url
stringlengths 14
2.42k
| text
stringlengths 100
1.02M
| date
stringlengths 19
19
| metadata
stringlengths 1.06k
1.1k
|
|---|---|---|---|
https://testbook.com/objective-questions/mcq-on-production-management--5eea6a0e39140f30f369e4ee | Production Management MCQ Quiz - Objective Question with Answer for Production Management - Download Free PDF
Last updated on Jul 26, 2022
Latest Production Management MCQ Objective Questions
Production Management MCQ Question 1:
Which of the following statement is correct regarding Micro motion study?
1. Micro motion study is one of the most accurate techniques to enhance the production rate of the assembly line.
2. The micro-motion study technique is best suited for those operations or activities which are of short duration.
3. The motion time data from the film is transferred to SIMO chart which is further analyzed for the purpose of workplace layout or method improvement.
4. None of the above
Option :
Production Management MCQ Question 1 Detailed Solution
Explanation:
Definition:
• The micro-motion study technique is best suited for those operations or activities which are of short duration and which are repeated hundreds of times so statement 2 is correct.
• “Thus micro motion study is the technique of recording and analyzing the timing of basic elements of an operation with the objective of achieving the best method of performing the operation.” Such respective short-duration activities involve quick movement of limbs which cannot be accurately studied and timed using two-handed process charts.
• This is due to the fact that such record microscopic details such as different operations, Inspection and transport, etc. The study of such microscopic movements in short cycle repetitive jobs is not sufficient.
• Short cycle operations require to be studied for microscopic motions e.g., the operation of picking up a nut from the bin and its fixing consists of three hand motions namely reach for the nut, grasp the nut, and move the hand back to assembly position.
• Such detailed analysis help to develop the best possible pattern of movements and hence enabling the operator to perform various operations repeatedly with minimum effort and fatigue.
• Micro motion study is one of the most accurate techniques of work analysis used for work improvement. so statement 1 is incorrect.
• It makes use of motion pictures of different activities movement, so with the help of the camera. small time up to 0.0005 minutes can be measured and recorded by this system.
• When a picture camera is utilized, the procedure is known as “Micro motion study”.
• The motion time data from the film is transferred to SIMO chart.
• The SIMO chart data can be further analyzed for the purpose of workplace layout or method improvement so statement 3 is correct.
Purpose of Micro Motion Study:
1. It can be used for the following purposes:
2. To study the nature and path of movements for obtaining the elements of an operation.
3. To study the activities of the machine and the operator.
4. To impart training to the workers or operators regarding motion; economy so that unnecessary movement by the workers may be avoided.
5. To study the relationship between the activities of the operator and the machine.
6. To keep a permanent record of the most efficient way of performing a task for future reference.
7. To obtain motion time data for developing synthetic time standards for various elements.
Advantages of Micro Motion Study:
1. It has the following important advantages:
2. It provides a permanent record of motion study on films.
3. A large number of operators can see the procedure at any time even after the completion of motion study work.
4. Films can easily reveal the difference between the present and the proposed technique.
5. Films can be demonstrated to large workforce at any desired speed.
6. It provides very accurate time for each operation or motion in comparison to stopwatch time study.
7. It helps in making detailed and accurate analyses of the prevailing technique.
Production Management MCQ Question 2:
Which of the following industries will consume maximum power per tonne of product?
1. Zinc
2. Aluminium
3. Alloy steel
4. Cement
Answer (Detailed Solution Below)
Option 2 : Aluminium
Production Management MCQ Question 2 Detailed Solution
Aluminum Production: About 17,000 kWh of electricity is required to produce 1 tonne of aluminum.
Zinc Production: The amount of energy required for casting one tonne of zinc is 208 kWh, and for remelting, the same amount of metal is 155 kWh.
Alloy steel Production: With an average of 770 kWh/ton and a typical EAF consumption under 500 kWh/ton.
Cement Production: While total electrical energy consumption for cement production is about 100 kWh/ton of cement.
Production Management MCQ Question 3:
Safety in a workshop can be classified as
1. General safety
2. Personal safety
3. Machine safety
4. All of the above
5. None of these
Answer (Detailed Solution Below)
Option 4 : All of the above
Production Management MCQ Question 3 Detailed Solution
Generally, accidents do not happen; they are caused. Most accidents are avoidable. To achieve this, it is essential that every person should follow the safety procedure.
Safety in a workshop can be broadly classified into 3 categories.
1. General safety
2. Personal safety
3. Machine safety
General safety
• Keep the floor and gangways clean and clear.
• Don't leave the machine which is in motion.
• Don't touch or handle any equipment/ machine unless authorized to do so.
• Use the correct tools for the job and keep at their proper place.
• Wipeout spilt oil immediately.
• Replace worn out or damaged tools immediately.
• Never direct compressed air at yourself or at your co-worker.
• Ensure adequate light in the workshop.
• Sweep away the metal cuttings.
Personal safety
• Wear a one-piece overall or boiler suit.
• Keep the overall buttons fastened.
• Don't use ties and scarves.
• Roll up the sleeves tightly above the elbow.
• Wear safety shoes or boots.
• Cut the hair short.
• Don't wear a ring, watch or chain.
• Never lean on the machine.
• Don't clean hands in the coolant fluid.
• Don't use cracked or chipped tools.
• Don't start the machine until
• − the workpiece is securely mounted
• − the feed machinery is in the neutral
• − the work area is clear.
• Don't adjust clamps or holding devices while the machine is in motion.
• Never touch the electrical equipment with wet hands.
• Don't use any faulty electrical equipment.
• Don't engage yourself in conversation with others while concentrating on your job.
Machine safety
• Switch off the machine immediately if something goes wrong.
• Keep the machine clean.
• Replace any worn out or damaged accessories, holding devices, nuts, bolts etc as soon as possible.
• Do not attempt operating the machine until you know how to operate it properly.
• Do not adjust tool or the workpiece unless the power is off.
• Stop the machine before changing the speed.
• Disengage the automatic feeds before switching off.
• Check the oil level before starting the machine.
• Never start a machine unless all the safety guards are in position.
• Take measurements only after stopping the machine
• Use wooden planks over the bed while loading and unloading heavy jobs.
Production Management MCQ Question 4:
A system that uses the minimal amount of resources to produce a high volume of high-quality goods with some variety is known as:
1. Repetitive Production system
2. Mass Production system
3. Continuous Production system
4. Lean Production system
5. Just-in Time System
Answer (Detailed Solution Below)
Option 4 : Lean Production system
Production Management MCQ Question 4 Detailed Solution
The correct answer is Lean Production system
Important Points
Explanation:
Lean Production system:
• The system uses minimal amounts of resources to produce a high volume of high-quality goods with some variety.
• Lean production system uses a highly skilled workforce and flexible types of equipment.
Additional Information
Mass production:
• Mass production means the production of items on large scale.
• In this type of production, there is a continuous and steady flow of materials.
• It offers the lowest production cost per unit.
• Material handling is reduced to a minimum since very little time is spent on the resetting of the machine.
• Work cycles are short and of a repetitive nature.
• Plant layout and facilities are designed to suit production requirements.
• Mass production is preferred where there is a continuous and regular demand for products.
Continuous Production system:
• The same product is produced continuously in the same sequence of operations, e.g. chemical and power plant
Repetitive Production system:
• A repetitive Production system refers to the fabrication, machining, assembly and testing of discrete, standard units, which are produced in volume.
• It also refers to the fabrication, machining, assembly and testing of products assembled in volume from standard options.
• A typical repetitive manufacturing system is characterized by long runs of complex products produced in lower volume quantities.
Just-in Time System:
• A just-in-time (JIT) inventory system is a business approach that involves receiving products as close to when they are needed as possible.
• As a result, if a car assembly facility wants to install airbags, it does not have a supply of airbags on hand, but instead orders them as the cars come off the assembly line.
Production Management MCQ Question 5:
_______ is considered as an adult according to the Factory Act, 1948.
1. A person completed 15 years of age
2. A person completed 18 years of age
3. A person completed 21 years of age
4. A person completed 25 years of age
Answer (Detailed Solution Below)
Option 2 : A person completed 18 years of age
Production Management MCQ Question 5 Detailed Solution
Explanation:
• Adult means a person who has completed his eighteenth year of age.
• Adolescent means a person, who has completed his fifteenth year of age but has not completed his eighteenth year.
Top Production Management MCQ Objective Questions
Production Management MCQ Question 6
The original cost of equipment is rupees 1,00,000. Its salvage value at the end of its useful life of five years is 40,000. Its book value at the end of two years of its useful life as per straight line method of evaluation of depreciation will be
1. 68000
2. 76000
3. 58000
4. 940000
Option 2 : 76000
Production Management MCQ Question 6 Detailed Solution
Concept
By Straight line method
Depreciation is given by
$${D_m} = \frac{{{C_i} - {C_s}}}{n}$$
Where Dm is depreciation
Ci is Initial cost of an asset
Cs is Salvage or scrap value (Estimated at the end of utility period)
Book value (Bm) at the end of 'm ' years life is given by
$${B_m} = {C_i} - m × {D_m}$$
Calculation
Given,
Ci = 1,00,000 /-, Cs = 40,000 /-
n = 5 years
Depreciation is
$${D_m} = \frac{{1,00,000 - 40000}}{5} = 12,000$$
Book value at the end of 2 years of lifetime is given by
$${B_m} = {C_i} - m × {D_m}$$
= 1,00,000 - 2 × 12,000 = 76,000 /-
Production Management MCQ Question 7
Which of the following layout is useful when the product being processed is very big, heavy or difficult to move?
1. Fixed position
2. Process layout
3. Product layout
4. Cellular layout
Answer (Detailed Solution Below)
Option 1 : Fixed position
Production Management MCQ Question 7 Detailed Solution
Concept:
Layout planning in manufacturing and service organisations involves the physical arrangement of various resources available in the system to improve the performance of the operating system, thereby providing better customer services.
Layouts can be classified into the following four categories:
1. Process layout
2. Product layout
3. Group layout (Combination layout)
4. Fixed-position layout
Fixed-position layout:
In a fixed position layout, the service is performed around a customer that remains stationary while the work is being done.
For example, surgery is performed on a patient, where a patient remains stationary in operation theatre. Different doctors or specialist perform different activities in a sequence on a stationary patient.
Example: Manufacturers of aeroplanes, ships, locomotives, large turbines, heavy machinery, pressure vessels and others which involve heavy materials and sub-assemblies.
This type of layout is suitable:
• When one or a few pieces of an identical product are to be manufactured
• When the assembly consists of a large number of heavy parts, the cost of transportation of which is very high
Product Layout: In product layout, the workstations are located according to the processing sequence for the service.
Process Layout: In process layout, similar activities are grouped together according to the process or function they perform.
Production Management MCQ Question 8
The process of taking measures to check the quality, performance or reliability of the equipment is called:
1. testing
2. inspecting
3. servicing
4. analysing
Answer (Detailed Solution Below)
Option 1 : testing
Production Management MCQ Question 8 Detailed Solution
Testing: It is the process of taking measures to check the quality, performance, or reliability of the equipment.
Inspections: It includes measuring, testing, examining, or gauging the features of a process or product.
Servicing: It is the kind of timely maintenance of the equipment for better performance.
Analyzing: It provides information on the composition of a sample of matter. They are employed to obtain qualitative and quantitative information about the presence or absence of one or more components of the equipment.
Production Management MCQ Question 9
Acceptance sampling is normally used for
1. Job-shop production
2. Batch production
3. Mass production
4. Just-in -time production
Answer (Detailed Solution Below)
Option 3 : Mass production
Production Management MCQ Question 9 Detailed Solution
Explanation:
Acceptance sampling is a quality control procedure, which uses the inspection of small samples instead of 100 percent inspection in making the decision to accept or reject much larger quantities, called a lot. This is a statistical procedure, which uses random samples, that is, each item in the lot has an equal chance of being a part of the sample that is inspected.
• Acceptance Sampling is normally used for Mass production.
• In its simplest form, If the sample from a larger lot has an acceptable level of defects, it will be accepted. If not, the entire lot will be rejected.
• If acceptance sampling is used prior to accepting goods from a supplier (i.e., incoming inspection), then the acceptable level of defects must be agreed between the supplier and customer because the supplier may have to take back the entire lot if it fails acceptance sampling.
• A sampling plan establishes the rules guiding the sampling and the criteria for accepting or rejecting the lot.
Mass production:
• The manufacture of discrete parts or assemblies using a continuous process is called mass production. The machineries are arranged in a line or product layout. It is also called as continuous production.
• Mass manufacturing is characterized by very high production rates and low variety or flexibility
Advantages of Mass Production
• Higher rate of production with reduced cycle time.
• Higher capacity.
• A less skilled operator can manage the process.
• Low in process inventory.
• Production cost per unit item will come down due to economies of scale.
Batch Production:
• It is a common type of production. In this type of production, the flow of material is intermittent.
• Articles are manufactured in batch as per the specific order procured.
• A good production control system must be developed. Product planning is done for each batch.
• Proper maintenance of equipment and machinery is essential.
• The plant layout best suitable for this type of production is the Process Layout.
Job – shop production:
• Job – shop production is characterized by the manufacturing of a large variety of products in small quantities that are designed and produced as per specifications are given by customers
• The main feature of this production system is that it is highly flexible
• A shop – shop comprises general-purpose machines arranged in different departments
• Example – Manufacture of aeroplanes and oil field equipment, machine tools, giant hydro turbine, rolling mills, and other heavy equipment
Just In Time (JIT) Production system:
• Just in time (JIT) manufacturing is a workflow methodology aimed at reducing flow times within production systems, as well as response times from suppliers and to customers.
Important characteristics of JIT Production system
• Daily or hourly deliveries of small quantities of parts from suppliers.
• Certification of supplier quality, so that no receiving inspections are needed.
• The use of Kanbans to drive the demand at each workstation.
• Production stops as soon as the immediate demand level has been fulfilled.
• On-site inspection of each in-process product from the preceding work center, so that flaws are discovered at once.
• The use of rapid machine setups, so that production runs can be as short as one unit.
• The cross-training of employees, so that they are certified to work on multiple tasks.
• Immediate shipment of completed goods to customers as soon as an order has been fulfilled.
Mass production or continuous production High manufacturing rate with low flexibility or variety Job shop production Low volume and high flexibility or variety Batch production medium volume and medium variety
Production Management MCQ Question 10
Technology S-curves are helpful in
1. Product platform planning
2. Concept generation
3. Product architecture
4. Product design
Answer (Detailed Solution Below)
Option 4 : Product design
Production Management MCQ Question 10 Detailed Solution
Explanation:
S-curve:
• In the course of Product Design and Development, the S-Curve determines the performance in regards to time and effort.
• It assists in determining the level of maturity of the product.
• Technological innovation time cycle and market behaviour are well characterized by the ‘S’ curve.
• Technological innovation typically manifests themselves into a market along the ‘S’ curve.
• The Technology S-Curve of Innovation/product life cycle is a robust framework that can be used to analyze various products at their different stages and to explain their successes and failures.
It is divided into 3 portions namely Lower, Middle and Top portions.
Lower Portions
• This phase is at the beginning of the S-Curve pattern of innovation.
• It is when the product/ industry is completely new.
• As a result, a dominant design in the market hasn’t been established yet.
• Therefore, the competition between the various players in the industry is fierce.
• As a result, usually at this stage, most of the resources are spent on research and development.
• There will be apparently not much innovation.
• Changes are less and widely spaced.
Middle Portion
• In this phase, due to the ability to overcome a major technical obstacle or the ability to satisfy a demand of the market.
• The product/industry has been adopted by the early majority and managed to cross the differences and a dominant design has been established already.
• Hence, the market will be characterized by rapid growth in production, and the product will move quickly towards a full market acceptance.
• So, A rapid profusion of innovation (slope) will take place during this period.
• Many products are launched and many competitors join the market.
Top Portion
• Here, the product is adopted almost completely by society and is usually approaching a physical limit.
• Due to the strong competition among the major players in the market which is clearly defined at this stage, most of the resources at this point are spent on improving the production processes and making them cheaper.
• Therefore, oftentimes the products at this stage become completely standardized and the innovations at this stage are considered incremental.
Production Management MCQ Question 11
A system that uses the minimal amount of resources to produce a high volume of high-quality goods with some variety is known as:
1. Repetitive Production system
2. Mass Production system
3. Continuous Production system
4. Lean Production system
Answer (Detailed Solution Below)
Option 4 : Lean Production system
Production Management MCQ Question 11 Detailed Solution
Explanation:
Lean Production system:
• The system uses minimal amounts of resources to produce a high volume of high-quality goods with some variety.
• Lean production system uses a highly skilled workforce and flexible types of equipment.
Mass production:
• Mass production means the production of items on large scale.
• In this type of production, there is a continuous and steady flow of materials.
• It offers the lowest production cost per unit.
• Material handling is reduced to a minimum since very little time is spent on the resetting of the machine.
• Work cycles are short and of a repetitive nature.
• Plant layout and facilities are designed to suit production requirements.
• Mass production is preferred where there is a continuous and regular demand for products.
Continuous Production system:
• The same product is produced continuously in the same sequence of operations, e.g. chemical and power plant
Repetitive Production system:
• A repetitive Production system refers to the fabrication, machining, assembly and testing of discrete, standard units, which are produced in volume.
• It also refers to the fabrication, machining, assembly and testing of products assembled in volume from standard options.
• A typical repetitive manufacturing system is characterised by long runs of complex products produced in lower volume quantities.
Production Management MCQ Question 12
Which of the following Characteristics is not usually associated with batch production?
1. products made to customer order
2. low volume
3. stable, predictable demand
4. general purpose equipment
Answer (Detailed Solution Below)
Option 1 : products made to customer order
Production Management MCQ Question 12 Detailed Solution
Batch production
• Batch production is a type of production in which the job passes through the functional departments in batches and each batch may have a different routing.
• Batch production is characterized by the manufacture and stocking of a limited number of products at regular intervals, awaiting sales.
• Batch production is used for stable and predictable demand and for general purpose equipments.
• Example – Machine tools, pumps, compressors, stationary IC engines, etc.
Additional Information
Job – shop production
• Job – shop production is characterized by the manufacturing of a large variety of products in small quantities that are designed and produced as per specifications are given by customers
• The main feature of this production system is that it is highly flexible
• A shop – shop comprises general-purpose machines arranged in different departments
• Example – Manufacture of aeroplanes and oil field equipment, machine tools, giant hydro turbine, rolling mills, and other heavy equipment
Mass production
• In mass production, the same type of product is manufactured to meet the continuous demand for the product
• Manufacturing of discrete components or assemblies in a very large volume is called mass production
• Machines are arranged in a line according to the sequence of operations on the product
• Example – Nuts, bolts, screws, washers, pencils, matches, engine blocks, bicycles, electric motors, sewing machines, tractors, etc.
Production Management MCQ Question 13
Mass production is characterized by
1. low-volume items with maximum flexibility in their design
2. high-volume items with maximum flexibility in their design
3. high-volume items with minimum flexibility in their design
4. low-volume items with minimum flexibility in their design
Answer (Detailed Solution Below)
Option 3 : high-volume items with minimum flexibility in their design
Production Management MCQ Question 13 Detailed Solution
Explanation:
Mass production:
• Manufacture of discrete parts or assemblies using a continuous process are called mass production. The machineries are arranged in a line or product layout. It is also called as continuous production.
• Mass manufacturing is characterized by very high production rates and low variety or flexibility
Advantages of Mass Production
• Higher rate of production with reduced cycle time.
