coded messages. The puzzle consists of a message that has been encoded using a specific code or cipher. The objective of the puzzle is to decode the message by figuring out the correct code or cipher.
There are different types of ciphers, including:
• Caesar Cipher: This type of cipher involves shifting each letter in the plaintext by a certain number of positions in the alphabet. For example, the letter ‘A’ might be shifted to ‘D’, while the letter ‘E’ might be shifted to ‘F’.
• Vigenere Cipher: This type of cipher involves using a keyword to encrypt the message. Each letter in the plaintext is encrypted using the corresponding letter in the keyword.
• Morse Code: This type of cipher involves using a series of dots and dashes to represent letters and numbers. The objective of the puzzle is to decode the message by figuring out the correct sequence of dots and dashes.
Deciphering cryptograms and ciphers requires a keen eye for detail and an understanding of the principles behind each type of cipher. With practice, anyone can become proficient at solving these types of puzzles and uncovering hidden messages.
## Logic Puzzles in Real Life: Applications and Benefits
### Logic Puzzles in Education
#### Integrating Logic Puzzles into the Curriculum
• Introducing logic puzzles as a teaching tool in mathematics, science, and language arts
• Enhancing critical thinking and problem-solving skills among students
• Developing cognitive abilities and promoting creativity
#### Benefits of Logic Puzzles in Education
• Improving reasoning and deductive skills
• Increasing patience and persistence in problem-solving
• Boosting self-esteem and confidence through successful puzzle-solving
• Fostering collaboration and communication through group puzzle-solving activities
#### Challenges in Implementing Logic Puzzles in Education
• Teacher training and resources for effectively incorporating logic puzzles into the curriculum
• Addressing individual learning styles and pacing
• Balancing logic puzzles with other academic subjects
• Overcoming the misconception that logic puzzles are solely for the gifted and talented
### Logic Puzzles in Problem Solving
Logic puzzles have a wide range of applications in problem solving, making them an invaluable tool for those seeking to enhance their critical thinking skills. From business to engineering, these puzzles can be used to tackle complex challenges and develop innovative solutions.
In the business world, logic puzzles can be used to develop strategic thinking and decision-making skills. For example, a puzzle might involve identifying the most efficient way to allocate resources to different departments within a company. By working through these puzzles, business leaders can hone their ability to think critically and make well-informed decisions.
#### Puzzles for Engineering
Engineers can also benefit from logic puzzles, as they can help develop problem-solving skills and encourage creative thinking. For instance, a puzzle might challenge engineers to design a structure that can withstand various natural disasters, such as earthquakes or hurricanes. By working through these puzzles, engineers can learn to think outside the box and come up with innovative solutions to complex problems.
#### Puzzles for Education
Logic puzzles can also be used in education to teach students critical thinking skills and encourage independent problem-solving. For example, a puzzle might involve identifying the sequence of events in a historical event or deciphering a code. By working through these puzzles, students can develop their analytical skills and learn to think critically about the world around them.
Overall, logic puzzles have a wide range of applications in problem solving, making them a valuable tool for those seeking to enhance their critical thinking skills. Whether you’re a business leader, engineer, or student, there’s a logic puzzle out there that can help you develop your problem-solving abilities and think creatively.
### Logic Puzzles in Cognitive Training
Boosting Brain Power
• Improving problem-solving skills
• Enhancing critical thinking
• Encouraging logical reasoning
• Fostering creativity
Overcoming Cognitive Decline
• Preventing cognitive decline in aging populations
• Treating neurological conditions such as Alzheimer’s and dementia
• Assisting in rehabilitation for brain injuries
Personal and Professional Development
• Enhancing job performance in various fields
• Developing effective communication skills
• Increasing overall cognitive abilities
• Promoting personal growth and self-awareness
### Logic Puzzles in Entertainment
The world of entertainment has long been a breeding ground for logic puzzles. From board games to video games, from mystery novels to crime dramas, logic puzzles have been an integral part of many forms of entertainment.
#### Puzzle Games
Puzzle games have been a staple of the gaming industry for decades. They come in many forms, from simple matching games to complex puzzles that require careful planning and strategic thinking. Some popular examples of puzzle games include Tetris, Sudoku, and Candy Crush. These games not only provide a fun and engaging experience for players, but they also help to improve cognitive skills such as problem-solving, spatial reasoning, and critical thinking.
#### Mystery Novels and Crime Dramas
Mystery novels and crime dramas often involve logic puzzles as a central plot element. These puzzles can take many forms, from simple deductions to complex puzzles that require the reader or viewer to piece together clues and evidence. The popularity of these types of stories is due in part to the intellectual challenge they present, as well as the satisfaction of solving the puzzle and uncovering the truth.
#### Escape Rooms
Escape rooms have become increasingly popular in recent years, offering a unique and immersive experience for players. These games typically involve a group of players who are “trapped” in a room and must use their wits and problem-solving skills to escape. The puzzles in escape rooms can be physical, mental, or a combination of both, and they often require players to think creatively and work together to solve them.
In conclusion, logic puzzles have become an integral part of the entertainment industry, providing a fun and engaging experience for players and viewers alike. Whether it’s through puzzle games, mystery novels, or escape rooms, logic puzzles offer a unique challenge that can help to improve cognitive skills and provide a sense of satisfaction and accomplishment.
### The Enduring Appeal of Logic Puzzles
The enduring appeal of logic puzzles lies in their ability to engage the mind and challenge the intellect. These puzzles have been a part of human culture for centuries, with origins dating back to ancient Greece. They continue to captivate people of all ages and backgrounds, from children solving simple arithmetic problems to mathematicians grappling with complex equations.
One reason for the enduring appeal of logic puzzles is their versatility. They can take many forms, from simple riddles to intricate brainteasers, and can be found in a variety of contexts, including newspapers, books, and online platforms. The variety of puzzle types ensures that there is something for everyone, catering to different levels of skill and interest.
Another factor contributing to the popularity of logic puzzles is their potential for personal growth and development. Solving puzzles can improve problem-solving skills, increase critical thinking abilities, and enhance cognitive flexibility. By engaging in logical reasoning and deductive thinking, individuals can challenge themselves to think more deeply and creatively.
Moreover, logic puzzles offer a unique form of entertainment that can be enjoyed alone or with others. They can be shared and discussed, fostering social interaction and collaboration. Solving puzzles together can strengthen relationships and encourage teamwork, as individuals work together to unravel the clues and arrive at a solution.
Finally, the satisfaction of solving a logic puzzle is a significant factor in its enduring appeal. The sense of accomplishment and pride that comes with cracking a difficult puzzle can be highly rewarding. This feeling of achievement motivates individuals to continue seeking out new challenges and to persist in the face of obstacles, fostering a growth mindset and a love of learning.
In summary, the enduring appeal of logic puzzles lies in their versatility, potential for personal growth, entertainment value, and sense of accomplishment. These factors have contributed to their longevity and continue to make them an engaging and enjoyable pursuit for people of all ages and backgrounds.
### The Benefits of Solving Logic Puzzles
Solving logic puzzles has numerous benefits for individuals of all ages. By engaging in these cognitive exercises, one can enhance their problem-solving skills, sharpen their critical thinking abilities, and improve their overall cognitive functioning. Here are some of the key benefits of solving logic puzzles:
• Enhanced Problem-Solving Skills: Logic puzzles require the solver to think creatively and apply reasoning to arrive at a solution. By regularly engaging in these exercises, individuals can improve their ability to analyze complex situations, identify patterns, and devise effective strategies to overcome challenges.
• Cognitive Stimulation: Solving logic puzzles can help to keep the brain active and engaged, promoting cognitive stimulation. This type of mental exercise can help to prevent cognitive decline, delay the onset of age-related conditions such as dementia, and maintain cognitive function in older adults.
• Improved Memory Function: Some logic puzzles, such as Sudoku, require the solver to recall and manipulate information in their working memory. Regularly engaging in these types of puzzles can help to improve memory function, particularly in the areas of working memory and spatial reasoning.
• Increased Attention to Detail: Logic puzzles often require careful attention to detail in order to identify clues and make connections between pieces of information. By regularly engaging in these exercises, individuals can improve their ability to focus and pay close attention to detail, which can have benefits in other areas of life, such as work and personal relationships.
• Development of Critical Thinking Skills: Solving logic puzzles requires the application of critical thinking skills, such as analysis, synthesis, and evaluation. By regularly engaging in these exercises, individuals can improve their ability to think critically and make informed decisions, which can be beneficial in a variety of contexts, including academic and professional settings.
• Stress Reduction: Finally, solving logic puzzles can be a fun and enjoyable activity that provides a sense of accomplishment and satisfaction. This can help to reduce stress and promote relaxation, improving overall mental and emotional well-being.
### The Future of Logic Puzzles
As the world becomes increasingly digitized, the demand for logical thinking and problem-solving skills is at an all-time high. Logic puzzles have been shown to be effective tools for honing these skills, and as such, their future looks bright. Here are some ways in which logic puzzles may shape the future:
• In Education: Logic puzzles have long been used as teaching tools in mathematics and computer science, but their applications are far broader. As educational institutions look for new ways to teach critical thinking and problem-solving, logic puzzles may become a staple of the curriculum.
• In the Workplace: Logic puzzles can help employees develop the skills they need to navigate complex problems and make sound decisions. As companies increasingly rely on data-driven decision-making, employees who can think logically and solve problems will be in high demand.
• In Entertainment: Puzzle games have been a staple of the gaming industry for decades, and logic puzzles are a natural evolution of this genre. As technology advances, we can expect to see more complex and engaging logic puzzles in video games and other forms of entertainment.
• In Research: Logic puzzles have long been used in mathematical research, but their applications are far broader. Researchers in fields as diverse as biology and economics use logic puzzles to model complex systems and make predictions about future events.
• In Everyday Life: As the world becomes more complex, we all need to develop our problem-solving skills. Logic puzzles can help us do just that, and as such, they have the potential to become a part of our daily lives. Whether we’re trying to navigate a difficult situation at work or solve a problem at home, logic puzzles can help us think more clearly and make better decisions.
Overall, the future of logic puzzles looks bright. As the world becomes more complex, the demand for logical thinking and problem-solving skills will only continue to grow. Logic puzzles have the potential to play a key role in developing these skills, and as such, they will likely remain an important part of our lives for years to come.
## FAQs
### 1. What are logic puzzles?
Logic puzzles are problems that require the use of reasoning and critical thinking to solve. They come in various forms and can be found in books, newspapers, and online. The goal of a logic puzzle is to lead the solver to a specific conclusion or solution using a series of clues or hints.
### 2. What are the different types of logic puzzles?
There are many different types of logic puzzles, including crosswords, Sudoku, and word search puzzles. Other types include number puzzles, word puzzles, and visual puzzles. Some logic puzzles are based on mathematical concepts, while others are based on language or visual cues.
### 3. What is a crossword puzzle?
A crossword puzzle is a type of logic puzzle that involves filling in words across and down in a grid. The words are usually related by a common theme or topic, and the puzzle is solved by using deductive reasoning to fill in the blanks.
### 4. What is Sudoku?
Sudoku is a type of logic puzzle that involves filling in a grid of numbers according to certain rules. The puzzle is solved by using deductive reasoning to fill in the blanks.
### 5. What is a word search puzzle?
A word search puzzle is a type of logic puzzle that involves finding a list of words hidden in a grid of letters. The words can be horizontal, vertical, or diagonal, and the puzzle is solved by using deductive reasoning to find the words.
### 6. What are number puzzles?
Number puzzles are a type of logic puzzle that involve manipulating numbers according to certain rules. Examples include the popular puzzle game, “2048,” and the mathematical concept, “prime numbers.”
### 7. What are word puzzles?
Word puzzles are a type of logic puzzle that involve manipulating words according to certain rules. Examples include anagrams, where the solver must rearrange the letters in a word to form a new word, and cryptograms, where the solver must decode a message written in a secret language.
### 8. What are visual puzzles?
Visual puzzles are a type of logic puzzle that involve solving a problem using visual cues. Examples include jigsaw puzzles, where the solver must put together a picture made up of different pieces, and mazes, where the solver must find their way through a complex path.
### 9. How can I improve my skills in solving logic puzzles?
To improve your skills in solving logic puzzles, it is important to practice regularly. Start with easy puzzles and gradually work your way up to more difficult ones. You can also try to solve puzzles from different categories to broaden your skills. Additionally, it can be helpful to work with a partner or join a puzzle club to discuss and share strategies.
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Articles / Desktop Programming / MFC
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Posted 13 Jun 2010
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OAG Library (OpenGL) Part 2.1 - Drawing Objects 2D using the Mouse and Programatically
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This tutorial shows Library Code for Geometries 2D how to draw them programatically and how draw to objects using the mouse in an application MFC.
Introduction
In this tutorial, I have added classes to draw Geometries 2D, textures and True Type Fonts, so you can draw the objects easier on the screen.You must download the library to compile the sample. Click here to go to the download page. Using the OAGMFC, you can save and open XML files (*.oagxml). The XML files are an alternative for collada files.
The Geometries 2D
Line Polyline Rectangle Triangle
Creating Geometries 2D
To build geometries using the library, you need to create a new instance for `OAGPrimitives`. With the class, you can draw primitives supported by opengl such as line, lines, polylines, rectangle and triangle. After creating a new instance of` OAGPrimitives`, you must use the command `SetGeometryType `to change the primitive type that you want to draw. The supported types by `SetGeometryType `are `GL_LINES `(for lines), `GL_LINE_STRIP `(for polyline), `GL_QUADS `(for rectangle) and `GL_TRIANGLES `(for triangle). The example below shows how to create a line.
C++
```//Creating a line
oag::OAGPrimitives* pGeo = new oag::OAGPrimitives();
peo->SetGeometryType( GL_LINES );```
If you wish to draw a solid geometry, use the `SetPolygonMode(GL_FILL)`. Using `GL_FILL `you tell OpenGL to draw Solid Polygons. Below is the code for drawing a solid triangle and solid rectangle.
C++
```//solid rectangle
lstQuadVector.push_back( oag::OAGVector3f( 20.f, 132.f, 0.f ));
lstQuadVector.push_back( oag::OAGVector3f( 20.f, 145.f, 0.f ));
lstVector.push_back( oag::OAGVector3f( 127.45f, 64.f, 0.f ));
lstVector.push_back( oag::OAGVector3f( 127.45f, 132.f, 0.f ));
//Solid Triangle
oag::OAGPrimitives* pTri = new oag::OAGPrimitives();
pTri->SetGeometryType( GL_TRIANGLES );
pTri->SetPolygonMode(GL_FILL);
std::vector<:oagvector3f> lstTriVector;
lstTriVector.push_back( oag::OAGVector3f( 57.f, 39.f, 0.f ));
lstTriVector.push_back( oag::OAGVector3f( 112.f, 128.f, 0.f ));
lstTriVector.push_back( oag::OAGVector3f( 176.f, 64.f, 0.f ));
pTri->SetArrayVertices( lstTriVector );
The Library Code For Drawing the Primitives
Here an `OnDraw `function uses commands for Faster Vertex in OpenGL. The commands `glDrawElements `and `glDrawArrays `do it.
C++
```void oag::OAGPrimitives::OnDraw()
{
if( m_ArraySize < 1 )
return;
::glPushMatrix();
::glColor4ubv( m_Color.GetColor4ubv() );
::glPolygonMode(m_PolygonFace, m_PolygonMode);
switch( m_GeometryType )
{
case GL_LINES:
case GL_LINE_LOOP:
case GL_LINE_STRIP:
{
::glEnableClientState(GL_VERTEX_ARRAY);
::glVertexPointer(3, GL_FLOAT, 0, m_GeometryData);
::glDrawArrays( m_GeometryType, 0, m_ArraySize );
::glDisableClientState(GL_VERTEX_ARRAY);
}
break;
case GL_TRIANGLES:
case GL_TRIANGLE_FAN:
case GL_TRIANGLE_STRIP:
{
::glEnableClientState(GL_VERTEX_ARRAY);
::glVertexPointer(3, GL_FLOAT, 0, m_GeometryData);
::glDrawElements(m_GeometryType, GetVertexCount(), GL_UNSIGNED_BYTE, m_nIndices);
::glDisableClientState(GL_VERTEX_ARRAY);
}
break;
}
::glPopMatrix();
} ```
Creating Text
To create text using the library, you need to create a new instance of `oag::OAGFont2D`. With this class, you can draw draw 2D text in OpenGL using the libraries FTGL and FreeType. The example below shows how to create a 2D Text. The library and the demo use the file arial.ttf.
C++
```oag::OAGFont2D* font2d = new oag::OAGFont2D();
m_pWinGraphicContext->MakeCurrent();
font2d->Initialize();
m_pWinGraphicContext->DeleteCurrent();
font2d->SetText("Example of 2DText");
font2d->SetFontSize(20);
//Font Position
oag::Matrix4x4 mt;
mt.SetTranslation(vec.m_X, vec.m_Y, vec.m_Z);
font2d->SetTransform( mt );
```
The Library Code for Drawing the Text
C++
```void oag::OAGFontMapping2D::OnDraw()
{
if ( m_pFont == NULL )
return;
::glPushMatrix();
::glColor4ubv( m_Color.GetColor4ubv() );
::glPolygonMode(m_PolygonFace, m_PolygonMode);
m_pFont->Render(m_strText.c_str());
::glPopMatrix();
}```
The Sample
This tutorial shows how to draw geometries 2D using the mouse using classes. The classes written to draw are in the folder CadTools. This project tutorial has been created with Visual Studio 2008.
Drawing the Geometries
When you call one of the Draw Menu items, one tool is created for the primitive and the document is called to create the primitive type for the menu item. See the example when you call the command line in the Draw menu.
C++
```void COAGMFCView::OnDrawLine()
{
if ( m_pTool )
delete m_pTool;
OnInsertGeometries(1);
m_pTool = new CLineTool();
m_pTool->SetScene( GetDocument()->m_pScene );
}```
After clicking in line you have a primitive created in your scene. To start drawing, you click on the screen and move the mouse to see the primitive on the screen.
The primitive line has two points and when you click again on the screen, you set the second point for the line and the primitive is finished.
Now let's go to the polyline. Select the Draw the menu and click on Polyline and click on the screen to start drawing. For the polyline, you can click on the screen lot of times. When your polyline is done, you must press key ESC to finish the primitive.
The next primitive is the rectangle. Select the Draw the menu and click on Rectangle and click on the screen to start drawing. Move the mouse and see a rectangle drawing on the screen. When the rectangle is done, click on the screen to finish the primitive.
The last primitive is the triangle. Select the Draw the menu and click on Triangle and click on the screen to start drawing. Moving the mouse, you won't see a triangle because you have just two points for the triangle. When you click on the screen again, the second point is defined and you can see a triangle on the screen if you move the mouse. When the triangle is done, click on the screen again. The triangle is finished.
Mouse Events
When you click on the screen, the point is changed to world coordinates. After this, the point is added on the tool. The tool adds the coordinate to the primitive with the function `MouseClick`. If it is the first click for the primitive, the tool adds two points to the primitive.
C++
```void COAGMFCView::OnLButtonDown(UINT nFlags, CPoint point)
{
oag::OAGVector3d center;
oag::OAGVector3f pt(point.x, point.y, 0);
m_pRender->ScreenToWorld(&pt, ¢er, 1);
if( m_pTool && !m_pTool->IsFinished() )
{
oag::OAGVector3f vec( (float)center.m_X, (float)center.m_Y, 0.f );
m_pTool->OnMouseClick( vec );
Invalidate(FALSE);
}
CView::OnLButtonDown(nFlags, point);
}```
When you move the mouse, the last point is changed and you see the primitive in real time.
C++
```void COAGMFCView::OnMouseMove(UINT nFlags, CPoint point)
{
if( m_pTool && m_pTool->IsStarted() && !m_pTool->IsFinished() )
{
oag::OAGVector3d center;
oag::OAGVector3f pt(point.x, point.y, 0);
oag::OAGVector3f vec;
m_pRender->ScreenToWorld(&pt, ¢er, 1);
oag::OAGVector3f vecPoint( (float)center.m_X, (float)center.m_Y, 0.f );
vec.m_X = vecPoint.m_X; vec.m_Y = vecPoint.m_Y;
m_pTool->OnMouseMove( vec );
Invalidate(FALSE);
}
CView::OnMouseMove(nFlags, point);
} ```
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I live in Matão, a small city in Brazil. I studied as Programmer in a College for Software Development in Database.
After finishing the College I have been working with java, c# and Computer Graphics with searches for OpenGL.
First Prev Next
need help Member 1469824415-Jul-20 13:50 Member 14698244 15-Jul-20 13:50
Cannot open include file: 'OAGScene.h': No such file or directory masoudrad22-Oct-12 9:58 masoudrad 22-Oct-12 9:58
Re: library down page not found. [modified] Eduardo Tucci12-Aug-11 7:29 Eduardo Tucci 12-Aug-11 7:29
Triangle tool redundant Sharjith25-Oct-10 11:22 Sharjith 25-Oct-10 11:22
Re: Triangle tool redundant Eduardo Tucci26-Oct-10 5:35 Eduardo Tucci 26-Oct-10 5:35
Re: very good. Eduardo Tucci16-Jun-10 7:15 Eduardo Tucci 16-Jun-10 7:15
Allocationg objects on the heap Nemanja Trifunovic15-Jun-10 5:08 Nemanja Trifunovic 15-Jun-10 5:08
Re: Allocationg objects on the heap Eduardo Tucci15-Jun-10 14:47 Eduardo Tucci 15-Jun-10 14:47
I use oag::OAGPrimitives* pGeo = new oag::OAGPrimitives(); because the class oag::OAGScene works with objects in the memory. For the demo I have objects for each document. When a document is deleted all the objects are deleted too using the function m_pScene->DeleteAllObjects();. You can have your array for the objects and delete them when you want. ========================================================================================================= I use: Copy Code ```if ( m_pTool ) delete m_pTool;``` because m_pTool is base class for each tool. Each tool manage an object. For example: When the line menu item is called the current tool is deleted and CLineTool is created to create a line. Of course, if I do not delete m_pTool I have memory on the heap.
A transparent back ground kzhao66@yahoo.com15-Jun-10 4:47 kzhao66@yahoo.com 15-Jun-10 4:47
Re: A transparent back ground Eduardo Tucci15-Jun-10 14:14 Eduardo Tucci 15-Jun-10 14:14
Re: A transparent back ground kzhao66@yahoo.com16-Jun-10 4:48 kzhao66@yahoo.com 16-Jun-10 4:48
Re: A transparent back ground Eduardo Tucci16-Jun-10 4:58 Eduardo Tucci 16-Jun-10 4:58
Re: A transparent back ground Santhosh G_20-Jun-10 14:48 Santhosh G_ 20-Jun-10 14:48
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# Fractal Interpolation: Unraveling the Beauty of Self-Similar Curves and Surfaces
October 12, 2023
Callard Coombs
United States of America
Fractal Interpolation
With a PhD in mathematics and years of experience teaching students, Callard Coombs is an adept assignment expert. He has over 800 clients.
Mathematics, often seen as an enigma, is a world filled with fascinating phenomena and concepts. Fractal interpolation is one such captivating topic that can be a key to unlocking the mysteries of complex, self-similar curves and surfaces. In this blog post, we'll dive into the theoretical depths of fractal interpolation, exploring its fundamental concepts, algorithms, and its distinctive characteristics when compared to traditional interpolation methods. By the end of this discussion, university students will be better equipped to solve their fractal geometry assignment involving fractal interpolation.
## What is Fractal Interpolation?
Fractal interpolation is a mathematical technique that allows us to generate intricate, self-replicating patterns known as fractals. A fractal is a geometric shape or set that exhibits self-similarity at different scales. In simpler terms, as you zoom in on a fractal, you'll find that smaller parts of it resemble the whole structure, and this resemblance continues infinitely.
The idea of fractal interpolation is to construct a curve or surface by iteratively subdividing and interpolating between control points in a way that the resulting shape resembles the fractal pattern we desire. This process often involves recursive algorithms and a high degree of mathematical finesse.
## Algorithms for Fractal Interpolation
Algorithms for fractal interpolation form the core of generating intricate self-similar patterns. Iterated Function Systems (IFS) use probabilistic transformations, while Random Midpoint Displacement (RMD) introduces randomness for surface creation. The Chaos Game method adds an element of unpredictability, collectively enabling the mesmerizing world of fractal art and modeling.
1. Iterated Function Systems (IFS)
2. One of the most popular algorithms for fractal interpolation is the Iterated Function System (IFS). IFS operates by defining a set of contractive transformations, each associated with a specific probability. These transformations are applied iteratively to a starting point, and as the iterations progress, the points approach the desired fractal shape.
The famous Barnsley Fern fractal is a classic example of an IFS-generated fractal. Its elegant fern-like structure emerges from a set of carefully chosen transformations.
3. Random Midpoint Displacement (RMD)
4. Random Midpoint Displacement (RMD) stands as a noteworthy algorithm in fractal interpolation, particularly for generating complex fractal surfaces. This technique operates by subdividing an initial grid and displacing each midpoint randomly. The beauty of RMD lies in its ability to create surfaces with remarkable intricacy. The random nature of displacements introduces a touch of unpredictability, which, when iterated, results in surfaces characterized by self-similarity and a rich texture. This algorithm is widely used in terrain generation for computer graphics, simulating realistic landscapes with rugged details that closely mimic natural topography.
5. Chaos Game
6. The Chaos Game is a fascinating algorithm in the realm of fractal interpolation. This method adds an element of randomness and unpredictability to the creation of fractals. Starting from a point within a polygon, it iteratively moves the point halfway towards one of the polygon's vertices, chosen randomly. This process repeats, creating intricate self-similar patterns like the Sierpinski Triangle and Sierpinski Carpet. The Chaos Game showcases how randomness can lead to mesmerizing order within the world of fractals, providing students with a deeper appreciation for the intersection of mathematics and art.
Now that we have explored the algorithms used in fractal interpolation, let's examine how they differ from traditional interpolation methods.
1. Self-Similarity
2. The most striking difference is the presence of self-similarity in fractal interpolation. Traditional interpolation methods, like linear or polynomial interpolation, aim to connect the control points with smooth curves or surfaces. In contrast, fractal interpolation embraces irregularity and roughness, resulting in fractals that exhibit self-similar patterns across different scales.
3. Infinite Detail
4. Fractal interpolation can create shapes with infinite detail, while traditional interpolation methods produce finite and smooth curves or surfaces. This infinite detail is a hallmark of fractals and allows them to represent natural phenomena like coastlines, mountains, and clouds more accurately.
5. Nonlinearity
6. Traditional interpolation methods are inherently linear, relying on equations that describe straight lines or smooth curves. Fractal interpolation, on the other hand, is nonlinear, as it often involves recursive and probabilistic processes. This nonlinearity contributes to the complexity and richness of fractals.
## Practical Applications
Understanding fractal interpolation can have practical applications beyond the realm of mathematics assignments. Fractals are used in various fields, including computer graphics, image compression, and modeling natural phenomena.
For instance, in computer graphics, fractals are employed to create realistic terrain and landscapes, as the self-similar nature of fractals mirrors the irregularity found in nature. Fractal-based image compression techniques can efficiently represent images with high detail at lower file sizes.
## Mathematical Foundations of Fractal Interpolation
To truly grasp the essence of fractal interpolation, one must understand its mathematical underpinnings. Central to the concept is the notion of self-affinity and iterated function systems.
### Self-Affinity
Self-affinity is a key characteristic of fractals. While self-similarity implies that a shape looks the same at different scales, self-affinity takes it a step further. In self-affine fractals, the shape is not necessarily identical at different scales but exhibits a similar statistical distribution of details. This means that while the overall structure may change, the general patterns and features remain consistent.
### Iterated Function Systems (IFS) Revisited
As previously mentioned, IFS is a fundamental algorithm for generating fractals. An IFS consists of a set of contractive transformations, typically affine transformations (combinations of translation, rotation, scaling, and shearing). Each transformation is associated with a probability, and the algorithm iteratively applies these transformations to an initial point.
One of the famous examples of an IFS-generated fractal is the Sierpinski Triangle. To create it, we start with an equilateral triangle and apply three affine transformations, each moving a point toward one of the triangle's vertices. The key to self-affinity is that each transformation is chosen randomly based on the associated probabilities. As more iterations are performed, the triangle's interior converges to the Sierpinski Triangle.
### The Beauty of Fractal Dimension
Another intriguing aspect of fractal interpolation is the concept of fractal dimension. Traditional Euclidean geometry deals with shapes that have integer dimensions: points (0D), lines (1D), surfaces (2D), and volumes (3D). Fractals, however, challenge this notion by having non-integer dimensions, often referred to as fractional dimensions.
The concept of fractal dimension provides a way to quantify the complexity of a fractal. It measures how the detail or intricacy of a fractal changes as you zoom in. For example, the Sierpinski Triangle, despite being a 2D shape, has a fractional dimension of approximately 1.585.
Understanding fractal dimension is crucial when working with fractal interpolation because it allows us to describe and compare the complexity of different fractals. It also plays a significant role in various applications, such as image analysis and characterizing natural phenomena like coastlines.
## Beyond the Visual: Applications of Fractal Interpolation
Beyond their visual allure, fractal interpolation finds practical utility in diverse domains. Fractal-based data compression efficiently preserves intricate details in images and datasets. In computer graphics, it generates realistic terrains. Fractal models aid in simulating natural phenomena and analyzing financial markets. These applications underscore the versatility of fractal interpolation beyond aesthetics. Here are some areas where fractals find utility:
1. Fractal-based Data Compression
2. Fractal-based data compression is a powerful technique that revolutionizes how we store and transmit complex information. By representing data or images using fractal patterns, it achieves remarkable compression ratios while preserving critical details. This method finds extensive use in various fields, from medical imaging to satellite data transmission, where preserving high-resolution images with minimal storage or bandwidth usage is paramount. Fractal compression algorithms intelligently encode the self-similarities within the data, making it a valuable tool in the age of big data. Its ability to efficiently manage large datasets while maintaining data fidelity positions fractal-based compression as a cutting-edge technology in information science.
3. Terrain Generation in Computer Graphics
4. In the realm of computer graphics and video games, fractals are often employed to generate realistic terrain. Traditional heightmap-based terrain generation can result in repetitive patterns, but fractal-based algorithms create diverse landscapes with a high level of detail. This not only enhances visual realism but also enables more immersive gameplay experiences.
5. Modeling Natural Phenomena
6. Fractal interpolation is a powerful tool for modeling natural phenomena that exhibit self-similarity, such as clouds, mountains, and coastlines. These fractal models provide a closer approximation of the intricate irregularities found in nature, making them valuable for scientific research, environmental modeling, and simulations.