• Higher capacity.
• Less skilled operator can manage the process.
• Low in process inventory.
• Production cost per unit item will come down due to economies of scale.
Additional Information
Mass production or continuous production High manufacturing rate with low flexibility or variety Job shop production Low volume and high flexibility or variety Batch production medium volume and medium variety
Production Management MCQ Question 14
Which of the following is not the part of products planning competitive strategy?
1. Technology leadership
2. Cost leadership
3. Customer focus
4. Endurance
Answer (Detailed Solution Below)
Option 4 : Endurance
Production Management MCQ Question 14 Detailed Solution
Explanation:
Four parts of products planning competitive strategy are
1. Cost Leadership Strategy or Low-cost strategy or Technology leadership.
2. Differentiation strategy.
3. Best-cost strategy or Cost leadership.
4. Market-niche or focus strategy or Customer focus.
Hence, endurance is something unrelated here, so it will be the answer.
Production Management MCQ Question 15
Which of the following production system is characterised by the low production volume?
1. Project Production System
2. Job Shop Production System
3. Batch Production System
4. Mass Production System
Answer (Detailed Solution Below)
Option 2 : Job Shop Production System
Production Management MCQ Question 15 Detailed Solution
Explanation:
Production System is classified as Job Shop, Batch Production, Mass and Continuous Production systems.
Job Shop Production System
• High Variety and Low volume
• General-purpose machines
• Highly skilled operators
• Large inventory of materials
Batch Production
• Shorter production runs
• When plant and machinery are flexible
Mass Production
• Standard product and large volume of products
• Large Volume of products
Continuous Production
• Dedicated plant and equipment with zero flexibility
Production Management MCQ Question 16
Which of the following is the objective of MIS?
1. To keep the information to up to data
2. To increase the quantity of the data
3. To filter the data
4. To interpret data
Answer (Detailed Solution Below)
Option 2 : To increase the quantity of the data
Production Management MCQ Question 16 Detailed Solution
Explanation:
Management Information system or ‘MIS’ is a planned system of collecting, storing and disseminating (spreading or increasing) data in the form of information needed to carry out the function of management
Objectives of MIS:
• Capturing data
• Processing data
• Information storage
• Information Retrieval
• Information Propagation
Production Management MCQ Question 17
Lean production is related to
1. mass production
2. batch production
3. customized production
4. stock-driven production
Answer (Detailed Solution Below)
Option 3 : customized production
Production Management MCQ Question 17 Detailed Solution
Explanation:
Lean production:
• Lean production means supplying the customer with exactly what the customer wants, when the customer wants it, without waste, through continual improvement.
• Waste is anything that does not add value to the product.
• Lean production is driven by the "pull" system of the customer's order.
• These waste reduction efforts improve productivity and processes.
Customized Production:
• In this flow of material and parts from one location to another is intermittent or discontinuous.
• Each job order is different from the previous as regards its type, specifications, quality, and quantity.
• Product design takes a lot of time.
• Prior planning becomes difficult.
• General-purpose machinery and a flexible layout are preferred.
• The number of items to be manufactured is very small.
Mass production:
• Mass production means the production of items on large scale.
• In this type of production, there is a continuous and steady flow of materials.
• It offers the lowest production cost per unit.
• Material handling is reduced to a minimum since very little time is spent on the resetting of the machine.
• Work cycles are short and of a repetitive nature.
• Plant layout and facilities are designed to suit production requirements.
• Mass production is preferred where there is a continuous and regular demand for products.
Batch Production:
• It is a common type of production.
• The flow of material is intermittent.
• The plant layout is of the process type.
• Product planning is done for each batch.
• Proper maintenance of equipment and machinery is essential.
• A good production control system must be developed.
• Articles are manufactured in batch as per the specific order procured.
Production Management MCQ Question 18
In the context of factors utilized in FMEA, choose the odd one out.
1. Occurrence
2. Severity
3. Serviceability
4. Detection
Answer (Detailed Solution Below)
Option 3 : Serviceability
Production Management MCQ Question 18 Detailed Solution
Explanation:
Failure Mode and Effects Analysis (FMEA) is product recalls resulting from poorly designed products and/or processes.
Failure Mode and Effects Analysis, or FMEA, is a methodology aimed at allowing organizations to anticipate failure during the design stage by identifying all of the possible failures in a design or manufacturing process.
Here’s an overview of the 10 steps to a Process FMEA.
1. Review the process
• Use a process flowchart to identify each process component.
• List each process component in the FMEA table.
• If it starts feeling like the scope is too big, it probably is. This is a good time to break the Process Failure Mode and Effects Analysis into more manageable chunks.
2. Brainstorm potential failure modes
• Review existing documentation and data for clues about all of the ways each component can fail.
• The list should be exhaustive – it can be paired down and items can be combined after this initial list is generated.
• There will likely be several potential failures for each component.
3. List the potential effects of each failure
• The effect is the impact the failure has on the end product or on subsequent steps in the process.
• There will likely be more than one effect for each failure.
4. Assign Severity rankings
• Based on the severity of the consequences of failure.
5. Assign Occurrence rankings
• Rate the severity of each effect using customized ranking scales as a guide.
6. Assign Detection rankings
• The chances o failure will be detected prior to it occurring.
7. Calculate the RPN: Severity × Occurrence × Detection
8. Develop the action plan
• Decide which failures will be worked on based on the Risk Priority Numbers. Focus on the highest RPNs.
• Define who will do what by when.
9. Take action
• Implement the improvements identified by your Process Failure Mode and Effects Analysis team.
10. Calculate the resulting RPN
• Re-evaluate each of the potential failures once improvements have been made and determine the impact of the improvements.
Production Management MCQ Question 19
Line balancing is related to
1. product layout
2. process layout
3. hybrid layout
4. cellular layout
Answer (Detailed Solution Below)
Option 1 : product layout
Production Management MCQ Question 19 Detailed Solution
Explanation:
Line balancing:
Line balancing is a production strategy that involves balancing operator and machine time to match the production rate to the Takt time. Takt time is the rate at which parts or products must be produced in order to meet customer demand.
Benefits of Line Balancing
1. Reduce waiting waste.
2. Reduce inventory waste.
3. Absorb internal and external irregularities.
4. Reduce production costs and increase profits.
Line balancing is used in Product Layout as in this layout rate of production is very high and requires proper balancing between machine and operator for better efficiency.
Production Management MCQ Question 20
An assembly line is an example of which one of the following processes?
1. Product focused process
2. Customized process
3. Repetitive process
4. Specialized process
Answer (Detailed Solution Below)
Option 3 : Repetitive process
Production Management MCQ Question 20 Detailed Solution
Explanation:
In a manufacturing shop floor the production line is arranged in a manner that the product is moved sequentially along the line and stops at the work centers where operations are performed. The production line is designed to optimize the utilization of movement of work-piece during manufacturing. The line is designed according to the products volume and variety.
Assembly line:
• An assembly line is a manufacturing process in which interchangeable parts are added to a product in a sequential manner to create an end product.
• At each station along the line some part of the production process takes place.
• The workers and machinery used to produce the item are stationary along the line and the product moves through the cycle, from start to finish, therefore the assembly line have number of process which are repetitive in nature.
Customized Process:
• Customized processes are used when the company manufactures a wide range of products that can be modified according to the customer’s requirements.
• For customized products one kind of machinery are grouped together. The customized process are suitable for low volume and high variety of products.
Product Focused Process:
• In a product focused process the various operations on raw material are performed in a sequence and the machines are placed along the product flow line, i.e. machines are arranged in a sequence in which raw material will be operated upon. This is suitable for continuous production.
Specialized process:
• In a specialized process, the operations are very product specific and the special purpose machines are utilized in this process
Production Management MCQ Question 21
Which of the following is an example of Batch type of production system?
1. Automobiles production unit
2. Petrol refinery
3. Clothes factory
4. Aircraft manufacturing unit
Answer (Detailed Solution Below)
Option 3 : Clothes factory
Production Management MCQ Question 21 Detailed Solution
Explanation:
A flexible manufacturing system (FMS) is a manufacturing cell or system consisting of one or more CNC machines, connected by an automated material handling system, pick-and-place robots, and all operated under the control of a central computer. It also has auxiliary sub-systems like component load/unload station, automatic tool handling system, tool pre-setter, component measuring station, wash station, etc.
Job – shop production
• Job – shop production is characterized by the manufacturing of a large variety of products in small quantities that are designed and produced as per specifications are given by customers
• The main feature of this production system is that it is highly flexible
• A shop – shop comprises general-purpose machines arranged in different departments
• Example – Manufacture of aeroplanes and oil field equipment, machine tools, giant hydro turbine, rolling mills, and other heavy equipment
Batch production
• Batch production is a type of production in which the job passes through the functional departments in batches and each batch may have a different routing
• Batch production is characterized by the manufacture and stocking of a limited number of products at regular intervals, awaiting sales
• Example – Machine tools, pumps, compressors, stationary IC engines, Clothes factory, etc.
Mass production
• In mass production, the same type of product is manufactured to meet the continuous demand for the product
• Manufacturing of discrete components or assemblies in a very large volume is called mass production
• Machines are arranged in a line according to the sequence of operations on the product
• Example – Nuts, bolts, screws, washers, pencils, matches, engine blocks, bicycles, electric motors, sewing machines, tractors, etc.
The FMS is most suited for the mid-variety, mid-volume production range
Production Management MCQ Question 22
Fully computer-automated plants can be achieved for which of the following production systems ?
1. Batch production
2. Job shop production
3. Mass production of discrete product
4. Continuous flow processes
Answer (Detailed Solution Below)
Option 4 : Continuous flow processes
Production Management MCQ Question 22 Detailed Solution
Explanation:
Continuous Flow Process:
• The continuous flow process involves moving one work unit at a time between each step of the process — with no breaks in time, sequence, substance, or extent. For most applications, continuous flow saves time, energy, and costs and when implemented correctly, it can:
• Reduce waste
• Save money by reducing inventory and transportation costs
• Increase productivity by completing more units in less time
• Improve quality by making it easier to spot and correct errors
• Cut down on overhead via increased stability and reduced lead time
• Adapt to customer needs more effectively than batch processing
• Fully computer-automated plants can be achieved by Continuous Flow processes.
Additional InformationBatch production:
• batch production is a method used to produce similar items in groups, stage by stage. In batch production, the product goes through each stage of the process together before moving on to the next stage.
• The degree to which workers are involved in this type of production depends on the type of product. It is common for machinery to be used for the actual production and workers participate only at the beginning and end of the process.
Job shop production:
• A job shop is a type of manufacturing process in which small batches of a variety of custom products are made. In the job shop process flow, most of the products produced require a unique setup and sequencing of process steps.
Production Management MCQ Question 23
Which of the following is not basic element of the JIT production system?
1. cellular layouts
2. push production system
3. small-lot production
4. flexible resources
Answer (Detailed Solution Below)
Option 2 : push production system
Production Management MCQ Question 23 Detailed Solution
Explanation:
Just In Time (JIT) Production system:
• Just in time (JIT) manufacturing is a workflow methodology aimed at reducing flow times within production systems, as well as response times from suppliers and to customers.
Important characteristics of JIT Production system
• Daily or hourly deliveries of small quantities of parts from suppliers.
• Certification of supplier quality, so that no receiving inspections are needed.
• The use of Kanbans to drive the demand at each workstation.
• Just-in-Time manufacturing also uses a pull system to move materials through the production cycle. This leads to a more efficient manufacturing layout that can significantly reduce lead time.
• Another main objective of Just-In-Time Production systems is to improve the production process in the company hence increases productivity.
• Production stops as soon as the immediate demand level has been fulfilled.
• On-site inspection of each in-process product from the preceding work center, so that flaws are discovered at once.
• The use of rapid machine setups, so that production runs can be as short as one unit.
• The cross-training of employees, so that they are certified to work on multiple tasks.
• Immediate shipment of completed goods to customers as soon as an order has been fulfilled.
Production Management MCQ Question 24
MIS is not used in which of the following applications?
1. Forecasting
2. Scheduling problem
3. Inventory
4. Heat transfer
Answer (Detailed Solution Below)
Option 4 : Heat transfer
Production Management MCQ Question 24 Detailed Solution
Explanation:
MIS stands for "Management Information Systems".
The management information system (MIS) provides the required information for the effective functioning of business organizations such as forecasting, scheduling and inventory data.
It provides the information to the management as per the level in the organization.
• Top management seeks this information for policymaking, goal setting and strategic decision making.
• Middle management needs these data in the problem solving, monitoring progress towards achievement of goals, planning and scheduling.
Therefore, MIS is an essential field of study in business management and is useful for a manager, entrepreneur or a business professional.
Hence the answer will be Heat Transfer.
Production Management MCQ Question 25
Which of the following industries will consume maximum power per tonne of product?
1. Zinc
2. Aluminium
3. Alloy steel
4. Cement
Answer (Detailed Solution Below)
Option 2 : Aluminium
Production Management MCQ Question 25 Detailed Solution
Aluminum Production: About 17,000 kWh of electricity is required to produce 1 tonne of aluminum.
Zinc Production: The amount of energy required for casting one tonne of zinc is 208 kWh, and for remelting, the same amount of metal is 155 kWh.
Alloy steel Production: With an average of 770 kWh/ton and a typical EAF consumption under 500 kWh/ton.
Cement Production: While total electrical energy consumption for cement production is about 100 kWh/ton of cement. | 2022-10-05 15:18:11 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.31672993302345276, "perplexity": 4694.368542428524}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030337631.84/warc/CC-MAIN-20221005140739-20221005170739-00425.warc.gz"} |
https://www.yourdictionary.com/unelectrified | #### Sentence Examples
• If W is the weight required to depress the attracted disk into the same sighted position when the plates are unelectrified and g is the acceleration of gravity, then the difference of potentials of the conductors tested is expressed by the formula V - V'=(d - d') /87 W where S denotes the area of the attracted disk.
• If the gold-leaf is unelectrified, it is not acted upon by the two plates placed at equal distances on either side of it, but if its potential is raised or lowered it is attracted by one disk and repelled by the other, and the displacement becomes a measure of its potential.
• Instead of rebounding after collision, as the unelectrified drops of clean water generally, or always, do, the electrified drops coalesce, and then the jet is no longer scattered about. | 2019-01-22 19:31:20 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8031685948371887, "perplexity": 962.0441143363416}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-04/segments/1547583867214.54/warc/CC-MAIN-20190122182019-20190122204019-00259.warc.gz"} |
https://math.stackexchange.com/questions/3906628/zero-kl-divergence-rightarrow-same-distribution | # Zero KL-divergence $\Rightarrow$ same distribution?
I am looking at relative entropy=KL-divergence and trying to prove, at least in case of probability mass functions $$q,\,p$$ with possible states $$x\in A$$ that if:
$$D(p \,\lVert\,q)=\sum_{x\in A}p\left(x\right)\,\log\left[\frac{p\left(x\right)}{q\left(x\right)}\right] = 0$$
then $$p\left(x\right)=q\left(x\right)\: \forall x \in A$$.
I can see that the converse is true, and my book Cover, Thomas Elements of Information Theory does state that the above result is also valid, but the proof there is, in my opinion, incomplete. The authors use concavity of $$\log$$ to prove that $$D(p \,\lVert\,q)>0$$ (happy there). And then state, that the only way to have equality, as opposed to inequality, is to have $$,\log\left[\frac{p\left(x\right)}{q\left(x\right)}\right]$$ being not strictly concave. Sort of ok with that (not too happy), but then the authors state that the only way to get that is to have $$\left[\frac{p\left(x\right)}{q\left(x\right)}\right]=const$$. I am not sure how to make that last jump.
Can anyone help? Or suggest an alternative proof? Thanks
"The only way to have equality is to have $$\log \frac{p(x)}{q(x)}$$ being not strictly concave" is a weird way to say something here, so you may have misread the proof, or it may be badly worded. However, the idea is to figure out the equality case of the inequality we used to prove that KL-divergence is nonnegative.
To begin with, here's a proof that KL-divergence is nonnegative, which is probably essentially the same as the one you read. Let $$f(x) = -\log x = \log \frac1x$$; this is a strictly convex function on the positive reals. Then by Jensen's inequality $$D(p\,\|\,q) = \sum_{x \in A} p(x) f\left(\frac{q(x)}{p(x)}\right) \ge f\left(\sum_{x \in A} p(x) \frac{q(x)}{p(x)}\right) = f(1) = 0.$$ For a strictly convex function like $$f(x)$$, assuming that the weights $$p(x)$$ are all positive, equality holds if and only if the inputs to $$f$$ are all equal, which directly implies $$\frac{q(x)}{p(x)}$$ is constant and therefore $$p(x)=q(x)$$ for all $$x$$.
We have to be careful if $$p(x)=0$$ for some inputs $$x$$. Such values of $$x$$ are defined to contribute nothing to the KL-divergence, so essentially we have a sum over a different set $$A' = \{x \in A : p(x) > 0\}$$. Then on the right-hand side of the inequality, we get $$f\left(\sum_{x \in A'} p(x) \frac{q(x)}{p(x)}\right) = f\left(\sum_{x \in A'} q(x)\right).$$ But now, if the sum over $$A'$$ is some probability $$q<1$$, then $$f(q) > 0$$, and we conclude $$D(p\,\|\,q) > 0$$. So the sum of $$q(x)$$ over $$A'$$ should be $$1$$. We conclude that when the KL-divergence is $$0$$, $$p(x)=0 \implies q(x)=0$$ for all $$x \in A$$. In order for the KL-divergence to even be defined, $$q(x)=0 \implies p(x)=0$$ for all $$x\in A$$. So we can throw away any values where $$p(x)=q(x)=0$$, and now we're back in the simple case, where all the weights in Jensen's inequality are positive.