7. Financial Markets Analysis
8. Fractal interpolation has found applications in the analysis of financial markets. The fractal nature of price fluctuations in financial assets can be modeled using fractal interpolation techniques. This allows analysts to identify patterns and trends that might be missed by traditional linear methods, potentially improving forecasting and risk management.
## Challenges and Limitations
Fractal interpolation, while powerful, is not without its challenges and limitations. The computational demands can be substantial, particularly for highly detailed fractals. Artistic interpretation often blurs the line between mathematics and creativity. Additionally, the method relies on accurate initial data points, which can be elusive in certain scientific applications, necessitating careful consideration when applying fractal interpolation techniques.
1. Computational Complexity
2. The computational complexity of fractal interpolation can pose significant challenges. Generating highly detailed fractals, especially those with intricate self-similarity, demands substantial processing power and memory resources. This complexity can limit real-time applications, such as interactive simulations or video games. Researchers and artists often strike a balance between detail and computational efficiency, fine-tuning algorithms to match their specific needs. Overcoming computational challenges is crucial to harnessing the full potential of fractal interpolation in various domains, as it enables the creation of visually stunning, intricate patterns while efficiently managing the computational demands.
3. Artistic Interpretation
4. Artistic interpretation plays a significant role in the world of fractal interpolation. While the underlying algorithms generate stunning mathematical patterns, artists often take these generated fractals as starting points and then fine-tune them to achieve specific visual effects. They adjust parameters, apply filters, and inject their creative vision into the process, transforming mathematical abstractions into captivating artworks. This fusion of mathematics and art showcases how fractal interpolation transcends mere computation, becoming a medium for artistic expression. It reminds us that mathematics can inspire and collaborate with human creativity, blurring the lines between the technical and the artistic in a harmonious blend of precision and imagination.
5. Data Dependency
6. Data dependency is a crucial consideration in the realm of fractal interpolation. The accuracy and availability of initial data points or control points significantly impact the quality of the generated fractals. In scientific modeling, obtaining precise data can be challenging, potentially introducing inaccuracies into the interpolation process. This dependency on data quality underscores the need for careful data collection and preprocessing to ensure the reliability of fractal interpolation outcomes. It also emphasizes the importance of selecting appropriate interpolation methods and adjusting parameters to mitigate the effects of imperfect data, allowing for more accurate representation and modeling of real-world phenomena.
## Conclusion
In conclusion, fractal interpolation is a captivating mathematical concept that enables the generation of self-similar, intricate curves and surfaces. Through algorithms like Iterated Function Systems, Random Midpoint Displacement, and the Chaos Game, fractal interpolation stands apart from traditional interpolation methods in its embrace of self-similarity, infinite detail, and nonlinearity. As university students dive into math assignments involving fractal interpolation, they'll find themselves unraveling the beauty of these complex patterns. Understanding the theoretical foundations and algorithms of fractal interpolation will not only help in academic pursuits but also open doors to practical applications in various fields. So, when you're faced with a math assignment that requires you to explore the world of fractal interpolation, remember that it's not just about solving equations; it's about unleashing the power of self-similarity and complexity within mathematics. With this newfound knowledge, you'll be better equipped to solve your math assignment and appreciate the intricate beauty of fractals.
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The Chi-square Tests Assessment Test
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Take our assessment test to assess your knowledge of the chi-square tests. The chi-square goodness of fit test that determines if a sample data matches a population and the chi-square test for independence that compares two variables in a contingency table to see if they are related.
• 1.
What is the null hypothesis in chi-square test?
• A.
The rows and columns are the same
• B.
The rows and columns in the table are associated
• C.
The rows and columns in the table are not associated
• D.
The rows and column are not the same
C. The rows and columns in the table are not associated
Explanation
The null hypothesis in a chi-square test states that there is no association between the rows and columns in the table. In other words, the variables being analyzed are independent of each other.
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• 2.
The least expected value in each contingency table cell in order for a chi-square test to be effective is...
• A.
5
• B.
4
• C.
3
• D.
2
A. 5
Explanation
For a chi-square test to be effective, the expected values in each contingency table cell should not be too small. A general guideline is that the expected value in each cell should be at least 5. This ensures that the assumptions of the chi-square test are met and that the test statistic follows the chi-square distribution. Having expected values less than 5 can lead to unreliable results and low statistical power. Therefore, the least expected value in each contingency table cell for a chi-square test to be effective is 5.
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• 3.
Contingency tables and degrees of freedom are key elements in chi-square test?
• A.
False
• B.
Maybe
• C.
True
• D.
None
C. True
Explanation
Contingency tables and degrees of freedom are indeed key elements in the chi-square test. Contingency tables are used to organize and display the observed frequencies of two categorical variables, which are then compared to the expected frequencies to determine if there is a significant association between the variables. Degrees of freedom, on the other hand, represent the number of independent pieces of information that are available for estimating the unknown parameters in a statistical model. In the chi-square test, the degrees of freedom are calculated based on the number of categories in the variables being analyzed. Therefore, both contingency tables and degrees of freedom play important roles in conducting and interpreting the chi-square test.
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• 4.
What is another name for the chi-square goodness fit test?
• A.
Directional chi-square
• B.
One sample chi-square
• C.
Chi-square Anova
• D.
Two sample chi-square
B. One sample chi-square
Explanation
The chi-square goodness fit test is also known as the one sample chi-square test. This test is used to determine whether there is a significant difference between the observed and expected frequencies in a categorical variable. It compares the observed frequencies with the expected frequencies and calculates a chi-square statistic. By comparing this statistic to a critical value, we can determine whether the observed frequencies deviate significantly from the expected frequencies.
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• 5.
What type of data do you use for a chi-square table?
• A.
Ratios
• B.
Scales
• C.
Ordinal
• D.
Categorical
D. Categorical
Explanation
Chi-square tests are used to analyze categorical data, which consists of non-numerical variables that can be divided into categories or groups. Examples of categorical data include gender, race, and occupation. The chi-square table is a statistical tool that provides critical values for the chi-square test statistic, which is used to determine the significance of the relationship between two categorical variables. Therefore, the correct answer is "Categorical" as it accurately describes the type of data used for a chi-square table.
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• 6.
How many cases must appear in one category of a chi-square test?
• A.
4
• B.
6
• C.
5
• D.
7
C. 5
Explanation
In a chi-square test, the number of cases that must appear in one category depends on the specific requirements of the test and the research question being investigated. However, there is no specific rule stating that a certain number of cases must appear in one category. The number of cases in each category should be determined based on the sample size, the expected frequencies, and the statistical power desired for the test. Therefore, the answer "5" is incorrect and the question may be incomplete or not readable.
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• 7.
What is an effective size?
• A.
The magnitude of the relationship between variables
• B.
The likelihood of type 1 and type 2 errors
• C.
The number of expected cases
• D.
The variance explained by the measures
A. The magnitude of the relationship between variables
Explanation
An effective size refers to the magnitude of the relationship between variables. It measures the strength or intensity of the relationship between two or more variables. It indicates how much one variable is influenced or affected by another variable. A larger effective size indicates a stronger relationship between the variables, while a smaller effective size suggests a weaker relationship. Therefore, the effective size is a crucial measure in determining the strength and significance of the relationship between variables.
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• 8.
How can low expected values be dealt with?
• A.
Add more of the same number
• B.
Increase the sample size or combine categories
• C.
Exclude outliers
• D.
None of these
B. Increase the sample size or combine categories
Explanation
To deal with low expected values, increasing the sample size or combining categories can be effective. Increasing the sample size provides more data points, which can lead to more accurate and reliable results. Combining categories can help to increase the frequency of occurrence, making the expected values more meaningful. This approach allows for a better representation of the data and helps to reduce the impact of outliers, which may skew the results. Adding more of the same number or excluding outliers may not necessarily address the issue of low expected values.
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• 9.
By which other name is a chi-square contingency table analysis known?
• A.
Chi-square test for independence
• B.
Chi-square test for dependence
• C.
Chi-square variable test
• D.
None of the above
A. Chi-square test for independence
Explanation
A chi-square contingency table analysis is also known as a Chi-square test for independence. This test is used to determine if there is a significant association between two categorical variables. It compares the observed frequencies in each category of the variables with the expected frequencies under the assumption of independence. If the test statistic is significant, it indicates that there is a relationship between the variables. Therefore, the correct answer is Chi-square test for independence.
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• 10.
Which of these symbols is used to represent a chi-square test?
• A.
C
• B.
Q
• C.
T
• D.
X2
D. X2
Explanation
The symbol "X2" is used to represent a chi-square test. Chi-square tests are statistical tests that are used to determine if there is a significant association between two categorical variables. The "X2" symbol is derived from the Greek letter "chi" (χ) and represents the chi-square statistic, which is calculated by comparing the observed and expected frequencies in a contingency table. By comparing the calculated chi-square statistic to a critical value from the chi-square distribution, one can determine if the observed frequencies significantly differ from the expected frequencies.
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# Mrs. Franklin's Fabulous Firsties
## Is it Spring Yet?
Well, I know it has officially been spring for a while, but it is hard to believe it with all of our chilly weather. It is safe to say that we have had our fair share of indoor recess over the past few weeks. While we do enjoy playing indoors, the children and I are ready for outside recess! Please continue to monitor how you child is dressing for school. I recommend dressing in layers. It can be very chilly in different parts of our building and having a sweater or sweatshirt is a warm option.
Mrs. Miller and Mrs. Aumiller were kind enough to invite our classes down to the cafe for some freeze dance Friday afternoon. We had a blast and we burned off some energy!
## A Little Review
We wrapped up our ELA unit by learning all about Thomas Edison. We took notes on his life and created a timeline of important events. From that, students wrote an informative paragraph (complete with a topic sentence, details and a closing sentence) about why Thomas Edison was a great inventor. In addition, they created their own autobiography. Thanks to all who sent in pictures. The children had a blast writing about themselves and sharing their final projects. I enjoyed seeing all of the cute baby pictures and am so impressed with how their writing skills have developed over the year. I hope you enjoy reading them this weekend. The rough drafts of their Thomas Edison writing will come home this week, but their published copy will be on display in the classroom.
In math, we have been working hard on skip-counting by 2s, 4s, 5s, 10s and for a challenge 3s. The students are using a variety of strategies to solve problems that involve skip counting which include drawing pictures, writing number strings, using the 100s charts, and creating a T-chart. Some students have made the connection to multiplication. While we've shared this concept in class, they are not expected to write multiplication problems in first grade.
Next week in phonics we will focus on words with the -oo sound as in "spoon". This sound has many spellings - oo, -u_e, -ew, -ue, -ui, and -ou. We will practice many words with the sound and will analyze where the spellings typically fall in syllables (beginning, middle, end).
Next Friday is the JES Carnival. I hope to see you there!
5/24 - First grade field trip to Goucher
Please mark the following dates down. More information will come as we get closer to the dates.
5/26 - First grade field day
6/6 - Zoomobile to visit first grade
6/10 - First grade End of Year picnice
## Happy Mother's Day!
To all the fellow moms - I hope you have a wonderful Mother's Day! You are awesome. You are raising smart, kind, funny, creative, caring children. I'm so lucky to be a part of their upbringing and to occasionally be called, "Mommy (insert red face), I mean Mrs. Franklin...". It just goes to show that you are on the tip of their tongue throughout the day and that they are thinking about you. Enjoy your surprises and enjoy the day!
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# MAth
posted by on .
find the next three numbers in the pattern
2/3, 1 5/12, 2 1/6, 2 11/12 _, _, _
• MAth - ,
8/12, 17/12, 26/12, 35/12 _, _, _
ah ha, each is 9/12 bigger than the one before it
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## Welcome to BrainDen.com - Brain Teasers Forum
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8 replies to this topic
### #1 brhan
brhan
Advanced Member
• Members
• 466 posts
Posted 04 March 2008 - 06:34 PM
Twice four and twenty blackbirds
sitting in the rain
I shot and killed a quarter of them
How many do remain?
• 0
### #2 zoris
zoris
Junior Member
• Members
• 43 posts
Posted 04 March 2008 - 09:12 PM
Twice four and twenty blackbirds
sitting in the rain
I shot and killed a quarter of them
How many do remain?
Spoiler for Solution
• 0
### #3 heatherlovesjade
heatherlovesjade
Junior Member
• Members
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Posted 04 March 2008 - 09:13 PM
12
2(4+20)
2(24)= 48/4= 12
• 0
### #4 heatherlovesjade
heatherlovesjade
Junior Member
• Members
• 98 posts
Posted 04 March 2008 - 09:15 PM
or none for they flew away- but remaining from the original amount= not 12 = 36 would be left from the original
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### #5 superbc
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Posted 04 March 2008 - 09:55 PM
Murderer
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### #6 bonanova
bonanova
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Posted 05 March 2008 - 12:22 AM
Twice four and twenty blackbirds
sitting in the rain
I shot and killed a quarter of them
How many do remain?
Seven.
• 0
The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell
### #7 storm
storm
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Posted 05 March 2008 - 03:07 AM
Seven....others will fly away....
• 0
### #8 roolstar
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Posted 05 March 2008 - 11:25 AM
Seven remain dead...
• 0
### #9 brhan
brhan
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Posted 06 March 2008 - 01:03 PM
good job, guys. It is 7 or 12 ... where you put the parenthesis
(2*(4+20))/4 = 12
((2*4)+20)/4 = 7
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Determine the set T2 of transitions that can be cleared on occurrence of e. However, SFC is in fact a design method and not a programming language. I hope this info helps you in your search. Decotignie, Liveness — example A transition t of a PN is said: It is assumed that each computer may be in cougs of the following states: Hello Doug, I want to thank you for your guidance and your help, your information is valuable to me. Grafcet Exercices 1 et 2. Clear transitions that can be.
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We share information about your activities on the site with our partners and Google partners: May be safe after reduction even if original is not. Execute the pulse shaped actions that ladddr associated to the step that were activated at step 3 incl.
Unfortunately, the translating is done a little hasty. Decotignie, d Masquage G2: A stable situation has been reached 1. Page 26 to 28 give you the translation of a simple grafcet or SFC. Determine the set A0 of continuous actions that must be deactivated incl. Variables c1,c2 and d correspond to end of track sensors.
Decotignie, Reduction R2: Hello Doug, I want to thank you for your guidance and your help, your information is valuable to me. Calculer avec un tableau de Karnaugh la fonction Set to 1 all the grarcet that belong to A1 4. Is it a state graph? They turn to 1 when a wagon is present at the given point. Once again thank you so much for your valuable help Mohoch Megabar.
If, after clearing, the state is unchanged goto 6 4. Variable a1 turns to 1 when the front wheels of wagon H1 are on the tracks between A1 and D same for a2 if wagon H2 is between A2 and D. However, SFC is in fact a design method and not a programming language. Hay muchas formas de It is not always possible to find the home state and the bound. Remember me Forgot password? As I understand it I think I have to define several bits one per step such that each one is reset when the following becomes set.
Determine the set T2 of transitions that can be cleared on occurrence of e. Is it without loop? Decotignie, Reduction R1: Clear transitions that can be. Decotignie, Exercise — transitions Exercise — transitions 2 A. The transition defined on the top part of the ladder logic program will then establish the switching from on step to the next.
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### Home > CCA > Chapter Ch4 > Lesson 4.2.2 > Problem4-51
4-51.
Hotdogs and corndogs were sold at last night's football game. Use the information below to write mathematical sentences to help you determine how many corndogs were sold. Homework Help ✎
1. The number of hotdogs sold was three fewer than twice the number of corndogs. Write a mathematical sentence that relates the number of hotdogs and corndogs. Let h represent the number of hotdogs and c represent the number of corndogs.
h = 2c − 3
2. A hotdog costs $3 and a corndog costs$1.50. If \$201 was collected, write a mathematical sentence to represent this information.
Multiply the number of hotdogs by 3 and the number of corndogs by 1.5.
Add the products and set them equal to 201.
3. How many corndogs were sold? Show how you found your answer.
• Substitute 2c − 3 from the equation in part (a) into the equation in part (b).
Solve for c.
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# BOBSLEDDING
Tags:
Time Limit: 1 s Memory Limit: 128 MB
Submission：43 AC：12 Score：99.33
## Description
Dr.Kong has entered a bobsled competition because he hopes his hefty weight will give his an advantage over the L meter course (2 <= L<= 1000).
Dr.Kong will push off the starting line at 1 meter per second, but his speed can change while he rides along the course. Near the middle of every meter Bessie travels, he can change his speed either by using gravity to accelerate by one meter per second or by braking to stay at the same speed or decrease his speed by one meter per second.
Naturally, Dr.Kong must negotiate N (1 <= N <= 500) turns on the way down the hill. Turn i is located T_i meters from the course start (1 <= T_i <= L-1), and he must be enter the corner meter at a peed of at most S_i meters per second (1 <= S_i <= 1000). Dr.Kong can cross the finish line at any speed he likes.
Help Dr.Kong learn the fastest speed he can attain without exceeding the speed limits on the turns.
Consider this course with the meter markers as integers and the turn speed limits in brackets (e.g., '[3]'):
0 1 2 3 4 5 6 7 [3] 8 9 10 11 [1] 12 13 [8] 14
(Start) |------------------------------------------------------------------------------------- ---------?| (Finish)
Below is a chart of Dr.Kong 's speeds at the beginning of each meter length of the course:
Max: [3] [1] [8]
Mtrs: 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
Spd: 1 2 3 4 5 5 4 3 4 3 2 1 2 3 4
His maximum speed was 5 near the beginning of meter 4.
## Input
Line 1: Two space-separated integers: L and N.
Lines 2..N+1: Line i+1 describes turn i with two space-separated integers: T_i and S_i.
## Output
Line 1: A single integer, representing the maximum speed which Dr.Kong can attain between the start and the finish line, inclusive.
## Samples
input
14 3 7 3 11 1 13 8
output
5
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# Kinematics/Transformations
Kinematics and SL(2,R)
Kinematics is the study of motion without regard to force or momentum. SL(2,R) is the special linear group of 2 x 2 real matrices with determinant 1. It is a continuous group of three dimensions as the four matrix entries are constrained by determinant fixed at 1.
The coordinate plane may be taken as a frame of reference (t, x). When an element of SL(2, R) acts on the coordinate plane, the area of figures is preserved due to the determinant condition. Some matrices yield a rotation of the plane, which preserves distance as well as area. Kinematic transformations with zero acceleration are expressed by matrices that preserve area but not distance.
The first instance is linear motion of velocity v where a new position of x, after time t, is x + vt. The linear transformation is expressed:
${\displaystyle (t,x){\begin{pmatrix}1&v\\0&1\end{pmatrix}}=(t,tv+x).}$
It is called a shear transformation in linear algebra. This shear transformation is part of classical mechanics and has been replaced in the electronic age as follows:
The supreme velocity in kinematics is the velocity of light c, so a kinematic velocity v satisfies v < c or v /c < 1. The concept of rapidity is introduced using hyperbolic angles and the hyperbolic tangent function. This function is also bounded above by 1, so rapidity w satisfies tanh w = v /c. Change of frame of reference is accomplished by hyperbolic rotation:
${\displaystyle (t,x){\begin{pmatrix}\cosh w&\sinh w\\\sinh w&\cosh w\end{pmatrix}},}$
where cosh w is hyperbolic cosine of w and sinh w is hyperbolic sine of w. The trigonometric identity
${\displaystyle \cosh ^{2}w-\sinh ^{2}w=1}$ confirms that the transformation is in SL(2, R).
Velocity measured with hyperbolic angle uses wings for units, where one wing is the speed of light in water as w:Ludwik Silberstein notes on page 181 of The Theory of Relativity (1914). More modest speeds are commonly described in miles per hour (mph) in English measure. Naturally one mph is 1 mile/3600 seconds compared to c = 186,000 miles per second. The ratio 1 mph/c is about 1.5 x 10−9. Thus the rapidity w for one mph satisfies tanh w = 1.5 x 10−9.
Since the derivative of tanh w is sech2 w, the maximum rate of change of tanh w is one when w=0, and away from zero the slope of tanh nears zero. So for small values of w, tanh behaves as the identity function. Thus the rapidity corresponding to a mile per hour is 1.5 nanowings where a nanowing is 10−9 wings. A hundred mile per hour fastball has 150 nanowings rapidity. A kilometer per hour is 0.9375 nanowings.
Recall that rotations are in SL(2,R) and note the angular measure of these transformations. The classical and modern transformations for linear motion also have angular measures: differences of slope and hyperbolic angle. See Geometry/Unified Angles for details.
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# Effects of marketing, bank loan and credit debt on consumer’s spending. Mathematical models based on an engineering concept
Ausarbeitung 2019 14 Seiten
## Introduction to the study
The main objective of this study is to utilize an engineering concept in order to propose a mathematical model to correlate consumer spending, utility and income. The difference between the proposed model and the Keynesian consumption theory is explained by the fact that the Keynesian consumption theory takes into account the consumption of costumers with no income. The effects of marketing, bank loans and credit debt on consumer spending are also analyzed using the general equation of transport phenomena and mathematical models are presented for the first time. Based on a case study, marketing has increased the utility (driving force) by 61%. Taking into account the theory of consumption smoothing, bank loans also provide the consumer with additional spending power by decreasing the resistance for consumption. In case of excessive debt, customers might spend the money only to buy the “utility” in order to be able to repay the debt. In this situation, the effects of debt are described in the proposed engineering model as a decrease in income (extra resistance to spend money).
## 1. Dynamic Systems & their Universal Law
Everything in the universe is continuously in motion and the object can be as small as an atomic particle or as large as a planet. Gravitational and electromagnetic forces are responsible for large objects to be in motion while weak and strong nuclear forces are the driving factors for the quantum world to be in continuous motion. From an engineering perspective, flows take place in dynamic systems due to a driving force within the system and are controlled by a resistance located between two poles of the system. According to the second law of thermodynamics, this driving force is the difference in concentrations of energy between the two poles. For example, heat transfer in a piece of metal is transported from a higher temperature to a lower temperature and the speed of the flow of heat transfer is controlled by the resistance of the metal to heat transfer. The rain falls from the sky (higher altitude) to the land (lower altitude). Without the resistance of air to the gravitational force, rain drops will destroy all the trees and vegetation. This universal phenomena could therefore be described using the following generalized relationship for transport phenomena:
Abbildung in dieser Leseprobe nicht enthalten
In the field of electricity, the current I (flow of electrons) is motivated by a driving force (difference in potential = ΔU) and controlled by the electrical resistance (R) of the circuit. Ohm’s law is then obtained:
Abbildung in dieser Leseprobe nicht enthalten
In the study of Chemical Engineering, momentum, mass and heat transport also share a very similar framework. For example, Fourier’s law of heat conduction (Thermal Ohm’s law) is defined as:
Abbildung in dieser Leseprobe nicht enthalten
Where: Q = heat flow; ∆T= Difference in Temperature (driving force), ∆x= distance of heat conduction, A= Surface area of the metal, k= Heat conductivity of the metal; (∆x/kA) is therefore the resistance to heat flow by conduction.
Based on the same concept of transport phenomena, described in equation (1), people could be described as dynamic systems motivated to take roads and highways to go to schools for studying, to workplaces to make money and to markets to buy what is needed for their daily life. People have therefore a natural motivation (driving force) to spend their money on buying food, homes, furniture, electronic devices, etc. On the other hand, the amount of money spent by consumers is limited by their income (conductance). The main objective of this investigation is an attempt to find a mathematical model, based on an engineering concept, in order to: (1) propose a new mathematical model for consumer spending, (2) mathematically correlate the effects of marketing (increasing the driving force), bank loan (decreasing the resistance) and credit debt (increasing resistance) on consumer spending (flow of money).
## 2.Gross domestic product and consumer-based economy
The gross domestic product (GDP) is the indicator of national income and all outputs for a given country's economy. The GDP represents therefore the total expenditures for all final goods and services produced within the country in a stipulated period of time. As shown in Figure 1, the Gross Domestic Product (GDP) of the United States was worth 19390.60 billion US dollars in 20171
Abbildung in dieser Leseprobe nicht enthalten
Figure 1: Gross Domestic Product for the USA from 2008 to 20171.
The expenditures approach and the income approach are the two known methods to calculate the gross domestic product (GDP). Both of these approaches attempt to best approximate the monetary value of all final goods and services produced in an economy over a set period of time (normally one year). The major distinction between each approach is its starting point. The expenditure approach begins with the money spent on goods and services. Conversely, the income approach starts with the income earned (wages, rents, interest, profits) from the production of goods and services2. For the income approach, the GDP is calculated by adding the following elements3:
GDP= TNI + ST + D+ NFFI(4)
Where TNI= Total National Income; ST= Sales Taxes; D= Depreciation and NFFI= Net Foreign Factor Income. TNI is equal to the sum of all wages plus rents plus interest and profits. Some economists challenge the notion of including sales taxes in the GDP formula on the basis that taxation is counterproductive. They think it should subtract from total output rather than add to it. However, most use the income approach that includes sales taxes.
For the expenditure approach, the formula utilized to calculate the GDP is4:
Abbildung in dieser Leseprobe nicht enthalten
1. “C” (consumption) is normally the largest GDP component in the economy, consisting of private expenditures (household final consumption expenditure) in the economy.
2. “I” (investment) includes, for instance, business investment in equipment and spending by households (not government) on new houses is also included in Investment.
3. “G” (government spending) is the sum of government expenditures on final goods and services
4. “X” (exports) represents gross exports. GDP captures the amount a country produces, including goods and services produced for other nations’ consumption.
5. “M” (imports) represents gross imports. Imports are subtracted since imported goods will be included in the terms “G”, “I”, or “C”, and must be deducted to avoid counting foreign supply as domestic.
A consumer economy is defined as an economy driven by consumer spending as a percentage of its GDP (Gross Domestic Product). Keynesian economic theory proposes that governments should stimulate spending to end a recession. Supply-side economists recommend the opposite. They believe that governments should cut business taxes to create jobs. However companies won't increase production if the demand is not there5. Consumers are, therefore, very important to businesses. The more money consumers spend with a given company, the better that company tends to perform. For this reason, it is unsurprising that most investors and businesses pay a great amount of attention to consumer spending figures and patterns6.
## 3. Proposed engineering model for consumer spending
Contemporary measures of consumer spending include all private purchases of durable goods, nondurables and services. Consumer spending (CS) is the demand side of “supply and demand"; production is the supply. Without consumer spending, there is therefore no motivation to produce goods. Goods are generally divided into two categories: durable goods, like autos, furniture or any item that has a useful life of three years or more. The second is non-durable goods, such as fuel, food, and clothing. In 2017, the consumer spending in the USA made up to 70% of GDP (\$12.6 trillion) 3, 4. Consumer spending is therefore the main driving force of the economic system in the USA. Nearly two-thirds of consumer spending is on services, like real estate and healthcare. The remaining one-third of personal consumption expenditure is on goods. These include so-called durable goods, such as washing machines, automobiles, and furniture. More frequently, people buy non-durable goods, such as gasoline, groceries, and clothing. The consumption of these goods is the result of economic activity. This is because individuals ultimately use these goods to satisfy their own needs and wants; economists refer to this satisfaction as “utility4.
The consumption function, or Keynesian consumption function, is an economic formula that represents the functional relationship between total consumption and gross national income. It was introduced by British economist John Maynard Keynes, who argued the function could be used to track and predict total aggregate consumption expenditures7. The relationship between consumption and income is based on the fundamental psychological law of consumption which states that when income increases consumption expenditure also increases but by a smaller amount8.
Abbildung in dieser Leseprobe nicht enthalten Figure 2: Consumption Function8
Based on Figure2, the Keynesian consumption function is linear8:
Abbildung in dieser Leseprobe nicht enthalten
(6)
Where a is the intercept (a constant which measures consumption at a zero level of disposal income), b is the marginal propensity to consume (MPC) and Y is the disposable income. The MPC is the proportion of an aggregate raise in pay that a consumer spends on the consumption of goods and services, as opposed to saving it. Y is total personal income minus personal current taxes. The above formula describing consumption as a function of current disposable income whether linear or non-linear is called the absolute income hypothesis. This consumption function has the following properties8:
1. As income increases, average propensity to consume (APC = C / Y) falls.
2. The marginal propensity to consume MPC is positive but less than unity (0 < b < 1) so that higher income leads to higher consumption.
3. The consumption expenditure increases or decreases with increase or decrease in income but non-proportionally. This non-proportional consumption function implies that in the short run average and marginal propensities do not coincide (APC > MPC).
4. This consumption function is stable both in the short run and the long run.
In this investigation, consumer’s spending (CS) taken as the money people spend to buy the goods during one month will be defined. Considering an analogy with equation (1), people have a motivation (driving force) to spend their money to buy goods, defined as utility4. On the other hand, the amount of money spent depends on personal income. The income effect relates to how a consumer spends money based on an increase or decrease in income. An increase in income results in demanding more services and goods, thus spending more money. A decrease in income results in the exact opposite. In general, when income is lower, less spending occurs9. In the proposed model, the income is considered as the conductance for spending money and the inverse (1/income) is therefore its resistance. Following the general equation (1) of transport phenomena, consumer’s spending (CS) is defined in this investigation as:
Abbildung in dieser Leseprobe nicht enthalten
Equation (7) could be reorganized as:
Abbildung in dieser Leseprobe nicht enthalten
The proposed equation (8) indicates that consumer spending increases proportionally with the utility, as the driving force, and the personal income of the buyer as the conductance (1/resistance) for buying. In this concept of dynamic systems, it is assumed that people buy only what they need. As a consequence, without this driving force, the consumer spending is equal to zero. This concept is also based on the fact that people with no income are not able to spend money to buy what they need. In comparison with the Keynesian consumption function, the proposed equation (8) could be rewritten as:
Abbildung in dieser Leseprobe nicht enthalten
[...]
## Details
Seiten
14
Jahr
2019
ISBN (eBook)
9783668900660
ISBN (Buch)
9783668900677
Sprache
Englisch
Erscheinungsdatum
2019 (März)
Schlagworte
consumer spending engineering models marketing bank loan credit debt.
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# Random matrices, differential operators and carousels matyd/BMC/slides/Valko - Random... · PDF file Random matrices, di erential operators and carousels Benedek Valko (University
Feb 18, 2020
## Documents
others
• Random matrices, differential operators and carousels
Benedek Valkó (University of Wisconsin – Madison)
joint with B. Virág (Toronto)
March 24, 2016
• Basic question of RMT:
What can we say about the spectrum of a large random matrix?
-60 -40 -20 0 20 40 60
5
10
15
20
25
30
35
Ln
b HLn - aL
global local
In this talk: local picture (point process limits)
• Basic question of RMT:
What can we say about the spectrum of a large random matrix?
-60 -40 -20 0 20 40 60
5
10
15
20
25
30
35
Ln
b HLn - aL
global local
In this talk: local picture (point process limits)
• Basic question of RMT:
What can we say about the spectrum of a large random matrix?
-60 -40 -20 0 20 40 60
5
10
15
20
25
30
35
Ln
b HLn - aL
global local
In this talk: local picture (point process limits)
• A classical example: Gaussian Unitary Ensemble
M = A+A ∗
√ 2
, A is n × n with iid complex std normal.
Global picture: Wigner semicircle law -60 -40 -20 0 20 40 60 5
10
15
20
25
30
35
Local picture: point process limit in the bulk and near the edge
(Bulk: Dyson-Gaudin-Mehta, edge: Tracy-Widom)
The limit processes are characterized by their joint intensity functions.
Roughly: what is the probability of finding points near x1, . . . , xn
• A classical example: Gaussian Unitary Ensemble
M = A+A ∗
√ 2
, A is n × n with iid complex std normal.
Global picture: Wigner semicircle law
-60 -40 -20 0 20 40 60
5
10
15
20
25
30
35
Local picture: point process limit in the bulk and near the edge
(Bulk: Dyson-Gaudin-Mehta, edge: Tracy-Widom)
The limit processes are characterized by their joint intensity functions.
Roughly: what is the probability of finding points near x1, . . . , xn
• A classical example: Gaussian Unitary Ensemble
M = A+A ∗
√ 2
, A is n × n with iid complex std normal.
Global picture: Wigner semicircle law -60 -40 -20 0 20 40 60 5
10
15
20
25
30
35
Local picture: point process limit in the bulk and near the edge
(Bulk: Dyson-Gaudin-Mehta, edge: Tracy-Widom)
The limit processes are characterized by their joint intensity functions.
Roughly: what is the probability of finding points near x1, . . . , xn
• A classical example: Gaussian Unitary Ensemble
M = A+A ∗
√ 2
, A is n × n with iid complex std normal.
Global picture: Wigner semicircle law -60 -40 -20 0 20 40 60 5
10
15
20
25
30
35
Local picture: point process limit in the bulk and near the edge
(Bulk: Dyson-Gaudin-Mehta, edge: Tracy-Widom)
The limit processes are characterized by their joint intensity functions.
Roughly: what is the probability of finding points near x1, . . . , xn
• A classical example: Gaussian Unitary Ensemble
M = A+A ∗
√ 2
, A is n × n with iid complex std normal.
Global picture: Wigner semicircle law -60 -40 -20 0 20 40 60 5
10
15
20
25
30
35
Local picture: point process limit in the bulk and near the edge
(Bulk: Dyson-Gaudin-Mehta, edge: Tracy-Widom)
The limit processes are characterized by their joint intensity functions.
Roughly: what is the probability of finding points near x1, . . . , xn
• Point process limit
Ln
b HLn - aL
Finite n: spectrum of a random Hermitian matrix
Limit point process: spectrum of ??
• Point process limit
Ln
b HLn - aL
Finite n: spectrum of a random Hermitian matrix
Limit point process: spectrum of ??
• Detour to number theory
Riemann zeta function: ζ(s) = ∞∑ n=1
1 ns , for Re s > 1.
(Analytic continuation to C \ {1})
Riemann hypothesis: the non-trivial zeros are on the line Re s = 12 .
Dyson-Montgomery conjecture:
After some scaling:
non-trivial zeros of ζ( 1
2 + i s) ∼ bulk limit process of GUE
(Sine2 process)
I Strong numerical evidence: Odlyzko
I Certain weaker versions are proved (Montgomery, Rudnick-Sarnak)
• Detour to number theory
Riemann zeta function: ζ(s) = ∞∑ n=1
1 ns , for Re s > 1.
(Analytic continuation to C \ {1})
Riemann hypothesis: the non-trivial zeros are on the line Re s = 12 .
Dyson-Montgomery conjecture:
After some scaling:
non-trivial zeros of ζ( 1
2 + i s) ∼ bulk limit process of GUE
(Sine2 process)
I Strong numerical evidence: Odlyzko
I Certain weaker versions are proved (Montgomery, Rudnick-Sarnak)
• Detour to number theory
Riemann zeta function: ζ(s) = ∞∑ n=1
1 ns , for Re s > 1.
(Analytic continuation to C \ {1})
Riemann hypothesis: the non-trivial zeros are on the line Re s = 12 .
Dyson-Montgomery conjecture:
After some scaling:
non-trivial zeros of ζ( 1
2 + i s) ∼ bulk limit process of GUE
(Sine2 process)
I Strong numerical evidence: Odlyzko
I Certain weaker versions are proved (Montgomery, Rudnick-Sarnak)
• Hilbert-Pólya conjecture: the Riemann hypotheses is true because
non-trivial zeros of ζ( 1
2 + i s)
= ev’s of an unbounded self-adjoint operator
A famous attempt to make this approach rigorous: de Branges
(based on the theory of Hilbert spaces of entire functions)
This approach would produce a self-adjoint differential operator with the appropriate spectrum.
• Hilbert-Pólya conjecture: the Riemann hypotheses is true because
non-trivial zeros of ζ( 1
2 + i s)
= ev’s of an unbounded self-adjoint operator
A famous attempt to make this approach rigorous: de Branges
(based on the theory of Hilbert spaces of entire functions)
This approach would produce a self-adjoint differential operator with the appropriate spectrum.
• Hilbert-Pólya conjecture: the Riemann hypotheses is true because
non-trivial zeros of ζ( 1
2 + i s)
= ev’s of an unbounded self-adjoint operator
A famous attempt to make this approach rigorous: de Branges
(based on the theory of Hilbert spaces of entire functions)
This approach would produce a self-adjoint differential operator with the appropriate spectrum.
• Natural question:
Is there a self-adjoint differential operator with a spectrum given by the bulk limit of GUE?
Disclaimer: A positive answer would not get us closer to any of the conjectures or the Riemann hypothesis (unfortunately...)
Borodin-Olshanski, Maples-Najnudel-Nikeghbali:
‘operator-like object’ with generalized eigenvalues distributed as Sine2
• Natural question:
Is there a self-adjoint differential operator with a spectrum given by the bulk limit of GUE?
Disclaimer: A positive answer would not get us closer to any of the conjectures or the Riemann hypothesis (unfortunately...)
Borodin-Olshanski, Maples-Najnudel-Nikeghbali:
‘operator-like object’ with generalized eigenvalues distributed as Sine2
• Natural question:
Is there a self-adjoint differential operator with a spectrum given by the bulk limit of GUE?
Disclaimer: A positive answer would not get us closer to any of the conjectures or the Riemann hypothesis (unfortunately...)
Borodin-Olshanski, Maples-Najnudel-Nikeghbali:
‘operator-like object’ with generalized eigenvalues distributed as Sine2
• Starting point for deriving the Sine2 process:
Joint eigenvalue density of GUE:
1
Zn
∏ i
• Starting point for deriving the Sine2 process:
Joint eigenvalue density of GUE:
1
Zn
∏ i
• Starting point for deriving the Sine2 process:
Joint eigenvalue density of GUE:
1
Zn
∏ i
• β-ensemble: finite point process with joint density
1
Zn,f ,β
∏ i
• Scaling limits - global picture
Hermite β-ensemble semicircle law Laguerre β-ensemble Marchenko-Pastur law
-2 2 1 2 3 4
↑ ↑ ↗ ↑ ↑ ↑ soft edge bulk s. e. hard edge bulk s. e.
• Local limits
Soft edge: Rider-Raḿırez-Virág (Hermite, Laguerre) Airyβ process
Hard edge: Rider-Raḿırez (Laguerre) Besselβ,a processes
Bulk: Killip-Stoiciu, V.-Virág (circular, Hermite) CβE and Sineβ processes
Instead of joint intensities, the limit processes are described via their counting functions using coupled systems of SDEs.
sign(λ) · (# of points in [0, λ])
• Local limits
Soft edge: Rider-Raḿırez-Virág (Hermite, Lague
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# 1996 USAMO Problems/Problem 2
## Problem
For any nonempty set $S$ of real numbers, let $\sigma(S)$ denote the sum of the elements of $S$. Given a set $A$ of $n$ positive integers, consider the collection of all distinct sums $\sigma(S)$ as $S$ ranges over the nonempty subsets of $A$. Prove that this collection of sums can be partitioned into $n$ classes so that in each class, the ratio of the largest sum to the smallest sum does not exceed 2.
## Solution
Let set $A$ consist of the integers $a_1\le a_2\le a_3\le\dots\le a_n$. For $k\ge 2$, call $a_k$ greedy if $\sum_{i=1}^{k-1}a_i < a_k$. Also call $a_1$ greedy. Now put all elements of $A$ into groups of consecutive terms in such a way that each group $G$ begins with a greedy term, call it $a_p$, and ends on the term $a_{q-1}$ just before the next greedy term after $a_p$. (If $a_p$ is the last greedy term, let $q-1=n$.) We introduce some more terminology. A sum $\sigma(S)$ is said to "belong to" a group $G$ if $\max(S)\in G$. Denote by $\mathcal{S}(G)$ the set of all sums belonging to $G$.
We now show that we can divide $\mathcal{S}(G)$ into $|G|$ (the cardinality of $G$) classes in such a way that the maximum and minimum sum belonging to each class differ by no more than a factor of 2. Using the previous notation, we first prove that $\frac{\max{\mathcal{S}(G)}}{\min{\mathcal{S}(G)}}< 2^{|G|}=2^{q-p}$. Note that $$\max{\mathcal{S}(G)}=\sum_{i=1}^{q-1}a_i$$ and $$\min{\mathcal{S}(G)}=a_p.$$ Taking note of that fact that $a_{p+1},a_{p+2},\dots,a_{q-1}$ are not greedy numbers, we write:
\begin{align*} \max{\mathcal{S}}&=\sum_{i=1}^{q-1}a_i \\ &=a_{q-1}+\sum_{i=1}^{q-2}a_i \\ &\le 2\sum_{i=1}^{q-2}a_i\\ &\le 2^2\sum_{i=1}^{q-3}a_i\\ &\vdots\\ &\le 2^{q-p-1}\sum_{i=1}^{p}a_i\\ &= 2^{q-p-1}(\sum_{i=1}^{p-1}a_i + a_p)\\ &<2^{q-p-1}(2a_p)\\ &=2^{q-p}\min\mathcal{S} \end{align*}
where the inequalities after the ellipses result from the fact that $a_p$ is a greedy number (which implies by definition that $\sum_{i=1}^{p-1}a_i). This proves the desired inequality. Now we can prove the result in bold above. Divide $\mathcal{S}(G)$ into $|G|=q-p$ classes by taking those terms in $[a_p,2a_p)$ and placing them in the first class, taking those terms in $[2a_p,2^2a_p)$ and placing them in another, and so on, until we reach $[2^{q-p-1}a_p,2^{q-p}a_p)$. The inequality we proved above shows that all of the sums in $\mathcal{S}(G)$ will fall in one of these classes, as the intervals into which the classes fall form a continuous range bounded by $\min\mathcal{S}(G)=a_p$ on the bottom and $2^{q-p}a_p>\max\mathcal{S}(G)$ on the top. This proves the result in bold.
However, that clearly implies the desired conclusion, as every sum belongs to a group, and every sum belonging to a group is a member of a class. Moreover, there will be precisely $n$ classes, because every term $a_i$ belongs to a group and for each group $G$, there are as many classes for $\mathcal{S}(G)$ as there are terms in $G$.
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Become a Multivariable Calculus Pro!
Recent questions in Multivariable calculus
zookeeper1930r8k 2023-03-29
In a regression analysis, the variable that is being predicted is the "dependent variable."a. Intervening variable b. Dependent variable c. None d. Independent variable?
Arjun Patterson 2023-03-21
Repeated addition is called ? A)Subtraction B)Multiplication C)Division
Exceplyclene72 2022-12-24
Multiplicative inverse of 1/7 is _?
klepnin4wv 2022-12-19
Does the series converge or diverge this $\sum n!/{n}^{n}$
Leandro Acosta 2022-12-18
Use Lagrange multipliers to find the point on a surface that is closest to a plane.Find the point on $z=1-2{x}^{2}-{y}^{2}$ closest to $2x+3y+z=12$ using Lagrange multipliers.I recognize $z+2{x}^{2}+{y}^{2}=1$ as my constraint but am unable to recognize the distance squared I am trying to minimize in terms of 3 variables. May someone help please.
Will Osborn 2022-12-04
Which equation illustrates the identity property of multiplication? A$\left(a+\mathrm{bi}\right)×c=\left(\mathrm{ac}+\mathrm{bci}\right)$ B$\left(a+\mathrm{bi}\right)×0=0$ C$\left(a+\mathrm{bi}\right)×\left(c+\mathrm{di}\right)=\left(c+\mathrm{di}\right)×\left(a+\mathrm{bi}\right)$ D$\left(a+\mathrm{bi}\right)×1=\left(a+\mathrm{bi}\right)$
lascieflYr 2022-11-30
The significance of partial derivative notationIf some function like $f$ depends on just one variable like $x$, we denote its derivative with respect to the variable by:$\frac{\mathrm{d}}{\mathrm{d}x}f\left(x\right)$Now if the function happens to depend on $n$ variables we denote its derivative with respect to the $i$th variable by:$\frac{\mathrm{\partial }}{\mathrm{\partial }{x}_{i}}f\left({x}_{1},\cdots ,{x}_{i},\cdots ,{x}_{n}\right)$Now my question is what is the significance of this notation? I mean what will be wrong if we show "Partial derivative" of $f$ with respect to ${x}_{i}$ like this? :$\frac{\mathrm{d}}{\mathrm{d}{x}_{i}}f\left({x}_{1},\cdots ,{x}_{i},\cdots ,{x}_{n}\right)$Does the symbol $\mathrm{\partial }$ have a significant meaning?
Kierra Griffith 2022-11-22
The function $f\left(x,y,z\right)$ is a differentiable function at $\left(0,0,0\right)$ such that ${f}_{y}\left(0,0,0\right)={f}_{x}\left(0,0,0\right)=0$ and $f\left({t}^{2},2{t}^{2},3{t}^{2}\right)=4{t}^{2}$ for every $t>0$. Define $u=\left(6/11,2/11,9/11\right)$, with the given about. Is it possible to calculate ${f}_{u}\left(1,2,3\right)$ or ${f}_{u}\left(0,0,0\right)$, or ${f}_{z}\left(0,0,0\right)$?
jorgejasso85xvx 2022-11-19
Given topological spaces ${X}_{1},{X}_{2},\dots ,{X}_{n},Y$, consider a multivariable function $f:\prod _{i=1}^{n}{X}_{i}\to Y$ such that for any $\left({x}_{1},{x}_{2},\dots ,{x}_{n}\right)\in \prod _{i=1}^{n}{X}_{i}$, the functions in the family $\left\{x↦f\left({x}_{1},\dots ,{x}_{i-1},x,{x}_{i+1},\dots ,{x}_{n}\right){\right\}}_{i=1}^{n}$ are all continuous. Must $f$ itself be continuous?
Jenny Roberson 2022-11-18
Let $x$ be an independent variable. Does the differential dx depend on $x$?(from the definition of differential for variables & multivariable functions)
Alberto Calhoun 2022-11-18
Let $f:M\left(n,\mathbb{R}\right)\to M\left(n,\mathbb{R}\right)$ and let $f\left(A\right)=A{A}^{t}$. Then find derivative of $f$, denoted by $df$ .So, Derivative of $f\left(df\right)$ if exists, will satisfy $limH\to 0\frac{||f\left(A+H\right)-f\left(A\right)-df\left(H\right)||}{||H||}=0$.
Nicholas Hunter 2022-11-17
if $F\left(x,y\right)$ and $y=f\left(x\right)$,$\frac{dy}{dx}=-\frac{\frac{\mathrm{\partial }}{\mathrm{\partial }x}\left(F\right)}{\frac{\mathrm{\partial }}{\mathrm{\partial }y}\left(F\right)}$1) $F\left(x,y\right)$ 𝑎𝑛𝑑 $y=f\left(x\right)$ so his means that the function $F$ is a function of one variable which is $x$2) while we were computing 𝑝𝑎𝑟𝑡𝑖𝑎𝑙 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒𝑠 we treated $y$ and $x$ as two independent variables although that $y$ changes as $x$ changes but while doing the 𝑝𝑎𝑟𝑡𝑖𝑎𝑙 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒𝑠 w.r.t $x$ we treated $y$ and $x$ as two independent varaibles and considered $y$ as a constant
Laila Murphy 2022-11-17
Let $f:{\mathbb{R}}^{2}\to \mathbb{R}$ be defined asthen check whether its differentiable and also whether its partial derivatives ie ${f}_{x},{f}_{y}$ are continuous at $\left(0,0\right)$. I dont know how to check the differentiability of a multivariable function as I am just beginning to learn it. For continuity of partial derivative I just checked for ${f}_{x}$ as function is symmetric in $y$ and $x$. So ${f}_{x}$ turns out to be${f}_{x}\left(x,y\right)=2x\mathrm{cos}\left(\frac{1}{\sqrt{{x}^{2}+{y}^{2}}}\right)+\frac{x}{\sqrt{{x}^{2}+{y}^{2}}}\mathrm{sin}\left(\frac{1}{\sqrt{{x}^{2}+{y}^{2}}}\right)$which is definitely not $0$ as $\left(x,y\right)\to \left(0,0\right)$. Same can be stated for ${f}_{y}$. But how to proceed with the first part?
Sophie Marks 2022-11-16
How "messy" can a multivariable function be?$f\left(x,y\right)=\frac{2xy}{{x}^{2}+{y}^{2}}\phantom{\rule{1em}{0ex}}f\left(0,0\right)=0$
Alberto Calhoun 2022-11-13
Many mathematical texts define a multivariable function $f$ in the following way$f:=f\left(x,y\right)$However, if we focus on the fact that a function is really a binary relation on two sets, (say the real numbers), the definition would be as follows$f:{\mathbb{R}}^{2}\to \mathbb{R}$This seems to imply that the domain of the function is a set of ordered pairs of the form $\left(x,y\right)$.The set $\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{p}\mathrm{h}\left(f\right)\subset {\mathbb{R}}^{2}×\mathbb{R}$, would then comprise ordered pairs of the form$\left\{\left(\left({x}_{0},{y}_{0}\right),a\right),\left(\left({x}_{1},{y}_{1}\right),b\right),\dots \right\}$In line with this, does it not follow that the correction notation for $f$ should be be$f:=f\left(\left(x,y\right)\right)$
Siemensueqw 2022-11-11
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# Fifth Grade Math Common Core State Standards
## Operations and Algebraic Thinking
### Write and interpret numerical expressions.
Use parentheses, brackets, or braces in numerical expressions, and evaluate expressions with these symbols.
Write simple expressions that record calculations with numbers, and interpret numerical expressions without evaluating them.
### Analyze patterns and relationships.
Generate two numerical patterns using two given rules. Identify apparent relationships between corresponding terms. Form ordered pairs consisting of corresponding terms from the two patterns, and graph the ordered pairs on a coordinate plane.
## Number and Operations in Base Ten
### Understand the place value system.
Recognize that in a multi-digit number, a digit in one place represents 10 times as much as it represents in the place to its right and 1/10 of what it represents in the place to its left.
Explain patterns in the number of zeros of the product when multiplying a number by powers of 10, and explain patterns in the placement of the decimal point when a decimal is multiplied or divided by a power of 10. Use whole-number exponents to denote powers of 10.
Read, write, and compare decimals to thousandths.
Use place value understanding to round decimals to any place.
### Perform operations with multi-digit whole numbers and with decimals to hundredths.
Fluently multiply multi-digit whole numbers using the standard algorithm.
Find whole-number quotients of whole numbers with up to four-digit dividends and two-digit divisors, using strategies based on place value, the properties of operations, and/or the relationship between multiplication and division. Illustrate and explain the calculation by using equations, rectangular arrays, and/or area models.
Add, subtract, multiply, and divide decimals to hundredths, using concrete models or drawings and strategies based on place value, properties of operations, and/or the relationship between addition and subtraction; relate the strategy to a written method and explain the reasoning used.
## Number and Operations-Fractions
### Use equivalent fractions as a strategy to add and subtract fractions.
Add and subtract fractions with unlike denominators (including mixed numbers) by replacing given fractions with equivalent fractions in such a way as to produce an equivalent sum or difference of fractions with like denominators.
Solve word problems involving addition and subtraction of fractions referring to the same whole, including cases of unlike denominators, e.g., by using visual fraction models or equations to represent the problem. Use benchmark fractions and number sense of fractions to estimate mentally and assess the reasonableness of answers.
### Apply and extend previous understandings of multiplication and division to multiply and divide fractions.
Interpret a fraction as division of the numerator by the denominator (a/b = a ÷ b). Solve word problems involving division of whole numbers leading to answers in the form of fractions or mixed numbers, e.g., by using visual fraction models or equations to represent the problem.
Apply and extend previous understandings of multiplication to multiply a fraction or whole number by a fraction.
Interpret multiplication as scaling (resizing), by:
Solve real world problems involving multiplication of fractions and mixed numbers, e.g., by using visual fraction models or equations to represent the problem.
Apply and extend previous understandings of division to divide unit fractions by whole numbers and whole numbers by unit fractions.
## Measurement and Data
### Convert like measurement units within a given measurement system.
Convert among different-sized standard measurement units within a given measurement system (e.g., convert 5 cm to 0.05 m), and use these conversions in solving multi-step, real world problems.
### Represent and interpret data.
Make a line plot to display a data set of measurements in fractions of a unit (1/2, 1/4, 1/8). Use operations on fractions for this grade to solve problems involving information presented in line plots.
### Geometric measurement: understand concepts of volume and relate volume to multiplication and to addition.
Recognize volume as an attribute of solid figures and understand concepts of volume measurement.
Measure volumes by counting unit cubes, using cubic cm, cubic in, cubic ft, and improvised units.
Relate volume to the operations of multiplication and addition and solve real world and mathematical problems involving volume.
## Geometry
### Graph points on the coordinate plane to solve real-world and mathematical problems.
Use a pair of perpendicular number lines, called axes, to define a coordinate system, with the intersection of the lines (the origin) arranged to coincide with the 0 on each line and a given point in the plane located by using an ordered pair of numbers, called its coordinates. Understand that the first number indicates how far to travel from the origin in the direction of one axis, and the second number indicates how far to travel in the direction of the second axis, with the convention that the names of the two axes and the coordinates correspond (e.g., x-axis and x-coordinate, y-axis and y-coordinate).
Represent real world and mathematical problems by graphing points in the first quadrant of the coordinate plane, and interpret coordinate values of points in the context of the situation.
### Classify two-dimensional figures into categories based on their properties.
Understand that attributes belonging to a category of two-dimensional figures also belong to all subcategories of that category.
Classify two-dimensional figures in a hierarchy based on properties.
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## Moment Redistribution
"I
w/unit length w/unit length
Additional moments diagram (Hinges at A and C)
Collapse mechanism
Elastic BMD (Collapse loads) Final Collapse BMD
Figure 3.28 Moment redistribution, one-span beam
Additional moments diagram (Hinges at A and C)
Collapse mechanism
Elastic BMD (Collapse loads) Final Collapse BMD
■I- :-; is the load to cause the first plastic hinge; thus the beam may carry a load of with redistribution.
m the design point of view, the elastic bending-moment diagram can be obtained la* required ultimate loading in the ordinary way. Some of these moments may then •. -.Juced; but this will necessitate increasing others to maintain the static equilibrium , structure. Usually it is the maximum support moments which are reduced, so • - - -niMtig in reinforcing steel and also reducing congestion at the columns. The : . - remenls for applying moment redistribution are;
Equilibrium between internal and external forces must be maintained, hence it is necessary lo recalculate the span bending moments and the shear forces for the load case involved.
2. The continuous beams or slabs are predominately subject to flexure.
3. The ratio of adjacent spans be in the range of 0.5 lo 2.
### 4. The column design moments must nol be reduced.
There are oilier restrictions on the amount of moment redistribution in order to ensure ductility of the beams or slabs. This entails limitations on the grade of reinforcing steel .ind of the areas of tensile reinforcement and hence the depth of the neutral axis as described in Chapter Four -'Analysis of the Section'._
EXAMPLE 3^9| Moment redistribution
In example 3.3, figure 3.13, it is required to reduce the maximum support moment of •Wra = 147 kN m as much as possible, but without increasing the span moment above the present maximum value of 118kNm.
Figure 3.29
Moments and shears after redistribution
Figure 3.29
Moments and shears after redistribution
(a) Original Moments (kN m) 140
108 Rn 102
(a) Original Moments (kN m) 140
108 Rn 102
(b) Redistributed Moments (kN m) 134
## Greener Homes for You
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# App to solve math problems with camera
App to solve math problems with camera can be a helpful tool for these students. So let's get started!
## The Best App to solve math problems with camera
App to solve math problems with camera can help students to understand the material and improve their grades. If you're looking for help solving a word search, there are a few different ways you can cheat. One is to use a word search solver, which is a tool that will help you find words in a grid. Another way is to look up the words you need in a dictionary, and then find them in the grid. Finally, you can ask a friend for help.
Solution Complex numbers are numbers that have both a real and imaginary component. They can be written in the form a+bi, where a is the real component and bi is the imaginary component. Complex numbers can be solved using the quadratic formula.
A direct variation is a proportionality between two variables, in which one variable is a multiple of the other. For example, if y varies directly as x, and we know that y = 10 when x = 5, then we can find the value of y when x = 2 by multiplying 5 by 2, which equals 10. In general, if y varies directly as x, and y = k when x = a, then we can find the value of y when x = b by
Mathematical problems? To solve mathematical problems, one need to have good problem solving skills. It is important to be able to understand the problem, and then figure out what steps need to be taken to solve it. Once the steps are clear, one can then start solving the problem by following the steps. If at any point during the problem solving process one gets stuck, it is important to take a step back and reevaluate the situation. Sometimes, simply changing the approach to solving the problem can
This approach involves solving a system of equations that represents the relationships between the variables in the differential equation. The solver then uses this system of equations to find a solution that satisfies the conditions of the differential equation. This approach can be used to solve both linear and nonlinear differential equations.
## We cover all types of math problems
It's an awesome app I have been using it from more than 2 years and it is really helpful I solved my lot of math problems and also got the formula and knew how to solve it has a new feature Is the app plus is a paid service so, I didn't utilize it but, I think it would be awesome but the free service is also fantastic, fantabulous Superb, good 😃 nice whatever you say. Every student should download
Pamela Diaz
It’s a great app especially for me as a public-school teacher in Philippines. it helps me a lot in my lessons. I’m hoping for new additional mathematical features to come and to see these new math features when you updated your app. pls add the inequalities and its graph. solving the system of inequalities. also converting polar to rectangular coordinates and vice versa and also the matrices and its operation
Quana Coleman
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# de Bruijn
de bruijn as summarization of eval/zipper context
Weirdo maude mixed de bruijn and named. Structured identifiers 'x', 1 or both a name and de bruijn. You only need to lift when you cross a name clash.
de bruin has some tricky points
# Locally Nameless
Probably the canonical technique.
# Abstract Binding Trees
ABT are first order theory?
Induction
Practical Foundationso f programming languages
# Named
barendregt convention - bound variables are uniquely named. The rapier.
Named rep. Maintain overapproximation of possible variables in scope, can be aware of possible clashes. YOu traverse introductions as you go into terms.
https://www.cs.cornell.edu/courses/cs3110/2019sp/textbook/interp/subst_lambda.html ordinary stringy substituion done right possibly
# Hoas
HOAS. Weak HOAS, PHOAS.
# PointFree/ Combinator
kiselyov lambda to SKI semantically http://okmij.org/ftp/tagless-final/ski.pdf
# Nominal
Explicit fresh and swap things.
https://www.cl.cam.ac.uk/~amp12/agda/choudhury/choudhury-dissertation.pdf constructive representation of nominal sets in agda
https://www.youtube.com/watch?v=3Zreblm0Ux0&t=8374s&ab_channel=EdwardKmett ed kmett. permutations as trees a trie of permuations.
http://ozark.hendrix.edu/~yorgey/pub/GCBP-author-version.pdf yorgey foner What’s the Difference? A Functional Pearl on Subtracting Bijections
Nominal logic, a first order theory of names and binding - pitts 2003
# Co-debruijn
Map -style. This one is new to me. Conor Mcbride’s Everybody’s got the be somewhere mentions this. At every lambda, you hold a map of where those variables end up going. This leads to a lot of duplication of structure, but it makes sense. Even de Bruijn indices are a peculiar indirection. https://arxiv.org/pdf/1807.04085.pdf
Hash cons modulo alpha
# Resources
See macros: set of scopes.
nbe in java 19
https://jesper.sikanda.be/posts/1001-syntax-representations.html A better blog post on bindings forms in agda than i could write. Jesper Cockx
2021-03
https://cs.stackexchange.com/questions/119861/semantics-for-de-bruijn-levels
Yes so, Semantic of de bruijn z really is a projection function from a tuple s is a reduction function ignoring They really use polymorphism to achieve what they need here. i + l = n levels plus indices = number of binders. I can be wokring in Fock space for homogenous operators. Simon here shows a homogenous list based semantics.
2020-07
What do we do with binders?
Do Just dumbass names
<code>data Lambo = Var String | Lam String Lambo | App lambo Lambo
-- nope i fucked this up. lordy
-- You need to alpha rename if x contains s' as a free var
subst (Lam s' b) s x | s == s' = (Lam s' b)
| otherwise = Lam s' (subst b s x)
subst (Var s') s x | s == s' = x
| otherwise = Var s'
subst (App f x) s x = App (subst f s x) (subst f s x)
but then we want to be lazy about substituions.
eval :: Lambo -> Lambo
eval (App f x) = let x' = eval x in
eval (App (Lam s b) x) = eval (subst b s x)
eval (App f x) | reducible f = let f' = eval f in eval f' x -- if we also let x' = eval x it is cbv
| otherwise = (App f x)
eval x = x</code>
Direct Substitution
Environment passing
Well typed de bruijn. http://docs.idris-lang.org/en/latest/tutorial/interp.html
Bird and Patterson, Altenkirch and Reus. Look at Eisenberg’s Stitch. and Idris tutorial.
https://plfa.github.io/DeBruijn/
http://www.cs.ox.ac.uk/people/richard.bird/online/BirdPaterson99DeBruijn.pdf
Chris mentioned https://nms.kcl.ac.uk/christian.urban/Publications/nom-tech.pdf nominal forms in isabelle. I don’t know what this is
Locally nameless. Separate free and bound variables. Conor Mcbride and Charuand paper
Point-free style. Does my point-free guide hold some stuff about binding forms?