• Thanks! I think I can fix this in my head by saying that strict convexity means $$\lambda f\left(x_1\right)+\left(1-\lambda\right)f\left(x_2\right)>f\left(\lambda x_1+(1-\lambda) x_2\right),\: \forall x_1\neq x_2,\,\lambda\in \left(0,\,1\right)$$ Therefore equality for a strictly convex function implies that one of the conditions is violated, and since the probabilities are not zero or one, the $x_{\dots}$ s must be equal | 2022-01-16 22:02:18 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 35, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9699946045875549, "perplexity": 78.95195612171025}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320300244.42/warc/CC-MAIN-20220116210734-20220117000734-00533.warc.gz"} |
https://www.semanticscholar.org/paper/The-Gravitational-Bending-of-Acoustic-Schwarzschild-Qiao-Zhou/2146799f52519c6918db8ce341563cfef4bdae03 | Corpus ID: 237490311
# The Gravitational Bending of Acoustic Schwarzschild Black Hole
@inproceedings{Qiao2021TheGB,
title={The Gravitational Bending of Acoustic Schwarzschild Black Hole},
author={C. K. Qiao and Mi Zhou},
year={2021}
}
• C. Qiao, Mi Zhou
• Published 13 September 2021
• Physics
Acoustic black hole is becoming an attractive topic in recent years, for it open-up new direction for experimental explorations of black holes in laboratories. In this work, the gravitational bending of acoustic Schwarzschild black hole is investigated. We resort to the approach developed by Gibbons and Werner, in which the gravitational bending is calculated using the Gauss-Bonnet theorem in geometrical topology. In this approach, the gravitational bending is directly connected with the… Expand
1 Citations
#### Figures and Tables from this paper
Deriving Weak Deflection Angle by Black Holes or Wormholes using Gauss-Bonnet Theorem
• Physics
• TURKISH JOURNAL OF PHYSICS
• 2021
In this review, various researches on finding the bending angle of light deflected by a massive gravitating object which regard the Gauss-Bonnet theorem as the premise have been revised. Primarily,Expand
#### References
SHOWING 1-10 OF 65 REFERENCES
Note on acoustic black holes from black D3-brane
• Physics
• International Journal of Modern Physics D
• 2019
Black D3-branes are known to admit an effective hydrodynamic description when low frequency and long wavelength perturbations are introduced into the system. We use this perturbed nonextremal blackExpand
Deflection of light by rotating regular black holes using the Gauss-Bonnet theorem
• Physics
• Physical Review D
• 2018
In this paper, we study the weak gravitational lensing in the spacetime of rotating regular black hole geometries such as Ayon-Beato-Garc\'ia (ABG), Bardeen, and Hayward black holes. We calculate theExpand
Acoustic black holes in curved spacetime and the emergence of analogue Minkowski spacetime
• Physics
• Physical Review D
• 2019
Gravity is not only able to be mimicked in flat spacetimes, but also in curved spacetimes. We study analogue gravity models in curved spacetime by considering the relativistic Gross-Pitaevskii theoryExpand
Deflection angle of photon from magnetized black hole and effect of nonlinear electrodynamics
• Physics
• The European Physical Journal C
• 2019
In this paper, we analyze deflection angle of photon from magnetized black hole within non-linear electrodynamics with parameter $$\beta$$ . In doing so, we find the corresponding optical spacetimeExpand
Acoustic black holes for relativistic fluids
• Physics
• 2010
We derive a new acoustic black hole metric from the Abelian Higgs model. In the non-relativistic limit, while the Abelian Higgs model becomes the Ginzburg-Landau model, the metric reduces to anExpand
Gravitational bending angle of light for finite distance and the Gauss-Bonnet theorem
• Physics
• 2016
We discuss a possible extension of calculations of the bending angle of light in a static, spherically symmetric and asymptotically flat spacetime to a nonasymptotically flat case. We examine aExpand
Gravitational deflection of relativistic massive particles by wormholes
• Physics
• 2020
In this paper, the gravitational deflection of relativistic massive particles up to the second post-Minkowskian order by static and spherically symmetric wormholes is investigated in the weak-fieldExpand | 2021-11-27 00:02:08 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6533370614051819, "perplexity": 2220.7328369851193}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964358074.14/warc/CC-MAIN-20211126224056-20211127014056-00437.warc.gz"} |
http://openstudy.com/updates/556766fae4b050a18e82c2d1 | ## anonymous one year ago How do I find the length of an arc that subtends a central angle??
1. anonymous
@ganeshie8
2. anonymous
@Here_to_Help15
3. anonymous
The ratio of the arc's length to the circumference of the circle is the same as the ratio of the angle subtended by the arc to an entire revolution of the circle. In other words, $\frac{L}{2\pi r}=\frac{\theta}{2\pi}~~\implies~~L=r\theta$ where $$r$$ is the radius and $$\theta$$ is the central angle. | 2017-01-18 10:36:34 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9164093732833862, "perplexity": 130.4341241732561}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-04/segments/1484560280266.9/warc/CC-MAIN-20170116095120-00044-ip-10-171-10-70.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/how-to-prove-x-is-irrational-number.384075/ | # How to prove √X is irrational number
1. Mar 5, 2010
### SOHAWONG
when X is even number,it's easy to prove
but how about the condition which X is odd number?
I have no idea of this
2. Mar 5, 2010
### Hurkyl
Staff Emeritus
$\sqrt{4}$ is irrational?
3. Mar 5, 2010
4. Mar 5, 2010
### Char. Limit
So in other words...
$$\sqrt{x}$$ is irrational iff x=/=n^2 for n belonging to the integer set.
5. Mar 5, 2010
### SOHAWONG
yes, but how to prove?
6. Mar 6, 2010
### Tinyboss
Fundamental theorem of arithmetic. Assume p^2/q^2=x with gcd(p,q)=1, and see what has to divide what.
7. Mar 6, 2010
### SOHAWONG
what does gcd mean?
8. Mar 6, 2010
### Tinyboss
Greatest common divisor. If gcd(p,q)=1, it means the fraction p/q is in lowest terms.
Look at the proof for sqrt(2), and adapt it. Remember that "even" just means "is divisible by 2", so that if you're checking a number other than 2, you won't be thinking about "even" anymore.
9. Mar 6, 2010 | 2018-09-24 11:06:02 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5185443758964539, "perplexity": 10516.188665189547}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-39/segments/1537267160400.74/warc/CC-MAIN-20180924110050-20180924130450-00425.warc.gz"} |
https://www.gradesaver.com/textbooks/math/algebra/elementary-algebra/chapter-11-additional-topics-chapter-11-test-page-521/6 | ## Elementary Algebra
$5i\sqrt 3$
Recall, $i=\sqrt{-1}$. Thus, we obtain: $\sqrt {-75}$ $=\sqrt {-1\times3\times25}$ $=\sqrt {-1}\times \sqrt 3\times \sqrt {25}$ $=i\times \sqrt 3\times \sqrt {5^{2}}$ $=i\times \sqrt 3\times 5$ $=5i\sqrt 3$ | 2021-04-12 22:33:44 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5566425323486328, "perplexity": 432.2746758434706}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038069267.22/warc/CC-MAIN-20210412210312-20210413000312-00472.warc.gz"} |
https://leetcode.ca/2016-05-17-169-Majority-Element/ | # Question
Formatted question description: https://leetcode.ca/all/169.html
169. Majority Element
Given an array of size n, find the majority element. The majority element is the element that appears more than ⌊ n/2 ⌋ times.
You may assume that the array is non-empty and the majority element always exist in the array.
Example 1:
Input: [3,2,3]
Output: 3
Example 2:
Input: [2,2,1,1,1,2,2]
Output: 2
@tag-array
# Algorithm
First of all, there is a prerequisite, there must be a number that appears more than half of the number, then if the counter is reduced to 0, it means that the number of numbers that are not candidates currently is the same as the number of candidates, then this candidate It is already very weak, and it may not appear more than half. At this time, choose to replace the current candidate.
Or, after sorting, mid is always the majority.
Java | 2022-07-05 09:23:59 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5808831453323364, "perplexity": 830.4791482141309}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656104542759.82/warc/CC-MAIN-20220705083545-20220705113545-00113.warc.gz"} |
http://bib-pubdb1.desy.de/collection/PUB_OLYMP-20120731?ln=en&as=1 | # OLYMP
2017-06-1912:23 [PUBDB-2017-05437] Journal Article et al Composite bottlebrush mechanics: ?-internexin fine-tunes neurofilament network properties Soft matter 11(29), 5839 - 5849 (2015) [10.1039/C5SM00662G] Neuronal cytoplasmic intermediate filaments are principal structural and mechanical elements of the axon. Their expression during embryonic development follows a differential pattern, while their unregulated expression is correlated to neurodegenerative diseases. [...] OpenAccess: PDF PDF (PDFA); 2016-11-2414:11 [PUBDB-2016-05585] Preprint/Report et al Hard Two-Photon Contribution to Elastic Lepton-Proton Scattering Determined by the OLYMPUS Experiment [DESY-16-217; arXiv:1611.04685] The OLYMPUS collaboration reports on a precision measurement of the positron-proton to electron-proton elastic cross section ratio, $\it R_{2 \gamma}$, a direct measure of the contribution of hard two-photon exchange to the elastic cross section. In the OLYMPUS measurement, 2.01 GeV electron and positron beams were directed through a hydrogen gas target internal to the DORIS storage ring at DESY. [...] OpenAccess: 2016_paper Physical review letters - PDF PDF (PDFA); 1611.04685v1(1) - PDF PDF (PDFA); External link: Fulltext 2016-11-2411:53 [PUBDB-2016-05579] Report/Journal Article et al Hard Two-Photon Contribution to Elastic Lepton-Proton Scattering Determined by the OLYMPUS Experiment [DESY-16-217; arXiv:1611.04685] Physical review letters 118(9), 092501 (2017) [10.1103/PhysRevLett.118.092501] The OLYMPUS collaboration reports on a precision measurement of the positron-proton to electron-proton elastic cross section ratio, $\it R_{2 \gamma}$, a direct measure of the contribution of hard two-photon exchange to the elastic cross section. In the OLYMPUS measurement, 2.01 GeV electron and positron beams were directed through a hydrogen gas target internal to the DORIS storage ring at DESY [...] OpenAccess: PDF PDF (PDFA); External link: Fulltext 2016-08-1116:12 [PUBDB-2016-02987] Journal Article Kohl, M. The OLYMPUS Experiment at DESY INTERNATIONAL WORKSHOP ON POSITRONS AT JEFFERSON LAB, JPOS09, Meeting locationNewport News (Virginia), US, 25 Mar 2009 - 27 Mar 2009 AIP conference proceedings 1160, 19 (2009) [10.1063/1.3232027] Recent determinations of the proton electric to magnetic elastic form factor ratio from polarization transfer measurements at Jefferson Lab indicate an unexpected and dramatic discrepancy with the elastic form factor ratio obtained using the Rosenbluth separation technique in unpolarized cross section measurements. This discrepancy has been explained as the effect of two-photon exchange beyond the usual one-photon exchange approximation in the calculation of the elastic electron-proton scattering cross section. [...] OpenAccess: PDF PDF (PDFA); External link: Fulltext 2016-08-1115:37 [PUBDB-2016-02983] Journal Article/Contribution to a conference proceedings Kohl, M. EM Form Factors and OLYMPUS 13th International Workshop on Meson Production, Properties and Interaction, MESON2014, Meeting locationKRAKÓW, Poland, 29 May 2014 - 3 Jun 2014 The elastic form factors of the nucleon characterize the distributions of charge and magnetization in momentum space and are important input for calculations of strong interaction phenomena and nuclear structure. The dramatic discrepancy in the observed ratio of elastic proton form factors between the Rosenbluth separation and polarization transfer methods has invoked numerous theoretical and experimental investigations. [...] OpenAccess: PDF PDF (PDFA); External link: Fulltext 2016-08-1115:20 [PUBDB-2016-02981] Journal Article et al The OLYMPUS Internal Hydrogen Target An internal hydrogen target system was developed for the OLYMPUS experiment at DESY, in Hamburg, Germany. The target consisted of a long, thin-walled, tubular cell within an aluminum scattering chamber. [...] OpenAccess: PDF PDF (PDFA); Restricted: PDF PDF (PDFA); External link: Fulltext 2016-08-1114:56 [PUBDB-2016-02977] Journal Article et al Measurement and tricubic interpolation of the magnetic field for the OLYMPUS experiment The OLYMPUS experiment used a 0.3T toroidal magnetic spectrometer to measure the momenta of outgoing charged particles. In order to accurately determine particle trajectories, knowledge of the magnetic field was needed throughout the spectrometer volume. [...] OpenAccess: PDF PDF (PDFA); Restricted: PDF PDF (PDFA); External link: Fulltext 2016-08-1114:21 [PUBDB-2016-02975] Journal Article et al Design and Performance of a Lead Fluoride Detector as a Luminosity Monitor Precise luminosity measurements for the OLYMPUS two-photon exchange experiment at DESY were performed by counting scattering events with alternating beams of electrons and positrons incident on atomic electrons in a gaseous hydrogen target. Final products of M$\phi$ller, Bhabha, and pair annihilation interactions were observed using a pair of lead fluoride ($PbF_{2}$) Cherenkov calorimeters with custom housings and electronics, adapted from a system used by the A4 parity violation experiment at MAMI. [...] Published on 2016-04-20. Available in OpenAccess from 2017-04-20.: PDF; Restricted: PDF PDF (PDFA); External link: Fulltext 2016-08-1114:02 [PUBDB-2016-02974] Conference Presentation Schmidt, A. OLYMPUS Lepton-Nucleus Scattering XIV, ELBA XIV, Meeting locationIsola d'Elba, MIT, Italy, 27 Jun 2016 - 1 Jul 2016 OpenAccess: PDF PDF (PDFA); External link: Fulltext 2016-08-1113:54 [PUBDB-2016-02972] Poster Kohl, M. The OLYMPUS Experiment at DESY International conference on Precision Physics of Simple Atomic Systems, PSAS'2016, Meeting locationJerusalem, Israel, 22 May 2016 - 27 May 2016 OpenAccess: PDF PDF (PDFA); External link: Fulltext | 2017-12-12 12:19:12 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7375229001045227, "perplexity": 12566.131833148393}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948516843.8/warc/CC-MAIN-20171212114902-20171212134902-00222.warc.gz"} |
https://gmatclub.com/forum/the-rectangular-garden-represented-in-the-figure-above-with-dimension-250186.html | It is currently 18 Nov 2017, 09:06
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# The rectangular garden represented in the figure above, with dimension
Author Message
TAGS:
### Hide Tags
Math Expert
Joined: 02 Sep 2009
Posts: 42249
Kudos [?]: 132588 [0], given: 12326
The rectangular garden represented in the figure above, with dimension [#permalink]
### Show Tags
26 Sep 2017, 23:24
00:00
Difficulty:
(N/A)
Question Stats:
86% (00:30) correct 14% (01:08) wrong based on 14 sessions
### HideShow timer Statistics
The rectangular garden represented in the figure above, with dimensions x feet by y feet is surrounded by a walkway 2 feet wide. Which of the following represents the area of the
walkway, in square feet?
(A) 2x + 2y + 4
(B) 2x + 2y + 16
(C) 4x + 4y + 8
(D) 4x + 4y + 16
(E) 4x + 4y + 32
[Reveal] Spoiler:
Attachment:
2017-09-27_1015_002.png [ 961 Bytes | Viewed 320 times ]
[Reveal] Spoiler: OA
_________________
Kudos [?]: 132588 [0], given: 12326
BSchool Forum Moderator
Joined: 26 Feb 2016
Posts: 1596
Kudos [?]: 670 [0], given: 17
Location: India
WE: Sales (Retail)
The rectangular garden represented in the figure above, with dimension [#permalink]
### Show Tags
26 Sep 2017, 23:40
The outer rectangle has dimensions x+4 and y+4,
when the dimensions of the inner rectangle have been given to be x and y.
The area of the walkway, will be the difference of the area of the outer rectangle and the area of the inner rectangle.
Mathematically, this can be represented as follows,
$$(x+4)(y+4) - xy = xy + 4x + 4y + 16 - xy = 4x + 4y + 16$$
Therefore, 4x+4y+16(Option D) is the area of the walkway
_________________
Stay hungry, Stay foolish
Kudos [?]: 670 [0], given: 17
The rectangular garden represented in the figure above, with dimension [#permalink] 26 Sep 2017, 23:40
Display posts from previous: Sort by | 2017-11-18 16:06:26 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6190468668937683, "perplexity": 6537.529843004814}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-47/segments/1510934804976.22/warc/CC-MAIN-20171118151819-20171118171819-00614.warc.gz"} |
https://mathoverflow.net/questions/138523/is-there-a-proof-that-the-c-algebras-dont-see-the-invariant-subspace-prob/143223 | # Is there a proof that the $C^{*}$-algebras don't see the invariant subspace problem?
This post is an appendix of this one.
Let $H$ be an infinite dimensional separable Hilbert space and $B(H)$ the algebra of bounded operators.
Invariant subspace problem: Let $T \in B(H)$. Is there a non-trivial closed $T$-invariant subspace?
Hypothesis : The ISP admits a negative answer, i.e., there are ISP counter-examples.
Definition : A category $\mathcal{S}$ of operator algebras see the ISP if $\forall T, T' \in B(H)$ with $\mathcal{S}(T) \simeq \mathcal{S}(T')$: $$T \text{ is an ISP counter-example} \Leftrightarrow T' \text{ is an ISP counter-example }$$
Proposition: The category $W^{*}$ of von Neumann algebras, doesn't see the ISP.
proof: Under the previous hypothesis, let $T \in B(H)$ be an ISP counter-example. Then $T$ is irreducible, i.e., $W^{*}(T) = B(H)$. But there are many irreducible operators checking the ISP, for example, the unilateral shift $S$. So $W^{*}(T) \simeq W^{*}(S)$, $S$ checks the ISP and $T$ not. $\square$
This post asks about an equivalent result for the category of $C^{*}$-algebras :
Is there a proof that the category of $C^{*}$-algebras doesn't see the ISP ?
• Is $\mathcal{S}(T)$ the operator algebra generated by $T$? – Ulrich Pennig Aug 4 '13 at 12:57
• Yes, $\mathcal{S}(T)$ is the operator algebra (of category $\mathcal{S}$) generated by $T \in B(H)$. Just a precision, the $C^{∗}$-algebras and von Neumann algebras are here separable (the categories $C^{∗}$ and $W^{∗}$). If we can prove that $C^{∗}(T)$ is a Cuntz algebra (with $T\in B(H)$ an ISP counter-example), the result should follow. – Sebastien Palcoux Aug 4 '13 at 14:00
C*-algebras don't see the ISP. The operators $T\in B(H)$ and $T\oplus T\in B(H\oplus H)$ generate isomorphic C*-algebras, but the latter clearly has non-trivial invariant subspaces. To have both operators in the same Hilbert space, pick isometries $v_1,v_2\in B(H)$ with orthogonal ranges that add up to $H$. Then $$T\mapsto v_1Tv_1^*+v_2Tv_2^*$$ is an injective *-endomorphism of $B(H)$ that maps $T$ to an operator with non-trivial invariant subspaces.
• Thank you Leonel ! This map is continuous for the main topologies of operators algebras (norm-topology, weak-topology, strong-topology...), and this argument runs also without a $\star$-structure (on the algebra), so that it shows that no category of operator algebras see the ISP. Is it right ? – Sebastien Palcoux Sep 26 '13 at 13:09
• A generic way for tweaking a question, is to improve it by excluding the counter-examples. Here we can improve the definition of "see the ISP" by : $\forall T, T' \in B(H)$ with $\mathcal{S}(T) \simeq \mathcal{S}(T')$ and $T' \ne v_{1} T v^{*}_{1} + v_{2} T v^{*}_{2}$ (with $v_{1}$, $v_{2}$ as in your answer), then : "$T$ is an ISP counter-example" $\Leftrightarrow$ "$T'$ is an ISP counter-example". Is this what you thought? Else what do you suggest ? – Sebastien Palcoux Sep 27 '13 at 8:09 | 2020-12-02 03:32:18 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9459930062294006, "perplexity": 401.0662911212132}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141686635.62/warc/CC-MAIN-20201202021743-20201202051743-00229.warc.gz"} |
http://math.stackexchange.com/questions/203179/origin-of-the-notion-of-a-well-formed-formula | # Origin of the Notion of a Well-Formed Formula
When was the idea of a well-formed formula first stated or can get inferred as such under another name?