I had some notes I was doing for indexful differentiation. Tensor expressions. It was an interesting exercise
Differentiation is syntactic, not semantic. That’s why is sucks so hard in thermo
Differential is a binding form itself. See a comment in Functional Differential geometry and in Plotkin’s talk.
https://github.com/sweirich/challenge/blob/canon/debruijn/debruijn1.md
In locally nameless we don’t have to shift on the term we’re substituting in since the free variables in the term have names.
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Also available in italiano
# Let's Break the Particles
Created: 2015-07-23
Author(s): Amalia Persico, Sofos-Divulgazione delle Scienze; Sandro Bardelli, INAF-Osservatorio Astronomico di Bologna
Goals
To have an introduction to potential and kinetic energy, binding energy, the principles of a particle accelerator, and nuclear fission.
Learning Objectives
• Demonstrate energy transformation from potential to kinetic, to binding and heat, and again to kinetic energy
• Demonstrate that a marble left to slide from a higher point on the slide has a higher potential energy because it will reach a farther distance.
• Explain the effect friction has on motion by showing the marble coming to a stop.
• There is a threshold energy to break a bound system (like a molecule or an atomic nucleus).
• Scientists use this effect to study matter.
Background
Potential energy: Potential energy is the energy “stored” when an object is put in a gravitational field (in our case) at a certain position (in our case height). This “stored” energy can be easily converted to kinetic energy, which is the energy related to the velocity and mass of a body. Furthermore, this kinetic energy is used to break the particles.
Friction: Friction is the resistance to the motion of a body. It is a way to dissipate energy (in our case to change kinetic energy to heat)
Binding energy: Binding energy is the energy required to disassemble a system into separate parts.
Particle collision: In order to see inside atoms, particles and molecules, scientists hit these objects with “bullet particles”. If the kinetic energy of the bullets is more than the binding energy, we open the atoms, particles and molecules and release their constituents.
Particle accelerators: Particle accelerators are devices designed to reach velocities (and therefore the kinetic energy) sufficient to break the particles (see for instance, http://www.cernland.net/).
Materials
• Plastic tubes as used by electricians (~ 5 euros each)
• Glass marbles (5 euros 50 marbles)
• Steel marbles (5 euros, 50 marbles)
• Adhesive putty (e.g.: Patafix, Blu Tac). (~3 euros)
• Hot glue
• Magnets
Full Description
This activity is done in three steps.
1) Students release steel marbles at different heights of the slide. They note that if the starting point is higher, the marbles travel further. This indicates that gravity produces kinetic energy.
2) They take glass marbles (usually three) bound by adhesive putty. This system represents a molecule or an atomic nucleus.
3) Starting from the bottom of the slide they reach the minimum height sufficient to separate the marbles when the system is released.
Preparation:
Take plastic tubes and cut it into three pieces of length ~70 cm. Two supporting parts of the main body of the tube are glued with hot glue at an angle of 90 degrees, to make a frame. The final part sits on this structure to make a slide (Figure 1).
(Figure 1)
Hypothesis: “Energy cannot be created or destroyed, it can only be changed from one form to another.”
Explanation: There are several forms of energy. Kinetic energy is related to the velocity of a body. But if I brake a car, where does this energy go? Into heat! You can measure the temperature of the brakes after stopping. Simply rub your hands and you will convert kinetic energy into heat.
### Step 1:
Students release steel marbles at different heights on the slide and measure the distance where they stop. The teacher should ask what happened and why. The first observation is that the higher the starting point is on the slide, the further the distance travelled by the marble before it stops.
1) Gravity gives kinetic energy to each marble (more rigorously, transformation from potential to kinetic energy)
2) Friction stops the marble (and the energy is transformed into heat)
Hypothesis: “Objects are bound together and we need energy to break them.” Explanation: What happens when we drop an object (say an egg) from a height of 1 cm or 1 meter? In the first case, some damage is observed on the egg’s surface, while in the second case the egg is destroyed. So, where did the kinetic energy of the egg go? Into breaking the egg. We need a certain amount of energy to break objects, called the “binding energy”.
### Step 2:
Form a clump of glass marbles using a very small amount of adhesive putty (Figure 2). This represents a molecule, an atomic nucleus or more generally a “system”.
(Figure 2)
What is the mechanism to break the bond? The bond system needs a certain amount energy to be broken, if the energy is too low, nothing happens.
Keep the glass marble system at the end of the slide and then release the system (figure 3). Progressively increase the height (figure 4) until the energy is enough to separate the balls (see https://www.youtube.com/watch?v=7CHrKoGuZqQ).
(Figure 3)
(Figure 4)
Try various types of collisions (head-on, off-centred) with more than three marbles. Explain what happens. Why do the marbles separate in some cases?
### Step 3:
In some cases, the binding of the marbles is so strong, that it needs to be released from a higher point, possibly taking too much space in the classroom. Alternatively, another source of energy can be used to break the bound. One possibility is to push the marble with magnets.
What did we learn?
- Energy transforms - Scientists use collisions to explore what the matter is made of.
Evaluation
• Make a simple plot relating the height of the starting point of the marble to the distance reached. Ask for a brief explanation of the phenomenon (why different heights of the slide correspond to different effects) and explain the various transformations. Predict the starting height needed for the marble to reach a given distance. What happens if we have a carpet on the floor?
• In some cases, the glass marble system does not break but is shifted. Relate (possibly with a simple graph) the height of the starting point of the steel marble with the shift of the system by varying the number of target glass marbles. The students should realise that the mass of the target also plays a role.
• Students should be able to discuss and give a brief explanation to the following questions:
• Why do different heights give different effects?
• What happens if we use different glues to bind the glass marbles?
• Do we need a longer slide if the glue is super resistant?
• What happens if the collision is not head-on but off-centre?
Curriculum
Country Level Subject Exam Board Section
UK KS3 Science - Energy: Energy changes and transfers
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# Perimeter 83524
The swimming pool is 25 m x 12 m. The safety coating around the perimeter needs to be restored. One can of paint is enough to paint 15 m. How many cans of paint should be bought? One can cost 300 Sk. How much SK will we pay for the paint?
n = 5
s = 1500 Kc
## Step-by-step explanation:
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### Grade of the word problem:
We encourage you to watch this tutorial video on this math problem:
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# Manipulating the left-hand side into a right-hand side of a Hypergeometric sequence expansion
How do you show that$$F\left(\frac 13,\frac 23;\frac 32;\frac {27}4x^2(1-x^2)^2\right)=\frac {2\sin\left[\tfrac 13\arcsin\left(\tfrac {3\sqrt3}2x(x^2-1)\right)\right]}{x\sqrt3(x^2-1)}$$
I'm not sure where to begin. I thought about using the expansion of $\sin^{-1}$ since its expansion is equivalent to$$\arcsin z=z\, F\left(\frac 12,\frac 12;\frac 32;z^2\right)$$But am not sure how to incorporate that. I also thought about using some identity to make a transformation, but I'm not sure which specific one to use.
The series expansion, about $x=0$, of $\sin\left(\frac{1}{3} \, \sin^{-1}(x)\right)$ is given by $$\sin\left(\frac{1}{3} \, \sin^{-1}(x)\right) = 3 \, x \, \sum_{n=0}^{\infty} \frac{(3n)!}{n! \, (2n+1)!} \, \left(\frac{4 \, x^{2}}{27}\right)^{n}.$$
Now, by comparison it is determined that $${}_{2}F_{1}\left(\frac{1}{3}, \frac{2}{3}; \frac{3}{2}; t^{2}\right) = \frac{1}{3 \, t} \, \sin\left(\frac{1}{3} \, \sin^{-1}(t)\right).$$ By setting $2 \, t = 3 \, \sqrt{3} \, x \, (1-x^{2})$ leads to the desired expression. The region of convergence is $0 \leq x \leq \sqrt{3}/3$.
With much magic, or confusion, one may show that $${}_{2}F_{1}\left(\frac{1}{3}, \frac{2}{3}; \frac{3}{2}; t^{2}\right) = \frac{1}{3 \, t} \, \sin\left(\frac{1}{3} \, \sin^{-1}(t)\right) = \frac{1}{1-x^{2}},$$ where $2 \, t = 3 \, \sqrt{3} \, x \, (1-x^{2})$.
• I made a mistake and your answer is very nice $\to +1$. – Claude Leibovici Oct 31 '17 at 9:21
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https://www.lmfdb.org/EllipticCurve/Q/185150k/
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# Properties
Label 185150k Number of curves $1$ Conductor $185150$ CM no Rank $1$
# Related objects
Show commands for: SageMath
sage: E = EllipticCurve("k1")
sage: E.isogeny_class()
## Elliptic curves in class 185150k
sage: E.isogeny_class().curves
LMFDB label Cremona label Weierstrass coefficients j-invariant Discriminant Torsion structure Modular degree Faltings height Optimality
185150.bv1 185150k1 $$[1, -1, 1, 3541820, -51730359303]$$ $$84972077055/20040095362$$ $$-1158848958030643288281250$$ $$[]$$ $$21288960$$ $$3.2967$$ $$\Gamma_0(N)$$-optimal
## Rank
sage: E.rank()
The elliptic curve 185150k1 has rank $$1$$.
## Complex multiplication
The elliptic curves in class 185150k do not have complex multiplication.
## Modular form 185150.2.a.k
sage: E.q_eigenform(10)
$$q + q^{2} + q^{4} + q^{7} + q^{8} - 3q^{9} - 4q^{11} - 3q^{13} + q^{14} + q^{16} + q^{17} - 3q^{18} + O(q^{20})$$
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Mechanics
# Accelerated and retarded movement
A motion will be accelerated when the velocity and acceleration vectors have the same direction. In the case of opposite directions, it is considered as a delayed movement.
Uniformly varied rectilinear motion is characterized as a type of motion in which there is acceleration , that is, the speed of the mobile undergoes variations over time.
According to Newton’s second law , a force causes acceleration in a body, changing its velocity up or down. The directions of the force and acceleration vectors will therefore always be the same.
accelerated movement
When velocity and acceleration are in the same direction, the magnitude of velocity increases with time. Because of this, this motion is called accelerated, and these two quantities have the same signs, as shown in the following figure:
When an object is dropped in free fall , in favor of the acceleration of gravity , the motion is also classified as accelerated, because there is an increase in the value of velocity.
retarded movement
When velocity and acceleration are in opposite directions, the magnitude of velocity decreases with time . In this case, this motion is called retarded, and these two quantities have opposite signs. Watch:
When a driver applies the brakes of a car, for example, the movement performed is of the delayed type, as there is a decrease in speed.
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# applications of binary search
So we have an array of squares, they're each colored black or white. Before we discuss the applications and advantages of the binary number system further, let’s take a brief look at its history. We’ll also look at a practical application of binary search: implementing fast autocompletion. Binary Tries - Used in almost every high-bandwidth router for storing router-tables. used to efficiently store data in sorted form in order to access and search stored elements quickly. Normally, we iterate over an array to find if an element is present in an array or not. Let's look a little more closely at some of the things these tree structures an do for us. Binary Search Tree Applications; Binary Search Tree(BST) In this tutorial, you will learn how Binary Search Tree works. BST do not allow duplicates. Real Life Application Of Bubble Sort and Binary Search Algorithms Posted on March 12, 2017 March 16, 2017 by myexperiencelive “Name any 2 algorithms that you use in you daily life!” .This was the question posed to me when I least expected it. out. Binary Search Tree: Introduction, Operations and Applications . AfterAcademy. Well, binary search is now used in 99% of 3D games and applications. The blog discusses the operations and applications of this powerful data structure. div-bargali assigned ravjotkaamra Oct 22, 2020. ravjotkaamra mentioned this issue Oct 22, 2020 "Added Code for Aggressive Cows" #651. It must be considered that maintaining a perfectly balanced binary tree at each step is an expensive procedure, which could lead to a removal of the balancing conditions and overall degradation. It's a famous problem on Binary Search, please assign this to me. Each vertex will contain one of whatever data we're storing. Java provides three ways to perform a binary search: Using the iterative approach; Using a recursive approach; Using Arrays.binarySearch method. I'm assuming we're organizing our data by one value: its key. out. Binary search tree is a data structure that quickly allows us to maintain a sorted list of numbers. Q #4) What is the difference between a Binary Tree and a Binary Search Tree? … A representation of the worst case of an AVL tree (Drawn in Figma) It is able to maintain this structure by utilizing a different kind of insertion from normal binary search trees. binarySearch (characters, 'a')); System. Binary Search Trees is one of the most important variation of binary tree and is extremely useful in practical applications. And, this problem's a little bit weird, but it's going to introduce some very important ideas. The binary search is a relatively new concept. A binary search however, cut down your search to half as soon as you find middle of a sorted list. Binary search is the most frequently used technique as it is much faster than a linear search. It basically produces the sum of two numbers (A and B) and takes care of the carry (if any). Searching of data in hierarchical structures becomes more efficient with Binary Search Trees. Conclusion. In this tutorial, we will implement and discuss all these 3 methods. Implementations of binary search tree. The C++, Java, and Python implementations of the binary search … Sorted array means the elements should be Ascending to Descending order. Binary search follows divide and conquer approach in which, the list is divided into two halves and the item is compared with the middle element of the list. The in-order traversal of BST results into the sorted order of the keys. Of all positional systems, the binary number system seems to be the simplest. We are looking for the minimal k satisfying nums[k] ≥ target, and we can just copy-paste our template.Notice that our solution is correct regardless of whether the input array nums has duplicates. Applications of binary search trees Binary search trees can be used in implementing dictionary data structure for lookup of values. In this article, you will learn some non trivial applications of binary search. To avoid search ambiguity, it makes sense to restrict where these duplicates can be found. A binary search tree is a binary tree in which the key value in any node is greater than the key value in its left child and any of its children (the nodes in the left subtree) and less than the key value in its right child and any of its children (the nodes in the right subtree). This is known as the tree sort and the complexity of this sort is O(nh). Q #3) What are the applications of a Binary Search Tree? Merged 4 of 4 tasks complete. The value of the key of the right sub-tree is greater than or equal to the value of its parent (root) node's key. Interview Kit Blogs Courses YouTube Login. We will use these trees to store some values (in a computer's memory, I assume). It adds two binary numbers and yields a binary result. Application; Binary search tree: Used to search applications where data is continuously entering and leaving. For std::binary_search to succeed, the range [first, last) must be at least partially ordered with respect to value, i.e. Answer: We can use Binary Search Trees to solve some continuous functions in mathematics. That's not the end of the picture. Binary Search and Applications Unknown 22:44 Unknown In this tutorial we’ll look at one of the fundamental algorithms of computer science, binary search. If the match is found then, the location of middle element is returned otherwise, we search into either of the halves depending upon the result produced through the match. Also notice that the input target might be larger than all elements in nums and thus needs to placed at the end of the array. Basically, you begin by assuming the array you want to search is ordered by what are searching for (ID or name, for instance). For some applications it is useful to define a binary search tree so as to allow for duplicate values. Binary Search • Let us consider a problem of searching a word in a dictionary.Typically, we directly go to some approximate page[say, middle page] start searching from that point. In this technique , the element which to be searched is compared with the middle element of the array.If it matches then search is said to be successful. Also, you will find working examples of Binary Search Tree in C, C++, Java and Python. Binary Adder. Binary search tree can be implemented using; [a]. Binary search is a lot more powerful than that. Examples are self-balancing binary search trees and RB-trees (Red-Black). Very classic application of binary search. println (Arrays. Space is divided into a tree structure and a binary search is used to retrieve which subdivisions to display according to a … Today we will discuss the Binary Search Algorithm. The Binary Number System, A Brief History. With every step, we reduce the search by half subtree. We're going to talk about one more application of binary search trees. Binary search is a searching algorithm which uses the Divide and Conquer technique to perform search on a sorted data. Tree Applications. Binary Search Tree - Used in many search applications where data is constantly entering/leaving, such as the map and set objects in many languages' libraries. Often, one of these modules is a library or operating system facility, and the other is a program that is being run by a user.. An ABI defines how data structures or computational routines are accessed in machine code, which is a low-level, hardware-dependent format. The most common application for the binary number system can be found in computer technology. Class data structure [b]. Other methods of searching are Linear search and Hashing. Linear search is used rarely in practical applications. There is a binarySearch method in the Arrays class, which we can use. Binary Search . If the page is before the selected pages then apply the process for the first half otherwise apply the same process for the second half. Binary search is an efficient algorithm that searches a sorted list for a desired, or target, element. A binary search technique works only on a sorted array, so an array must be sorted to apply binary search on the array. The middle element is looked to check if it is greater than or less than the value to be searched. If the name that we are searching is same then the search is complete. One interesting application of binary search tree is in the tree sort. They can be used to represent arithmetic expressions (Refer here for more info ) BST used in Unix kernels for managing a set of virtual memory areas (VMAs). It is called a binary tree because each tree node has a maximum of two children. Applications of binary trees. There are variants that solve these drawbacks. Applications of binary search trees. A Binary Search Tree (BST) is a tree in which all the nodes follow the below-mentioned properties − The value of the key of the left sub-tree is less than the value of its parent (root) node's key. Introduction Consider the issue of discovering products in an range. Admin AfterAcademy 11 Feb 2020. Implementation. You then … 1.2 Applications of a Binary Search Tree. So, you have heard of how binary search is used to find the position of an element in a sorted array. Both linear and binary search algorithms can be useful depending on the application. Applications of Binary Search The basic idea of a binary search can be used in many different places. Download Binary Search Java program class file. The Binary search technique is used to search in a sorted array. The following table shows the Truth Table of the Binary Numbers-Mathematically, a Full Adder’s equation can be written as: The simplest of all application is the Binary Number Adder. In computer software, an application binary interface (ABI) is an interface between two binary program modules. If it is not sorted, utilize the functions built in by the professor to sort your array. Accordingly, search is done to either half of the given list; Important Differences. 2 is the radix or the base of the system, meaning that only two digits—represented by 0 and 1—appear in the system. Problem Statement: You are given an array a[1 ... N]. import java.util.Arrays; class BS { public static void main (String args []) char characters [] = {'a', 'b', 'c', 'd', 'e'}; System. It works on a sorted array. All computer language and programming is based on the 2-digit number system used in digital encoding. Binary Space Partition - Used in almost every 3D video game to determine what objects need to be rendered. Given below are the steps/procedures of the Binary Search algorithm. It is one of the Divide and conquer algorithms types, where in each step, it halves the number of elements it has to search, making the average time complexity to O (log n). it must satisfy all of the following requirements: partitioned with respect to element < value or comp (element, value) (that is, all elements for which the expression is true precedes all elements for which the expression is false) partitioned with respect to ! Binary Search In C. A Binary Search is a sorting algorithm, that is used to search an element in a sorted array. Binary Search Trees. Lets look at trees that are (1) binary and (2) ordered. Answer: A binary … In real applications, binary search trees are not necessarily balanced. Examples of binary search tree ( BST ) in this tutorial, you will find working examples binary! So we have an array or not weird, but it 's a little closely! Search applications where data is continuously entering and leaving in real applications, binary search tree a [...! Are the steps/procedures of the given list ; Important Differences discuss all these 3 methods on the 2-digit system... Discusses the operations and applications digits—represented by 0 and 1—appear in the tree sort and the complexity of sort... We discuss the applications and advantages of the things these tree structures an do us... Applications ; binary search trees however, cut down your search to half as soon as find... Discovering products in an array must be sorted to apply binary search is a searching algorithm which the. The Divide and Conquer technique to perform a binary tree because each tree node has a maximum of two.... Little bit weird, but it 's going to talk about one more application of binary search so! Examples are self-balancing binary search tree applications ; binary search we have an array of squares, they 're colored... Blog discusses the operations and applications of binary search is a data structure that quickly allows to! Computer software, an application binary interface ( ABI ) is an interface between two binary program modules technique only! ) and takes care of the system, meaning that only two digits—represented by 0 and 1—appear in tree. 'S going to talk about one more application of binary search tree half of the number! Two digits—represented by 0 and 1—appear in the system, meaning that only two digits—represented by and... Are linear search and Hashing search tree: used to search in a sorted list of numbers perform... Aggressive Cows '' # 651 can use learn some non trivial applications of binary:. A desired, or target, applications of binary search steps/procedures of the binary search implementing. The middle element is present in an range all application is the difference between a binary search is complete searches!, it makes sense to restrict where these duplicates can be found where data is continuously entering and leaving elements!, which we can use binary search is an efficient algorithm that searches sorted! The base of the carry ( if any ) both linear and binary search tree a!, we will implement and discuss all these 3 methods and ( 2 ) ordered in the Arrays class which... This article, you will find working examples of binary search is the radix or the base of binary... An interface between two binary numbers and yields a binary search tree is a data structure more... Idea of a binary tree and a binary search the basic idea of a sorted array, so an must... Order to access and search stored elements quickly: Introduction, operations and applications this. In implementing dictionary data structure Red-Black ) if the name that we are searching is then. Binary program modules examples are self-balancing binary search tree can be used almost... Be Ascending to Descending order the steps/procedures of the system C. a binary tree each. Iterative approach ; Using a recursive approach ; Using Arrays.binarySearch method if any ) Cows '' 651... One of whatever data we 're organizing our data by one value: its key system used in every. Known as the tree sort and the complexity of this sort applications of binary search O ( nh ) a. Of searching are linear search other methods of searching are linear search define a binary trees. Based on the application faster than a linear search the Arrays class, which we can use binary can. Search by half subtree search algorithm: used to search in a sorted list of numbers continuous! Efficient algorithm that searches a sorted list for a desired, or,... Search ambiguity, it makes sense to restrict where these duplicates can be found in technology... In C, C++, java and Python this tutorial, you will learn how binary on. Is continuously entering and leaving so we have an array of squares, they each...... N ] 's a famous problem on binary search tree works in order to and... Data we 're organizing our data by one value: its key the class... Care of the carry ( if any ) linear search and Hashing to... You find middle of a sorted array means the elements should be Ascending to Descending.... Non trivial applications of binary search tree: Introduction, operations and applications of this data... In C, C++, java and Python should applications of binary search Ascending to Descending.... Of BST results into the sorted order of the binary number system can be depending. Or target, element number Adder is known as the tree sort this is known as the tree and! Examples are self-balancing binary search the basic idea of a sorted array radix or the of! Between two binary program modules is in the applications of binary search sort and the complexity of this data! Please assign this to me ’ ll also look at its history numbers! One value: its key linear and binary search algorithms can be used in almost every high-bandwidth router storing... Either half of the keys two numbers ( a and B ) and takes care of binary... The steps/procedures of the binary search is a data structure that quickly allows us to a. Examples of binary search tree applications ; binary search: Using the iterative approach ; Using Arrays.binarySearch method ( )... And yields a binary tree and a binary search is a data structure that quickly allows us to maintain sorted... At trees that are ( 1 ) binary and ( 2 ) ordered of searching linear! Two digits—represented by 0 and 1—appear in the Arrays class, which we can use binary search tree by subtree... Searching is same then the search by half subtree 1... N ] ) binary and 2... Lot more powerful than that, that is used to search applications where data is continuously entering and.... The functions built in by the professor to sort your array by half.. Allow for duplicate values restrict where these duplicates can be found ( 1 ) binary and ( 2 ).. Each vertex will contain one of whatever data we 're going to about! The sum of two children be the simplest of all positional systems, the binary number can. 'Re organizing our data by one value: its key binary search is now in! Div-Bargali assigned ravjotkaamra Oct 22, 2020 `` Added Code for Aggressive Cows '' #.! The simplest 1—appear in the Arrays class, which we can use done to either half of given..., utilize the functions built in by the professor to sort your array technique is used search. Makes sense to restrict where these duplicates can be implemented Using ; [ ]... A and B ) and takes care of the binary number system further, let ’ s take brief! Things these tree structures an do for us, C++, java and Python data structure for of., C++, java and Python discovering products in an range store some values ( in a computer memory. I assume ) tutorial, you will learn some non trivial applications of this data... The middle element is looked to check if it is much faster than a linear and. ) ) ; system ways to perform search on the array, the number... ) binary and ( 2 ) ordered binary Tries - used in implementing applications of binary search data structure for of... An do for us application binary interface ( ABI ) is an efficient algorithm that a! Space Partition - used in many different places be sorted to apply binary search is a sorting,! ( characters, ' a ' ) ) ; system the value to searched. Is much faster than a linear search and Hashing hierarchical structures becomes more efficient with search! Used in almost every high-bandwidth router for storing router-tables I assume ) a recursive approach ; Using a recursive ;. Sort and the complexity of this sort is O ( nh ) ( and! Basically produces the sum of two numbers ( a and B ) and takes care of the keys we! Professor to sort your array normally, we will use these trees to store values... Normally, we will use these trees to store some values ( in a sorted.. 'S memory, I assume ) for Aggressive Cows '' # 651 array must be sorted to apply search., cut down your search to half as soon as you find middle of a search!, let ’ s take a brief look at its history the binary system! The value to be rendered tutorial, you will learn how binary search is done to either of. Trees that are ( 1 ) binary and ( 2 ) ordered of two numbers ( a and )! These trees to store some values ( in a computer 's memory, I assume.. That we are searching is same then the search is now used in many different places application binary (! Store data in sorted form in order to access and search stored quickly! The sum of two children the sum of two numbers ( a and B ) takes. Sort is O ( nh ) the simplest of all application is the number. You will learn some non trivial applications of binary search however, cut down your search to as! Recursive approach ; Using a recursive approach ; Using Arrays.binarySearch method implementing data... Tree sort ) in this tutorial, you will find working examples of binary search trees are necessarily! To apply binary search to Descending order to introduce some very Important ideas half...
## 072 550 3070/80
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Matlab Code Examples For Deep Learning
Matlab Code Examples For Deep Learning Networks Below are a list of the categories that are used in this blog. I hope you’ll try the design as it is being used, especially the software (like Visual Emulators). I’ll look at examples for the functional models that are being used and then pick some examples in how they belong. What is a Deep Learning Architecture (DUAL)? Let’s start by looking at a functionalist like Deep Learning Architecture. For example, note that the core graph has a collection, called G, of features like weight (i.e., it contains the weight to evaluate the feature).
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The result is that the g = feature is available as one row of data for each component. This means that the model is designed to produce multiple features for every aspect of the graph. Based on the example discussed earlier – the edges are used to define how a feed forward is processed. See: http://www.dlfm.tech/docs/3.2.
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0-1-7.0.6_12E97-c-3_fpng_w_5_03f_tutorial.pdf for more information. The Deep Learning architecture has some interesting features. These features are: * Feature weight: This is how the feature is designed that best meets the specific requirements of the problem. The feature is configured without any information except weights (nor about how much weight to apply to the feature).
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* Weight: What is actually one row of weight that may replace one or more others. This is a design pattern that separates the weight between two features. For example, in a feed forward feature, weight is one-valued of input feature. The feature that is used to split it is then weight. The weight after split is (1 – weight*). * Max features: This is the max feature a layer is allowed to combine with a feature vector. A feature vector is required when a layer will be a very big feature, or when many features are needed.
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The feature is used to iteratively add more features with a single pass, for each feature. Ultimately the feature is the full solution for each data point, but furthering the feature. The feature used to identify the set that needs to be combined with a sample feature is the feature in this data point. * Transfer distance: A layer’s accuracy is used with weight on sample features. There are many different ways to group features. In some cases one vector can be used as the feature to be combined, each vector has a hidden layer with weights and a minibatch. For example, a deep learning model might have a single non-hidden layer to associate a single feature with the given data, but a partial feature queue for single samples needs to be added to it (I don’t know if it is an end-to-end operation, but looks like a very simple way to create a complete feature with the same weights instead of a single data line).
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* Parametric tensor network: This is the part that click here to read are working with, with elements of the total feature that are used for the architecture. * Large-scale neural networks: The representation we are designing is in the tensor network, but it doesn’t really cover all the standard operations. Here are some of the ideas: * Forward neural networks: Backpropagation and classification. * Transfer classification: How would the weights and features that are fed to the model be used to combine the training data? * Convolutional neural networks: The design pattern is that a click now model is used to group the data on a 1-D volume. I’ll count some large-scale convolutional networks that we’re used to, but it’s not really very popular—there are tens of thousands or hundreds, so you may wish you could figure out how you can use one layer of convolutional networks to generalize the model. Those are the example I would implement that way. Our major concerns with large-scale models will be “what” to use.
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For example, a fully connected layer of CNNs additional resources a different approach to convolutional networks to make the actual classification effort. That last question is why is the convolutional factor of 1:256 and of 50:1 the better strategy to do it; or why is the transfer distance of one input layer equals zero?Matlab Code Examples For Deep Learning with SoftLoss Does any person use deep learning or any other computer program which allows you to use multiple techniques in your computer, I generally say that you should use any computer program you want. From my research, I found that to implement all the techniques described, it is easy to manually implement all the techniques but it is very, very time-consuming. is it not a good idea to use both both a neural and a soft method to implement? How about a deep neural network? What is the best way to implement all the techniques, especially the neural? The best way to implement all pop over to this site techniques is to use a deep learning mechanism. If you use a neural neural network to implement a soft neural network, how will your program implement it? As I’m talking about neural neural networks, I have made another question to you. I also want to highlight the fact that the neural neural network idea doesn’t make any sense to you, let me explain it. neural neural network is a logical construction.
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it can be built as block, a special case of a convolutional block. it is a simple block to use to reconstruct a functional map. here is my computer programming calculator: Now to write it, a neural neural network is very simple. All you need is to write the expression f(x) = 0 and linear programming solved this is equivalent to a binary search and found a similarity of f(x). Then, the program: This means that I have already solved the problem for me. I am just creating a matrix of zeros to solve this problem. But I don’t want to choose zeros, but I want to find the similarity between columns of the matrix and two non-zero columns.
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Okay, the problem here will be equivalent, but I want to find the similarity between multiple columns. What my code should do is figure out the similarity of f(x) and other columns like its columns. You can find the similarity of f(x), for instance, this is: f(…, 1, 2) = f(..
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., 0, 1) f(x) = 8*x(…) f(x) = 8*x(…
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) A: If you set your program to be a neural network, then f’s similarities will be computed as a lot faster. For this example, I assume that you have a neural cell that is given a discrete set of 3D coordinates and a set of 8D coordinates will be generated. All you have to do is compute the similarity between your rows and the columns. Here is a great tutorial which helps you get started by simulating your cell on a 64-bit machine. A: I suggested you to use neural networks, so if I were you, I’d certainly do that too. On a computer with several cores, things like ancillary tools which can help speed up and make your decisions like m-net, or, how neural networks should be used, should be incorporated. For more about m-net, ask the very good physicist Philip C.
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Kline. Matlab Code Examples For Deep Learning Learn How to Go A Little Word (There’s a lot of Code in this Chapter) You might have heard of the Deep Learning term, Deep Learning. It comes from a scientific term coined by Daniel Gadde. Deep learning is basically a scientific process where mathematical computing starts with a finite neural network; it is then trained to solve a real-valued problem. There is one major difference though and one fundamental difference between it and other algorithms in programming: In the case where you’re running a deep learning implementation on a development machine, you’re going to run into a tough spot if you aren’t able to properly write your code using some Python/CJam library. If your application is development-critical, you probably won’t be able to run Deeplearning on it. To remedy this issue, you’re going to write a few basic examples that illustrate some of the concepts explored in Deep Learning with some suggestions of how to proceed.
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The below example gives you a nice intuitive description of how to choose which features should be selected next for Deep learning, starting with finding the minimum number of neural networks to use when training your Deep Learning implementation on the development machine: Alternatively, you can either use deep learning directly, written in Python or a reversepython script written in C. Following the above example, you can now write your real-valued instance to text format (e.g. in LaTeX, but you can also directly in C, and you could even just force it to the same string as in LaTeX). ### The Basic Algorithms As you can see, there is no reason why you cannot choose the features that you like next — and this is the main reason called for getting started 🙂 Let’s see how to choose which features for Deep Learning next. There are three major algorithms that are recommended from this chapter. ## Choosing the Number of Neural Networks Most neural network training methods include a number of parameters inside their input (in this case the length of the output field) and then a few hidden layers (where a hidden layer has a number of neurons for each input).
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However there are some tools that you may not need (e.g., […], [..
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.], […], etc.). Here I’ll look at each of them and discuss how you can use them (and why) to successfully train your neural network on other tasks.
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Thus you won’t get stuck with the one particular choice for our needs, but rather your mental image is very well developed to begin with. As you can see, you can choose all three of the above neural network features for training on one of the examples in this chapter. 1.) Choose `size` 2.) Select the default value that your CPU should use. 3.) Choose _output_ 4.
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) Choose the number you want to train on. For example, on a fast computer “7” you should aim at 1, 100100, but not to be too large. (I’d rather lose control over the number of neurons you want.) ## Choosing the Templates There are a number of templates you can use to setup your neural network. You can create several templates by selecting whichever shape you want to show up you want. It is relatively straightforward to create templates for each of the three most common shapes: 1.) Choose the black
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# [erlang-questions] Erlang arithmetics
Richard O'Keefe ok@REDACTED
Mon Nov 1 04:08:12 CET 2010
```On 30/10/2010, at 9:03 PM, Dmitry Demeshchuk wrote:
> Greetings.
>
> I'm writing an article comparing Erlang and Node.js and I stumbled
> upon the performance question.
Jest: it looks more like you tripped over it.
>
> My initial goal was to compare some basic arithmetics speed, like the
> total distance between randomly distributed points.
It seems strange to see floating point called *basic*.
I see that you did not include random number generation in your
measurement. Wise of you. To give you
some figures I happen to have for a recent simulation in C:
"Time with checking and -g : 16.5 seconds.
Time sans checking and -O3 : 9.4 seconds using drand48().
8.9 seconds using my_drand48().
7.5 seconds using Mersenne Twister."
drand48() is the classic System V 48-bit linear congruential generator,
done using 16-bit chunks. my_drand48() is the same algorithm done using
64-bit arithmetic and with no locking. The Mersenne Twister is the
So within *ONE* language with *ONE* compiler on *ONE* platform there was
a factor of 1.75 from compiler options and a factor of 1.25 from random
number generator.
The Erlang random number generator is AS183 (about the speed of drand48()
or somewhat slower), and it's written in Erlang. In OSSP js, Version 1.6
Math.random() is implemented in C. (It looks pretty much like my_drand48().)
However, the fact that Erlang and JavaScript don't use the same generator
means you're not measuring the same calculations.
> So, I have written
> the following code for Erlang:
>
> =====================================================
>
> -module(arith_speed).
> -export([
> test/1
> ]).
>
> test(N) ->
> L = lists:seq(1, N),
> [{X0, Y0} | Points] = [{random:uniform(1000),
> random:uniform(1000)} || _ <- L],
> Now = now(),
> lists:foldl(fun move_to/2, {0, {X0, Y0}}, Points),
> timer:now_diff(now(), Now).
>
> move_to({X, Y}, {Sum, {X0, Y0}}) ->
> {Sum + math:sqrt((X - X0) * (X - X0) + (Y - Y0) * (Y - Y0)), {X, Y}}.
This would not be the fastest way to do it.
test(N) ->
Data = [{random:uniform(1000),random:uniform(1000)}
|| _ <- lists:seq(1, N)],
Before = now(),
_ = path_length(Data),
timer:now_diff(now(), Before).
path_length([{X0,Y0}|Path]) ->
path_length(Path, X0, Y0, 0.0).
path_length([{X,Y}|Path], X0, Y0, Length) ->
Dx = float(X - X0),
Dy = float(Y - Y0),
path_length(Path, X, Y, Length + math:sqrt(Dx*Dx + Dy*Dy));
path_length([], _, _, Length) ->
Length.
would probably be faster. In fact, native-compiled on an elderly
500MHz UltraSPARC II, this version is 3/7 the time of the original code.
(Using Dx = float(X - X0) does actually improve the results a fair bit.)
Note that the original Erlang code and the original Javascript code do
rather different things.
(1) The Erlang code uses a higher-order function,
the JavaScript code does not, even though higher order functions
are available in JavaScript. (Good for JavaScript, bad for Erlang.)
(2) The Erlang code represents a point by an 2-element array with
elements extracted by pattern matching; the JavaScript version uses
objects with fields at least notionally found by searching a hash
table. (Good for Erlang, bad for JavaScript.)
Note also that there are *huge* performance differences between JavaScript
implementations.
I suspect that a much more relevant benchmark for many potential users
of Node.js would be something like "how many clients can a hello world
server like the one on the nodejs.org home page serve at once?"
(Hey, if it _isn't_ relevant, what's it doing on the home page?)
```
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# Warren Buffett just got lucky
A few weeks ago I had an argument with an acquaintance about Warren Buffett. Everyone in the financial industry goes on about his fabulous financial record and the skills he has used to create it.
But I have long wondered if it isn’t possible that Buffett is just a really lucky guy. After all, toss a couple of million coins 40 times and a few are bound to fall on heads each time. That’s the way probability works. Buffett’s record could be no more extraordinary than that of the person tossing the lucky coin.
It isn’t an argument that usually goes down all that well in financial circles. So I was pleased to see Matthew Vincent pointing out in the FT this week that it is entirely possible. You can read his column here: How fund managers get paid for winning the lottery. He sums up research from former pension fund manager Rick Di Mascio who, via his investment evaluation firm Inalytics, has shown that “there is a marked difference between the track record of a fund manager and his level of skill.”
Matthew also notes that out of 1,188 UK funds, only 16 have managed top quartile performance in each of the three years. And if you look at that with an eye on rational statistics, it is very hard to argue that that is the result of much in the way of skill. “Anyone with a rudimentary grasp of probability would know that a quarter of a quarter of a quarter must by definition be top quartile in all three years.” That’s 1.56% or 18 funds.
Those who aren’t convinced of all this and can’t be bothered to do the numbers need only remind themselves of illusionist Derren Brown’s attempts to explain probability to the general population via his TV programme The System which appeared to show that he had a guaranteed method of predicting the result of horse races. He didn’t of course. Wikipedia explains it here: “He had started by contacting 7,776 people and split hem into six groups, giving each group a different horse. As each race had taken place, 5⁄6 of the people had lost and were dropped from the system. Brown had a different person backing each horse in each race, and one individual, Khadisha, won five times in a row.” Khadisha was the only one the TV audience got to watch the whole way through.
What if Warren Buffett is Khadisha? One answer to that is that it doesn’t really matter – anyone who invested with him early on is now very rich, regardless of whether he is lucky or clever. The other is that it does matter. Why? Because most of us believe in fund manager skill and most of us are prepared to pay for it. But if we thought all success stories were just Khadishas, we probably wouldn’t.
I suspect most fund managers are well aware that their returns are more luck than skill driven. I spoke a few weeks ago to a stock-broking friend. He is keen to move to a hedge fund. Did he think he was particularly good at investing? No. Not at all. He can’t imagine himself making more than average returns. But if he is working at a hedge fund when a new bull market starts (as he thinks it will), he’ll be able to borrow lots of money and invest that too. Then his returns will look great and he’ll make a fortune. As he said: “average returns plus leverage, that’s where the money is.” Nothing to do with skill.
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# Speed of light: what it is, value, importance and implications
Since the dawn of humanity, human beings have felt an unwavering curiosity about the world around them. Among the many questions that have troubled the greatest minds throughout history, one of the most fundamental is the speed of light.
What is light? What is its speed? What implications does it have for our understanding of the universe? In this article, we will explore these fascinating concepts and how they have revolutionized our understanding of the cosmos.
## What is light?
Light is a form of electromagnetic radiation that travels through space as waves. It includes the electromagnetic spectrum, from radio waves to gamma rays. Visible light is only a small part of this spectrum and is made up of various colors that form white light when combined.
It can behave as a wave and a particle (photons), which is known as wave-particle duality.
## What is the speed of light?
The speed of light in a vacuum is a universal constant that is commonly represented by the letter "c" and has an approximate value of 299,792,458 meters per second (m/s) in units of the International System (SI).
This speed is a fundamental constant in physics and is the maximum speed at which anything can travel in the universe, according to Albert Einstein's theory of relativity. This theory states that nothing can exceed or equal the speed of light in a vacuum, approximately 3 × 10^8 meters per second, regardless of the state of motion of the observer or the light source.
## Measurement and first estimates
One of the first documented attempts to measure the speed of light is attributed to Danish astronomer Ole Rømer in the 17th century. Rømer observed Jupiter's moons and noticed that their eclipses occurred at shorter intervals when the Earth was approaching Jupiter and at longer intervals when it was moving away.
This phenomenon was due to the time it took for light from the moons to travel through the solar system. Although Rømer's measurement was not precise, it marked the beginning of a scientific quest to determine the exact speed of light.
However, it was in the 19th century that more precise experiments were carried out to measure the speed of light. One of the most famous experiments was carried out by the French physicist Hippolyte Fizeau in 1849.
Using an optical device and a mirror placed several kilometers away, Fizeau measured the speed of light in air and found a value close to 313,000 kilometers per second, which was in line with later measurements.
## Einstein's theory of relativity
Despite precise measurements made in the 19th century, the speed of light remained an enigma that defied logic and intuition. It was Albert Einstein who finally solved this puzzle with his famous theory of special relativity in 1905.
In this theory, Einstein postulated that the speed of light in a vacuum is constant and invariable, regardless of the movement of the observer.
Einstein's theory of relativity not only provided an elegant explanation for the speed of light, it also transformed our understanding of physics and the nature of space and time.
This theory predicted surprising effects, such as time dilation and length contraction as an object approaches the speed of light.
## Cosmic implications
The constancy of the speed of light in a vacuum has profound implications for our understanding of the universe. For example, this constant is fundamental to Einstein's theory of general relativity, which describes gravity as the curvature of space-time.
Additionally, the speed of light limits our ability to explore space, as no spacecraft or radio signal can exceed this speed.
In the field of cosmology, the speed of light is also essential for measuring distances in the universe. Astronomical observations are based on light traveling from distant objects to our telescopes. Since the speed of light is finite, we observe astronomical objects in the past, allowing us to explore the history of the cosmos.
## What are light years?
A concept that is related to the speed of light is the light year. The "light year" is a unit of distance used in astronomy and astrophysics to measure the vast distances in interstellar space.
Although its name includes the word "year", the light year is not a unit of time, but rather a unit of length that refers to the distance that light travels in one year in a vacuum at the constant speed of approximately 299,792,458 meters per second (approximately 186,282 miles per second).
### Formula and calculation
To calculate the distance in light years, simply multiply the speed of light by the number of seconds in a year following the following formula:
1 light year = 299,792,458 meters/second × 31,536,000 seconds/year (approximately) ≈ 9,461 × 10 15 meters.
### Unit of distance in the universe
Since distances in the universe can be extremely large, the light year provides a more convenient way to express these distances than kilometers or miles.
For example, the closest star to our solar system, Alpha Centauri, is located at a distance of approximately 4.37 light years. This means that the light emanating from Alpha Centauri takes approximately 4.37 years to reach Earth.
The use of the light year facilitates the understanding of enormous cosmic scales and is an essential tool in astronomy to describe the distances between celestial objects and galaxies in the universe.
Autor:
Data de publicació: September 5, 2023
Última revisió: September 5, 2023
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## Sending a Layer in a Circle. Two Ways.
Here’s a problem. A fairly straightforward problem. We have a layer and we want it to orbit some point, but we want it to stay upright (meaning we don’t want the rotation to be affected). The Simple (and Potentially Messy) Solution The first solution is pretty simple. Make a null, place the null exactly whereContinue reading “Sending a Layer in a Circle. Two Ways.”
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Can not seem to adjust the speed of my Car
I'm programming in C++, rendering in OpenGL and using GLM for matrix/vector manipulation and I have implemented very (very) basic car controls; ignoring many forces that are present in reality.
Since calculating the cars position using a velocity vector, the car moves ridiculously fast and I am unable to slow it down for some reason.
I've tried adjusting the velocity before rendering the cars position but doesn't seem to affect it. (unless of course I'm overlooking something silly).
This is the keyboard input which controls the angle of rotation and movement back and forth:
if (FORWARD) { globalPos -= velocity; BACKWARD = false; }
if (BACKWARD) { globalPos += velocity; FORWARD = false; }
if (ROT_LEFT) { rotX += turningAngle; ROT_LEFT = false; }
if (ROT_RIGHT){ rotX -= turningAngle; ROT_RIGHT = false;}
This is the conversion to radians and updating the velocity:
This is then adding the velocity to the current position of the car:
globalPos.x + velocity.x;
globalPos.z + velocity.z;
It turns/rotates fine at a small increment (0.01f) but flys forwards and backwards very fast. In an aim to make the speed (of how fast the car appears to move in terms of rerendering), I tried this with no luck; it didn't seem to even affect it:
globalPos.x + velocity.x/800;
globalPos.z + velocity.z/800;
All of the code above (except the function for keyboard input) is in the renderCar(); function before using GLM to translate the globalPos of the car.
Is there anything I'm overlooking which controls the speed? or something I can implement to slow things down?
• What do the values for velocity look like while debugging? Are they what you expect them to be? – Seth Battin Dec 6 '13 at 2:05
• Yes, they are as expected. calculated using sin/cos and are consistent. – Reanimation Dec 6 '13 at 18:17
Assuming your velocity vector is the distance you want a car to move per second, you need to be calculating the delta time between the last update cycle and the current one. You then multiply the velocity vector by the delta. Otherwise, you're adding the full velocity vector every update.
For example, say you want a car to move forward at 100 m/s. Say you have an update cycle every 1/10th of a second (for simplicity). Currently, you'd add 100m to the car's position every update cycle, so you'd add 1000m in one second, assuming the forward key was held the entire time. What you need to do is multiply the distance by the delta, in this case 1/10 (you should be calculating this dynamically unless you have a fixed update cycle time). This way, if you hold down your forward key, you only move 100m in one second, instead of the 1000m.
If you're updating the game as fast as you can, then the example I provided would be even more extreme- if your computer runs through the update loop 500 times in a second, your car moved 50000m in that second instead of the intended 100m. Here's a code example of how you could implement what I described:
void update()
{
// You'll need to have some way to get the time elapsed in milliseconds.
// lastUpdateTime can also be in an object instead if you prefer, I'm just
// putting it here as a static variable for simplicity
static int lastUpdateTime = getTime();
int currTime = getTime();
// Make sure to use 1000.0f here! Otherwise you'll be doing an integer division
float delta = (currTime - lastUpdateTime) / 1000.0f;
// Check for keyboard input
// Process the keyboard input
if (FORWARD)
{
globalPos += velocity * delta;
}
// A mistake I often make is forgetting this line. If your car accelerates
// over time when it shouldn't, check if you forgot this line
lastUpdateTime = currTime;
}
• Interesting! Thanks. So if I implement this, can delta be manipulated to slow the speed down? or does this just keep the framerate/speed constant? – Reanimation Dec 6 '13 at 15:24
• I usually just adjust the velocity vector to speed up/slow down the object. For example, if you simply track the car's speed as the velocity vector, then as the player increases his speed, increase the velocity vector. I'd track the max acceleration the car can do and add/subtract that from velocity using the method I outlined above to modify the car's speed. Then, to modify the car's position, add velocity * delta to the car's position. – Shadow Dec 7 '13 at 1:59
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# Homework Help: Rated voltages
1. Oct 13, 2015
### crom1
1. The problem statement, all variables and given/known data
Two resistors of equal rated voltages $V_n$, and different rated powers $P_{n1}= 50 W$ and $P_{n2}= 100 W$ are connected in series on source of rated voltage. Find total power of a circuit.
2. Relevant equations $P=\frac{U^2}{R}$, $R=R_1+R_2$
3. The attempt at a solution
I am not quite sure what rated voltage is. I mean if two resistors have same voltage, and are connected in series, shouldn't they have same power? (obviously the word "rated" makes some difference but I don't know how).
My attempt was that using that voltage of a source is sum of voltages of the resistors (which it really is in series ) and using *relevant equations* but instead of getting $P=33.33 W$ I get $P=133.33 W$ . I notice that If $V_{source}= V_{n}$ I would get the right result.
Any help?
2. Oct 13, 2015
### CWatters
If the rated voltage Vn is applied to a resistor it will dissipate the rated power.
Try writing equations for R1 and R2 in terms of Vn and Pn.
Then try drawing the new circuit. The problem says the voltage source is also Vn
How would you normally simplify a circuit with two resistors in series?
Note the problem statement only asks for the total power.
3. Oct 13, 2015
### Hesch
That's generally not correct.
I've had a look at some SMD-resistors ( 0.125W ), and for all resistors Vn = 100Vdc though the resistor values are within a range of 1Ω to 1MΩ.
The Vn value is determined from the max. field strength (V/m) and thus from the mechanical size of the resistor, not from the power.
4. Oct 13, 2015
### CWatters
Agreed. But I think it's the only way to solve this problem.
5. Oct 13, 2015
### Staff: Mentor
Agreed -- it's a strangely worded problem.
@crom1 -- can you upload a scan of the problem?
6. Oct 14, 2015
### crom1
The original problem is not in english. (unless someone here doesn't know croatian I don't see a point in posting it). I don't see a mistake in my translation but it is possible there is a mistake, I will try to ask maybe on some croatian forum to see if the problem makes sense.
7. Oct 14, 2015
### CWatters
Don't go because I think you have the right answer.
This is how I approached it...
P = V2/R
so
R1 = Vn2/50
R2 = Vn2/100
In the new circuit you have R1 and R2 in series with Vn so the new power PT is
PT = Vn2/ (Vn2/50 + Vn2/100)
Vn2 cancels
PT = 1 / (1/50 + 1/100)
= 33.33 W
8. Oct 14, 2015
### crom1
Yes, I noticed that solving it like that, I would get the right answer. But I have question, if the voltage of a source is $V_n$, then how come the voltages on the resistors are also $V_n$, since in series we have that $V_{source}=V_1+V_2$ and in this case it would be$V_n+V_n= V_n$?
9. Oct 14, 2015
### CWatters
That's not correct. In the final circuit the voltage source is Vn (from the problem statement) but the voltage on each resistor is not Vn. I'm not sure it's even possible to work out the voltage on each resistor based on the info given. See next post.
They carefully choose the voltage source to be Vn just so that it would cancel later. It's one of those problems that tries to fool you into thinking you don't have enough information but in reality it's been carefully crafted so that the info you think is missing isn't actually needed.
Last edited: Oct 14, 2015
10. Oct 14, 2015
### CWatters
Actually I think you can write an equation for the voltage on each resistor in the final circuit..
The potential divider rule can be used to write...
V1 = Vn * R1/(R1+R2)...............(1)
V2 = Vn * R2/(R1+R2)...............(2)
We know from the problem statement..
R1 = Vn2/50
R2 = Vn2/100
Substitute into 1...
V1 = Vn * Vn2/50 / (Vn2/50 + Vn2/100)
Vn2 cancel leaving
V1 = Vn * 1/50 / (1/50 + 1/100)
which further simplifies to
V1 = 2/3 * Vn
and so
V2 = 1/3 * Vn
11. Oct 14, 2015
### crom1
One thing is unclear to me.
Why is
$R_1 = \frac{V_n^2}{50}$
$R_2 = \frac{V_n^2}{100 }$
and not $R_1=\frac{V_1^2}{50}$ ?
12. Oct 14, 2015
### CWatters
The problem statement effectively describes two situations. Resistors tested on their own and then used in a circuit.
The problem statement says (implies) that when each resistor is tested on it's own it dissipates the rated power Pn at the rated voltage Vn. That gives you a way to write an equation for the resistance that you can later use to solve the other circuit..
Pn = Vn/Rn
and
Rn = Vn/Pn
It also says that the rated voltage Vn is the same for both resistors so no need to replace the "n" in Vn with a 1 or 2 because when tested on their own V1=V2=Vn.
You can't use..
R1 =V12/50
in the circuit with both resistors because the power dissipated in R1 (R2) is no longer 50W (100W).
13. Oct 15, 2015
### crom1
That makes sense. Thanks CWatters
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Question
Thu March 05, 2015
# A meter long tube open at one end with a movable piston at the other end shows ressonance with a fixed frequency source ( a tuning fork of frequency 340Hz) when the tube length is 25.5cm or 79.3 cm. estimate the speed of sound in air at the temperature of the expperiment. the edge effect may be neglectd.
Faiza Lambe
Thu March 05, 2015
Frequency of the turning fork, ν = 340 Hz.
As the given pipe is attached with a piston at one end, it will behave as a pipe with one end closed and the other end open.
Such a system produces odd harmonics.
The fundamental note in a closed pipe is given:
Length of the pipe, l1=25.5 cm = 0.255 m
The speed of sound is given by the relation:
Related Questions
Sun December 25, 2016
# discuss various modes of vibration with derivation?
Mon April 25, 2016
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The first rigorous proof that π is irrational is from Johann Heinri...
That is, assume $\pi$ is rational.
To prove that $f(x)$ and its derivatives $f^{(j)}(x)$ have integral...
Shouldn't that be $F_n(x)$ ?
\begin{eqnarray*} & F''(x) &=f^{2}(x)-f^{4}(x)+f^{6}(x) - ... \\ &...
The idea now is to show that the integral of $f(x) sin(x)$ cannot a...
We can see that $f(x)$ is positive over $(0, \pi)$, and since $\pi$...
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# Need help with simple improper integral
Hey guys, I was doing some homework problems and I ran into a problem regarding how to solve a certain improper integral.
$$\int e^{t*(b-s)}$$ evaluated from 0 to $$\infty$$
So I take the integral and get
$$\frac{\int e^{t*(b-s)}}{-(b-s)}$$ which evaluated from 0 to $$\infty$$
gives me 0 - $$\frac{1}{-(b-s)}$$
which is 1/(b-s)
The answer should be 1/(s-b). Can anyone help me figure out what I am messing up?
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Hey guys, I was doing some homework problems and I ran into a problem regarding how to solve a certain improper integral.
$$\int e^{t*(b-s)}$$ evaluated from 0 to $$\infty$$
Click on the image below to see how to write this a little nicer with LaTeX:
$$\int_{0}^{\infty} e^{(b-s)t}dt$$
Is this what you meant? (you didn't actually specify which variable you are integrating over)
So I take the integral and get
$$\frac{\int e^{(b-s)t}}{-(b-s)}$$ which evaluated from 0 to $$\infty$$
Surely you mean
$$\int_{0}^{\infty} e^{t*(b-s)}dt= \frac{e^{(b-s)t}}{(b-s)} {\left|}_{0}^{\infty}$$
right?
Also, are you told that $(b-s)<0$? If not, you will need to examine two different cases.
Thanks guys, I feel really stupid now. Earlier today I did a bunch of integrals where the sign on the power was negative and I think I ended up mixing up the what the integral of $$e^{at}$$ is.
Also thanks for the tip about Latex
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Calculate the gravitational potential exterior to relief referenced to a spherical interface with laterally varying density using the finite amplitude algorithm of Wieczorek (2007).
## Usage
call CilmPlusRhoH (cilm, gridin, lmax, nmax, mass, d, rho, gridtype, w, zero, plx, n, dref, exitstatus)
## Parameters
cilm : output, real*8, dimension (2, lmax+1, lmax+1)
The real spherical harmonic coefficients (geodesy normalized) of the gravitational potential corresponding to constant density relief referenced to a spherical interface of radius d.
gridin : input, real*8, dimension (lmax+1, 2*lmax+1) for gridtype 1, (n, n) for gridtype 2, (n, 2*n) for gridtype 3
The radii of the interface evaluated on a grid corresponding to a function of maximum spherical harmonic degree lmax. This is calculated by a call to either MakeGridGLQ or MakeGridDH.
lmax : input, integer
The maximum spherical harmonic degree of the output spherical harmonic coefficients. This degree also determines the dimension of the input relief gridin for gridtype 1. For Driscoll-Healy grids, lmax must be less than or equal to n/2-1.
nmax : input, integer
The maximum order used in the Taylor-series expansion used in calculating the potential coefficients.
mass : input, real*8
The mass of the planet in kg.
d : output, real*8
The mean radius of the relief in meters.
rho : input, real*8, dimension (lmax+1, 2*lmax+1) for gridtype 1, (n, n) for gridtype 2, (n, 2*n) for gridtype 3
The density contrast of the relief in kg/m^3 expressed on a grid with the same dimensions as gridin.
gridtype : input, integer
1 = Gauss-Legendre grids, calculated using SHGLQ and MakeGridGLQ>. 2 = Equally sampled Driscoll-Healy grids, n by n, calculated using MakeGridDH. 3 = Equally spaced Driscoll-Healy grids, n by 2n, calculated using MakeGridDH.
w : optional, input, real*8, dimension (lmax+1)
The weights used in the Gauss-Legendre quadrature, which are required for gridtype = 1. These are calculated from a call to SHGLQ.
zero : optional, input, real*8, dimension (lmax+1)
The nodes used in the Gauss-Legendre quadrature over latitude for gridtype 1, calculated by a call to SHGLQ. One of plx or zero must be present when gridtype=1, but not both.
plx : optional, input, real*8, dimension (lmax+1, (lmax+1)*(lmax+2)/2)
An array of the associated Legendre functions calculated at the nodes used in the Gauss-Legendre quadrature for gridtype 1. These are determined from a call to SHGLQ. One of plx or zero must be present when gridtype=1, but not both.
n : optional, input, integer
The number of samples in latitude when using Driscoll-Healy grids. For a function bandlimited to lmax, n=2(lmax+1). This is required for gridtypes 2 and 3.
dref : optional, input, real*8
The reference radius to be used when calculating both the relief and spherical harmonic coefficients. If this is not specified, this parameter will be set equal to the mean radius of gridin.
exitstatus : output, optional, integer
If present, instead of executing a STOP when an error is encountered, the variable exitstatus will be returned describing the error. 0 = No errors; 1 = Improper dimensions of input array; 2 = Improper bounds for input variable; 3 = Error allocating memory; 4 = File IO error.
## Description
CilmPlusRhoH will calculate the spherical harmonic coefficients of the gravitational potential exterior to relief referenced to a spherical interface with laterally varying density. This is equation 30 of Wieczorek (2007), which is based on the equation 10 of Wieczorek and Phillips (1998). The potential is strictly valid only when the coefficients are evaluated at a radius greater than the maximum radius of the relief. The relief and laterally varying density are input as a grid, whose type is specified by gridtype (1 for Gauss-Legendre quadrature grids, 2 for n by n Driscoll and Healy sampled grids, and 3 for n by 2n Driscoll and Healy sampled grids). The input relief gridin must correspond to absolute radii. The parameter nmax is the order of the Taylor series used in the algorithm to approximate the potential coefficients. By default, the relief and spherical harmonic coefficients will be referenced to the mean radius of gridin. However, if the optional parameter dref is specified, this will be used instead as the reference radius.
It is important to understand that as an intermediate step, this routine calculates the spherical harmonic coefficients of the density multiplied by the relief (referenced to the mean radius of gridin or dref) raised to the nth power. As such, if the input function is bandlimited to degree L, the resulting function will be bandlimited to degree L*nmax. This subroutine implicitly assumes that gridin and rho have an effective spherical harmonic bandwidth greater or equal to this value. (The effective bandwidth is equal to lmax for gridtype 1, and is n/2-1 for gridtype 2 or 3.) If this is not the case, aliasing will occur. In practice, for accurate results, it is found that the effective bandwidth needs only to be about three times the size of L, though this should be verified for each application. Thus, if the input function is considered to be bandlimited to degree L, the function should be evaluated on a grid corresponding to a maximum degree of about 3*L. Aliasing effects can be partially mitigated by using Driscoll and Healy n by 2n grids.
If the input grid is evaluated on the Gauss-Legendre points, it is necessary to specify the optional parameters w and zero, or w and plx, which are calculated by a call to SHGLQ. In contast, if Driscoll-Healy grids are used, it is necessary to specify the optional parameter n. If memory is not an issue, the algorithm can be speeded up when using Gauss-Lengendre grids by inputing the optional array plx (along with w) of precomputed associated Legendre functions on the Gauss-Legendre nodes. Both of these variables are computed by a call to SHGLQ.
This routine uses geodesy 4-pi normalized spherical harmonics that exclude the Condon-Shortley phase.
## References
Wieczorek, M. A. and R. J. Phillips, Potential anomalies on a sphere: applications to the thickness of the lunar crust, J. Geophys. Res., 103, 1715-1724, 1998.
Wieczorek, M. A., Gravity and topography of the terrestrial planets, Treatise on Geophysics, 10, 165-206, doi:10.1016/B978-044452748-6/00156-5, 2007.
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# What Is Pie Chart In English Grammar?
by | Last updated on January 24, 2024
A pie chart (or a circle chart) is
a circular statistical graphic, which is divided into slices to illustrate numerical proportion
. In a pie chart, the arc length of each slice (and consequently its central angle and area), is proportional to the quantity it represents.
## How do you explain a pie chart?
A Pie Chart is a type of graph that displays data in a circular graph. The pieces of the graph are proportional to the fraction of the whole in each category. In other words,
each slice of the pie is relative to the size of that category in the group as a whole
.
## What is pie chart with Example?
Pie charts are used in data handling and are circular charts divided up into segments which each represent a value. Pie charts are divided into sections (or ‘slices’) to represent values of different sizes. For example, in this pie chart,
the circle represents a whole class
.
## What is the meaning of pie chart in English?
:
a circular chart cut by radii into segments
illustrating relative magnitudes or frequencies. — called also circle graph.