-
at en.wikipedia.org/wiki/Gottlob_Frege they give links to translations. I cannot be positive it was Frege, of course. – Will Jagy Sep 27 '12 at 1:33 | 2015-04-19 04:52:23 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.880774974822998, "perplexity": 1337.4957769278856}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-18/segments/1429246637445.19/warc/CC-MAIN-20150417045717-00284-ip-10-235-10-82.ec2.internal.warc.gz"} |
https://tel.archives-ouvertes.fr/tel-00985181 | # Structures de Poisson Logarithmiques : invariants cohomologiques et préquantification
Abstract : The main objective of this thesis is to propose a criteria of prequantization of singular Poisson structures with singularities carried by a free divisor of a fi nite dimensional complex manifold. For this, we start from an algebraic construction of formal logarithmic di fferentials along a fi nitely generated non trivial ideal of a commutative and unitary algebra. We introduce the concept of logarithmic Poisson algebra. Then, we show that these Poisson structures induce a new cohomological invariant, this is dow via the Lie-Rinehart algebra structure, that they induced on the module of formal logarithmic di fferentials. With the latter, we study the integrale conditions of such Poisson structures. First, we show that the Hamiltonian map of logarithmic Poisson structure extends to the module of formal logarithmic diff erential and induces a structure of Lie-Rinehart algebra on it. Furthermore, we show that its image is contained in the module of logarithmic derivations. We called logaruthmic Poisson cohomologie, the cohomologie induced by this representation. Subsequently, we show on some examples that Poisson cohomologies groups and Poisson logarithmic cohomologies groups are diff erent in general, although they coincide in the case of logsymplectic Poisson structures. We conclude with a study the prequantization conditions of all such structures by means of this cohomology.
Mots-clés :
Document type :
Theses
https://tel.archives-ouvertes.fr/tel-00985181
Contributor : Anne-Marie Plé <>
Submitted on : Tuesday, April 29, 2014 - 12:31:43 PM
Last modification on : Friday, May 10, 2019 - 12:14:02 PM
Long-term archiving on : Tuesday, July 29, 2014 - 12:15:29 PM
### Identifiers
• HAL Id : tel-00985181, version 1
### Citation
Joseph Dongho. Structures de Poisson Logarithmiques : invariants cohomologiques et préquantification. Analyse classique [math.CA]. Université d'Angers, 2012. Français. ⟨tel-00985181⟩
Record views | 2019-06-25 09:46:24 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.838259220123291, "perplexity": 2846.5895887312545}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560627999817.30/warc/CC-MAIN-20190625092324-20190625114324-00217.warc.gz"} |
https://nips.cc/Conferences/2020/ScheduleMultitrack?event=16910 | `
Timezone: »
Poster
Kernel Based Progressive Distillation for Adder Neural Networks
Yixing Xu · Chang Xu · Xinghao Chen · Wei Zhang · Chunjing XU · Yunhe Wang
Thu Dec 10 09:00 PM -- 11:00 PM (PST) @ Poster Session 6 #1891
Adder Neural Networks (ANNs) which only contain additions bring us a new way of developing deep neural networks with low energy consumption. Unfortunately, there is an accuracy drop when replacing all convolution filters by adder filters. The main reason here is the optimization difficulty of ANNs using $\ell_1$-norm, in which the estimation of gradient in back propagation is inaccurate. In this paper, we present a novel method for further improving the performance of ANNs without increasing the trainable parameters via a progressive kernel based knowledge distillation (PKKD) method. A convolutional neural network (CNN) with the same architecture is simultaneously initialized and trained as a teacher network, features and weights of ANN and CNN will be transformed to a new space to eliminate the accuracy drop. The similarity is conducted in a higher-dimensional space to disentangle the difference of their distributions using a kernel based method. Finally, the desired ANN is learned based on the information from both the ground-truth and teacher, progressively. The effectiveness of the proposed method for learning ANN with higher performance is then well-verified on several benchmarks. For instance, the ANN-50 trained using the proposed PKKD method obtains a 76.8\% top-1 accuracy on ImageNet dataset, which is 0.6\% higher than that of the ResNet-50. | 2021-09-22 08:01:59 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.38609468936920166, "perplexity": 1414.242719888872}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780057337.81/warc/CC-MAIN-20210922072047-20210922102047-00358.warc.gz"} |
http://mathoverflow.net/questions/84826/connected-parts-of-hyperbola | ## connected parts of hyperbola [closed]
I have the coordinates of an hyperbola how can I separate the two connected arcs?
-
mathoverflow.net/faq#whatnot – Yemon Choi Jan 3 2012 at 22:14 | 2013-05-22 16:53:18 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9620221257209778, "perplexity": 2514.7880198846924}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368702019913/warc/CC-MAIN-20130516110019-00095-ip-10-60-113-184.ec2.internal.warc.gz"} |
https://mathematica.stackexchange.com/questions/128593/i-want-to-use-mathematica-reference-pages-for-an-appendix-in-a-latex-report | # I want to use Mathematica Reference pages for an Appendix in a LaTeX report
I want to put the aforementioned information in an appendix(obviously I will Cite it) of a LaTeX report, however I get a variety of Errors when I attempt to do so.
I was trying the Module page, I've opened it in mathematica and tried copying certain information in LaTeX form and saving it as a LaTeX file but it doesn't work. I've also Tried ?Module in Mathematica and I have the same Issues.
Is there any way to do this?
Print it to PDF. Use the appropriate page size during printing, and in File -> Printing settings adjust the header and footer as necessary. Insert the PDF pages directly into your document. The pdfpages LaTeX package should make this possible. | 2020-02-19 03:19:00 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5809478759765625, "perplexity": 1354.62960340778}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875144027.33/warc/CC-MAIN-20200219030731-20200219060731-00030.warc.gz"} |
https://www.gamedev.net/forums/topic/397530-question-about-what-to-consider-when-making-c-wrapper-classes-for-win32/ | • 13
• 18
• 19
• 27
• 10
# Question about what to consider when making C++ wrapper classes for Win32
This topic is 4299 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.
## Recommended Posts
I have a question regarding C++ Wrapper classes. I have some experience with C and classes seem kind of odd to me. What things should I consider in order to create a wrapper, say for Win32 GUI code? Everytime I create a class Win32 code seems to not want to compile and complains about WINAPI and other Win32 objects not being defined. Also is it even possible to truly make a wrapper class since Win32 is older than dirt? Thank you in advance for any replies!
##### Share on other sites
Quote:
Also is it even possible to truly make a wrapper class since Win32 is older than dirt?
C is older than the Win32 API. What's your point?
Yes, it is entirely possible to write a wrapper around Win32 API functions. If you are getting errors, its because you are doing something wrong. Possibly missing an #include, not linking with the right libraries, or writing illegal C++ (dispite what you might think, C++ is quite a different language from C).
The other question is, why bother? There are plenty of existing wrappers or, quite frankly, better languages (C# + .NET) for GUI development.
##### Share on other sites
Well no need to be rude or a jerk. I am trying to better understand the methodology behind wrapping a class around the Win32 interface, and what things I need to understand to do it. I don't necessarily want to blindly use a class thats already been created because I am not going to understand why it works if I don't try. I find most tutorials "show you how to do it", they don't explain the "why's" regarding why you do it. To me, the class is very odd considering I have only done coding in languages that are non-OOP based and most C++ books do that lame Critter caretaker program which is just confusing as hell.
My point is that the Win32 is antique, it was mainly used for C programming and was not written with a perspective on OOP languages from what I can tell, so is it hard to wrap in a C++ class or nearly impossible to wrap the API completely? As I said, C++ is an enigma and I am trying to understand it since I can program somewhat decently with the Win32 API. I was hoping for rational info if anyone can provide some.
##### Share on other sites
If you don't understand OOP, I would reccommend using an OOP library before you try to write one. Perhaps using MFC or wxWidgets. Or better yet giving Java or C# a try.
##### Share on other sites
I'm just being blunt.
However, I'm confused. If the "concept" of a class seems foreign or confusing to you (it seems likes that's what you are saying), you aren't going to want to jump into something like trying to build a collection of classes for handling Windows GUI programming. The OO paradigm is much, much more than just using the "class" keyword in C++ instead of "struct," and sticking some member functions in your class.
The desire to want to understand how something works rather than just blindly using it is commendable. However, attempting reimplementation doesn't always lead to understanding, and can very often lead to the acquision of incorrect information, especially when you bite of more than you can chew.
You might want to start with something a little less ambitious than OO Win32 wrappers (which will be fraught with not only the perils of sane object-oriented design, but also the perils of Win32 and the perils of C++). But perhaps more ambitious than whatever this "critter caretaker" example you mentioned is (I have never heard of it before).
There are many reasons why the Win32 API is in C. Some good, some bad. Among them, however, is the fact that C is very, very good at interoperating with other languages (and its pretty easy to do).
##### Share on other sites
Learn the language you're going to use first. Wrapping Win32 to get basic functionality up should take 2-4 hours and it does go well with OO. It's all in how you design your classes. I'm not trying to be rude, I'm trying to help.
I was in your shoes a few years ago when I moved from C to C++. Get a good book on C++ and know that language inside and out (it's QUITE different then C compared to what most people think I find).
Start with a simpler project and work with the console to leave out the Windows issue to make life easier too. Walk before you run :)
##### Share on other sites
I don't like posting here because of the numerous ignorant/immature users like sgalland who are running on empty when it comes to the facts and are quick to accuse others of being rude or a "jerk." I find that anything you say that is contrary to their opinion and that isn't qualified with platitudes is automatically interpreted as a flame by these nitwits.
##### Share on other sites
Thank you for the responses, and sorry for the attitude, I read what I wrote and I sounded like a jerk. Anyhoo, what are some good C++ books, I have 3 books, C++ for game programming for beginners, C++ for dummies (worst c++ book written), and C++ in 21 days and frankly the C++ class section are all critter programs or classes that seem impossible to decipher. I for some reason just don't understand the class paradigm although they seem simple in theory, they seem to be a complex subject. Maybe I should just stick with C :)
##### Share on other sites
A class is a datatype, in the same way that a C struct is. You can create instances (also referred to as objects in OO parlance; C structs normally just have "instances" when spoken of by C programmers, if they even bother to distinguish concept of the type from the concept of the instantiations) of it in basically the same ways as in C. However, in C++ you can also provide member functions in addition to data members, 'protection' of members (public/private/protected access levels), inheritance (explained below), and some "auto-run" functionality (constructors and destructors): special functions that are run when an object of the class is instantiated and when it is cleaned up (either as a result of delete'ing an object allocated with new, delete[]'ing an array of objects allocated with new[], or because the object was stack-allocated and fell out of scope).
The 'class' keyword in C++ is syntactic sugar; a C++ compiler allows you to add all of these things to 'struct's as well. By typing 'class' instead, you default any inherited bases to inherit privately (you usually want public inheritance, so you have to then specify this), and members to be private (you usually want all or almost all data to be private, and most functions to be public). However, the term also carries semantic baggage: it is idiomatic to use 'struct' for mere collections of data (so typically they only have no member functions except for constructors and a destructor, and don't contain any pointers), and 'class' for types that provide behaviour and thus actually model something.
Inheritance allows for subtyping relationships: you can say that class X IS-A class Y, which is to say, an X object can do anything a Y object can. This is implemented by the compiler by adding all the Y data members to the X "object layout", almost as if you had a data member of the X type - except that you can call X functions on (and access X data members from) the Y directly, rather than having to access the X member. In exchange for this, you have somewhat less flexibility than you get with real "composition" (modelling a HAS-A relationship, by making an actual member of type X). It is quite unusual for it to be a good idea for X to both "be" and "have" a Y; and beware that when you define X, you should NOT redefine members of Y (unless you want *extra* members that happen to have the same name, and even then, things get difficult to manage).
Inheritance also facilitates polymorphism. I could describe this in some detail, but I think the best approach for you would be to read here and here (the latter is a fairly random Google hit that I selected because there's actual code that's fairly illustrative; you might want to just google for 'C "pointer cast polymorphism"' yourself) for a description of how similar functionality gets brutally hacked into C, and then here to see how C++ does the work for you (in a much safer way which is still as efficient as could sanely be expected).
Constructors and destructors are used for setup and cleanup of objects. This facilitates the very powerful C++ idiom of RAII (Resource Acquisition Is Initialization). The idea is that every object is responsible for its own "toys", and puts them away when it's done. The C++ standard library provides several classes that respect these rules. When everyone pitches in like that, it becomes easier for everyone - for example, if you want to make your own class do RAII properly, then it makes good sense to prefer standard library "container" objects over raw arrays or hand-rolled list/graph structures, because the constructors and destructors of the members will be called as well as those for your containing object, greatly reducing your workload. See more here and here.
Actually, hell, just read the whole C++ Faq Lite. It's quite worthwhile.
##### Share on other sites
Maybe a code example:
#include "windows.h"#define WIN32_LEAN_AND_MEANclass cWindowsGui{ private: int WINAPI WinMain(HINSTANCE hInstance, HINSTANCE hPrevInstance, LPSTR lCmdLine, int nCmdShow) ; public: cWindowsGui();};cWindowsGui::cWindowsGui(){ this.WinMain(HINSTANCE h, HINSTANCE p, LPSTR c, int cs);}int WINAPI cWindowsGui::WinMain(HINSTANCE hInstance, HINSTANCE hPrevInstance, LPSTR lCmdLine, int nCmdShow){ MessageBox(NULL,"Test Message","Test",MB_OK);}
When I use this code it tells me that WINAPI, LPSTR, and other Windows.h defined variables are invalid once I initialise them. The problem isn't that I have a book, I guess its that in my 3 C++ books they all have a lame critter example that just doesn't show me what I would call a real world class. I know that Java works differnetly about C++, but this is the only thing I can come up with to write self contained code that follows the OOP paradigm since Java is the only OOP language that I know. I have heard that C++ is not a true OOP from many people and I don't know if this is the cause of the issue, nor the fact I am using MinGW as my compiler, but from what I have seen (and that's not saying much) it should work. As I keep stating, how would I do something like this in the real world to understand C++ classes better? | 2018-03-19 13:09:41 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3003985583782196, "perplexity": 1588.0170436005037}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-13/segments/1521257646914.32/warc/CC-MAIN-20180319120712-20180319140712-00137.warc.gz"} |
https://cseducators.stackexchange.com/questions/1463/unit-review-for-boolean-algebra | # Background:
This is the first unit in junior year for a course in theoretical computer science. Prior to this, the students have already had AP CS A and a mish-mash of other topics. They have, by this point, studied:
1. Python (as their introduction to programming)
2. Java for AP CS A (loops, arrays, object design, recursion, polymorphism)
3. 6502 Assembly (for stack operation, including recursion again)
4. C (largely for pointers)
5. Additional things in Java (e.g. trees, linked lists, tries, Hashmaps, generics) and some algorithms (such as BFS, DFS, Huffman Encoding, Conway's Game of Life, Prim's, etc)
This course, then, is a year of theoretical computer science. The first unit is background mathematics that they will need during the rest of the course, and it focuses on boolean algebra and sets. The requested Unit Review here is for my boolean algebra opening.
The general goals are to help the students gain serious familiarity with algebraic manipulations and with the symbols themselves. After these lessons, we will move onto Conjunctive and Disjunctive Normal Forms, fairly substantial algebraic manipulations involving 5-10 steps, and then into logic gates for a gentle introduction to computer circuitry, so it is important that they come out prepared for that.
Word of warning: The unit below has some serious weaknesses, as you will see. Please be gentle! This year will be my second time giving these lessons, and while I have made a series of improvements from what I did this last September, there is still a long way to go. There is a lot of lecture, and I would really like to make it more engaging when I do it again27.
# The Curriculum
Lesson one:
We begin our foray into boolean algebra by converting our Java boolean symbols to formal math symbology: && to $\land$, || to $\lor$, and ! to $\lnot$. We then discuss how != is essentially $\oplus$, and begin our discussion of $\Longleftrightarrow$ as being a near approximation of == (with a promise that we will come back to this again, because we are not done with it yet).
We then introduce tautologies (such as $B \lor \lnot B$), and go over their truth tables. I finally provide them with an extremely short homework worksheet where they have to circle the boolean statements that are tautologies. (It also provides practice with all of the symbols we have discussed)
Lesson two:
We review lesson 1, and then introduce the concept of contradictions as the opposite of tautologies, go over their truth tables, and look at the obvious contradiction $B \land \lnot B$.
I ask them to create two tautologies with a partner, the more creative the better, walk around, and ask for a number of students to write interesting ones on the board.
We move on to $A \Rightarrow B$, discuss its meaning and truth table, and then ask what in Java is similar? I first propose:
if (A)
B = true;
We talk about this for awhile, and talk about why, though it may seem like a reasonable choice at first, it is not really the same idea. We then spend some time on the hardest idea of implies: what we mean when we say that $False \Rightarrow True$ is $True$.
At this point, I take a break to introduce the way I approach proofs in this class. Over the course of the year, certain very important proofs must be reproduced by a student chosen at random during the following period. This forces the students not to ignore these proofs, and helps them to internalize both the mathematical symbols that I need them to gain fluency with, and some very clever proof techniques.
I spend a substantial chunk of time then doing a formal proof that $False \Rightarrow True \equiv True$. I finish the class with a second, informal, and intuitive proof of the same (which they will not need to reproduce.)
Lesson 3:
We review from the prior day, and a student is called up to reproduce the proof.
We then spend about 15 minutes going over necessary and sufficient in depth, and re-examine both $\Longleftrightarrow$ and $\Rightarrow$ through this new lense. During this process, I introduce $\Longleftrightarrow$ more properly as $(A \Rightarrow B) \land (A \Leftarrow B)$ and as iff, and we talk about the similarities that $\Longleftrightarrow$ shares with $\equiv$.
I go over DeMorgan's Laws, and briefly cover 5 more symbols: ∀, ∃, |, ∈, and ∴.
At this point, I give them English statements to translate to mathematical statements, and mathematical statements to translate to English statements, one at a time. We use the following format: I provide an exercise, they work on it with a buddy or on their own while I bounce around the room offering help, and then I go over how we do it. The final exercise is to translate Goldbach's Conjecture (which I do not identify to them until after the translation work is done) from boolean symbology into English:
$\forall n|n \gt 2 \wedge (\frac{n}{2}) \in \mathbb{N}$,
$\exists (m,ℓ)\,\,|\,\, m \in P \wedge ℓ \in P \wedge m+ℓ=n$
As they leave, I give them a homework assignment with practice problems in very simple 1- or 2-step boolean algebraic reductions (including applying DeMorgan's Law), translating statements from English back and forth into symbols, creating truth tables from algebraic statements, and creating algebraic statements from truth tables.
# The Request
This is my opener for the year, and it is so. very. dry. There is also very little activity! I am already aware that I've got a real snoozer here.
This material must to come first in the year for large-scale organizational reasons, but I would really like ideas to make it more engaging. I am also particularly seeking out ideas for active learning and ways to utilize pair partners to improve both engagement and mastery.
I don't know if I have a solid answer to the main thrust of your question, but I do have some one-off suggestions that may or may not be helpful.
First, something you could try integrating into your lessons is some kind of meta-narrative about why proofs are useful in the first place. This seems to be a very common concern from students in the discrete maths course that I help teach (especially since they've been mostly programming up to that point) -- this post made earlier might be applicable (though idk if I fully agree with all of the answers).
The second thing you could try doing is squashing your first and second lessons into one. If your students are already familiar with that many programming languages, I think they'll also be pretty familiar with boolean symbols, and won't find truth tables terribly hard (or interesting).
It might then be more fruitful to cover tautologies, contradictions, and contingencies around the same time -- might as well cover all of the related definitions at once.
I think focusing on implications for some time would be useful, though. That seems to be a very challenging concepts for students to fully grasp -- see How to teach logical implication for more details. Making students do challenging exercises that the Wason selection task to see if they really grok implication may also help.
You mentioned you briefly cover first-order logic (symbols like ∀, ∃, |, ∈, and ∴). I think those merit more time -- those also seem to be challenging concepts for students to grasp (in particular, what they mean, how variable scoping works, how to correctly use rules like ∀-introduction, ∀-elimination, and the like). One good challenge question for your students is to ask them whether or not $\forall x \, \exists \, y P(x, y)$ and $\exists y \, \forall x \, P(x, y)$ have same or differing truth values.