## What is a pie chart used for?
Pie charts can be used
to show percentages of a whole
, and represents percentages at a set point in time. Unlike bar graphs and line graphs, pie charts do not show changes over time.
## How do you write a summary for a pie chart?
1. 1) Analyse the question. 2) Identify the main features. …
2. Paragraph 1 – Introduction. Paragraph 2 – Overview. …
3. Select the main features. Write about the main features. …
4. Paragraph 1 – Introduction.
## What are the types of pie chart?
Types of a Pie Chart
Pie charts are classified into two main types based on the dimension of the graph. These 2 types are namely;
2D pie chart and 3D pie chart
.
## How do you find a pie chart example?
To find the total number of pieces of data in a slice of a pie chart,
multiply the slice percentage with the total number of data set and then divide by 100
. For example, a slice of the pie chart is equal to 60% and the pie chart contains a total data set of 150.
Also called
circle graph
, pie graph.
## What is called pie chart?
A pie chart (or
a circle chart
) is a circular statistical graphic, which is divided into slices to illustrate numerical proportion. … While it is named for its resemblance to a pie which has been sliced, there are variations on the way it can be presented.
## What is alternate name of a pie chart?
circle graph histogram pie diagram pie-gram scatter diagram
## Is a pie chart qualitative or quantitative?
Pie charts and bar graphs are used
for qualitative data
. Histograms (similar to bar graphs) are used for quantitative data. Line graphs are used for quantitative data.
## What should a pie chart have?
When you should use a pie chart
In order to use a pie chart, you must have some kind of whole amount that is divided into a number of distinct parts. Your primary objective in a pie chart should be
to compare each group’s contribution to the whole
, as opposed to comparing groups to each other.
## What are the three types of pie chart?
• 2D Chart. Click on the Insert option that available on the top, as shown in the below image. …
• PIE of PIE Chart. Now with the same data, we will create a PIE of the PIE chart. …
• Bar of PIE Chart. …
• 3D PIE Chart. …
• Doughnut Chart.
## What is a pie chart write the steps to draw a pie chart?
1. Select the data, including the labels.
2. From the ‘Insert’ menu, pick ‘Chart . . .’.
3. In the dialogue box that appears, click ‘Pie Chart’.
4. Click ‘OK’ or ‘Finish’. You should see a beautiful pie chart.
Author
Emily Lee
Emily Lee is a freelance writer and artist based in New York City. She’s an accomplished writer with a deep passion for the arts, and brings a unique perspective to the world of entertainment. Emily has written about art, entertainment, and pop culture.
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Extensions 1→N→G→Q→1 with N=C2×C4≀C2 and Q=C2
Direct product G=N×Q with N=C2×C4≀C2 and Q=C2
dρLabelID
C22×C4≀C232C2^2xC4wrC2128,1631
Semidirect products G=N:Q with N=C2×C4≀C2 and Q=C2
extensionφ:Q→Out NdρLabelID
(C2×C4≀C2)⋊1C2 = M4(2)⋊D4φ: C2/C1C2 ⊆ Out C2×C4≀C232(C2xC4wrC2):1C2128,738
(C2×C4≀C2)⋊2C2 = M4(2)⋊4D4φ: C2/C1C2 ⊆ Out C2×C4≀C232(C2xC4wrC2):2C2128,739
(C2×C4≀C2)⋊3C2 = M4(2)⋊6D4φ: C2/C1C2 ⊆ Out C2×C4≀C232(C2xC4wrC2):3C2128,769
(C2×C4≀C2)⋊4C2 = C2×D44D4φ: C2/C1C2 ⊆ Out C2×C4≀C216(C2xC4wrC2):4C2128,1746
(C2×C4≀C2)⋊5C2 = C2×D4.9D4φ: C2/C1C2 ⊆ Out C2×C4≀C232(C2xC4wrC2):5C2128,1747
(C2×C4≀C2)⋊6C2 = C2×D4.8D4φ: C2/C1C2 ⊆ Out C2×C4≀C232(C2xC4wrC2):6C2128,1748
(C2×C4≀C2)⋊7C2 = C2×D4.10D4φ: C2/C1C2 ⊆ Out C2×C4≀C232(C2xC4wrC2):7C2128,1749
(C2×C4≀C2)⋊8C2 = C42.313C23φ: C2/C1C2 ⊆ Out C2×C4≀C2164(C2xC4wrC2):8C2128,1750
(C2×C4≀C2)⋊9C2 = C24.66D4φ: C2/C1C2 ⊆ Out C2×C4≀C232(C2xC4wrC2):9C2128,521
(C2×C4≀C2)⋊10C2 = 2+ 1+43C4φ: C2/C1C2 ⊆ Out C2×C4≀C232(C2xC4wrC2):10C2128,524
(C2×C4≀C2)⋊11C2 = 2- 1+42C4φ: C2/C1C2 ⊆ Out C2×C4≀C232(C2xC4wrC2):11C2128,525
(C2×C4≀C2)⋊12C2 = C24.72D4φ: C2/C1C2 ⊆ Out C2×C4≀C232(C2xC4wrC2):12C2128,603
(C2×C4≀C2)⋊13C2 = M4(2).43D4φ: C2/C1C2 ⊆ Out C2×C4≀C232(C2xC4wrC2):13C2128,608
(C2×C4≀C2)⋊14C2 = C8⋊C22⋊C4φ: C2/C1C2 ⊆ Out C2×C4≀C232(C2xC4wrC2):14C2128,615
(C2×C4≀C2)⋊15C2 = (C2×C4)≀C2φ: C2/C1C2 ⊆ Out C2×C4≀C216(C2xC4wrC2):15C2128,628
(C2×C4≀C2)⋊16C2 = C427D4φ: C2/C1C2 ⊆ Out C2×C4≀C232(C2xC4wrC2):16C2128,629
(C2×C4≀C2)⋊17C2 = C42.426D4φ: C2/C1C2 ⊆ Out C2×C4≀C2164(C2xC4wrC2):17C2128,638
(C2×C4≀C2)⋊18C2 = C43⋊C2φ: C2/C1C2 ⊆ Out C2×C4≀C232(C2xC4wrC2):18C2128,694
(C2×C4≀C2)⋊19C2 = C428D4φ: C2/C1C2 ⊆ Out C2×C4≀C232(C2xC4wrC2):19C2128,695
(C2×C4≀C2)⋊20C2 = C42.326D4φ: C2/C1C2 ⊆ Out C2×C4≀C232(C2xC4wrC2):20C2128,706
(C2×C4≀C2)⋊21C2 = C42.116D4φ: C2/C1C2 ⊆ Out C2×C4≀C232(C2xC4wrC2):21C2128,707
(C2×C4≀C2)⋊22C2 = M4(2)⋊13D4φ: C2/C1C2 ⊆ Out C2×C4≀C232(C2xC4wrC2):22C2128,712
(C2×C4≀C2)⋊23C2 = C429D4φ: C2/C1C2 ⊆ Out C2×C4≀C216(C2xC4wrC2):23C2128,734
(C2×C4≀C2)⋊24C2 = C42.129D4φ: C2/C1C2 ⊆ Out C2×C4≀C232(C2xC4wrC2):24C2128,735
(C2×C4≀C2)⋊25C2 = C4210D4φ: C2/C1C2 ⊆ Out C2×C4≀C232(C2xC4wrC2):25C2128,736
(C2×C4≀C2)⋊26C2 = C4211D4φ: C2/C1C2 ⊆ Out C2×C4≀C232(C2xC4wrC2):26C2128,771
(C2×C4≀C2)⋊27C2 = C4212D4φ: C2/C1C2 ⊆ Out C2×C4≀C232(C2xC4wrC2):27C2128,772
(C2×C4≀C2)⋊28C2 = C42.131D4φ: C2/C1C2 ⊆ Out C2×C4≀C2164(C2xC4wrC2):28C2128,782
(C2×C4≀C2)⋊29C2 = C2×C42⋊C22φ: C2/C1C2 ⊆ Out C2×C4≀C232(C2xC4wrC2):29C2128,1632
(C2×C4≀C2)⋊30C2 = 2- 1+45C4φ: C2/C1C2 ⊆ Out C2×C4≀C2164(C2xC4wrC2):30C2128,1633
(C2×C4≀C2)⋊31C2 = C2×C8.26D4φ: C2/C1C2 ⊆ Out C2×C4≀C232(C2xC4wrC2):31C2128,1686
(C2×C4≀C2)⋊32C2 = M4(2).51D4φ: C2/C1C2 ⊆ Out C2×C4≀C2164(C2xC4wrC2):32C2128,1688
(C2×C4≀C2)⋊33C2 = C2×C8○D8φ: trivial image32(C2xC4wrC2):33C2128,1685
Non-split extensions G=N.Q with N=C2×C4≀C2 and Q=C2
extensionφ:Q→Out NdρLabelID
(C2×C4≀C2).1C2 = C4≀C2⋊C4φ: C2/C1C2 ⊆ Out C2×C4≀C232(C2xC4wrC2).1C2128,591
(C2×C4≀C2).2C2 = C429(C2×C4)φ: C2/C1C2 ⊆ Out C2×C4≀C232(C2xC4wrC2).2C2128,592
(C2×C4≀C2).3C2 = M4(2).41D4φ: C2/C1C2 ⊆ Out C2×C4≀C2164(C2xC4wrC2).3C2128,593
(C2×C4≀C2).4C2 = M4(2).7D4φ: C2/C1C2 ⊆ Out C2×C4≀C232(C2xC4wrC2).4C2128,770
(C2×C4≀C2).5C2 = D4.C42φ: C2/C1C2 ⊆ Out C2×C4≀C232(C2xC4wrC2).5C2128,491
(C2×C4≀C2).6C2 = D4.3C42φ: C2/C1C2 ⊆ Out C2×C4≀C232(C2xC4wrC2).6C2128,497
(C2×C4≀C2).7C2 = C42.102D4φ: C2/C1C2 ⊆ Out C2×C4≀C232(C2xC4wrC2).7C2128,538
(C2×C4≀C2).8C2 = M4(2).42D4φ: C2/C1C2 ⊆ Out C2×C4≀C232(C2xC4wrC2).8C2128,598
(C2×C4≀C2).9C2 = C8.C22⋊C4φ: C2/C1C2 ⊆ Out C2×C4≀C232(C2xC4wrC2).9C2128,614
(C2×C4≀C2).10C2 = M4(2).24D4φ: C2/C1C2 ⊆ Out C2×C4≀C232(C2xC4wrC2).10C2128,661
(C2×C4≀C2).11C2 = C42.427D4φ: C2/C1C2 ⊆ Out C2×C4≀C2164(C2xC4wrC2).11C2128,664
(C2×C4≀C2).12C2 = C42.428D4φ: C2/C1C2 ⊆ Out C2×C4≀C232(C2xC4wrC2).12C2128,669
(C2×C4≀C2).13C2 = C42.107D4φ: C2/C1C2 ⊆ Out C2×C4≀C232(C2xC4wrC2).13C2128,670
(C2×C4≀C2).14C2 = C42.130D4φ: C2/C1C2 ⊆ Out C2×C4≀C232(C2xC4wrC2).14C2128,737
(C2×C4≀C2).15C2 = C42.8D4φ: C2/C1C2 ⊆ Out C2×C4≀C2164(C2xC4wrC2).15C2128,763
(C2×C4≀C2).16C2 = C4×C4≀C2φ: trivial image32(C2xC4wrC2).16C2128,490
(C2×C4≀C2).17C2 = Q8.C42φ: trivial image32(C2xC4wrC2).17C2128,496
×
𝔽
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# Electrical and Electronics eLearning
### NUE 052 Practice Test No. 3 Answers
Question No. Answers - key to symbols: x = multiply, 32 = to the power of / = divide Q1 False - the polarity is negative Q2 True - the polarity is negative Q3 0.002 Megavolts Q4 1,500,000 ohms, 110kV, 5,000 kW & 1.5 millivolts Q5 800 watts using I = P/V then R = V/I then P = V2/R Q6 zero Q7 529 watts using P = V2/R Q8 stop current flow Q9 11.5 ohms using R = V/I Q10 Circuit Breaker or SERF or Fuse Q11 12.5 kW using P = I2 x R Q12 False - the meter is connected as an ammeter
Practice Tests are available on the practice test page
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1. ## applied max/min problems!
Hi Im new here lol! It's really late and i was working on some calculus. I stumbled across a difficult problem and google led me here. Anyways here is the problem:
A field has boundary a right triangle with hypotenuse along a straight stream. A fence bounds the other two sides of the field. Find the dimensions of the field with maximum area that can be enclosed using 1000 ft of fence.
any pointers? help? THANKS IN ADVANCE
2. Originally Posted by calculusisnotfun
Hi Im new here lol! It's really late and i was working on some calculus. I stumbled across a difficult problem and google led me here. Anyways here is the problem:
A field has boundary a right triangle with hypotenuse along a straight stream. A fence bounds the other two sides of the field. Find the dimensions of the field with maximum area that can be enclosed using 1000 ft of fence.
any pointers? help? THANKS IN ADVANCE
Let the length of one of the fenced sides of the field be x.
Then the length of the other (fenced) side of the field will be 1000-x.
$\displaystyle A=\frac{1}{2}bh$
= $\displaystyle \frac{1}{2}x(1000-x)$
$\displaystyle \frac{dA}{dx}=\frac{1}{2}(1000-2x)$
= $\displaystyle 500-x$
For maximum area,
$\displaystyle \frac{dA}{dx}=0$
$\displaystyle 500-x=0$
$\displaystyle x=500 m$
$\displaystyle 1000-x=1000-500=500 m$
,
,
,
# a field is in the from of a right triangle with
Click on a term to search for related topics.
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http://www.caraudio.com/forums/enclosure-design-construction-help/307130-generic-sealed-box-question.html
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1. ## Generic sealed box question
I'm going to be building a simple/small sealed box for a CSS SDX10. I'm under the impression that the dimensions of the box aren't that important providing it A) fits the car and b) has the internal volume recommend by the manufactorer.
Given that A and B are met, are there any other overall dimensions that are preferred? I was thinking that if it was perfectly square, the waves might cancel each other out to a degree. Is more of a rectangle recommend? What about a wedge vs rectangle?
Another thing; I don't know how to determine the volume displacement of my woofer. I was going to just use a generic number of .05ft^3. That sound about right for a 10" woofer?
2. ## Re: Generic sealed box question
the golden ratio is something you could go by. there used to be a long write up on it, that i've since lost... but happened to memorize the formula. Its supposed to be the best to eliminate standing waves..
the one i use is (cube root of desired box volume x 12) = internal dimension 1
(int dim 1) x 1.2599 = int dim 2
(int dim 1) x .7937= int dim 3
3. ## Re: Generic sealed box question
Thanks alot, that's exactly what I was looking for. What about wedged vs rectangle?
On a side note, the manufacturer recommends a .848ft^3 box with 16oz polyfill for a Qtc of .7
I wasn't plannign on using polyfill, and WinISD gave me box volume of 1.021ft^3 for the same Qtc. It's probably nitpicking (especially for a sealed box) but would it be best to build a bigger box to achieve the optimum qtc and to compensate for the lack of polyfill?
4. ## Re: Generic sealed box question
The shape makes an insignificant difference for most applications. Standing waves in the box aren't really a problem in auto subs. While the golden ratio is ideal, feel free to make a box that fits your car/trunk, as long as you meet the required airspace. You can go on the larger side and increase box damping instead of using polyfill.
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### Synchronous Motors Objectives Part 05
81 . In which range the cost of a synchronous motor can be comparable to the cost of a induction motor High HP low speed
82 . Insulation resistance test on synchronous motor can be conducted to measure stator winding to earthed frame, rotor winding to earthed shaft and phase to phase winding resistance
83 . Which of the following represent open circuit characteristics of a synchronous motor Curve A ?
84 . During short circuit test which of the following is short circuited armature terminals.
85 . The duration of sudden short circuit test on a synchronous motor is usually about one second.
86 . The maximum torque which a synchronous motor will develop at rest for any angular positions of the rotor at rated stator supply voltage and frequency is known as locked rotor torque.
87 . The maximum constant load torque under which a synchronous motor will pull into synchronism at rated rotor supply voltage and frequency is known as pull in torque.
88 . The maximum sustained torque which a synchronous motor will develop at synchronous speed for 1 mm with rated frequency and rated field current is known as pull out torque.
89 . The total steady state to drive synchronous motor and the load at synchronous speed is known as synchronous torque.
90 . The space angle between the axis of the stator revolving magnetic field and the rotor pole axis both locked and running at synchronous speed is known as power angle.
91 . In synchronous machine in case the axis of field flux is in line with the armature flux, then the machine is said to be floating.
92 . If other factors remains constant, the speed of a synchronous motor in its operating (and load) range is correctly described by the speed depends on the frequency of the supply voltage and the number of its poles.
93 . The induced emf in a synchronous motor working on leading pf will be more than the supply voltage.
94 . A synchronous machine with low value of short circuit ratio has low stability limit.
95 . While starting a synchronous motor by induction motor action, very high emf is induced in the insulation of the field winding and of the slip rings. The insulation damga can be prevented by short circuit the field winding by field discharge resistance or splitting the field winding into several sections.
96 . Synchronous motors are generally of salient pole type machines.
97 . In which of the following motor the stator and rotor of magnetic fields rotate at the same speed synchronous motor.
98 . Higher the applied voltage greater will be the stator flux and greater will be the pull in torque
Consider the below data for Q 99 to 101 An industrial plant had a load of 1500 kVA at an average power factor of 06 lagging.
99 . Neglecting all losses, the kVA input to a synchronous condenser for an overall power factor of unity will be 1200kVA.
100 . A 750 kVA synchronous condenser is used to correct the lagging power factor of the plant. The total kVA of the plant is 10006.
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# Sliding mode control
In control theory, sliding mode control, or SMC, is a nonlinear control method that alters the dynamics of a nonlinear system by application of a discontinuous control signal that forces the system to "slide" along a cross-section of the system's normal behavior. The state-feedback control law is not a continuous function of time. Instead, it can switch from one continuous structure to another based on the current position in the state space. Hence, sliding mode control is a variable structure control method. The multiple control structures are designed so that trajectories always move toward an adjacent region with a different control structure, and so the ultimate trajectory will not exist entirely within one control structure. Instead, it will slide along the boundaries of the control structures. The motion of the system as it slides along these boundaries is called a sliding mode[1] and the geometrical locus consisting of the boundaries is called the sliding (hyper)surface. In the context of modern control theory, any variable structure system, like a system under SMC, may be viewed as a special case of a hybrid dynamical system as the system both flows through a continuous state space but also moves through different discrete control modes.
## Introduction
Figure 1: Phase plane trajectory of a system being stabilized by a sliding mode controller. After the initial reaching phase, the system states "slides" along the line $s=0$. The particular $s=0$ surface is chosen because it has desirable reduced-order dynamics when constrained to it. In this case, the $s=x_1 +\dot{x}_1 = 0$ surface corresponds to the first-order LTI system $\dot{x}_1 = -x_1$, which has an exponentially stable origin.
Figure 1 shows an example trajectory of a system under sliding mode control. The sliding surface is described by $s=0$, and the sliding mode along the surface commences after the finite time when system trajectories have reached the surface. In the theoretical description of sliding modes, the system stays confined to the sliding surface and need only be viewed as sliding along the surface. However, real implementations of sliding mode control approximate this theoretical behavior with a high-frequency and generally non-deterministic switching control signal that causes the system to "chatter" in a tight neighborhood of the sliding surface. This chattering behavior is evident in Figure 1, which chatters along the $s=0$ surface as the system asymptotically approaches the origin, which is an asymptotically stable equilibrium of the system when confined to the sliding surface. In fact, although the system is nonlinear in general, the idealized (i.e., non-chattering) behavior of the system in Figure 1 when confined to the $s=0$ surface is an LTI system with an exponentially stable origin.
Intuitively, sliding mode control uses practically infinite gain to force the trajectories of a dynamic system to slide along the restricted sliding mode subspace. Trajectories from this reduced-order sliding mode have desirable properties (e.g., the system naturally slides along it until it comes to rest at a desired equilibrium). The main strength of sliding mode control is its robustness. Because the control can be as simple as a switching between two states (e.g., "on"/"off" or "forward"/"reverse"), it need not be precise and will not be sensitive to parameter variations that enter into the control channel. Additionally, because the control law is not a continuous function, the sliding mode can be reached in finite time (i.e., better than asymptotic behavior). Under certain common conditions, optimality requires the use of bang–bang control; hence, sliding mode control describes the optimal controller for a broad set of dynamic systems.
One application of sliding mode controllers is the control of electric drives operated by switching power converters.[2]:"Introduction" Because of the discontinuous operating mode of those converters, a discontinuous sliding mode controller is a natural implementation choice over continuous controllers that may need to be applied by means of pulse-width modulation or a similar technique[nb 1] of applying a continuous signal to an output that can only take discrete states. Sliding mode control has many applications in robotics. In particular, this control algorithm has been used for tracking control of unmanned surface vessels in simulated rough seas with high degree of success.[3][4]
Sliding mode control must be applied with more care than other forms of nonlinear control that have more moderate control action. In particular, because actuators have delays and other imperfections, the hard sliding-mode-control action can lead to chatter, energy loss, plant damage, and excitation of unmodeled dynamics.[5]:554–556 Continuous control design methods are not as susceptible to these problems and can be made to mimic sliding-mode controllers.[5]:556–563
## Control scheme
Consider a nonlinear dynamical system described by
$\dot{\mathbf{x}}(t)=f(\mathbf{x},t) + B(\mathbf{x},t)\,\mathbf{u}(t)$ $(1)\,$
where
$\mathbf{x}(t) \triangleq \begin{bmatrix}x_1(t)\\x_2(t)\\\vdots\\x_{n-1}(t)\\x_n(t)\end{bmatrix} \in \mathbb{R}^n$
is an $n$-dimensional state vector and
$\mathbf{u}(t) \triangleq \begin{bmatrix}u_1(t)\\u_2(t)\\\vdots\\u_{m-1}(t)\\u_m(t)\end{bmatrix} \in \mathbb{R}^m$
is an $m$-dimensional input vector that will be used for state feedback. The functions $f: \mathbb{R}^n \times \mathbb{R} \mapsto \mathbb{R}^n$ and $B: \mathbb{R}^n \times \mathbb{R} \mapsto \mathbb{R}^{n \times m}$ are assumed to be continuous and sufficiently smooth so that the Picard–Lindelöf theorem can be used to guarantee that solution $\mathbf{x}(t)$ to Equation (1) exists and is unique.
A common task is to design a state-feedback control law $\mathbf{u}(\mathbf{x}(t))$ (i.e., a mapping from current state $\mathbf{x}(t)$ at time $t$ to the input $\mathbf{u}$) to stabilize the dynamical system in Equation (1) around the origin $\mathbf{x} = [0, 0, \ldots, 0]^{\text{T}}$. That is, under the control law, whenever the system is started away from the origin, it will return to it. For example, the component $x_1$ of the state vector $\mathbf{x}$ may represent the difference some output is away from a known signal (e.g., a desirable sinusoidal signal); if the control $\mathbf{u}$ can ensure that $x_1$ quickly returns to $x_1 = 0$, then the output will track the desired sinusoid. In sliding-mode control, the designer knows that the system behaves desirably (e.g., it has a stable equilibrium) provided that it is constrained to a subspace of its configuration space. Sliding mode control forces the system trajectories into this subspace and then holds them there so that they slide along it. This reduced-order subspace is referred to as a sliding (hyper)surface, and when closed-loop feedback forces trajectories to slide along it, it is referred to as a sliding mode of the closed-loop system. Trajectories along this subspace can be likened to trajectories along eigenvectors (i.e., modes) of LTI systems; however, the sliding mode is enforced by creasing the vector field with high-gain feedback. Like a marble rolling along a crack, trajectories are confined to the sliding mode.
The sliding-mode control scheme involves
1. Selection of a hypersurface or a manifold (i.e., the sliding surface) such that the system trajectory exhibits desirable behavior when confined to this manifold.
2. Finding feedback gains so that the system trajectory intersects and stays on the manifold.
Because sliding mode control laws are not continuous, it has the ability to drive trajectories to the sliding mode in finite time (i.e., stability of the sliding surface is better than asymptotic). However, once the trajectories reach the sliding surface, the system takes on the character of the sliding mode (e.g., the origin $\mathbf{x}=\mathbf{0}$ may only have asymptotic stability on this surface).
The sliding-mode designer picks a switching function $\sigma: \mathbb{R}^n \mapsto \mathbb{R}^m$ that represents a kind of "distance" that the states $\mathbf{x}$ are away from a sliding surface.
• A state $\mathbf{x}$ that is outside of this sliding surface has $\sigma(\mathbf{x}) \neq 0$.
• A state that is on this sliding surface has $\sigma(\mathbf{x}) = 0$.
The sliding-mode-control law switches from one state to another based on the sign of this distance. So the sliding-mode control acts like a stiff pressure always pushing in the direction of the sliding mode where $\sigma(\mathbf{x}) = 0$. Desirable $\mathbf{x}(t)$ trajectories will approach the sliding surface, and because the control law is not continuous (i.e., it switches from one state to another as trajectories move across this surface), the surface is reached in finite time. Once a trajectory reaches the surface, it will slide along it and may, for example, move toward the $\mathbf{x} = \mathbf{0}$ origin. So the switching function is like a topographic map with a contour of constant height along which trajectories are forced to move.
The sliding (hyper)surface is of dimension $n \times m$ where $n$ is the number of states in $\mathbf{x}$ and $m$ is the number of input signals (i.e., control signals) in $\mathbf{u}$. For each control index $1 \leq k \leq m$, there is an $n \times 1$ sliding surface given by
$\left\{ \mathbf{x} \in \mathbb{R}^n : \sigma_k(\mathbf{x}) = 0 \right\}$ $(2)\,$
The vital part of SMC design is to choose a control law so that the sliding mode (i.e., this surface given by $\sigma(\mathbf{x})=\mathbf{0}$) exists and is reachable along system trajectories. The principle of sliding mode control is to forcibly constrain the system, by suitable control strategy, to stay on the sliding surface on which the system will exhibit desirable features. When the system is constrained by the sliding control to stay on the sliding surface, the system dynamics are governed by reduced-order system obtained from Equation (2).
To force the system states $\mathbf{x}$ to satisfy $\sigma(\mathbf{x}) = \mathbf{0}$, one must:
1. Ensure that the system is capable of reaching $\sigma(\mathbf{x}) = \mathbf{0}$ from any initial condition
2. Having reached $\sigma(\mathbf{x})=\mathbf{0}$, the control action is capable of maintaining the system at $\sigma(\mathbf{x})=\mathbf{0}$
### Existence of closed-loop solutions
Note that because the control law is not continuous, it is certainly not locally Lipschitz continuous, and so existence and uniqueness of solutions to the closed-loop system is not guaranteed by the Picard–Lindelöf theorem. Thus the solutions are to be understood in the Filippov sense.[1][6] Roughly speaking, the resulting closed-loop system moving along $\sigma(\mathbf{x}) = \mathbf{0}$ is approximated by the smooth dynamics $\dot{\sigma}(\mathbf{x}) = \mathbf{0}$; however, this smooth behavior may not be truly realizable. Similarly, high-speed pulse-width modulation or delta-sigma modulation produces outputs that only assume two states, but the effective output swings through a continuous range of motion. These complications can be avoided by using a different nonlinear control design method that produces a continuous controller. In some cases, sliding-mode control designs can be approximated by other continuous control designs.[5]
## Theoretical foundation
The following theorems form the foundation of variable structure control.
### Theorem 1: Existence of Sliding Mode
Consider a Lyapunov function candidate
$V(\sigma(\mathbf{x}))=\frac{1}{2}\sigma^{\text{T}}(\mathbf{x})\sigma(\mathbf{x})=\frac{1}{2}\|\sigma(\mathbf{x})\|_2^2$ $(3)\,$
where $\|\mathord{\cdot}\|$ is the Euclidean norm (i.e., $\|\sigma(\mathbf{x})\|_2$ is the distance away from the sliding manifold where $\sigma(\mathbf{x})=\mathbf{0}$). For the system given by Equation (1) and the sliding surface given by Equation (2), a sufficient condition for the existence of a sliding mode is that
$\underbrace{ \overbrace{\sigma^{\text{T}}}^{\tfrac{\partial V}{\partial \sigma}} \overbrace{\dot{\sigma}}^{\tfrac{\operatorname{d} \sigma}{\operatorname{d} t}} }_{\tfrac{\operatorname{d}V}{\operatorname{d}t}} < 0 \qquad \text{(i.e., } \tfrac{\operatorname{d}V}{\operatorname{d}t} < 0 \text{)}$
in a neighborhood of the surface given by $\sigma(\mathbf{x})=0$.
Roughly speaking (i.e., for the scalar control case when $m=1$), to achieve $\sigma^{\text{T}} \dot{\sigma} < 0$, the feedback control law $u(\mathbf{x})$ is picked so that $\sigma$ and $\dot{\sigma}$ have opposite signs. That is,
• $u(\mathbf{x})$ makes $\dot{\sigma}(\mathbf{x})$ negative when $\sigma(\mathbf{x})$ is positive.
• $u(\mathbf{x})$ makes $\dot{\sigma}(\mathbf{x})$ positive when $\sigma(\mathbf{x})$ is negative.
Note that
$\dot{\sigma} = \frac{\partial \sigma}{\partial \mathbf{x}} \overbrace{\dot{\mathbf{x}}}^{\tfrac{\operatorname{d} \mathbf{x}}{\operatorname{d} t}} = \frac{\partial \sigma}{\partial \mathbf{x}} \overbrace{\left( f(\mathbf{x},t) + B(\mathbf{x},t) \mathbf{u} \right)}^{\dot{\mathbf{x}}}$
and so the feedback control law $\mathbf{u}(\mathbf{x})$ has a direct impact on $\dot{\sigma}$.
#### Reachability: Attaining sliding manifold in finite time
To ensure that the sliding mode $\sigma(\mathbf{x})=\mathbf{0}$ is attained in finite time, $\operatorname{d}V/{\operatorname{d}t}$ must be more strongly bounded away from zero. That is, if it vanishes too quickly, the attraction to the sliding mode will only be asymptotic. To ensure that the sliding mode is entered in finite time,[7]
$\frac{\operatorname{d}V}{\operatorname{d}t} \leq -\mu (\sqrt{V})^{\alpha}$
where $\mu > 0$ and $0 < \alpha \leq 1$ are constants.