I'd also save teaching what ∈ for when you formally introduce sets.
Finally, one way to perhaps make your lessons more exciting is to introduce homework/introduce problems that involve solving logic puzzles, particularly ones that are challenging to reason through normally. Have your students start by converting the constraints into logical propositions then have them simplify them until they're able to find an answer. This lets them practice their English to logic skills and their logic manipulation skills in a hopefully more fun setting.
If you want, you can even take this one step further and turn it into a programming assignment by asking them to generate constraints they then feed into a SAT solver like z3. (For example, take sudoku -- you can make each cell a variable, define 9 predicates that encode whether or not a cell contains a number or not, then define a whole bunch of implications encoding relationships between relevant pairs of cells). This lets them see more concretely how propositional logic is relevant to computer science, and also lets them play around with a cool piece of tech.
I've found a lot of success giving real world (often times very silly) examples of boolean algebra to give them a more intuitive understanding in addition to the pure algebraic laws. An example would be "If it rains tomorrow, I will bring an umbrella so I will stay dry".
This is a simple A -> B:
If it rains tomorrow then I will bring an umbrella, I will stay dry (T -> T = T)
If it rains tomorrow then I will not bring an umbrella, I will not stay dry (T -> F = F)
If it does not rain tomorrow then I will bring an umbrella, I will stay dry (F -> T = T)
If it does not rain tomorrow then I will not bring an umbrella, I will stay dry (F -> F = T)
Using DeMorgan's we know A -> B = !A V B. We can say A = it will rain tomorrow, B = bringing an umbrella and whether you stay dry or not is the equivalent of the resulting truth table value.
You can incorporate students in coming up with these silly examples, and having them figure out how the narrative would look like to reflect the truth table values. In addition, pairs can come up with scenarios and test each other's knowledge. (This was during 2nd year University too! So it's never too old to get silly)
Lastly as a remark, I did not see you mention some Laws of Boolean Algebra such as Associative, Commutative, Idempotent, Identity, and Distributive. I think it's worth while to introduce these laws during lesson 1 or 2 because solving boolean algebra down the road is built off of these fundamentals.
• Wikipedia has a very nice page showing all of the boolean operations. Then there is my personal favorite page of colorful... analogies: Allen's Interval Algebra. It's beautiful – user737 Jul 7 '17 at 13:43
• Thank you! With regards to the other laws, I cover those in my next mini-unit (CNF and DNF), so they're basically the next thing. Should I mention that in my question? – Ben I. Jul 7 '17 at 14:52
• @BenI. I don't think you need to mention it in your question, I was just being thorough. I think it's not too relevant with respect to the question asked :) – Kaneki Jul 7 '17 at 14:57
A couple of ideas:
1st day
Start by getting everybody up and moving. Have people form sets - say, a set of colors, with everybody wearing a different color, or a set of ages - but remind them that they can't have two people representing the same color in the set. Perform different operations with the human sets - intersections, unions, etc. After a couple of exercises like this, show them the formal notation for sets and match it to what exactly they were doing (i.e., when you created a new set of only the people that were in both sets, you were performing an operation called an intersection, which you can notate as $A\cap B$, etc).
Hopefully you can get far enough into set theory such that you can reach DeMorgan's Law. Another good thing to introduce would be proofs - you could, for example, do the proof of DeMorgan's Law, or a simpler topic in set theory. As you continue with set theory and set theory notation, let them experiment with Python's set type - maybe your exercises could have them solve simple set problems on paper and with Python, and also a little work with proofs.
2nd day
This day is talking about truth tables, logic gates, and boolean symbols. A couple of ideas for making this part less dry:
• Include some information about George Boole and why his work was such a success - I particularly like Martin Gardner's Logic Machines and Diagrams; it provides some entertaining history.
• Put logic gates (and all the notation that goes with them) in the context of the rest of the computer - I enjoy Crash Course's Computer Science series of videos; the relevant one can be found here.
After the history and the introduction of logic gates, perhaps let them mess around with logic gates trying to create their own "circuits" that do something. (If you wish, let them test it out with some transistors, batteries, LEDs, or whatever.) Obviously, this is hard, so as they start complaining about difficulty, introduce the notation as an easier way to keep track of it. For example, let them sketch out truth tables for different gates, (perhaps testing them). At this stage you can also show them how the symbols they are using connect to the notation in Java/Python. The exercises probably should get them fairly familiar with the notation and truth tables.
3rd day
This is where everything gets put together. The first part of the class could be devoted to putting English language and notation together, starting with an introduction of the notation, and then an activity with partners contributing phrases to each other to translate, or picking a phrase from a movie for the class to translate (easier phrases preferred, obviously). Cover tautologies and contradictions, and advance through to translating Goldbach's conjecture.
The second part of class would then be devoted to the one thing left in your plan of the first three days - the formal proof that $False\Rightarrow True\equiv True$. Since you've already done a formal proof, ask for suggestions, let students complete simple half-finished statements in the proof, and so on. The exercises would basically be the same as your original third-day exercises, along with some tautology practice, perhaps. | 2019-02-19 08:39:14 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5549652576446533, "perplexity": 767.0855932877892}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-09/segments/1550247489729.11/warc/CC-MAIN-20190219081639-20190219103639-00469.warc.gz"} |
https://www.cheenta.com/area-of-isosceles-triangle-amc-8-2005-problem-23/ | Select Page
Try this beautiful problem from AMC-8, 2005, Problem-23 based on the area of an isosceles triangle.
## Area of the Triangle- AMC- 8, 2005 – Problem 23
Isosceles right triangle ABC encloses a semicircle of area $2\pi$ . The circle has its center O on hypotenuse AB and is tangent to sides AC and BC . What is the area of triangle ABC ?
• $6$
• $8$
• $10$
### Key Concepts
Geometry
Triangle
Semi-circle
But try the problem first…
Answer:$8$
Source
AMC-8 (2005) Problem 23
Pre College Mathematics
## Try with Hints
First hint
Join Oand D
Can you now finish the problem ……….
Second Hint
This is an isosceles right triangle, the center is the midpoint of the hypotenuse. Radii drawn to the tangent points of the semicircle and the radii also divide the legs into two equal segments
can you finish the problem……..
Final Step
Given that AB and AC are the two sides of the Isosceles right triangle ABC and encloses a semicircle of area $2\pi$, center of the semicircle is O.
Let OD=r be the radius of the semi-circle
then area of semi-circle be $\frac{\pi r^2}{2}$
Now $\frac{\pi r^2}{2}$ = $2\pi$
$\Rightarrow r^2=4$
$\Rightarrow r=2$
this is an isosceles right triangle, the center is the midpoint of the hypotenuse. Radii drawn to the tangent points of the semicircle and the radii also divide the legs into two equal segments
They also create a square in the top left corner. From this, we can conclude the legs of the triangle are twice the length of the radii, 4
The area of the triangle is $\frac{1}{2} \times 4 \times 4$=8 | 2020-09-23 22:47:19 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7103619575500488, "perplexity": 738.25381667891}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600400212959.12/warc/CC-MAIN-20200923211300-20200924001300-00256.warc.gz"} |
https://labs.tib.eu/arxiv/?author=V.%20Vovchenko | • ### Puzzles in statistical hadron-gas treatment of systems created in proton-proton interactions at CERN SPS(1805.01901)
May 4, 2018 hep-ph, hep-ex, nucl-ex, nucl-th
We analyze the newest data from the NA61/SHINE collaboration which, in addition to previous results on pions and kaons, include mean multiplicities of $p$, $\Lambda$, and $\phi$-mesons produced in inelastic proton-proton (p+p) interactions at $\sqrt{s_{NN}}=6.3-17.3$~GeV. The canonical ensemble formulation of the ideal hadron resonance gas (HRG) model is used with exact conservation of net baryon number $B=2$, electric charge $Q=2$, and strangeness $S=0$. The chemical freeze-out parameters in p+p interactions are obtained and compared to those in central nucleus-nucleus collisions. We find several serious problems of the statistical model treatment of p+p interactions at the CERN SPS: 1) the inclusion of the $\phi$-meson yields in thermal fits worsens significantly the fit quality ({\it the $\phi$-meson puzzle}); 2) the data shows large event-by-event multiplicity fluctuations in inelastic p+p interactions which cannot be explained within a statistical model, and provide evidence for {\it large non-statistical fluctuations}; 3) the fits within the canonical ensemble formulation of HRG do not give any improvement over the fits within the grand canonical ensemble formulation, i.e., there are {\it no indications for existence of a single statistical system in} p+p interactions.
• ### Flavor-dependent eigenvolume interactions in a hadron resonance gas(1606.06542)
March 22, 2018 hep-ph, nucl-th
Eigenvolume effects in the hadron resonance gas (HRG) model are studied for experimental hadronic yields in nucleus-nucleus collisions. If particle eigenvolumes are different for different hadron species, the excluded volume HRG (EV-HRG) improves fits to multiplicity data. In particular, using different mass~-~volume relations for strange and non-strange hadrons we observe a remarkable improvement in the quality of the fits. This effect appears to be rather insensitive to other details in the schemes employed in the EV-HRG. We show that the parameters found from fitting the data of the ALICE Collaboration in central Pb+Pb collisions at the collision energy $\sqrt{s_{\rm NN}} = 2.76$~TeV entail the same improvement for all centralities at the same collision energy, and for the RHIC and SPS data at lower collision energies. Our findings are put in the context of recent fits of lattice QCD results.
• ### Quantum van der Waals and Walecka models of nuclear matter(1708.05605)
Oct. 12, 2017 nucl-th
A comparable study of the quantum van der Waals and Walecka models of nuclear matter is presented. Each model contains two parameters which characterize the repulsive and attractive interactions between nucleons. These parameters are fixed in order to reproduce the known properties of the nuclear ground state. Both models predict a first-order liquid-gas phase transition and a very similar behavior in the vicinity of the critical point. Critical exponents of the quantum van der Waals model are studied both analytically and numerically. There are important differences in the behavior of the thermodynamical functions of the considered models at large values of the nucleon number density. At the same time both models fall into the universality class of mean-field theory.
• ### Bose-Einstein condensation and liquid-gas phase transition in alpha-matter(1704.08039)
April 26, 2017 nucl-th
Systems of Bose particles with both repulsive and attractive interactions are studied using the Skyrme-like mean-field model. The phase diagram of such systems exhibits two special lines in the chemical potential-temperature plane: one line which represents the first-order liquid-gas phase transition with the critical end point, and another line which represents the onset of Bose-Einstein condensation. The calculations are made for strongly-interacting matter composed of alpha particles. The phase diagram of this matter is qualitatively similar to that observed for the atomic He4 liquid. The sensitivity of the results to the model parameters is studied. For weak interaction coupling the critical point is located at the Bose-condensation line.
• Substantial experimental and theoretical efforts worldwide are devoted to explore the phase diagram of strongly interacting matter. At LHC and top RHIC energies, QCD matter is studied at very high temperatures and nearly vanishing net-baryon densities. There is evidence that a Quark-Gluon-Plasma (QGP) was created at experiments at RHIC and LHC. The transition from the QGP back to the hadron gas is found to be a smooth cross over. For larger net-baryon densities and lower temperatures, it is expected that the QCD phase diagram exhibits a rich structure, such as a first-order phase transition between hadronic and partonic matter which terminates in a critical point, or exotic phases like quarkyonic matter. The discovery of these landmarks would be a breakthrough in our understanding of the strong interaction and is therefore in the focus of various high-energy heavy-ion research programs. The Compressed Baryonic Matter (CBM) experiment at FAIR will play a unique role in the exploration of the QCD phase diagram in the region of high net-baryon densities, because it is designed to run at unprecedented interaction rates. High-rate operation is the key prerequisite for high-precision measurements of multi-differential observables and of rare diagnostic probes which are sensitive to the dense phase of the nuclear fireball. The goal of the CBM experiment at SIS100 (sqrt(s_NN) = 2.7 - 4.9 GeV) is to discover fundamental properties of QCD matter: the phase structure at large baryon-chemical potentials (mu_B > 500 MeV), effects of chiral symmetry, and the equation-of-state at high density as it is expected to occur in the core of neutron stars. In this article, we review the motivation for and the physics programme of CBM, including activities before the start of data taking in 2022, in the context of the worldwide efforts to explore high-density QCD matter.
• ### New scenarios for hard-core interactions in a hadron resonance gas(1610.08753)
Oct. 27, 2016 hep-ph, nucl-th
The equation of state of a baryon-symmetric hadronic matter with hard-sphere interactions is studied. It is assumed that mesons are point-like, but baryons and antibaryons have the same hard-core radius rB. Three possibilities are considered: 1) the baryon-baryon and antibaryon-baryon interactions are the same; 2) baryons do not interact with antibaryons; 3) the baryon-antibaryon and meson-(anti)baryon interactions are negligible. By choosing the parameter rB=0.3-0.6 fm, we calculate the nucleon to pion ratio as a function of temperature and perform the fit of hadron yields measured in central Pb+Pb collisions at the bombarding energy Ecm=2.76 TeV per nucleon pair. New nontrivial effects in the interacting hadron resonance gas at temperatures 150-200 MeV are found.
• ### Critical fluctuations in models with van der Waals interactions(1610.01036)
Oct. 4, 2016 hep-ph, nucl-th
Particle number fluctuations are considered within the van der Waals (VDW) equation, which contains both attractive (mean-field) and repulsive (eigenvolume) interactions. The VDW equation is used to calculate the scaled variance of particle number fluctuations in generic Boltzmann VDW system and in nuclear matter. The strongly intensive measures $\Delta[E^*,N]$ and $\Sigma[E^*,N]$ of the particle number and excitation energy fluctuations are also considered, and, similarly, show singular behavior near the critical point. The $\Delta[E^*,N]$ measure is shown to attain both positive and negative values in the vicinity of critical point. Based on universality argument, similar behavior is expected to occur in the vicinity of the QCD critical point.
• ### Updates to the p+p and A+A chemical freeze-out lines from the new experimental data(1609.04827)
Sept. 15, 2016 hep-ph, hep-ex, nucl-ex, nucl-th
We show that the new data on mean multiplicities measured in p+p and A+A collisions together with the updated list of resonances lead to the significant changes of the obtained freeze-out lines. The new A+A line gives much smaller temperatures at high collision energies and agrees with the values obtained at the LHC. The newly obtained p+p line is much closer to the A+A line than previously expected, and even touches it in the region where the $K^+/\pi^+$ horn appears in the data. It indicates that the temperatures that will be obtained in the beam energy and system size scan by the NA61/SHINE Collaboration might be very close. However, our analysis shows that the chemical potentials could be very different for the same energies in A+A and p+p. It adds more puzzles to the set of surprising coincidences at the energies close to the possible onset of deconfinement.
• ### Limiting temperature of pion gas with the van der Waals equation of state(1508.04585)
Aug. 24, 2016 nucl-th
The grand canonical ensemble formulation of the van der Waals equation of state that includes the effects of Bose statistics is applied to an equilibrium system of interacting pions. If the attractive interaction between pions is large enough, a limiting temperature $T_0$ emerges, i.e., no thermodynamical equilibrium is possible at $T>T_0$. The system pressure $p$, particle number density $n$, and energy density $\varepsilon$ remain finite at $T=T_0$, whereas for $T$ near $T_0$ both the specific heat $C=d\varepsilon/dT$ and the scaled variance of particle number fluctuations $\omega[N]$ are proportional to $(T_0-T)^{-1/2}$ and, thus, go to infinity at $T\rightarrow T_0$. The limiting temperature corresponds also to the softest point of the equation of state, i.e., the speed of sound squared $c_s^2=dp/d\varepsilon$ goes to zero as $(T_0-T)^{1/2}$. Very similar thermodynamical behavior takes place in the Hagedorn model for the special choice of a power, namely $m^{-4}$, in the pre-exponential factor of the mass spectrum $\rho(m)$.
• ### Hydrodynamic modeling of a pure-glue initial scenario in high-energy hadron and heavy-ion collisions(1608.04318)
Aug. 15, 2016 hep-ph, nucl-th
Partonic matter produced in the early stage of ultrarelativistic nucleus-nucleus collisions is assumed to be composed mainly of gluons, and quarks and antiquarks are produced at later times. The comparable hydrodynamic simulations of heavy-ion collisions for (2+1)-flavor and Yang-Mills equations of state performed by using three different hydrodynamic codes are presented. Assuming slow chemical equilibration of quarks, the spectra and elliptic flows of thermal dileptons and photons are calculated for central Pb+Pb collisions at the LHC energy of $\sqrt{s_{_{\rm NN}}} = 2.76$ TeV. It is shown that a suppression of quarks at early times leads to a significant reduction of the yield of the thermal dileptons, but only to a rather modest suppression of the $p_T$-distribution of direct photons. It is demonstrated that an enhancement of photon and dilepton elliptic flows might serve as a promising signature of the pure-glue initial state. Calculations based on Bjorken hydrodynamics suggest that collisions of small systems at intermediate energies available at RHIC or future FAIR facilities may show stronger effects associated with initial pure gluodynamic evolution.
• ### Electromagnetic probes of a pure-glue initial state in nucleus-nucleus collisions at energies available at the CERN Large Hadron Collider(1604.06346)
Aug. 15, 2016 hep-ph, nucl-th
Partonic matter produced in the early stage of ultrarelativistic nucleus-nucleus collisions is assumed to be composed mainly of gluons, and quarks and antiquarks are produced at later times. To study the implications of such a scenario, the dynamical evolution of a chemically nonequilibrated system is described by the ideal (2+1)-dimensional hydrodynamics with a time dependent (anti)quark fugacity. The equation of state interpolates linearly between the lattice data for the pure gluonic matter and the lattice data for the chemically equilibrated quark-gluon plasma. The spectra and elliptic flows of thermal dileptons and photons are calculated for central Pb+Pb collisions at the CERN Large Hadron Collider energy of $\sqrt{s_{_{\rm NN}}} = 2.76$ TeV. We test the sensitivity of the results to the choice of equilibration times, including also the case where the complete chemical equilibrium of partons is reached already at the initial stage. It is shown that a suppression of quarks at early times leads to a significant reduction of the yield of the thermal dileptons, but only to a rather modest suppression of the $p_T$-distribution of direct photons. It is demonstrated that an enhancement of photon and dilepton elliptic flows might serve as a promising signature of the pure-glue initial state.
• ### Conserved charge fluctuations are not conserved during the hadronic phase(1608.03737)
Aug. 12, 2016 hep-ph, nucl-th
We study the correlation between the distributions of the net-charge, net-kaon, net-baryon and net-proton number at hadronization and after the final hadronic decoupling by simulating ultra relativistic heavy ion collisions with the hybrid version of the ultrarelativistic quantum molecular dynamics (UrQMD) model. We find that due to the hadronic rescattering these distributions are not strongly correlated. The calculated change of the correlation, during the hadronic expansion stage, does not support the recent paradigm, namely that the measured final moments of the experimentally observed distributions do give directly the values of those distributions at earlier times, when the system had been closer to the QCD crossover.
• ### Chemical freeze-out conditions in hadron resonance gas(1606.06350)
June 20, 2016 hep-ph
The hadron resonance gas model with hadron-type dependent eigenvolume corrections is employed to fit the hadron yield data of the NA49 collaboration for central Pb+Pb collisions at the c.m. energy per nucleon pair Ecm=6.3, 7.6, 8.8, 12.3, and 17.3 GeV, the hadron midrapidity yield data of the STAR collaboration for Au+Au collisions at Ecm=200 GeV, and the hadron midrapidity yield data of the ALICE collaboration for Pb+Pb collisions at Ecm=2760 GeV. The influence of the eigenvolume corrections is studied.
• ### Hadron multiplicities and chemical freeze-out conditions in proton-proton and nucleus-nucleus collisions(1512.08025)
June 20, 2016 hep-ph, nucl-th
New results of the NA61/SHINE Collaboration at the CERN SPS on mean hadron multiplicities in proton-proton (p+p) interactions are analyzed within the transport models and the hadron resonance gas (HRG) statistical model. The chemical freeze-out parameters in p+p interactions and central Pb+Pb (or Au+Au) collisions are found and compared with each other in the range of the center of mass energy of the nucleon pair $\sqrt{s_{NN}}=3.2-17.3$ GeV. The canonical ensemble formulation of the HRG model is used to describe mean hadron multiplicities in p+p interactions and the grand canonical ensemble in central Pb+Pb and Au+Au collisions. The chemical freeze-out temperatures in p+p interactions are found to be larger than the corresponding temperatures in central nucleus-nucleus collisions.
• ### Entropy production in chemically non-equilibrium quark-gluon plasma created in central Pb+Pb collisions at LHC energies(1510.01235)
Jan. 20, 2016 hep-ph, nucl-th
We study the possibility that partonic matter produced at early stage of ultrarelativistic heavy-ion collisions is out of chemical equilibrium. It is assumed that initially this matter is mostly composed of gluons, but quarks and antiquarks are produced at later times. The dynamical evolution of partonic system is described by the Bjorken-like ideal hydrodynamics with a time dependent quark fugacity. The results of this model are compared with those obtained by assuming the complete chemical equilibrium of partons already at the initial stage. It is shown that in a chemically non-equilibrium scenario the entropy gradually increases, and about 25% of the total final entropy is generated during the hydrodynamic evolution of deconfined matter. We argue that the (anti)quark suppression included in this approach may be responsible for reduced (anti)baryon to meson ratios observed in heavy-ion collisions at LHC energies.
• ### Non-Gaussian particle number fluctuations in vicinity of the critical point for van der Waals equation of state(1507.06537)
Dec. 21, 2015 nucl-th
The non-Gaussian measures of the particle number fluctuations -- skewness $S\sigma$ and kurtosis $\kappa \sigma^2$ -- are calculated in a vicinity of the critical point. This point corresponds to the end point of the first-order liquid-gas phase transition. The gaseous phase is characterized by the positive values of skewness while the liquid phase has negative skew. The kurtosis appears to be significantly negative at the critical density and supercritical temperatures. The skewness and kurtosis diverge at the critical point. The classical van der Waals equation of state in the grand canonical ensemble formulation is used in our studies. Neglecting effects of the quantum statistics we succeed to obtain the analytical expressions for the rich structures of the skewness and kurtosis in a wide region around the critical point. These results have universal form, i.e., they do not depend on particular values of the van der Waals parameters $a$ and $b$. The strongly intensive measures of particle number and energy fluctuations are also considered and show singular behavior in the vicinity of the critical point.
• ### Undersaturation of quarks at early stages of relativistic nuclear collisions: the hot glue initial scenario and its observable signatures(1509.07682)
Nov. 17, 2015 hep-ph, hep-ex, nucl-ex, nucl-th, hep-lat
The early stage of high multiplicity nuclear collisions is represented by a nearly quarkless, hot, deconfined pure gluon plasma. This new scenario should be characterized by a suppression of high $p_T$ photons and dileptons as well as by reduced baryon to meson ratios. We present the numerical results for central Pb+Pb collisions at the LHC energies by using the ideal Bjorken hydrodynamics with time-dependent quark fugacity. It is shown that about 25\% of final total entropy is generated during the hydrodynamic evolution of chemically undersaturated quark-gluon plasma.
• ### Scaled variance, skewness, and kurtosis near the critical point of nuclear matter(1506.05763)
Nov. 4, 2015 nucl-th
The van der Waals (VDW) equation of state predicts the existence of a first-order liquid-gas phase transition and contains a critical point. The VDW equation with Fermi statistics is applied to a description of the nuclear matter. The nucleon number fluctuations near the critical point of nuclear matter are studied. The scaled variance, skewness, and kurtosis diverge at the critical point. It is found that the crossover region of the phase diagram is characterized by the large values of the scaled variance, the almost zero skewness, and the significantly negative kurtosis. The rich structures of the skewness and kurtosis are observed in the phase diagram in the wide region around the critical point, namely, they both may attain large positive or negative values.
• ### Mean-field approach in the multi-component gas of interacting particles applied to relativistic heavy-ion collisions(1411.1444)
Aug. 13, 2015 nucl-th
Generalized mean-field approach for thermodynamic description of relativistic single- and multi-component gas in the grand canonical ensemble is formulated. In the framework of the proposed approach different phenomenological excluded-volume procedures are presented and compared to the existing ones. The mean-field approach is then used to effectively include hard-core repulsion in hadron-resonance gas model for description of chemical freeze-out in heavy-ion collisions. We calculate the collision energy dependence of several quantities for different values of hard-core hadron radius and for different excluded-volume procedures such as van der Waals and Carnahan-Starling models. It is shown that a choice of the excluded-volume model becomes important for large particle densities. For large enough values of hadron radii ($r\gtrsim0.9$ fm) there can be a sizable difference between different excluded-volume procedures used to describe the chemical freeze-out in heavy-ion collisions. At the same time, for the smaller and more commonly used values of hard-core hadron radii ($r\lesssim0.5$ fm), the precision of the van der Waals excluded-volume procedure is shown to be sufficient.
• ### Van der Waals Equation of State with Fermi Statistics for Nuclear Matter(1504.01363)
June 26, 2015 nucl-th
The van der Waals (VDW) equation of state is a simple and popular model to describe the pressure function in equilibrium systems of particles with both repulsive and attractive interactions. This equation predicts an existence of a first-order liquid-gas phase transition and contains a critical point. Two steps to extend the VDW equation and make it appropriate for new physical applications are carried out in this paper: 1) the grand canonical ensemble formulation; 2) an inclusion of the quantum statistics. The VDW equation with Fermi statistics is then applied to a description of the system of interacting nucleons. The VDW parameters $a$ and $b$ are fixed to reproduce the properties of nuclear matter at saturation density $n_0=0.16$ fm$^{-3}$ and zero temperature. The model predicts a location of the critical point for the symmetric nuclear matter at temperature $T_c\cong 19.7$ MeV and nucleon number density $n_c \cong 0.07$ fm$^{-3}$.
• ### Particle Number Fluctuations for van der Waals Equation of State(1501.03785)
Feb. 4, 2015 cond-mat.stat-mech, nucl-th
The van der Waals (VDW) equation of state describes a thermal equilibrium in system of particles, where both repulsive and attractive interactions between them are included. This equation predicts an existence of the 1st order liquid-gas phase transition and the critical point. The standard form of the VDW equation is given by the pressure function in the canonical ensemble (CE) with a fixed number of particles. In the present paper the VDW equation is transformed to the grand canonical ensemble (GCE). We argue that this procedure can be useful for new physical applications. Particularly, the fluctuations of number of particles, which are absent in the CE, can be studied in the GCE. For the VDW equation of state in the GCE the particle number fluctuations are calculated for the whole phase diagram, both outside and inside the liquid-gas mixed phase region. It is shown that the scaled variance of these fluctuations remains finite within the mixed phase and goes to infinity at the critical point. The GCE formulation of the VDW equation of state can be also an important step for its application to a statistical description of hadronic systems, where numbers of different particle species are usually not conserved.
• ### Hadron Resonance Gas Equation of State from Lattice QCD(1412.5478)
Jan. 23, 2015 hep-ph, nucl-th, hep-lat
The Monte Carlo results in lattice QCD for the pressure and energy density at small temperature $T < 155$ MeV and zero baryonic chemical potential are analyzed within the hadron resonance gas model. Two extensions of the ideal hadron resonance gas are considered: the excluded volume model which describes a repulsion of hadrons at short distances and Hagedorn model with the exponential mass spectrum. Considering both of these models one by one we do not find the conclusive evidences in favor of any of them. The controversial results appear because of rather different sensitivities of the pressure and energy density to both excluded volume and Hagedorn mass spectrum effects. On the other hand, we have found a clear evidence for a simultaneous presence of both of them. They lead to rather essential contributions: suppression effects for thermodynamical functions of the hadron resonance gas due to the excluded volume effects and enhancement due to the Hagedorn mass spectrum.
• ### Time dependence of partition into spectators and participants in relativistic heavy-ion collisions(1407.4644)
Oct. 27, 2014 hep-ph, nucl-th
The process of formation of the participant system in heavy-ion collisions is investigated in the framework of a simplified analytic Glauber-like model, which is based on the relativistic Boltzmann transport equation. The key point lies in the time-dependent partition of the nucleon system into two groups: nucleons, which did not take part in any interaction before a given time and nucleons, which already have interacted. In the framework of the proposed model we introduce a natural energy-dependent temporal scale $t_c$, which allows us to remove all dependencies of the model on the collision energy except for the energy dependence of the nucleon-nucleon cross-section. By investigating the time dependence of the total number of participants we conclude that the formation process of the participant system becomes complete at $t\simeq1.5 t_c$. Time dependencies of participant total angular momentum and vorticity are also considered and used to describe the emergence of rotation in the reaction plane.
• ### A new approach to time-dependent transport through an interacting quantum dot within Keldysh formalism(1311.4377)
The time-dependent transport through a nano-scale device, consisting of a single spin-degenerate orbital with on-site Coulomb interaction, coupled to two leads, is investigated. Various gate and bias voltage time-dependences are considered. The key and new point lies in the proposed way to avoid the difficulties of the usual heavy computation when dealing with two time Green's functions within Keldysh formalism. The time-dependent retarded dot Green's functions are evaluated, in an efficient manner within a non-canonical Hubbard I approximation. Calculations of the time-dependent current are then presented in the wide-band limit for different parameter sets. A comparison between the method and the Hartree-Fock approximation is performed as well. It is shown that the later cannot account reliably for dynamical aspects of transport phenomena.
• ### Longitudinal fluctuations of the center of mass of the participants in heavy-ion collisions(1306.5208)
July 18, 2013 hep-ph, nucl-ex, nucl-th
A model for computing the probability density of event-by-event participant center-of-mass rapidity y^{c.m.} is presented. The evaluations of the y^{c.m.} distribution are performed for different collision energies and different centralities. We show that for certain conditions the rapidity distribution is described by a Gaussian with a variance determined mostly by the collision centrality. It is found that the width of the y^{c.m.} distribution increases strongly for more peripheral collisions, while it depends weakly on the collision energy. Other theoretical estimates of rapidity distribution are presented and questions of interaction and separation between spectators and participants are discussed. | 2020-06-02 17:28:10 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7718610763549805, "perplexity": 952.9928134998567}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-24/segments/1590347425481.58/warc/CC-MAIN-20200602162157-20200602192157-00088.warc.gz"} |
http://gostudious.com/ | # Programming Continued
I’ve been thinking some more about programming in the classroom. I tweeted Fraser Speirs and asked him why Ruby over Python:
I’m realizing that this is the only type of response you can give when asked “why one language over another?” Ultimately, it doesn’t really matter which language you learn, and at the end of the day, the preference and comfort of the instructor is what matters most, at least in the environment we’re both trying to succeed in.
It was Fraser’s recent article that really started giving me some good ideas. A few people on the Python edu-sig list have recommended iPython Notebooks, but I like the idea of setting up a server somewhere and running it straight and simple for everyone.
Fraser does mention Coda, which I’ve just tried out and I think this is the direction I want to go, even if just for myself right now.1 It’s easy enough to run a text editor and a terminal as separate applications, but why clutter your desktop when an application like Coda has already put together these elements into one app? Coda does Python and Ruby syntax highlighting, and the Terminal window in the app connects to the localhost no problem. Plus, once you’re using Coda for Python or Ruby development, you can start to use it for web development as well by learning about the preview and CSS editor windows. It’s an IDE that’s not really an IDE. Plus, it’s lightweight and has an iPad version (not sure if that can use a local terminal though).
The iPad is great, but I agree with Fraser that you’ll probably want to use a keyboard. The Verge recently wrote a review on the best iPad keyboards, and they pretty much say the same thing about typing on the iPad:
No matter which keyboard you choose, typing for extended periods on an iPad is a pain. Why? For starters, the iPad is meant to be touched with your fingers, and as much command you might have over “next line,” “next word,” and “next paragraph” keyboard shortcuts, you don’t quite realize how often you move your cursor onscreen until you type for a while on an iPad. This means you’ll be touching the screen quite a bit as you type, a distracting and not always precise procedure.
As I mentioned in my previous post, I want to go straight ahead with Python. I will be attending PyCon Canada and I hope I’ll run into some fellow educators.
Meanwhile, I may start looking at Ruby. This article was pretty honest and clear about the differences, but Python still appeals to me, for some of the reasons the author admits. This article about Ruby in education is pretty basic, but makes an interesting argument. I think I could make as equally a convincing argument with Python, but it’s worth exploring anyway.
One last thing… Codecademy has about 4 units up in Python and it’s worth checking out. There are lots of ways to learn programming in interactive ways, as well as great books like Learn Python the Hard Way. Hopefully my students will be able to find one way or another to get hooked.
1. I’m still using version 1, not sure if I want to upgrade yet.
# Teaching Programming
I’m working on developing a programming workflow so that I can begin teaching programming to students. I’ve studied a lot of programming myself, and while I’ve never developed an ‘app’, I feel pretty confident in my ability to solve problems, especially those dealing with simple math operations, string manipulation, and basic logic.
When I learned programming it was always at the command line. We opened up pico or emacs, edited our code, saved, quit out of pico, compiled, and then ran. It was a simple workflow, and beyond including the standard libraries (in C/C++), there wasn’t much more to focus on than the actual program itself.
Fast forward a few years, and there I am with XCode installed on my MacBook, trying to follow along in an Integrated Development Environment. In XCode and Interface Builder, you can still work on your code (of course), but you also have to build window elements, and buttons, and there’s app delegates, and all sorts of different things to get confused by. Things get even more confusing when you realize that Apple is updated XCode almost constantly, and so not only is the language framework changing, but the development environment changes along with it. Apple updates things incessantly, constantly, and if you’re not right on top of that stuff, it can be very hard to follow along.
Sure, other languages and compilers and environments get updated. Python is in version 3 dot something now, but many people are still using 2 dot something. There are IDEs available with Python and there’s any number of modules and builds of Python. I don’t know exactly, but I’m sure this is the case with most languages and development platforms.
But, there’s no denying that at the command line, everything is simpler. If you’re developing simpler programs, then that’s probably the best thing to stick to. I eventually want to be able to whip up an iPad app in a weekend, but for demonstrating simple programming concepts to students, a simpler environment is going to be better each time.
I’ve decided upon Python as my direction for integrating programming into the math curriculum. Lots of people are using Python, and with languages, it’s wisest to go where there’s lots of support and documentation. Ruby is an option as well, but Python just seems more integrated to me.
So, in developing this workflow I’ve been experimenting with a bunch of different options. I was initially using TextWrangler and its own build window, but then I started needing the Terminal. I tried Coda, and a plug-in that I found that lets you build python right in the app. I tried doing it on the server where I host my websites, but they didn’t have very good support around python. I’ve tried Pythonista on the iPad, and Sublime Text 2 and its built in interpreter, as well as its terminal plug-in SublimeREPL.
There’s more traditional IDEs, like Eclipse, and NetBeans, and JetBrains. Of course, Python offers their own IDLE, but I found that to be fairly limited.
One of the things that has made this search slightly more difficult is my thought that I should figure out a way to have my students using PyLab. The truth of the matter is, we’re not really doing intensive stuff, it’s just that I want to have access to the various mathematics functions in those libraries. Plus, with matplotlib, I want them to be able to graph. Essentially, I’d like to replace the TI-83 altogether and have them using Python/PyLab instead.
I think it could work really well because Python runs live as well as through scripts. The live python could act as an in-the-moment calculator. You see the >>> prompt and you type in 100.4*6 and it spits out an answer, no programming knowledge necessary. Meanwhile, slowly, and awesomely we could start getting the computer to perform repeated operations for us, using programming. I want my students to walk out of the class knowing a new skill that can actually be used later for other things (Django anyone?). The TI-83 is a dead-end for students who see it as a tool only to be used in a math classroom, and even then they don’t know half of the features available.
Here’s the problem. Imagine, day one of class. “Class, I need everyone to come to the next class with PyLab installed on your machine.” Look, I’m pretty savvy with this stuff and I found it very tough to figure out how to install scipy, numpy, etc. You have to compile things on the command line, and it’s not easy.
It’s so difficult in fact that I wasn’t ever able to do it. If someone showed me how to do it, I could understand it, but even finding good tutorials is tough, and trust me, I looked. So, what’s the alternative? I found a few: The Enthought Python Distribution, SAGE, and the SciPy superpack. I found the EPD pretty easy to install, and everything you need is there. I haven’t played around with SAGE or the superpack yet.
Perhaps I could have my students install the EPD, and we’d all be happy. There’s one problem: not everyone has a computer, or wants to bring one to class everyday. Some students have iPads, and there should be a way to let them participate as well. Unfortunately, numpy/scipy/matplotlib isn’t yet available on the iPad (as far as I know). This makes me want to figure out a way to do this without using the scientific libraries.
There’s one more alternative that I recently read about. Fraser Speirs is attempting to do something similar, and what he’s done is set up a Linux server on the Amazon cloud services and have the students essentially SSH in. Cool. I would need to spend some time myself to get that all set up, and hopefully the server would support a full installation of PyLab, and I’m not sure if I could run the plotting software through that. Hmm. It may work with a little work, but if I want to get this started soon, I’m looking for the easiest solution first.
Right now, that’s going ahead without the scientific stuff and just using a plain vanilla Python. And it means going back to basics. I like Sublime Text 2, but students could find the text editor that works for them, BBEdit, TextWrangler, something on Windows. We’ll edit our code in the editor, and we’ll run it at the command line, just like how I learned. No IDEs, no fancy applications, just code and command line. For those on the iPad, Pythonista is the best option (it’s really nice).
I want to get more complex eventually, both for myself and my students. But, that might be a few years down the road. The first attempt has to be the simplest possible setup, so that any student can set it up and feel comfortable working with it from day one.
# Number Theory Class 25 May 1
Previous: Encryption with Addition
Now: Encryption with Multiplication, hints of beyond
We are going to use a little trick from our past to help us calculate multiplicative inverses.
If we have time, we will see a nice table that “covers” exponential cryptography.
//Classwork:
Page Problems
349 1,2,3,5
397 1,2
Next:
• Read
• Encryption using Modular Multiplication, 8.5
• Exponentiation Ciphers with a Prime Modulus, 9.4
• Study for Final
# Stats Class 25 May 1
Previous: Wrap up of Hypothesis Testing
Now: Correlation and Regression
There is a calculation to go along with correlation, but since we are time-constrained, we’ll use our calculators to figure out a value for r, which is known as Pearson’s correlation coefficient.
Once we have a value for r, we want to compare it to a critical value just as we did with hypothesis testing. The critical value chosen depends on the sample size and $$\alpha$$, which could be 0.05 or 0.01.
Lastly, once you have decided that there is a correlation, and that r is significant, then one might want to generate an equation for the line of best fit. It’s pretty easy to generate the equation for the line of best fit once you have r using the formulas below.