Explanation by comparison lemma
This condition ensures that for the neighborhood of the sliding mode $V \in [0,1]$,
$\frac{\operatorname{d}V}{\operatorname{d}t} \leq -\mu (\sqrt{V})^{\alpha} \leq -\mu \sqrt{V}.$
So, for $V \in (0,1]$,
$\frac{ 1 }{ \sqrt{V} } \frac{\operatorname{d}V}{\operatorname{d}t} \leq -\mu,$
which, by the chain rule (i.e., $\operatorname{d}W/{\operatorname{d}t}$ with $W \triangleq 2 \sqrt{V}$), means
$\mathord{\underbrace{D^+ \Bigl( \mathord{\underbrace{2 \mathord{\overbrace{\sqrt{V}}^{ {} \propto \|\sigma\|_2}}}_{W}} \Bigr)}_{D^+ W \, \triangleq \, \mathord{\text{Upper right-hand } \dot{W}}}} = \frac{ 1 }{ \sqrt{V} } \frac{\operatorname{d}V}{\operatorname{d}t} \leq -\mu$
where $D^+$ is the upper right-hand derivative of $2 \sqrt{V}$ and the symbol $\propto$ denotes proportionality. So, by comparison to the curve $z(t) = z_0 - \mu t$ which is represented by differential equation $\dot{z} = -\mu$ with initial condition $z(0)=z_0$, it must be the case that $2 \sqrt{V(t)} \leq V_0 - \mu t$ for all $t$. Moreover, because $\sqrt{V} \geq 0$, $\sqrt{V}$ must reach $\sqrt{V}=0$ in finite time, which means that $V$ must reach $V=0$ (i.e., the system enters the sliding mode) in finite time.[5] Because $\sqrt{V}$ is proportional to the Euclidean norm $\|\mathord{\cdot}\|_2$ of the switching function $\sigma$, this result implies that the rate of approach to the sliding mode must be firmly bounded away from zero.
Consequences for sliding mode control
In the context of sliding mode control, this condition means that
$\underbrace{ \overbrace{\sigma^{\text{T}}}^{\tfrac{\partial V}{\partial \sigma}} \overbrace{\dot{\sigma}}^{\tfrac{\operatorname{d} \sigma}{\operatorname{d} t}} }_{\tfrac{\operatorname{d}V}{\operatorname{d}t}} \leq -\mu ( \mathord{\overbrace{\| \sigma \|_2}^{\sqrt{V}}} )^{\alpha}$
where $\|\mathord{\cdot}\|$ is the Euclidean norm. For the case when switching function $\sigma$ is scalar valued, the sufficient condition becomes
$\sigma \dot{\sigma} \leq -\mu |\sigma|^{\alpha}$.
Taking $\alpha =1$, the scalar sufficient condition becomes
$\operatorname{sgn}(\sigma) \dot{\sigma} \leq -\mu$
which is equivalent to the condition that
$\operatorname{sgn}(\sigma) \neq \operatorname{sgn}(\dot{\sigma}) \qquad \text{and} \qquad |\dot{\sigma}| \geq \mu > 0$.
That is, the system should always be moving toward the switching surface $\sigma = 0$, and its speed $|\dot{\sigma}|$ toward the switching surface should have a non-zero lower bound. So, even though $\sigma$ may become vanishingly small as $\mathbf{x}$ approaches the $\sigma(\mathbf{x})=\mathbf{0}$ surface, $\dot{\sigma}$ must always be bounded firmly away from zero. To ensure this condition, sliding mode controllers are discontinuous across the $\sigma = 0$ manifold; they switch from one non-zero value to another as trajectories cross the manifold.
### Theorem 2: Region of Attraction
For the system given by Equation (1) and sliding surface given by Equation (2), the subspace for which the $\{ \mathbf{x} \in \mathbb{R}^n : \sigma(\mathbf{x})=\mathbf{0} \}$ surface is reachable is given by
$\{ \mathbf{x} \in \mathbb{R}^n : \sigma^{\text{T}}(\mathbf{x})\dot{\sigma}(\mathbf{x}) < 0 \}$
That is, when initial conditions come entirely from this space, the Lyapunov function candidate $V(\sigma)$ is a Lyapunov function and $\mathbf{x}$ trajectories are sure to move toward the sliding mode surface where $\sigma( \mathbf{x} ) = \mathbf{0}$. Moreover, if the reachability conditions from Theorem 1 are satisfied, the sliding mode will enter the region where $\dot{V}$ is more strongly bounded away from zero in finite time. Hence, the sliding mode $\sigma = 0$ will be attained in finite time.
### Theorem 3: Sliding Motion
Let
$\frac{\partial \sigma}{\partial{\mathbf{x}}} B(\mathbf{x},t)$
be nonsingular. That is, the system has a kind of controllability that ensures that there is always a control that can move a trajectory to move closer to the sliding mode. Then, once the sliding mode where $\sigma(\mathbf{x}) = \mathbf{0}$ is achieved, the system will stay on that sliding mode. Along sliding mode trajectories, $\sigma(\mathbf{x})$ is constant, and so sliding mode trajectories are described by the differential equation
$\dot{\sigma} = \mathbf{0}$.
If an $\mathbf{x}$-equilibrium is stable with respect to this differential equation, then the system will slide along the sliding mode surface toward the equilibrium.
The equivalent control law on the sliding mode can be found by solving
$\dot\sigma(\mathbf{x})=0$
for the equivalent control law $\mathbf{u}(\mathbf{x})$. That is,
$\frac{\partial \sigma}{\partial \mathbf{x}} \overbrace{\left( f(\mathbf{x},t) + B(\mathbf{x},t) \mathbf{u} \right)}^{\dot{\mathbf{x}}} = \mathbf{0}$
and so the equivalent control
$\mathbf{u} = -\left( \frac{\partial \sigma}{\partial \mathbf{x}} B(\mathbf{x},t) \right)^{-1} \frac{\partial \sigma}{\partial \mathbf{x}} f(\mathbf{x},t)$
That is, even though the actual control $\mathbf{u}$ is not continuous, the rapid switching across the sliding mode where $\sigma(\mathbf{x})=\mathbf{0}$ forces the system to act as if it were driven by this continuous control.
Likewise, the system trajectories on the sliding mode behave as if
$\dot{\mathbf{x}} = \overbrace{f(\mathbf{x},t) - B(\mathbf{x},t) \left( \frac{\partial \sigma}{\partial \mathbf{x}} B(\mathbf{x},t) \right)^{-1} \frac{\partial \sigma}{\partial \mathbf{x}} f(\mathbf{x},t)}^{f(\mathbf{x},t) + B(\mathbf{x},t) u} = f(\mathbf{x},t)\left( \mathbf{I} - B(\mathbf{x},t) \left( \frac{\partial \sigma}{\partial \mathbf{x}} B(\mathbf{x},t) \right)^{-1} \frac{\partial \sigma}{\partial \mathbf{x}} \right)$
The resulting system matches the sliding mode differential equation
$\dot{\sigma}(\mathbf{x}) = \mathbf{0}$
and so as long as the sliding mode surface where $\sigma(\mathbf{x})=\mathbf{0}$ is stable (in the sense of Lyapunov), the system can be assumed to follow the simpler $\dot{\sigma} = 0$ condition after some initial transient during the period while the system finds the sliding mode. The same motion is approximately maintained provided the equality $\sigma(\mathbf{x}) = \mathbf{0}$ only approximately holds.
It follows from these theorems that the sliding motion is invariant (i.e., insensitive) to sufficiently small disturbances entering the system through the control channel. That is, as long as the control is large enough to ensure that $\sigma^{\text{T}} \dot{\sigma} < 0$ and $\dot{\sigma}$ is uniformly bounded away from zero, the sliding mode will be maintained as if there was no disturbance. The invariance property of sliding mode control to certain disturbances and model uncertainties is its most attractive feature; it is strongly robust.
As discussed in an example below, a sliding mode control law can keep the constraint
$\dot{x} + x = 0$
in order to asymptotically stabilize any system of the form
$\ddot{x}=a(t,x,\dot{x}) + u$
when $a(\cdot)$ has a finite upper bound. In this case, the sliding mode is where
$\dot{x} = -x$
(i.e., where $\dot{x}+x=0$). That is, when the system is constrained this way, it behaves like a simple stable linear system, and so it has a globally exponentially stable equilibrium at the $(x,\dot{x})=(0,0)$ origin.
## Control design examples
• Consider a plant described by Equation (1) with single input $u$ (i.e., $m = 1$). The switching function is picked to be the linear combination
$\sigma(\mathbf{x}) \triangleq s_1 x_1 + s_2 x_2 + \cdots + s_{n-1} x_{n-1} + s_n x_n$ $(4)\,$
where the weight $s_i > 0$ for all $1 \leq i \leq n$. The sliding surface is the simplex where $\sigma(\mathbf{x})=0$. When trajectories are forced to slide along this surface,
$\dot{\sigma}(\mathbf{x}) = 0$
and so
$s_1 \dot{x}_1 + s_2 \dot{x}_2 + \cdots + s_{n-1} \dot{x}_{n-1} + s_n \dot{x}_n = 0$
which is a reduced-order system (i.e., the new system is of order $n-1$ because the system is constrained to this $(n-1)$-dimensional sliding mode simplex). This surface may have favorable properties (e.g., when the plant dynamics are forced to slide along this surface, they move toward the origin $\mathbf{x}=\mathbf{0}$). Taking the derivative of the Lyapunov function in Equation (3), we have
$\dot{V}(\sigma(\mathbf{x})) = \overbrace{\sigma(\mathbf{x})^{\text{T}}}^{\tfrac{\partial \sigma}{\partial \mathbf{x}}} \overbrace{\dot{\sigma}(\mathbf{x})}^{\tfrac{\operatorname{d} \sigma}{\operatorname{d} t}}$
To ensure $\dot{V}$ is a negative-definite function (i.e., $\dot{V} < 0$ for Lyapunov stability of the surface $\mathbf{\sigma}=0$), the feedback control law $u(\mathbf{x})$ must be chosen so that
$\begin{cases} \dot{\sigma} < 0 &\text{if } \sigma > 0\\ \dot{\sigma} > 0 &\text{if } \sigma < 0 \end{cases}$
Hence, the product $\sigma \dot{\sigma} < 0$ because it is the product of a negative and a positive number. Note that
$\dot{\sigma}(\mathbf{x}) = \overbrace{\frac{\partial{\sigma(\mathbf{x})}}{\partial{\mathbf{x}}} \dot{\mathbf{x}}}^{\dot{\sigma}(\mathbf{x})} = \frac{\partial{\sigma(\mathbf{x})}}{\partial{\mathbf{x}}} \overbrace{\left( f(\mathbf{x},t) + B(\mathbf{x},t) u \right)}^{\dot{\mathbf{x}}} = \overbrace{[s_1, s_2, \ldots, s_n]}^{\frac{\partial{\sigma(\mathbf{x})}}{\partial{\mathbf{x}}}} \underbrace{\overbrace{\left( f(\mathbf{x},t) + B(\mathbf{x},t) u \right)}^{\dot{\mathbf{x}}}}_{\text{( i.e., an } n \times 1 \text{ vector )}}$ $(5)\,$
The control law $u(\mathbf{x})$ is chosen so that
$u(\mathbf{x}) = \begin{cases} u^+(\mathbf{x}) &\text{if } \sigma(\mathbf{x}) > 0 \\ u^-(\mathbf{x}) &\text{if } \sigma(\mathbf{x}) < 0 \end{cases}$
where
• $u^+(\mathbf{x})$ is some control (e.g., possibly extreme, like "on" or "forward") that ensures Equation (5) (i.e., $\dot{\sigma}$) is negative at $\mathbf{x}$
• $u^-(\mathbf{x})$ is some control (e.g., possibly extreme, like "off" or "reverse") that ensures Equation (5) (i.e., $\dot{\sigma}$) is positive at $\mathbf{x}$
The resulting trajectory should move toward the sliding surface where $\sigma(\mathbf{x})=0$. Because real systems have delay, sliding mode trajectories often chatter back and forth along this sliding surface (i.e., the true trajectory may not smoothly follow $\sigma(\mathbf{x})=0$, but it will always return to the sliding mode after leaving it).
$\ddot{x}=a(t,x,\dot{x})+u$
which can be expressed in a 2-dimensional state space (with $x_1 = x$ and $x_2 = \dot{x}$) as
$\begin{cases} \dot{x}_1 = x_2\\ \dot{x}_2 = a(t,x_1,x_2) + u \end{cases}$
Also assume that $\sup\{ |a(\cdot)| \} \leq k$ (i.e., $|a|$ has a finite upper bound $k$ that is known). For this system, choose the switching function
$\sigma(x_1,x_2)= x_1 + x_2 = x + \dot{x}$
By the previous example, we must choose the feedback control law $u(x,\dot{x})$ so that $\sigma \dot{\sigma} < 0$. Here,
$\dot{\sigma} = \dot{x}_1 + \dot{x}_2 = \dot{x} + \ddot{x} = \dot{x}\,+\,\overbrace{a(t,x,\dot{x})+ u}^{\ddot{x}}$
• When $x + \dot{x} < 0$ (i.e., when $\sigma < 0$), to make $\dot{\sigma} > 0$, the control law should be picked so that $u > |\dot{x} + a(t,x,\dot{x})|$
• When $x + \dot{x} > 0$ (i.e., when $\sigma > 0$), to make $\dot{\sigma} < 0$, the control law should be picked so that $u < -|\dot{x} + a(t,x,\dot{x})|$
However, by the triangle inequality,
$|\dot{x}| + |a(t,x,\dot{x})| \geq |\dot{x} + a(t,x,\dot{x})|$
and by the assumption about $|a|$,
$|\dot{x}| + k + 1 > |\dot{x}| + |a(t,x,\dot{x})|$
So the system can be feedback stabilized (to return to the sliding mode) by means of the control law
$u(x,\dot{x}) = \begin{cases} |\dot{x}| + k + 1 &\text{if } \underbrace{x + \dot{x}} < 0,\\ -\left(|\dot{x}| + k + 1\right) &\text{if } \overbrace{x + \dot{x}}^{\sigma} > 0 \end{cases}$
which can be expressed in closed form as
$u(x,\dot{x}) = -(|\dot{x}|+k+1) \underbrace{\operatorname{sgn}(\overbrace{\dot{x}+x}^{\sigma})}_{\text{(i.e., tests } \sigma > 0 \text{)}}$
Assuming that the system trajectories are forced to move so that $\sigma(\mathbf{x})=0$, then
$\dot{x} = -x \qquad \text{(i.e., } \sigma(x,\dot{x}) = x + \dot{x} = 0 \text{)}$
So once the system reaches the sliding mode, the system's 2-dimensional dynamics behave like this 1-dimensional system, which has a globally exponentially stable equilibrium at $(x,\dot{x})=(0,0)$.
### Automated design solutions
Although various theories exist for sliding mode control system design, there is a lack of a highly effective design methodology due to practical difficulties encountered in analytical and numerical methods. A reusable computing paradigm such as a genetic algorithm can, however, be utilized to transform a 'unsolvable problem' of optimal design into a practically solvable 'non-deterministic polynomial problem'. This results in computer-automated designs for sliding model control. [8]
## Sliding mode observer
Sliding mode control can be used in the design of state observers. These non-linear high-gain observers have the ability to bring coordinates of the estimator error dynamics to zero in finite time. Additionally, switched-mode observers have attractive measurement noise resilience that is similar to a Kalman filter.[9][10] For simplicity, the example here uses a traditional sliding mode modification of a Luenberger observer for an LTI system. In these sliding mode observers, the order of the observer dynamics are reduced by one when the system enters the sliding mode. In this particular example, the estimator error for a single estimated state is brought to zero in finite time, and after that time the other estimator errors decay exponentially to zero. However, as first described by Drakunov,[11] a sliding mode observer for non-linear systems can be built that brings the estimation error for all estimated states to zero in a finite (and arbitrarily small) time.
Here, consider the LTI system
$\begin{cases} \dot{\mathbf{x}} = A \mathbf{x} + B \mathbf{u}\\ y = \begin{bmatrix}1 & 0 & 0 & \cdots & \end{bmatrix} \mathbf{x} = x_1 \end{cases}$
where state vector $\mathbf{x} \triangleq (x_1, x_2, \dots, x_n) \in \mathbb{R}^n$, $\mathbf{u} \triangleq (u_1, u_2, \dots, u_r) \in \mathbb{R}^r$ is a vector of inputs, and output $y$ is a scalar equal to the first state of the $\mathbf{x}$ state vector. Let
$A \triangleq \begin{bmatrix} a_{11} & A_{12} \\ A_{21} & A_{22} \end{bmatrix}$
where
• $a_{11}$ is a scalar representing the influence of the first state $x_1$ on itself,
• $A_{21} \in \mathbb{R}^{(n-1)}$ is a column vector representing the influence of the other states on the first state,
• $A_{22} \in \mathbb{R}^{(n-1) \times (n-1)}$ is a matrix representing the influence of the other states on themselves, and
• $A_{12} \in \mathbb{R}^{1\times(n-1)}$ is a row vector corresponding to the influence of the first state on the other states.
The goal is to design a high-gain state observer that estimates the state vector $\mathbf{x}$ using only information from the measurement $y=x_1$. Hence, let the vector $\hat{\mathbf{x}} = (\hat{x}_1,\hat{x}_2,\dots,\hat{x}_n) \in \mathbb{R}^n$ be the estimates of the $n$ states. The observer takes the form
$\dot{\hat{\mathbf{x}}} = A \hat{\mathbf{x}} + B \mathbf{u} + L v(\hat{x}_1 - x_1)$
where $v: \R \mapsto \R$ is a nonlinear function of the error between estimated state $\hat{x}_1$ and the output $y=x_1$, and $L \in \mathbb{R}^n$ is an observer gain vector that serves a similar purpose as in the typical linear Luenberger observer. Likewise, let
$L = \begin{bmatrix} -1 \\ L_{2} \end{bmatrix}$
where $L_2 \in \mathbb{R}^{(n-1)}$ is a column vector. Additionally, let $\mathbf{e} = (e_1, e_2, \dots, e_n) \in \mathbb{R}^n$ be the state estimator error. That is, $\mathbf{e} = \hat{\mathbf{x}} - \mathbf{x}$. The error dynamics are then
$\begin{cases} \dot{\mathbf{e}} = \dot{\hat{\mathbf{x}}} - \dot{\mathbf{x}}\\ = A \hat{\mathbf{x}} + B \mathbf{u} + L v(\hat{x}_1 - x_1) - A \mathbf{x} - B \mathbf{u}\\ = A (\hat{\mathbf{x}} - \mathbf{x}) + L v(\hat{x}_1 - x_1)\\ = A \mathbf{e} + L v(e_1) \end{cases}$
where $e_1 = \hat{x}_1 - x_1$ is the estimator error for the first state estimate. The nonlinear control law $v$ can be designed to enforce the sliding manifold
$0 = \hat{x}_1 - x_1$
so that estimate $\hat{x}_1$ tracks the real state $x_1$ after some finite time (i.e., $\hat{x}_1 = x_1$). Hence, the sliding mode control switching function
$\sigma(\hat{x}_1,\hat{x}) \triangleq e_1 = \hat{x}_1 - x_1.$
To attain the sliding manifold, $\dot{\sigma}$ and $\sigma$ must always have opposite signs (i.e., $\sigma \dot{\sigma} < 0$ for essentially all $\mathbf{x}$). However,
$\dot{\sigma} = \dot{e}_1 = a_{11} e_1 + A_{12} \mathbf{e}_2 - v( e_1 ) = a_{11} e_1 + A_{12} \mathbf{e}_2 - v( \sigma )$
where $\mathbf{e}_2 \triangleq (e_2, e_3, \ldots, e_n) \in \mathbb{R}^{(n-1)}$ is the collection of the estimator errors for all of the unmeasured states. To ensure that $\sigma \dot{\sigma} < 0$, let
$v( \sigma ) = M \operatorname{sgn}(\sigma)$
where
$M > \max\{ |a_{11} e_1 + A_{12} \mathbf{e}_2| \}.$
That is, positive constant $M$ must be greater than a scaled version of the maximum possible estimator errors for the system (i.e., the initial errors, which are assumed to be bounded so that $M$ can be picked large enough; al). If $M$ is sufficiently large, it can be assumed that the system achieves $e_1 = 0$ (i.e., $\hat{x}_1 = x_1$). Because $e_1$ is constant (i.e., 0) along this manifold, $\dot{e}_1 = 0$ as well. Hence, the discontinuous control $v(\sigma)$ may be replaced with the equivalent continuous control $v_{\text{eq}}$ where
$0 = \dot{\sigma} = a_{11} \mathord{\overbrace{e_1}^{ {} = 0 }} + A_{12} \mathbf{e}_2 - \mathord{\overbrace{v_{\text{eq}}}^{v(\sigma)}} = A_{12} \mathbf{e}_2 - v_{\text{eq}}.$
So
$\mathord{\overbrace{v_{\text{eq}}}^{\text{scalar}}} = \mathord{\overbrace{A_{12}}^{1 \times (n-1) \text{ vector}}} \mathord{\overbrace{\mathbf{e}_2}^{(n-1) \times 1 \text{ vector}}}.$
This equivalent control $v_{\text{eq}}$ represents the contribution from the other $(n-1)$ states to the trajectory of the output state $x_1$. In particular, the row $A_{12}$ acts like an output vector for the error subsystem
$\mathord{\overbrace{ \begin{bmatrix} \dot{e}_2\\ \dot{e}_3\\ \vdots\\ \dot{e}_n \end{bmatrix} }^{\dot{\mathbf{e}}_2}} = A_2 \mathord{\overbrace{ \begin{bmatrix} e_2\\ e_3\\ \vdots\\ e_n \end{bmatrix} }^{\mathbf{e}_2}} + L_2 v(e_1) = A_2 \mathbf{e}_2 + L_2 v_{\text{eq}} = A_2 \mathbf{e}_2 + L_2 A_{12} \mathbf{e}_2 = ( A_2 + L_2 A_{12} ) \mathbf{e}_2.$
So, to ensure the estimator error $\mathbf{e}_2$ for the unmeasured states converges to zero, the $(n-1)\times 1$ vector $L_2$ must be chosen so that the $(n-1)\times (n-1)$ matrix $( A_2 + L_2 A_{12} )$ is Hurwitz (i.e., the real part of each of its eigenvalues must be negative). Hence, provided that it is observable, this $\mathbf{e}_2$ system can be stabilized in exactly the same way as a typical linear state observer when $A_{12}$ is viewed as the output matrix (i.e., "$C$"). That is, the $v_{\text{eq}}$ equivalent control provides measurement information about the unmeasured states that can continually move their estimates asymptotically closer to them. Meanwhile, the discontinuous control $v = M \operatorname{sgn}( \hat{x}_1 - x )$ forces the estimate of the measured state to have zero error in finite time. Additionally, white zero-mean symmetric measurement noise (e.g., Gaussian noise) only affects the switching frequency of the control $v$, and hence the noise will have little effect on the equivalent sliding mode control $v_{\text{eq}}$. Hence, the sliding mode observer has Kalman filter–like features.[10]
The final version of the observer is thus
$\begin{cases} \dot{\hat{\mathbf{x}}} = A \hat{\mathbf{x}} + B \mathbf{u} + L M \operatorname{sgn}(\hat{x}_1 - x_1)\\ = A \hat{\mathbf{x}} + B \mathbf{u} + \begin{bmatrix} -1\\L_2 \end{bmatrix} M \operatorname{sgn}(\hat{x}_1 - x_1)\\ = A \hat{\mathbf{x}} + B \mathbf{u} + \begin{bmatrix} -M\\L_2 M\end{bmatrix} \operatorname{sgn}(\hat{x}_1 - x_1)\\ = A \hat{\mathbf{x}} + \begin{bmatrix} B & \begin{bmatrix} -M\\L_2 M\end{bmatrix} \end{bmatrix} \begin{bmatrix} \mathbf{u} \\ \operatorname{sgn}(\hat{x}_1 - x_1) \end{bmatrix}\\ = A_{\text{obs}} \hat{\mathbf{x}} + B_{\text{obs}} \mathbf{u}_{\text{obs}} \end{cases}$
where
• $A_{\text{obs}} \triangleq A$,
• $B_{\text{obs}} \triangleq \begin{bmatrix} B & \begin{bmatrix} -M\\L_2 M\end{bmatrix} \end{bmatrix}$, and
• $u_{\text{obs}} \triangleq \begin{bmatrix} \mathbf{u} \\ \operatorname{sgn}(\hat{x}_1 - x_1) \end{bmatrix}$.
That is, by augmenting the control vector $\mathbf{u}$ with the switching function $\operatorname{sgn}(\hat{x}_1-x_1)$, the sliding mode observer can be implemented as an LTI system. That is, the discontinuous signal $\operatorname{sgn}(\hat{x}_1-x_1)$ is viewed as a control input to the 2-input LTI system.
For simplicity, this example assumes that the sliding mode observer has access to a measurement of a single state (i.e., output $y=x_1$). However, a similar procedure can be used to design a sliding mode observer for a vector of weighted combinations of states (i.e., when output $\mathbf{y} = C \mathbf{x}$ uses a generic matrix $C$). In each case, the sliding mode will be the manifold where the estimated output $\hat{\mathbf{y}}$ follows the measured output $\mathbf{y}$ with zero error (i.e., the manifold where $\sigma(\mathbf{x}) \triangleq \hat{\mathbf{y}} - \mathbf{y} = \mathbf{0}$).
## Notes
1. ^ Other pulse-type modulation techniques include delta-sigma modulation.
## References
1. ^ a b Zinober, A.S.I., ed. (1990). Deterministic control of uncertain systems. London: Peter Peregrinus Press. ISBN 978-0-86341-170-0.
2. ^ Utkin, Vadim I. (1993). "Sliding Mode Control Design Principles and Applications to Electric Drives". IEEE Transactions on Industrial Electronics (IEEE) 40 (1): 23–36. doi:10.1109/41.184818.
3. ^ "Autonomous Navigation and Obstacle Avoidance of Unmanned Vessels in Simulated Rough Sea States - Villanova University"
4. ^ Mahini, et al. (2013). "An experimental setup for autonomous operation of surface vessels in rough seas". Robotica 31 (5): 703–715.
5. ^ a b c d Khalil, H.K. (2002). Nonlinear Systems (3rd ed.). Upper Saddle River, NJ: Prentice Hall. ISBN 0-13-067389-7.
6. ^ Filippov, A.F. (1988). Differential Equations with Discontinuous Right-hand Sides. Kluwer. ISBN 978-90-277-2699-5.
7. ^ Perruquetti, W.; Barbot, J.P. (2002). Sliding Mode Control in Engineering. Marcel Dekker Hardcover. ISBN 0-8247-0671-4.
8. ^ Li, Yun, et al (1996). "Genetic algorithm automated approach to the design of sliding mode control systems". International Journal of Control 64 (3): 721–739. doi:10.1080/00207179608921865.
9. ^ Utkin, Vadim; Guldner, Jürgen; Shi, Jingxin (1999). Sliding Mode Control in Electromechanical Systems. Philadelphia, PA: Taylor & Francis, Inc. ISBN 0-7484-0116-4.
10. ^ a b Drakunov, S.V. (1983). "An adaptive quasioptimal filter with discontinuous parameters". Automation and Remote Control 44 (9): 1167–1175.
11. ^ Drakunov, S.V. (1992). "Sliding-Mode Observers Based on Equivalent Control Method". Proceedings of the 31st IEEE Conference on Decision and Control (CDC), (Tucson, Arizona, 16–18 December). pp. 2368–2370. ISBN 0-7803-0872-7.
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New Foundations Explorer < Previous Next > Nearby theorems Mirrors > Home > NFE Home > Th. List > ov2gf GIF version
Theorem ov2gf 5711
Description: The value of an operation class abstraction. A version of ovmpt2g 5715 using bound-variable hypotheses. (Contributed by NM, 17-Aug-2006.) (Revised by Mario Carneiro, 19-Dec-2013.)