The book uses different letters than you are probably used to for a linear equation. Instead of $$y=mx+b$$, it’s written $$\hat{y}=a+bx$$ To keep it simple we will use the letters we know.
$m=r(\frac{S_y}{S_x})$
$b=\bar{y}-m\bar{x}$
//Classwork:
Page Problems
134 20
145 Verify the r value provided in 7-10 is correct, then calculate m, and b. Also, Do part f for each problem.
Next:
• Be Prepared
• Review on Thursday May 3
• Final on Thursday May 10, 8-10am
# QR Class 24 April 26
Previous: Exponential Models and Vampires
Now: Special Topics: Golden Ratio and the Euclidean Algorithm
1. Understanding the Fibonacci Series
2. Seeing how that turns into the golden ratio
3. Artistic Application Worksheet
4. Bjorkland Algorithm for Musical Rhythm
//Classwork:
• Lots of worksheets.
Page Problems
601 17
Next:
• Read
• Skim all of Chapter 11 and see if anything piques your interest
# Number Theory Class 24 April 26
Previous: Check-Digits
Now: Elementary Cryptography
Encryption, like check-digit schemes goes up in complexity giving more security.
Addition is the easiest code to crack Then, multiplication Then, exponential…prime modulus Then, exponential…composite modulus…The hardest code to crack
Basically all cryptography is is taking a string of numbers and shifting it in some way (add, multiply, power), and reducing it in some modular system. The modular reduction makes it that much harder to see a pattern.
Decrypting can be done by reversing what was done to encrypt it, or better, just inverse what was done. That’s why we want to understand additive and multiplicative inverses.
//Classwork:
Page Problems
343 1,4,5,6
349 1,2,3,5
Next:
• Read
• Encryption using Modular Multiplication, 8.5
# Stats Class 24 April 26
Previous: Estimation vs Hypothesis Testing
Now: Pearson’s Correlation Coefficient
//Classwork:
• Worksheets, “Tech Talk,” “Data Deluge”, and “Correlation and Regression”
Next:
• Read
• Linear Regression and the Coefficient of Determination, 4.2
# QR Class 23 April 24
Previous: Beginning Exponential Modeling
Now: Exponential Modeling
Just like in linear modeling, we had to learn to go between graphs, tables, and equations, we need to do the same thing with exponential models.
So far, we have learned how to plug in a growth rate and a starting value into a template to create an exponential model. But now, what if we encounter a graph, or a table with two points, how can we generate an equation from that?
//Classwork:
Supplement!
Next:
• Read
• Exponential Modeling, 9C | 2013-05-26 07:02:38 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3279914855957031, "perplexity": 1165.3108799268916}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368706635944/warc/CC-MAIN-20130516121715-00019-ip-10-60-113-184.ec2.internal.warc.gz"} |
https://www.bernardosulzbach.com/conservation-of-momentum-exercise/ | 2015-11-12
# Problem Statement
In a train yard, train cars are rolled down a long hill in order to link them up with other cars as shown. A car of mass 4000. kg starts to roll from rest at the top of a hill 5.0 m high, and inclined at an angle of 5.0° to the horizontal. The coefficient of rolling friction between the train and the track is 0.050. What velocity would the car have if it linked up with 3 identical cars sitting on flat ground at the bottom of the track?
Hint: The equation for rolling friction is just like the one for sliding friction.
# My Solution
The speed of the car after leaving the hill is equal to
v = sqrt(kinetic_energy * 2 / m)
And its kinetic energy will be equal to the work done on it:
kinetic_energy = F * d
Getting the distance is a matter of simple geometry:
d = csc (5°) * 5 m ~= 57.37 m
The force is a bit more complicated, note that I am using the net force, taking friction into account
F = tan (5°) * 4 * 10^4 N - 0.05 * sec (5°) * 4 * 10^4 N
= 4 * 10^4 N * (tan (5°) - 0.05 * sec (5°)) ~= 1,492 N
Throwing everything together
v = sqrt(F * d * 2 / m) ~= 6.542 m / s
As the velocity will be equally divided between the cars
v = (6.542 m / s) / 4 = 1.635 m / s
Thus, the answer is 1.64 m / s.
Note that I used 10 m s^-2 for the gravitational acceleration. If you want to be more precise, use 9.80665 m s^-2, the the standard acceleration due to gravity. | 2019-11-20 16:41:21 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.37572601437568665, "perplexity": 815.1562859963825}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496670597.74/warc/CC-MAIN-20191120162215-20191120190215-00286.warc.gz"} |
https://api-project-1022638073839.appspot.com/questions/how-do-you-solve-3-x-4-7-x-1 | # How do you solve 3^x + 4 = 7^x - 1 ?
Oct 8, 2016
$x \approx 1.08777$
#### Explanation:
Calling
$f \left(x\right) = {7}^{x} - {3}^{x} - 5$ Expanding in Taylor series around ${x}_{0}$
$f \left(x\right) \approx f \left({x}_{0}\right) + {\left(\frac{\mathrm{df}}{\mathrm{dx}}\right)}_{{x}_{0}} \left(x - {x}_{0}\right)$
or
$f \left({x}_{1}\right) \approx f \left({x}_{0}\right) + {\left(\frac{\mathrm{df}}{\mathrm{dx}}\right)}_{{x}_{0}} \left({x}_{1} - {x}_{0}\right)$
If ${x}_{0}$ is near a function zero, then $f \left({x}_{1}\right) \approx 0$ so
${x}_{k + 1} = {x}_{k} - f \frac{{x}_{k}}{\frac{\mathrm{df}}{\mathrm{dx}}} _ \left({x}_{k}\right)$
Here
${\left(\frac{\mathrm{df}}{\mathrm{dx}}\right)}_{{x}_{k}} = {7}^{{x}_{k}} L o g \left(7\right) - {3}^{{x}_{k}} L o g \left(3\right)$
Begining with ${x}_{0} = 1$ we have
${x}_{0} = 1$
${x}_{1} = 1.09685$
${x}_{2} = 1.08786$
${x}_{3} = 1.08777$
${x}_{4} = 1.08777$ | 2021-10-18 09:38:31 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 15, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9869969487190247, "perplexity": 2134.6663709795785}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323585201.94/warc/CC-MAIN-20211018093606-20211018123606-00143.warc.gz"} |
https://plainmath.net/algebra-i/63498-usually-regression-involves-variables-independent-variable-depende | Carole Juarez
2022-02-22
Usually linear regression involves two variables (x,y), i.e. an independent variable x and a dependent variable y, and they are related by the following expression
$y={a}_{0}+{a}_{1}x$
where ${a}_{0}$ and ${a}_{1}$ are parameters that define the linear model. In linear regression we have one equation of this form for each couple of observed variables $\left({x}_{i},{y}_{i}\right)$, thus we have a linear system and its solution gives us ${a}_{0}$ and ${a}_{1}$.
Let's consider that we have two set of independent-dependent variables, namely (x,y) and (w,z). The first two variables (x,y) are related by the previous equation, while the second two variables (w,z) are related by the following
$z={b}_{0}+{b}_{1}w$
where ${b}_{0}$ and ${b}_{1}$ are parameters that define the linear relation between z and w. Also in this case a set of observation $\left({w}_{j},{z}_{j}\right)$ leads to a linear system and its solution gives us ${b}_{0}$ and ${b}_{1}$.
In general, if ${a}_{0},{a}_{1}$ and ${b}_{1}$ are independent, then we can solve the two linear systems separately. But now, let's suppose that ${a}_{0}$ and ${b}_{0}$ are independent, while ${a}_{1}={b}_{1}$. In this case, the two linear systems should be solved simultaneously.
I've solved this problem just definying one linear system of equation involving both the two sets of equations, but I would like to know if this problem has a specific name and how to correctly approach it. In particular, I want to know how to assessing the fit quality (for example, with an equivalent of the ${R}^{2}$).
Gene Espinosa
$y={a}_{0}+{c}_{1}x$
$z={b}_{0}+{c}_{1}w$
you can minimize
$\sum {\left({a}_{0}+{c}_{1}x-y\right)}^{2}+\sum {\left({b}_{0}+{c}_{1}w-z\right)}^{2}$
giving the equations
$\sum {a}_{0}+{c}_{1}x-y=0$,
$\sum {b}_{0}+{c}_{1}w-z=0$,
$\sum x\left({a}_{0}+{c}_{1}x-y\right)+\sum w\left({b}_{0}+{c}_{1}w-z\right)=0$.
Now solve this $3×3$ system for ${a}_{0},{b}_{0},{c}_{1}$.
The fit quality is still given by the ratio of the explained variance over the total variance.
Do you have a similar question? | 2023-03-27 08:07:40 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 26, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9532970786094666, "perplexity": 97.26245136483895}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296948609.41/warc/CC-MAIN-20230327060940-20230327090940-00410.warc.gz"} |
https://studyib.net/physics/page/265/centripetal-force | # Centripetal force
If a body moves in a circle there must be a resultant force acting towards the centre. This is because the change in direction implies that the body is accelerating.
Deriving the equation is not so simple though...
Key Concepts
Perpendicular force
A force acting at right angles to the direction of motion causes the body to travel in a circular path.
Relationship between ω and v
ω is the angular velocity = $$2\pi \over T$$
v is the speed = $$2\pi r \over T$$
From this we can deduce that $$v = ωr$$
Essentials
When deriving the equation for centripetal acceleration we start by considering a small part of the motion we can deduce that
$$a = {mv^2\over r}$$
This is the acceleration.
Test Yourself
Use quizzes to practise application of theory.
Exam-style Questions | 2021-09-17 18:48:12 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7620406746864319, "perplexity": 433.11499434482425}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780055775.1/warc/CC-MAIN-20210917181500-20210917211500-00401.warc.gz"} |
http://mathoverflow.net/revisions/11253/list | 3 added 7 characters in body
1. For the moduli problem of a curve of genus $g$ with $n$ marked points, how large an $n$ is needed to ensure the existence of a fine moduli space? For this question, terminology is that of Mumford's GIT.
2. For the following three moduli problems, how big an $N$ is required for existence of a fine moduli space? The terminology is from the exposes of Deligne-Rapoport and Katz-Mazur, or Shimura. The first is in French, the second is too big, and the third is using old language and never mentions the modern terminology of universal elliptic curve, etc.. Therefore it is not possible for me to dig up the information myself.
i) Elliptic curves equipped with a cyclic subgroup of order $N$ -- this moduli problem corresponds to the modular group $\Gamma_0(N)$.
ii) Elliptic curves equipped with a point of order $N$ -- this moduli problem corresponds to the modular group $\Gamma_1(N)$.
ii) Elliptic curves equipped with a symplectic pairing on $N$-torsion points -- this moduli problem corresponds to the modular group $\Gamma (N)$.
References other than the above, will be appreciated.
2 edited tags
1
# Existence of fine moduli space for curves and elliptic curves
1. For the moduli problem of a curve of genus $g$ with $n$ marked points, how large an $n$ is needed to ensure the existence of a fine moduli space? For this question, terminology is that of Mumford's GIT.
2. For the following three moduli problems, how big an $N$ is required for existence of a fine moduli space? The terminology is from the exposes of Deligne-Rapoport and Katz-Mazur, or Shimura. The first is in French, the second is too big, and the third is using old language and never mentions the modern terminology of universal elliptic curve, etc.. Therefore it is not possible for me to dig up the information myself.
i) Elliptic curves equipped with a subgroup of order $N$ -- this moduli problem corresponds to the modular group $\Gamma_0(N)$.
ii) Elliptic curves equipped with a point of order $N$ -- this moduli problem corresponds to the modular group $\Gamma_1(N)$.
ii) Elliptic curves equipped with a symplectic pairing on $N$-torsion points -- this moduli problem corresponds to the modular group $\Gamma (N)$.
References other than the above, will be appreciated. | 2013-05-22 22:49:45 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5442774295806885, "perplexity": 281.9763471756038}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368702454815/warc/CC-MAIN-20130516110734-00009-ip-10-60-113-184.ec2.internal.warc.gz"} |
https://mran.revolutionanalytics.com/snapshot/2022-04-08/web/packages/geoFourierFDA/readme/README.html | geoFourierFDA
In this paper, we have observsed n curves χs1(t), …, χsn(t) in a region, where si = (θi, ηi), i = 1, …, n, and θi is the latitude and ηi is the longitude where the curve χsi was sampled. The goal of this package is to estimate an unsampled curve χs0(t) at s0 ∉ {s1, …, sn}. The Ideia proposed by Giraldo (2011) was simple: the curve χs0(t) is a linear combination of all curves χs1(t), …, χsn(t), i.e., $$\\widehat{\\chi\_{\\mathbf{s}\_0}}(t) = \\lambda\_1 \\chi\_{\\mathbf{s}\_1}(t) + \\lambda\_2 \\chi\_{\\mathbf{s}\_2}(t) + \\dots + \\lambda\_n \\chi\_{\\mathbf{s}\_1}(t)$$ where λ1, …, λn is solution of the linear system given by
where μ is an constant from the method of Lagrange’s multipliers and the function γ(h) = ∫γ(h; t)*dt* is called the trace-variogram, where, for each t, γ(h; t) is the semivariogram for the process χs1(t), …, χsn(t). More precisely, for each t, a weakly and isotropic spatial process is assumed for χs1(t), …, χs**n(t) and the integration of the semivariogram is carried out. Usually, the integration in the equation (1) is approximated using a modified version of the empirical semivariogram. In this pcackage, we have used the Legendre-Gauss quadrature, which is simple and it explicitly used the definition of the semivariogram.
Installation
This package can be installed using the devtools package.
devtools::install_github("gilberto-sassi/geoFourierFDA")
Examples
In this package, we have used the temperature dataset present in the package fda and in the package geofd. This dataset has temperature measurements from 35 weather stations from Canada. This data can be downloaded at weather.gov.ca. For illustration, we have separated the time series at The Pas station and used all others stations to estimate the curve temperature at The Pas.
How to interpolate a curve at an unmonitored location
# interpolating curve at Halifax using all remaining curves in the functional dataset
# Estimating the temperature at The Pas
Smoothed curve using Fourier series
# Coefficients of smoothing using Fourier series polynomial
# Coefficients of smoothing at The Pas
# coefficients of Fourier series polynomial
y_est <- fourier_b(coefs, x) | 2022-11-28 18:22:59 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8501181602478027, "perplexity": 970.7834459765088}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710534.53/warc/CC-MAIN-20221128171516-20221128201516-00671.warc.gz"} |
https://pos.sissa.it/256/210/ | Volume 256 - 34th annual International Symposium on Lattice Field Theory (LATTICE2016) - Physics Beyond the Standard Model
Numerical Analysis of Discretized ${\mathcal N}=(2,2)$ SYM on Polyhedra
S. Kamata,* S. Matsuura, T. Misumi, K. Ohta
*corresponding author
Full text: pdf
Pre-published on: 2016 December 27
Published on: 2017 March 24
Abstract
We perform a numerical simulation of the two-dimensional N = (2,2)
supersymmetric YangMills (SYM) theory on the discretized curved space. The U(1)A anomaly of the
continuum theory is maintained also in the discretized theory as an unbalance of the number of
the fermions. In the process, we propose a new phase-quenched approximation, which we call the
"anomaly-phasequenched (APQ) method", to make the partition function and observables well-defined by
U(1)A phase cancellation. By adopting APQ method, we estimate the Ward-Takahashi
identity for exact SUSY on lattice and clarify contribution of the pseudo zero-modes to the
pfaffian phase.
DOI: https://doi.org/10.22323/1.256.0210
Open Access | 2018-06-19 20:15:13 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3445547819137573, "perplexity": 7512.12334155241}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-26/segments/1529267863119.34/warc/CC-MAIN-20180619193031-20180619213031-00305.warc.gz"} |
http://mathhelpforum.com/calculus/42341-integrate-ln-tanx-1-a.html | # Math Help - integrate ln(tanx+1)
1. ## integrate ln(tanx+1)
I seen this problem a while back on the site and to my recollection it wasn't tackled. If it was, my apologies. I couldn't find it. I think Isomorphism had posted it. If I am a for doing this, so be it.
I am sure some of you have some clever show off methods, but here is something that kind of appeared to me rather quickly and turned into something familiar. I think it is cool to see the different innovative ways folks tackle these. It shows what a wonder the human brain can be.
You can easily run them through a computer, but they lack the human touch.
$\int_{0}^{\frac{\pi}{4}}ln(tan(x)+1)dx$
Let $u=1+tan(x)$
$u-1=tan(x)dx$
$du=sec^{2}(x)dx$
Make the subs and get:
$\int\frac{ln(u)}{sec^{2}(x)}dx$
An x mixed in with a u?. No.......to some that may seem like sacrilege.
But it's OK:
Since $tan(x)=\frac{u-1}{1}$
Then $sec^{2}(u)=(u-1)^{2}+1$
Then we have:
$\int_{1}^{2}\frac{ln(u)}{(u-1)^{2}+1}du$
Now, let $w=u-1, \;\ w+1=u, \;\ dw=du$
Then we have:
$\int_{0}^{1}\frac{ln(w+1)}{w^{2}+1}dw$
Now...this is a familiar integral that the Krizmeister done awhile back so I won't bother.
It's solution is $\frac{\pi}{8}ln(2)$
This kind of popped into my head when I seen it so I had to write it down.
2. Here. (That link will direct you into another one.) | 2014-03-08 08:49:01 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 11, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8049461841583252, "perplexity": 1063.7423168023208}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-10/segments/1393999654272/warc/CC-MAIN-20140305060734-00090-ip-10-183-142-35.ec2.internal.warc.gz"} |
https://motls.blogspot.com/2005/01/new-year-pictures.html | ## Sunday, January 02, 2005 ... //
### New Year - pictures
You can view a collection of random photographs if you click the thumbnail below:
Microsoft Internet Explorer is recommended. If your XP Service Pack 2 gives you a message that it protects you from Active-X content, you may click on this message and order the explorer to allow the slide show. It was safely created with Microsoft's own PowerToys.
The slide show allows you to browse through the thumbnails, much like in Windows XP Explorer. The collection includes ice sculptures, fireworks, and other things. The last 20 seconds of the fireworks are also available as a fireworks video.
Incidentally, there are other fireworks being planned, too. NASA is going to compete with the "atom smashers" at CERN and Fermilab. The guys can't wait to destroy their spacecraft and hit a comet on July 4th, much like in Deep Impact. Such things happen if adults behave as children - but to be honest, it should be a lot of fun! :-) | 2019-11-19 07:45:44 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.22789108753204346, "perplexity": 3178.162271642366}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496670036.23/warc/CC-MAIN-20191119070311-20191119094311-00374.warc.gz"} |
https://chemistry.stackexchange.com/tags/analytical-chemistry/new | # Tag Info
6
The Kjeldahl method has two versions, using $\ce{HCl}$ as trapping solution or a weak acid like boric acid. Depending which one is used, the nitrogen content is either determined by a back titration with $\ce{NaOH}$ (your case) or a direct titration with $\ce{HCl}$. The goal of the titration is to figure out how much ammonia (gas) has been trapped in the ...
2
There can be plenty. There is no definite list. I would suggest to look up the following terms in Skoog or Daniel Harris' Analytical Chemistry. Worth a trip to the library now :-) The following concepts give rise to systematic errors (i) buoyancy error (ii) temperature differences (iii) electrostatic charge build up on the sample (v) moisture (iv) above ...
2
I would imagine that you could just use one of the spectrophotometers you have available to determine the molar absorptivity constant experimentally. After all, the molar absortivity constant is simply the constant which equates the terms (those being Absorbance and Concentration) which are already proportional. So, using something like the beer-lambert law.....