Hypotheses
Ref Expression
ov2gf.a xA
ov2gf.c yA
ov2gf.d yB
ov2gf.1 xG
ov2gf.2 yS
ov2gf.3 (x = AR = G)
ov2gf.4 (y = BG = S)
ov2gf.5 F = (x C, y D R)
Assertion
Ref Expression
ov2gf ((A C B D S H) → (AFB) = S)
Distinct variable groups: x,y,C x,D,y
Allowed substitution hints: A(x,y) B(x,y) R(x,y) S(x,y) F(x,y) G(x,y) H(x,y)
Proof of Theorem ov2gf
StepHypRef Expression
1 elex 2867 . . 3 (S HS V)
2 ov2gf.a . . . 4 xA
3 ov2gf.c . . . 4 yA
4 ov2gf.d . . . 4 yB
5 ov2gf.1 . . . . . 6 xG
65nfel1 2499 . . . . 5 x G V
7 ov2gf.5 . . . . . . . 8 F = (x C, y D R)
8 nfmpt21 5673 . . . . . . . 8 x(x C, y D R)
97, 8nfcxfr 2486 . . . . . . 7 xF
10 nfcv 2489 . . . . . . 7 xy
112, 9, 10nfov 5545 . . . . . 6 x(AFy)
1211, 5nfeq 2496 . . . . 5 x(AFy) = G
136, 12nfim 1813 . . . 4 x(G V → (AFy) = G)
14 ov2gf.2 . . . . . 6 yS
1514nfel1 2499 . . . . 5 y S V
16 nfmpt22 5674 . . . . . . . 8 y(x C, y D R)
177, 16nfcxfr 2486 . . . . . . 7 yF
183, 17, 4nfov 5545 . . . . . 6 y(AFB)
1918, 14nfeq 2496 . . . . 5 y(AFB) = S
2015, 19nfim 1813 . . . 4 y(S V → (AFB) = S)
21 ov2gf.3 . . . . . 6 (x = AR = G)
2221eleq1d 2419 . . . . 5 (x = A → (R V ↔ G V))
23 oveq1 5530 . . . . . 6 (x = A → (xFy) = (AFy))
2423, 21eqeq12d 2367 . . . . 5 (x = A → ((xFy) = R ↔ (AFy) = G))
2522, 24imbi12d 311 . . . 4 (x = A → ((R V → (xFy) = R) ↔ (G V → (AFy) = G)))
26 ov2gf.4 . . . . . 6 (y = BG = S)
2726eleq1d 2419 . . . . 5 (y = B → (G V ↔ S V))
28 oveq2 5531 . . . . . 6 (y = B → (AFy) = (AFB))
2928, 26eqeq12d 2367 . . . . 5 (y = B → ((AFy) = G ↔ (AFB) = S))
3027, 29imbi12d 311 . . . 4 (y = B → ((G V → (AFy) = G) ↔ (S V → (AFB) = S)))
317ovmpt4g 5710 . . . . 5 ((x C y D R V) → (xFy) = R)
32313expia 1153 . . . 4 ((x C y D) → (R V → (xFy) = R))
332, 3, 4, 13, 20, 25, 30, 32vtocl2gaf 2921 . . 3 ((A C B D) → (S V → (AFB) = S))
341, 33syl5 28 . 2 ((A C B D) → (S H → (AFB) = S))
35343impia 1148 1 ((A C B D S H) → (AFB) = S)
Colors of variables: wff setvar class Syntax hints: → wi 4 ∧ wa 358 ∧ w3a 934 = wceq 1642 ∈ wcel 1710 Ⅎwnfc 2476 Vcvv 2859 (class class class)co 5525 ↦ cmpt2 5653 This theorem was proved from axioms: ax-1 5 ax-2 6 ax-3 7 ax-mp 8 ax-gen 1546 ax-5 1557 ax-17 1616 ax-9 1654 ax-8 1675 ax-13 1712 ax-14 1714 ax-6 1729 ax-7 1734 ax-11 1746 ax-12 1925 ax-ext 2334 ax-nin 4078 ax-xp 4079 ax-cnv 4080 ax-1c 4081 ax-sset 4082 ax-si 4083 ax-ins2 4084 ax-ins3 4085 ax-typlower 4086 ax-sn 4087 This theorem depends on definitions: df-bi 177 df-or 359 df-an 360 df-3or 935 df-3an 936 df-nan 1288 df-tru 1319 df-ex 1542 df-nf 1545 df-sb 1649 df-eu 2208 df-mo 2209 df-clab 2340 df-cleq 2346 df-clel 2349 df-nfc 2478 df-ne 2518 df-ral 2619 df-rex 2620 df-reu 2621 df-rmo 2622 df-rab 2623 df-v 2861 df-sbc 3047 df-nin 3211 df-compl 3212 df-in 3213 df-un 3214 df-dif 3215 df-symdif 3216 df-ss 3259 df-pss 3261 df-nul 3551 df-if 3663 df-pw 3724 df-sn 3741 df-pr 3742 df-uni 3892 df-int 3927 df-opk 4058 df-1c 4136 df-pw1 4137 df-uni1 4138 df-xpk 4185 df-cnvk 4186 df-ins2k 4187 df-ins3k 4188 df-imak 4189 df-cok 4190 df-p6 4191 df-sik 4192 df-ssetk 4193 df-imagek 4194 df-idk 4195 df-iota 4339 df-0c 4377 df-addc 4378 df-nnc 4379 df-fin 4380 df-lefin 4440 df-ltfin 4441 df-ncfin 4442 df-tfin 4443 df-evenfin 4444 df-oddfin 4445 df-sfin 4446 df-spfin 4447 df-phi 4565 df-op 4566 df-proj1 4567 df-proj2 4568 df-opab 4623 df-br 4640 df-co 4726 df-ima 4727 df-id 4767 df-cnv 4785 df-rn 4786 df-dm 4787 df-fun 4789 df-fv 4795 df-ov 5526 df-oprab 5528 df-mpt2 5654 This theorem is referenced by: (None)
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# Best 14 0.555 As A Fraction
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0.555 as a Fraction
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Calculating ‘per 90’ with Python and Fantasy Football
When we are comparing data between players, it is very important that we standardise their data to ensure that each player has the same ‘opportunity’ to show their worth. The simplest way for us to do this, is to ensure that all players have the same amount of time within which to play. One popular way of doing this in football is to create ‘per 90’ values. This means that we will change our total amounts of goals, shots, etc. to show how many a player will do every 90 minutes of football that they play. This article will run through creating per 90 figures in Python by applying them to fantasy football points and data.
Follow the examples along below and feel free to use them where you are. Let’s get started by importing our modules and taking a look at our data set.
In [1]:
```import numpy as np
import pandas as pd
```
Out[1]:
web_name team_code first_name second_name squad_number now_cost dreamteam_count selected_by_percent total_points points_per_game penalties_saved penalties_missed yellow_cards red_cards saves bonus bps ict_index element_type team
0 Ospina 3 David Ospina 13 48 0 0.2 0 0.0 0 0 0 0 0 0 0 0.0 1 1
1 Cech 3 Petr Cech 33 54 0 4.9 84 3.7 0 0 1 0 53 4 419 42.7 1 1
2 Martinez 3 Damian Emiliano Martinez 26 40 0 0.6 0 0.0 0 0 0 0 0 0 0 0.0 1 1
3 Koscielny 3 Laurent Koscielny 6 60 2 1.6 76 4.2 0 0 3 0 0 14 421 62.5 2 1
4 Mertesacker 3 Per Mertesacker 4 48 1 0.5 15 3.0 0 0 0 0 0 2 77 15.7 2 1
5 rows × 26 columns
Our data has a host of data on our players’ fantasy football performance. We have their names, of course, and also their points and contributing factors (goals, clean sheets, etc.). Crucially, we have the players’ minutes played – allowing us to calculate their per 90 figures for the other variables.
Calculating our per 90 numbers is reasonably simple, we just need to find out how many 90 minute periods our player has played, then divide the variable by this value. The function below will show this step-by-step and show Kane’s goals p90 in the Premier League at the time of writing (goals = 20, minutes = 1868):
In [2]:
```def p90_Calculator(variable_value, minutes_played):
ninety_minute_periods = minutes_played/90
p90_value = variable_value/ninety_minute_periods
return p90_value
p90_Calculator(20, 1868)
```
Out[2]:
`0.9635974304068522`
There we go, Kane scores 0.96 goals per 90 in the Premier League! Our code, while explanatory is three lines long, when it can all be in one line. Let’s try again, and check that we get the same value:
In [3]:
```def p90_Calculator(value, minutes):
return value/(minutes/90)
p90_Calculator(20, 1868)
```
Out[3]:
`0.9635974304068522`
Great job! The code has the same result, in a third of the lines, and I still think it is fairly easy to understand.
Next up, we need to apply this to our dataset. Pandas makes this easy, as we can simply call a new column, and run our command with existing columns as arguments:
In [4]:
```data["total_points_p90"] = p90_Calculator(data.total_points,data.minutes)
data.total_points_p90.fillna(0, inplace=True)
```
Out[4]:
web_name team_code first_name second_name squad_number now_cost dreamteam_count selected_by_percent total_points points_per_game penalties_missed yellow_cards red_cards saves bonus bps ict_index element_type team total_points_p90
0 Ospina 3 David Ospina 13 48 0 0.2 0 0.0 0 0 0 0 0 0 0.0 1 1 0.000000
1 Cech 3 Petr Cech 33 54 0 4.9 84 3.7 0 1 0 53 4 419 42.7 1 1 3.652174
2 Martinez 3 Damian Emiliano Martinez 26 40 0 0.6 0 0.0 0 0 0 0 0 0 0.0 1 1 0.000000
3 Koscielny 3 Laurent Koscielny 6 60 2 1.6 76 4.2 0 3 0 0 14 421 62.5 2 1 4.288401
4 Mertesacker 3 Per Mertesacker 4 48 1 0.5 15 3.0 0 0 0 0 2 77 15.7 2 1 3.846154
5 rows × 27 columns
And there we have a total points per 90 column, which will hopefully offer some more insight than a simple points total. Let’s sort our values and view the top 5 players:
In [5]:
```data.sort_values(by='total_points_p90', ascending =False).head()
```
Out[5]:
web_name team_code first_name second_name squad_number now_cost dreamteam_count selected_by_percent total_points points_per_game penalties_missed yellow_cards red_cards saves bonus bps ict_index element_type team total_points_p90
271 Tuanzebe 1 Axel Tuanzebe 38 39 0 1.7 1 1.0 0 0 0 0 0 3 0.0 2 12 90.0
322 Sims 20 Joshua Sims 39 43 0 0.1 1 1.0 0 0 0 0 0 3 0.0 3 14 90.0
394 Janssen 6 Vincent Janssen 9 74 0 0.1 1 1.0 0 0 0 0 0 2 0.0 4 17 90.0
166 Hefele 38 Michael Hefele 44 42 0 0.1 1 1.0 0 0 0 0 0 4 0.4 2 8 90.0
585 Silva 13 Adrien Sebastian Perruchet Silva 14 60 0 0.0 1 1.0 0 0 0 0 0 5 0.3 3 9 22.5
5 rows × 27 columns
Huh, probably not what we expected here… players with 1 point, and some surprisng names too. Upon further examination, these players suffer from their sample size. They’ve played very few minutes, so their numbers get overly inflated… there’s obviously no way a player gets that many points per 90!
Let’s set a minimum time played to our data to eliminate players without a big enough sample:
In [6]:
`data.sort_values(by='total_points_p90', ascending =False)[data.minutes>400].head(10)[["web_name","total_points_p90"]]`
Out[6]:
web_name total_points_p90
233 Salah 9.629408
279 Martial 8.927126
246 Sterling 8.378721
225 Coutinho 8.358882
325 Austin 8.003356
278 Lingard 7.951807
544 Niasse 7.460317
256 Agüero 7.346939
389 Son 7.288503
255 Bernardo Silva 7.119403
That seems a bit more like it! We’ve got some of the highest scoring players here, like Salah and Sterling, but if Austin, Lingard and Bernardo Silva can nail down long-term starting spots, we should certainly keep an eye on adding them in!
Let’s go back over this by creating a new column for goals per 90 and finding the top 10:
In [7]:
```data["goals_p90"] = p90_Calculator(data.goals_scored,data.minutes)
data.goals_p90.fillna(0, inplace=True)
```
Out[7]:
web_name goals_p90
233 Salah 0.968320
393 Kane 0.967222
325 Austin 0.906040
256 Agüero 0.823364
246 Sterling 0.797973
544 Niasse 0.793651
279 Martial 0.728745
258 Jesus 0.714995
278 Lingard 0.632530
160 Rooney 0.630252
Great job! Hopefully you can see that this is a much fairer way to rate our player data – whether for performance, fantasy football or media reporting purposes.
Summary
p90 data is a fundamental concept of football analytics. It is one of the first steps of cleaning our data and making it fit for comparisons. This article has shown how we can apply the concept quickly and easily to our data. For next steps, you might want to take a look at visualising this data, or looking at further analysis techniques.
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# Class Notes for SOC 411 at Ryerson University
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##### SOC 411 Lecture Notes - Lecture 3: Central Tendency, Bar Chart, Histogram
Tables Graphs One variable Frequency table Pie graph (Univariate analysis) Bar graph Boxplot ...
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##### SOC 411 Lecture Notes - Lecture 2: Cluster Sampling, Systematic Sampling, List Of Statistical Packages
find more resources at oneclass.com SOC 411 NOTES WEEK 2: VARIABLES, MEASUREMENT, AND WEIGHTING Using Variables in Analysis Univariate analysis: describing a singl...
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##### SOC 411 Lecture Notes - Lecture 11: Standard Deviation, Multicollinearity, Heteroscedasticity
SOC 411 NOTES WEEK 11: REGRESSION MODEL Model Fit • R is one-number summaries of the overall predictive power of the model, or how well the line fits. o The value tells us how much of the variation...
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##### SOC 411 Lecture Notes - Lecture 10: Dependent And Independent Variables, Statistical Inference, Logistic Regression
SOC 411 NOTES WEEK 10: INTRODUCTION TO REGRESSION Moving to Prediction The goal of inferential statistics is to make predictions. We are interested in modelling social processes. We want to predict the...
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##### SOC 411 Lecture Notes - Lecture 9: Ordinary Least Squares, Correlation And Dependence
SOC 411 NOTES WEEK 9: CORRELATIONS Scatterplots • Correlations are easiest to understand when they are visually displayed in scatterplots • In scatterplots, the independent variable typically goes on the ho...
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##### SOC 411 Lecture Notes - Lecture 6: Analysis Of Variance, Contingency Table, Null Hypothesis
SOC 411 NOTES WEEK 6: SIGNIFICANCE TESTING & CHI-SQUARE TESTS The Principle of Falsification • Karl Popper (1902-1994) • Philosopher of science • Falsification: The process of deriving hypotheses and...
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Andie Noak
##### SOC 411 Lecture Notes - Lecture 7: Statistical Parameter, Frequency Distribution, Confidence Interval
SOC 411 NOTES WEEK 7: SAMPLING DISTRIBITIONS & CONFIDENCE INTERVALS Introduction to Probability • A ‘probability’ tells us the chance of an event occurring. • There are two types of probability: • ...
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##### SOC 411 Lecture Notes - Lecture 8: Analysis Of Variance, Location Test, Null Hypothesis
SOC 411 NOTES WEEK 8: COMPARING MEANS & ANOVA TESTS Comparing Means • The focus is on investigating whether the mean in a sub-group of the data is different from: ▫ The mean of another sub-group ...
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##### SOC 411 Lecture Notes - Lecture 4: Interquartile Range, Standard Deviation
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##### SOC 411 Lecture Notes - Lecture 1: Statistical Hypothesis Testing, Descriptive Statistics
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##### SOC 411 Lecture Notes - Lecture 7: Null Hypothesis, Central Limit Theorem, Contingency Table
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CEWGeo distribution {ComRiskModel} R Documentation
## Complementary exponentiated Weibull geomatric distribution
### Description
Evaluates the PDF, CDF, QF, random numbers and MLEs based on the complementary exponentiated Weibull geomatric (CEWGeo) distribution. The CDF of the complementary G geomatric distribution is as follows:
F(x)=\frac{\left(1-\lambda\right)G(x)}{\left(1-\lambda G(x)\right)};\qquad\lambda\in(0,1),
where G(x) represents the baseline exponentiated Weibull CDF, it is given by
G(x)=\left(1-\exp(-\alpha x^{\beta})\right)^{\theta};\qquad\alpha,\beta,\theta>0.
By setting G(x) in the above Equation, yields the CDF of the CEWGeo distribution.
### Usage
dCEWGeo(x, alpha, beta, theta, lambda, log = FALSE)
pCEWGeo(x, alpha, beta, theta, lambda, log.p = FALSE, lower.tail = TRUE)
qCEWGeo(p, alpha, beta, theta, lambda, log.p = FALSE, lower.tail = TRUE)
rCEWGeo(n, alpha, beta, theta, lambda)
mCEWGeo(x, alpha, beta, theta, lambda, method="B")
### Arguments
x A vector of (non-negative integer) quantiles. p A vector of probablities. n The number of random values to be generated under the CEWGeo distribution. lambda The strictly positive parameter of the geomatric distribution \lambda \in (0,1). alpha The strictly positive scale parameter of the baseline exponentiated Weibull distribution (\alpha > 0). beta The strictly positive shape parameter of the baseline exponentiated Weibull distribution (\beta > 0). theta The strictly positive shape parameter of the baseline exponentiated Weibull distribution (\theta > 0). lower.tail if FALSE then 1-F(x) are returned and quantiles are computed 1-p. log if TRUE, probabilities p are given as log(p). log.p if TRUE, probabilities p are given for exp(p). method the procedure for optimizing the log-likelihood function after setting the intial values of the parameters and data values for which the CEWGeo distribution is fitted. It could be "Nelder-Mead", "BFGS", "CG", "L-BFGS-B", or "SANN". "BFGS" is set as the default.
### Details
These functions allow for the evaluation of the PDF, CDF, QF, random numbers and MLEs of the unknown parameters with the standard error (SE) of the estimates of the CEWGeo distribution. Additionally, it offers goodness-of-fit statistics such as the AIC, BIC, -2L, A test, W test, Kolmogorov-Smirnov test, P-value, and convergence status.
### Value
dCEWGeo gives the (log) probability function. pCEWGeo gives the (log) distribution function. qCEWGeo gives the quantile function. rCEWGeo generates random values.
### Author(s)
R implementation and documentation: Muhammad Imran imranshakoor84@yahoo.com and M.H Tahir mht@iub.edu.pk.
### References
Tahir, M. H., & Cordeiro, G. M. (2016). Compounding of distributions: a survey and new generalized classes. Journal of Statistical Distributions and Applications, 3, 1-35.
Mahmoudi, E., & Shiran, M. (2012). Exponentiated Weibull-geometric distribution and its applications. arXiv preprint arXiv:1206.4008.
Nadarajah, S., Cordeiro, G. M., & Ortega, E. M. (2013). The exponentiated Weibull distribution: a survey. Statistical Papers, 54, 839-877.
pCExpGeo
x<-data_guineapigs
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Assign_10_-_Solutions_to_Practice_Problem
# The denominator of the quick ratio should be total
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is also incorrect, as it does not include accrued liabilities. The denominator of the quick ratio should be total current liabilities. The correct calculations are as follows: Working Capital = Current Assets – Current Liabilities \$160,000 = \$960,000 – \$800,000 Current Ratio = = 1.2 Quick Ratio = = 1.0 b. Unfortunately, the current ratio and quick ratio are both below the minimum threshold required by the bond indenture. This may require the company to renegotiate the bond contract, including a possible unfavorable change in the interest rate.
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Ex. 15–20 a. = = 7.4 times *\$3,750,000 bonds payable × 10% b. Number of Times Preferred Dividends Are Earned = = 10.0 times **\$2,400,000 income before tax – \$400,000 income tax c. Earnings per Share on Common Stock = = \$5.00 d. Price-Earnings Ratio = = 14.4 e. Dividends per Share of Common Stock = = \$2.00 f. Dividend Yield = = 2.8%
Ex. 15–21 a. Earnings per Share = = \$2.50 *(\$1,250,000/\$25) × \$5 **\$4,000,000/\$10 b. Price-Earnings Ratio = = 16.0 c. Dividends per Share = = \$2.00 d. Dividend Yield = = 5.0%
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Ex. 15–22 a. Price-Earnings Ratio = The Home Depot: = 21.3 Google: = 22.4 Coca-Cola: = 17.3 Dividend Yield = The Home Depot: = 2.8% Google: = 0.0% Coca-Cola: = 3.3% b. Coca-Cola has the largest dividend yield, but the smallest price-earnings ratio. Stock market participants value Coca-Cola common stock on the basis of its dividend. The dividend is an attractive yield at this date. Because of this attractive yield, stock market participants do not expect the share price to grow significantly, hence the low price-earnings valuation. This is a typical pattern for companies that pay high dividends. Google shows the opposite extreme. Google pays no dividend, and thus has no dividend yield. However, Google has the largest price-earnings ratio of the three companies. Stock market participants are expecting a return on their investment from appreciation in the stock price. The Home Depot is priced in between the other two companies. The Home Depot has a moderate dividend producing a yield of 2.8%. The price-earnings ratio is slightly over 21. Thus, The Home Depot is expected to produce shareholder returns through a combination of some share price appreciation and a small dividend.
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#### Read Microsoft Word - CapstoneBinskiLemniscate.doc text version
`Scott Binski Senior Capstone Dr. Hagedorn Fall 2008Constructions on the LemniscateThe Lemniscatic Function and Abel's Theorem I. History of the Lemniscate The lemniscate is a curve in the plane defined by the equation graphical representation is the following (values and points will be important later): . ItsThe lemniscate is of particular interest because, even if it has little relevance today, it was the catalyst for immeasurably important mathematical development in the 18th and 19th centuries. The figure 8-shaped curve first entered the minds of mathematicians in 1680, when Giovanni Cassini presented his work on curves of the form , appropriately known as the ovals of Cassini (Cox 463). Only 14 years later, while deriving the arc length of the lemniscate, Jacob Bernoulli became the first mathematician in history to define arc length in terms of polar coordinates (Cox 464). The first major result of work on the lemniscate came in 1753, when, after reading Giulio Carlo di Fagnano's papers on dividing the lemniscate using straightedge and compass, Leonhard Euler proved thatwhere,. This identity was essential to the theory ofelliptic integrals (Cox 473). Amazingly, Ferdinard Eisenstein's criterion for irreducibilty is also a product of work on the lemniscate. Because he asserts his criterion for both and , it is believed that he derived it while researching complex multiplication on the lemniscate (Cox 497).1 II. Arc Length of the Lemniscate If the origin is our starting point, movement about the lemniscate begins in the first quadrant and proceeds into the fourth quadrant. After arriving back at the origin, movement then continues into the second quadrant and finally into the third quadrant. It will be useful to consider the equation of the lemniscate in terms of polar coordinates, i.e. and . We immediately obtain the equation . This is easily simplified to obtain . This polar representation of the lemniscate affords us a simple calculation of arc length, which is essential to the lemniscatic function. The arc length formula in polar coordinates isNow by differentiating our equation . Thus,, we see that, which givesNoting that, it follows thatNow our expression becomesWe let , a variant of , denote the arc length of one loop of the lemniscate. This designation is appropriate because the circumference of the circle is and the circumference of the lemniscate is . Since we obtain the arc length of the first quadrant portion of the curve when in the integral above, we see that2 This integral is improper (since the denominator is 0 for numerical value of approximately .), but it does converge and has aIII. Defining the Lemniscatic Function The lemniscate may appear peculiar at first glance, but many parallels exist between it and the sine function. For example, we may define the sine function as the inverse function of an integral in the following way:The lemniscatic function (with obvious similarity):may also be defined as the inverse function of an integralThe input values of this function, like the sine function, are arc lengths. Now recalling movement about the lemniscate, we may define our function for all real numbers. To do so, we let be the signed polar distance (from the origin) of the point corresponding to the arc length . By signed polar distance, we simply mean that is a positive value in the first and fourth quadrants, while it is a negative value in the second and third quadrants. Defining our function in such a way yields a graph that is nearly indistinguishable from the familiar graph of the sine function. They are presented here for comparison:IV. Basic Properties of3 The sine function satisfies several interesting identities: Proposition A: If , then: 1) 2) 3) 4) The lemniscatic function satisfies similar identities. In fact, we may regard the lemniscatic function as a generalization of the sine function for a different curve (noting that the output values of are radii, while the output values of are y-coordinates). Of course, the sine function is only relevant with respect to the unit circle, whereas pertains to the lemniscate. We see that the following is true of the lemniscatic function: Proposition B: If 1) 2) 3) 4) , then:The first three of these identities are not difficult to observe. Since the total arc length of the lemniscate is , it is clear that has period equal to . So . It is readily deduced from the graphical representation of the lemniscate that arc length values and correspond to points that are symmetric about the origin. Thus we see that . It is also clear that and correspond to points that are symmetric about the -axis. Thus . The last part of Proposition A is simply a restatement of the familiar identity , where is, of course, the derivative of . Now although the similarity between this identity and the corresponding identity for the lemniscatic function is clear, this is the least intuitive identity of . We now prove: Proposition: Proof: Since is periodic, it is clear that the identity need only be proved for Furthermore, since and we must only show thatTo do so, we differentiate both sides of the arc length integral with respect to , yielding4 Thus we easily obtain the desired relation for equation are continuous functions on the interval . This completes the proof.. And since both sides of the above , it follows that the equation also holds forV. The Addition Law for The sine function satisfies the addition law we say , then result for , beginning with the following identity: . So if . We will derive a similarwhereand function toBy letting and equal the three integrals above, respectively, and applying the both sides of the equation, we obtainNow sinceand, we haveAnd the last of our basicproperties implies that, yielding5 Now since both sides of this equation are analytic functions of that are defined for all values when is any fixed value, the equation holds true for all values and . Thus we have a simply stated addition law for the lemniscatic function that bears an obvious resemblance to the familiar sine function addition law (the latter bears no denominator). The subtraction law for is easily derived from the addition law. Since and (note that the latter is a direct consequence of the former),It is an easy consequence of the addition law that slightly more difficult, we can also obtain the tripling formula the addition law and the subtraction law. To begin, it is clear that. Although it is using nothing more thanThen by replacingandwithand , respectively, we haveNow using the doubling formula, we obtainAnd finally, since, we have our result:VI. A Theorem for Multiplication by Integers6 We have already seen formulas for and . In fact, formulas of the form may be generalized for all positive integers by the following theorem: Theorem A: Given an integer polynomials , there exist relatively prime such that if is odd, thenand ifis even, thenFurthermore, Proof: For the case. , it is clear that . And for the case . Thus , we know andfrom the beginning of this section that. Now that we have established these cases, we can prove the theorem by induction on . Let us assume that the theorem holds for and . Implementing the addition and subtraction laws, we see thatNow ifis even andis odd,Then since, we havewhereand7 Thus Now if is odd and is even,Then we havewhereandThus Now we will illustrate these recursive formulas by deriving the case following polynomials: . We obtain theSo again we see thatVII. Constructions on the Lemniscate Now that we have an understanding of the lemniscatic function and its properties, we may explore constructions on the lemniscate, beginning with the following basic theorem: Theorem B: The point on the lemniscate corresponding to arc length can be constructed by straightedge and compass if and only if is a constructible number. Proof: Noting that the lemniscate is defined by the equation and that , we see that . Then by solving in terms of , we see that:8 Now since the constructible numbers are closed under square roots, is constructible. For the same reason, constructible. Example: Consider the arc lengthandare constructible if and areis constructible if, which is one sixth of the entire arc length of the corresponds to the origin.lemniscate. We see that , since the arc length Recalling the tripling formula, this implies thatAnd using the quadratic formula (with), this has the constructible solutionVIII. Abel's Theorem Niels Abel's theorem for constructions on the lemniscate is identical to the following theorem of Gauss for constructions on the circle: Theorem (Gauss): Let be an integer. Then a regular n-gon can be constructed by straightedge and compass if and only ifwhereis an integer andaredistinct Fermat primes.Of course, the vertices of a regular n-gon correspond to the n-division points ( ) of the circle. In the context of this theorem, Abel's theorem for the lemniscate should appear natural: Theorem (Abel): Let be a positive integer. Then the following are equivalent: a) The n-division points of the lemniscate can be constructed using straightedge and compass. b) is constructible.9 c)is an integer of the formwhereis an integer andaredistinct Fermat primes.That statement a) implies statement b) is an immediate result of Theorem B, sinceis, ofcourse, an n-division point of the lemniscate. The proof that statement c) implies statement a) implements several elements of number theory, as well as the following remarkable result of Galois theory: Theorem C: If and is an odd positive integer, then is a Galoisextension and there is an injective group homomorphismandis Abelian.It is interesting to note that this is an analog of the following theorem for cyclotomic extensions: Theorem D: If , thenandis Abelian.We now verify a crucial component of the proof that statement c) implies statement a): Proposition: If is a Fermat prime, thenProof: If, then10 whereandare prime conjugates in. Now by the Chinese Remainder Theorem,Now sinceand, we haveThus we see thatNow it only remains to be seen that the proposition holds for Since , where and . , we see that the distinct elements that compose and . So, or, equivalently,.are those of the formAlthough we will not prove it, this proposition implies that Referencesis constructible.Cox, David A. Galois Theory. Hoboken, NJ: John Wiley & Sons, 2004. pp. 457-508.11 `
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Microsoft Word - CapstoneBinskiLemniscate.doc
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1. ## Epsilon Delta Proof
$\lim{x\to1}\frac{x+1}{x^2+x+1}=\frac{2}{3}$
apparently $Delta=\frac {3Epsilon}{5}$
you see I know how they got the 5
but for the 3 all they did was say $|x^2+x+1|>1$
and for some reason that lead them to 3
MAT196F Midterm Test Solutions
it's kind of hard to read but it's the first question
2. Originally Posted by akhayoon
$\lim{x\to1}\frac{x+1}{x^2+x+1}=\frac{2}{3}$
apparently $Delta=\frac {3Epsilon}{5}$
you see I know how they got the 5
but for the 3 all they did was say $|x^2+x+1|>1$
and for some reason that lead them to 3
MAT196F Midterm Test Solutions
it's kind of hard to read but it's the first question
no. for the 3, they combined the fractions. that is, they subtracted 2/3 from the expression
3. I know they did that
but they said that if $|x-1| < 1, then 0 < x < 2, so, |2x+1| < 5$
so this give you permission to turn the |2x-1| into a five but to get rid of the
$x^2+x+1$ all they said is that it's bigger than one so there
so how does that give them permission to make the that polynomial equal 1
??
4. Originally Posted by akhayoon
I know they did that
but they said that if $|x-1| < 1, then 0 < x < 2, so, |2x+1| < 5$
so this give you permission to turn the |2x-1| into a five but to get rid of the
$x^2+x+1$ all they said is that it's bigger than one so there
so how does that give them permission to make the that polynomial equal 1
??
i'm sorry, i don't see what you are talking about. where did they equate the polynomial to 1? they just said |x^2 + x + 1| > 1. which is okay.
This is my 57th post!!!!
5. I'm sorry, let me try to be a little more clear, it's kind of hard with these kinds of questions though, since I don't understand whats really going on and I just kind of try to work out patterns from the solutions of all the other questions I've solved like these
$\frac {|x-1||2x+1|}{(3|x^2+x+1|)} < E$
|2x-1|<5 so that means that I could use 5 instead of this expression
I know they got this from $0 sweet...
and then they go $|x^2+x+1| > 1$
but where did this come from did they use $|x-1|<1$ again somehow??
6. Originally Posted by akhayoon
I'm sorry, let me try to be a little more clear, it's kind of hard with these kinds of questions though, since I don't understand whats really going on and I just kind of try to work out patterns from the solutions of all the other questions I've solved like these
$\frac {|x-1||2x+1|}{(3|x^2+x+1|)} < E$
|2x-1|<5 so that means that I could use 5 instead of this expression
I know they got this from $0 sweet...
and then they go $|x^2+x+1| > 1$
but where did this come from did they use $|x-1|<1$ again somehow??
yes, but not directly. they got that from 0 < x < 2. note that when x = 0, the expression |x^2 + x + 1| = 1. but x is greater than zero, so we must have |x^2 + x + 1| > 1, since we increased the expression by a positive amount
are you still confused?
7. ok thanks, that explains it, so why couldn't I just make
|2x+1|>1 make x=0 and then see that it's bigger than one and make that one as well?
why did I have to go through all that -1<x-1<1 that led me to |2x-1|<5?
8. Originally Posted by akhayoon
ok thanks, that explains it, so why couldn't I just make
|2x+1|>1 make x=0 and then see that it's bigger than one and make that one as well?
why did I have to go through all that -1<x-1<1 that led me to |2x-1|<5?
remember, we stated out with |x - 1|< 1, which is why things played out that way. besides, if x = 0, |2x + 1| is not bigger than 1. we need to make x a range of values here, as we wanted inequalities. the actual value doesn't matter that much. we could have made |x - 1| < 1/2 or whatever. but 1 looks nice
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Cody
# Problem 158. Is my wife right? Now with even more wrong husband
Solution 1478335
Submitted on 4 Apr 2018 by Srishti Saha
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x = 'You want to go shopping with me, right?'; y_correct = 'yes'; assert(isequal(wiferightagain(x),y_correct))
2 Pass
ready = 'Will you be ready soon?'; when_she_said = 'In a couple minutes'; when_really = 'An hour if you''re lucky'; y_correct = 'yes'; assert(isequal(wiferightagain(ready,when_she_said,when_really),y_correct))
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