2
A spike recovery value of >100% does not mean that you added too much or the recovery process is too good. All it means is that the analytical method which is being followed "in letter and spirit" gives a positive error. There may be a bias in the method or the precision of the method is poor. We cannot say just by looking at one number. At this stage it is ...
2
The link above which I used for reference mentions that for the test for sulfate anions, hydrochloric acid (HCl) is used to ensure that there is no presence of carbonate ions. For the test for halide ions (F−, Cl−, Br−, I−, etc), nitric acid (HNO3) is used instead, for the same purpose. Good question, but this is not a universal practice. In reality,...
2
The logic is the same used for a monoprotic weak acid, considering the two ionization steps, as indicated by @MaxW. For a diprotic weak acid we have: The first ionization \begin{align*} \ce{H2A + H2O &<=> H3O+ + HA-} &K_{\ce{a}1}=\frac{\ce{[H3O+][HA-]}}{\ce{[H2A]}} \end{align*} and the second ionization \begin{align*} \ce{HA- + H2O &&...
11
It might contain $\ce{NaNO3}$ as well as $\ce{KNO3}$. This paper by Foong et. al. cites some thermal measurements, such as melting point and heat capacity, that you could make with a thermometer and heater. Note that there are inflections in the heat capacity curve as the different components ($\ce{NaNO3}$, $\ce{KNO3}$ and the eutectic of the two) melt. If ...
3
I see two possible answers. Not (A) and (B) but rather (A) and (D). Chakravarthy Kalyan is responsible for reminding us that (D) can be an answer too (refer to comments). For option (D): $\ce{HC ≡ CH + 2AgNO3 + 2NH4OH→ AgC ≡ CAg↓ + 2NH4NO3 + 2H2O}$ The greyish precipitate of silver acetylide will readily form when ethyne is passed through freshly ...
0
Very simple answer: IR radiation (in very first approximation) interacts with the individual bond. While methane is unpolar, the bonds in it are polar. Not terribly polar, but still.
1
The key point is estimation of standard deviation of estimation of measurement standard deviation on Stats SE. The exact formula is complicated with extensive Gamma function involvement. The essential info is the relative uncertainty of $s$ decreases very slowly with number of measurements. It can be approximated by involving Stirling approximation for ...
0
Your instinct about not rounding early is good. Before I get to that, I wanted to address one thing in another answer: "> I learned that the absolute and relative error have only 1 significant figure That really isn't true. In your example you showed 5.34532g ± 0.001428g which could result from some kind of experiment. The point being that the digits 1428 ...
1
These parameters arise from non-dimensionalization of the Mathieu equations for ion motion in exogenous, periodically varying electric fields (i.e. oscillating potentials). You can't calculate $m/z$ directly from $a$ and $q$. For a fixed $U$ and $V$, you can regard $a$ and $q$ as non-dimensional parameters that correspond to $m/z$. (Conversely, at fixed $... 2 I found one study that examined cerium(III) polyoxometalate salts and concluded that at least under the conditions of their study,$\ce{Ce(III)}$did not oxidize under electrospray conditions. The study is Bray et al. 2011 in the International Journal of Mass Spectrometry. They say: The presence of the redox active$\ce{Ce}\$ metal ion introduces ...
0
I'm looking for some way to calculate how much base I will need to add to an unbuffered solvent consisting of a mixture of ethanol and water in order to basify it to a specific reading (10) on an a pH probe calibrated with an aqueous standard. I think you have summarized the main issue yourself. The reading of a pH probe in a basic alcoholic solution is ...
6
If your supervisor is unconvinced by Brand's guidance on this (as supplied by Loong in his answer), you could always test the effect of ultrasonic cleaning directly: Take, say, six volumetric flasks, and weigh the amount of water each holds when filled to the line. Then subject three to several cycles of ultrasonic cleaning (the other three can be left to ...
21
According to technical information for volumetric measurement provided by Brand, it is acceptable to use an ultrasonic bath. Both glass and plastic labware may be cleaned in an ultrasonic bath. However, direct contact with the sonic membranes must be avoided. Nevertheless, For gentle treatment of labware, clean immediately after use – at low ...
Top 50 recent answers are included | 2019-07-19 20:33:16 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6925716996192932, "perplexity": 1436.4933811879835}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195526359.16/warc/CC-MAIN-20190719202605-20190719224605-00379.warc.gz"} |
https://neutrium.net/fluid_flow/pressure-drop-through-a-packed-bed/ | # Summary
As fluid flows through a packed bed it experiences a pressure loss due to friction. This article describes the use of the Carman-Kozeny and Ergun equations for the calculation of pressure drop through a randomly packed bed of spheres.
# Definitions
: Packed bed friction factor : Gravitational constant ( g = 9.81 m/s) : Height of the packed bed (m) : Fluidisation pressure drop (Pa) : Pressure drop through the packed bed (Pa) : Superficial fluid velocity (m/s) : Spherical equivalent particle diameter (m) : Bed voidage : Density of the fluid flowing through the packed bed (kg/m3) : Density of particles in the packed bed (kg/m3) : Viscosity of the fluid flowing through the packed bed (Pa.s)
# Introduction
As a fluid passes through a packed bed it experiences pressure loss due to factors such as friction. The relationships required to predict the pressure drop for a fluid flowing through a packed bed have been known for some time, with Darcy observing in 1896 that the laminar flow of water through a bed of sand was governed by the following relationship:
$\displaystyle \displaystyle \frac{-\Delta P}{H} \propto U$
This relationship was initially analysed in terms of the Hagen-Poiseuille equation for laminar flow through a tube and was later formulated as the Carman-Kozeny equation for pressure drop for laminar flow through a packed bed in 1937. The following sections present the Carman-Kozeny equation and subsequently Ergun's general equation for the pressure drop through a randomly packed bed of spheres.
# Laminar flow through a packed bed
The pressure drop for laminar fluid flow through a randomly packed bed of monosized spheres with diameter may be calculated using the Carman-Kozeny equation as follows:
# Turbulent Flow through a packed bed
The pressure drop for turbulent flow through a packed bed may be calculated from the turbulent component of the Ergun equation (discussed in section 5) as presented below:
# General Equation for pressure drop through a packed bed
Ergun (1952), using a extensive set of experimental data covering a wide range of particle size and shapes, presented a general equation to calculate the pressure drop across a packed bed for all flow conditions (laminar to turbulent). This equation is commonly referred to as the Ergun equation for flow through a randomly packed bed of spheres and takes the following form:
The Ergun equation combines both the laminar and turbulent components of the pressure loss across a packed bed. In laminar flow conditions the first component of the equation dominates with the Ergun equation essentially reducing to the Carman-Koreny equation presented in Section 3, although with a slight variation in the constants used due to variations in the experimental data with which the correlations was developed. In the laminar region the pressure drop through the packed bed is independent of fluid density and has a linear relationship with superficial velocity.
Under turbulent flow conditions the second component of the Ergun equation dominates. Here the pressure drop increases with the square of the superficial velocity and has a linear dependence on the density of the fluid passing through the bed.
## Calculating a Packed Bed Friction Factor
The Ergun equation may also be expressed through the use of a packed bed friction factor in a similar manner to how pressure drop is calculated for fluid flow in a pipe with the Darcy friction factor. The packed bed friction factor may be calculated using the packed bed Reynolds number as follows:
The Ergun equation may then be calculated using the packed bed friction factor as expressed below:
# Packed Bed of Non-Spherical Particles
Although the Ergun equation was constructed for mono-sized spherical particles, pressure drop can still be calculated for randomly packed non-spherical particles using the spherical equivalent particle diameter (the diameter of a sphere having the same surface area to volume ratio as the non-spherical particle). Here the Ergun equation becomes :
Alternatively if the particles in the packed bed are not mono-sized the surface-volume mean diameter , should be used in place of the spherical equivalent particle diameter . | 2017-03-26 11:06:03 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2997609078884125, "perplexity": 783.7082163458456}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-13/segments/1490218189214.2/warc/CC-MAIN-20170322212949-00660-ip-10-233-31-227.ec2.internal.warc.gz"} |
https://www.neetprep.com/question/56037-prism-angle--distance-screen-prism-b---mdistance-prism-source-amnm-Fresnal-biprism-findthe-value--fringe-width--m--mm---m----m/55-Physics--Wave-Optics/700-Wave-Optics | A star emitting light of wavelength 5896 Å is moving away from the earth with a speed of 3600 km/sec. The wavelength of light observed on earth will
(1) Decrease by 5825.25 Å
(2) Increase by 5966.75 Å
(3) Decrease by 70.75 Å
(4) Increase by 70.75 Å
(c = 3 × 108 m/sec is the speed of light)
To view Explanation, Please buy any of the course from below.
High Yielding Test Series + Question Bank - NEET 2020
Difficulty Level:
A heavenly body is receding away from the earth such that the fractional change in λ is 1, then its velocity is :
(1) C
(2) $\frac{3C}{5}$
(3) $\frac{C}{5}$
(4) $\frac{2C}{5}$
Concept Questions :-
Wave front
To view Explanation, Please buy any of the course from below.
High Yielding Test Series + Question Bank - NEET 2020
Difficulty Level:
A slit of width a is illuminated by white light. For red light (λ = 6500 Å), the first minima is obtained at θ = 30°. Then the value of a will be
(1) 3250 Å
(2) 6.5 × 10–4 mm
(3) 1.24 microns
(4) 2.6 × 10–4 cm
Concept Questions :-
Diffraction
To view Explanation, Please buy any of the course from below.
High Yielding Test Series + Question Bank - NEET 2020
Difficulty Level:
The radius of central zone of the circular zone plate is 2.3 mm. The wavelength of incident light is $5893\text{\hspace{0.17em}\hspace{0.17em}}Å.$ Source is at a distance of 6m. Then the distance of the first image will be
(1) 9 m
(2) 12 m
(3) 24 m
(4) 36 m
Concept Questions :-
Huygens' principle
To view Explanation, Please buy any of the course from below.
High Yielding Test Series + Question Bank - NEET 2020
Difficulty Level:
What will be the angular width of central maxima in Fraunhoffer diffraction when light of wavelength $6000\text{\hspace{0.17em}}Å$ is used and slit width is 12×10–5 cm
Concept Questions :-
Diffraction
To view Explanation, Please buy any of the course from below.
High Yielding Test Series + Question Bank - NEET 2020
Difficulty Level:
Direction of the first secondary maximum in the Fraunhofer diffraction pattern at a single slit is given by (a is the width of the slit)
(1) $a\mathrm{sin}\theta =\frac{\lambda }{2}$
(2) $a\mathrm{cos}\theta =\frac{3\lambda }{2}$
(3) $a\mathrm{sin}\theta =\lambda$
(4) $a\mathrm{sin}\theta =\frac{3\lambda }{2}$
Concept Questions :-
Diffraction
To view Explanation, Please buy any of the course from below.
High Yielding Test Series + Question Bank - NEET 2020
Difficulty Level:
A parallel monochromatic beam of light is incident normally on a narrow slit. A diffraction pattern is formed on a screen placed perpendicular to the direction of the incident beam. At the first maximum of the diffraction pattern, the phase difference between the rays coming from the edges of the slit is
(1) $0$
(2) $\frac{\pi }{2}$
(3) $\pi$
(4) 2$\pi$
Concept Questions :-
Diffraction
To view Explanation, Please buy any of the course from below.
High Yielding Test Series + Question Bank - NEET 2020
Difficulty Level:
A parallel beam of monochromatic light of wavelength 5000 Å is incident normally on a single narrow slit of width 0.001 mm. The light is focused by a convex lens on a screen placed on the focal plane. The first minimum will be formed for the angle of diffraction equal to
(1) 0o
(2) 15o
(3) 30o
(4) 60o
Concept Questions :-
Diffraction
To view Explanation, Please buy any of the course from below.
High Yielding Test Series + Question Bank - NEET 2020
Difficulty Level:
In the far field diffraction pattern of a single slit under polychromatic illumination, the first minimum with the wavelength ${\lambda }_{1}$ is found to be coincident with the third maximum at ${\lambda }_{2}$. So
(1) $3{\lambda }_{1}=0.3{\lambda }_{2}$
(2) $3{\lambda }_{1}={\lambda }_{2}$
(3) ${\lambda }_{1}=3.5{\lambda }_{2}$
(4) $0.3{\lambda }_{1}=3{\lambda }_{2}$
Concept Questions :-
Diffraction
To view Explanation, Please buy any of the course from below.
High Yielding Test Series + Question Bank - NEET 2020
Difficulty Level:
The angle of polarisation for any medium is 60o, what will be critical angle for this
(1) ${\mathrm{sin}}^{-1}\sqrt{3}$
(2) ${\mathrm{tan}}^{-1}\sqrt{3}$
(3) ${\mathrm{cos}}^{-1}\sqrt{3}$
(4) ${\mathrm{sin}}^{-1}\frac{1}{\sqrt{3}}$
Concept Questions :-
Polarization of light
To view Explanation, Please buy any of the course from below.
High Yielding Test Series + Question Bank - NEET 2020
Difficulty Level: | 2020-05-28 07:09:07 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 23, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3391810953617096, "perplexity": 2819.679577269696}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-24/segments/1590347398233.32/warc/CC-MAIN-20200528061845-20200528091845-00444.warc.gz"} |
https://www.techwhiff.com/issue/how-do-u-delete-a-question--220153 | # How do u delete a question
###### Question:
how do u delete a question
### Can someone help me with this worksheet?
can someone help me with this worksheet?...
### 1. If the point (-5, -3) is translated 4 units to the right and 3 units down, what are the coordinates of the new point? Write the algebraic expression that represents the translation.
1. If the point (-5, -3) is translated 4 units to the right and 3 units down, what are the coordinates of the new point? Write the algebraic expression that represents the translation....
### My auestion is ! Write the value and unit of gravitational constant G in SI units? Give me a answer for a short question !
My auestion is ! Write the value and unit of gravitational constant G in SI units? Give me a answer for a short question !...
### Determinar la cantidad de atomos que hay en 7.5g de S y 125g de Fe
Determinar la cantidad de atomos que hay en 7.5g de S y 125g de Fe...
### Problem Set 1. a. A person who weighs 100 pounds on Earth weighs 16.6 lb. on the moon. Which variable is the independent variable? Explain why. b. What is an equation that relates weight on Earth to weight on the moon? How much would a 185-pound astronaut weigh on the moon? Use an equation to explain how you know. d. How much would a man who weighs 50 pounds on the moon weigh on Earth?
Problem Set 1. a. A person who weighs 100 pounds on Earth weighs 16.6 lb. on the moon. Which variable is the independent variable? Explain why. b. What is an equation that relates weight on Earth to weight on the moon? How much would a 185-pound astronaut weigh on the moon? Use an equation to explai...
### Which verb correctly completes the following sentence? We __to go camping every summer. A.have try B. will tried C.try D.tries
Which verb correctly completes the following sentence? We __to go camping every summer. A.have try B. will tried C.try D.tries...
### Mycoplasma pneumoniae is a bacterium that lacks a cell wall. it causes atypical pneumonia. which antibiotic would be most effective in treating this infection, and why?
Mycoplasma pneumoniae is a bacterium that lacks a cell wall. it causes atypical pneumonia. which antibiotic would be most effective in treating this infection, and why?...
### Why is developing skills in racket sports like badminton and pickleball beneficial?
Why is developing skills in racket sports like badminton and pickleball beneficial?...
### The lunch at Dion's Place has choice of ham, turkey, or roast beef on rye or white bread white juice, milk, or tea. A. How many different lunches are possible. B. What is the probability that the lunch special of the day is ham on rye with tea? Please Help With Both Answers - I Will Give BRAINLIEST, and 50 points.
The lunch at Dion's Place has choice of ham, turkey, or roast beef on rye or white bread white juice, milk, or tea. A. How many different lunches are possible. B. What is the probability that the lunch special of the day is ham on rye with tea? Please Help With Both Answers - I Will Give BRAINLIES...
### According to Lapena, what is the importance of the oral tradition? To,what extend do you agree with his opinions and why?
According to Lapena, what is the importance of the oral tradition? To,what extend do you agree with his opinions and why?...
### 1/sec-1 + 1/sec+1 = 2cotcosec
1/sec-1 + 1/sec+1 = 2cotcosec...
### Using the table below showing median income by year, how many years did it take for the annual household income to double from its value in 1982? Year 2012 2010 2008 2006 2004 2002 2000 1998 1996 1994 1992 1990 Median Income $48,291$46,658 $47,624$45,618 $41,952$40,125 $39,772$36,802 $33,447$30,247 $28,424$27,522 Year 1988 1986 1984 1982 1980 1978 1976 1974 1972 Median Income $24.772$22,482 $20,196$18,347 $15,944$13,098 $10,947$9,554 $8,198 1970$7,368 1968 $6,421 Using the table below showing median income by year, how many years did it take for the annual household income to double from its value in 1982? Year 2012 2010 2008 2006 2004 2002 2000 1998 1996 1994 1992 1990 Median Income$48,291 $46,658$47,624 $45,618$41,952 $40,125$39,772 $36,802$33,447 $30... 1 answer ### ILL MARK AS BRAINLESS PLS HELP ILL MARK AS BRAINLESS PLS HELP... 1 answer ### How do you write: One third of the opposite of a number is less than 12? How do you write: One third of the opposite of a number is less than 12?... 1 answer ### 2y - x = 6 infinite No one 2y - x = 6 infinite No one... 1 answer ### Kelly can work for her dad and make$6 per hour, or she can work for Jana's Mowing Service and make $14 per hour. If she needs to make at least$84, and can only work 10 hours total. She can work for Jana's Mowing Service for at most 5 hours.
Kelly can work for her dad and make $6 per hour, or she can work for Jana's Mowing Service and make$14 per hour. If she needs to make at least \$84, and can only work 10 hours total. She can work for Jana's Mowing Service for at most 5 hours....
### Brian arrives to the College Ave student center bus stop at 5:00 PM. The waiting time for the bus to Busch campus is a random variable that satisfies roughly the exponential distribution (see below). If there is a traffic congestion on route 18, the average waiting time for the bus is 25 minutes. If there is no traffic congestion, the average waiting time is 7 minutes. Ordinarily there is a 35% chance of traffic congestion on route 18 around that time of the day. Brian waited 20 minutes until th
Brian arrives to the College Ave student center bus stop at 5:00 PM. The waiting time for the bus to Busch campus is a random variable that satisfies roughly the exponential distribution (see below). If there is a traffic congestion on route 18, the average waiting time for the bus is 25 minutes. If... | 2023-03-23 17:48:29 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.41317784786224365, "perplexity": 2064.795191979137}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296945182.12/warc/CC-MAIN-20230323163125-20230323193125-00269.warc.gz"} |
https://www.nextgurukul.in/wiki/concept/cbse/class-7/science/nutrition-in-animals/digestion-in-humans/3957254 | Notes On Digestion in Humans - CBSE Class 7 Science
Nutrition in animals Animals exhibit heterotrophic mode of nutrition. As animals cannot synthesise their own food, they depend on plants or other smaller animals for food. Types of animals Based on the food they consume, animals are classified into herbivores, carnivores and omnivores. Holozoic nutrition: It is a process by which animals take in their food. It involves different steps namely, ingestion, digestion, absorption, assimilation and egestion. Human beings exhibit holozoic mode of nutrition involving five basic steps. Digestion: Digestion is the process by which food is broken down into simple absorbable substances. Digestion of food takes place in the digestive system. Digestive system is made up of alimentary canal and associated glands. Digestion in man is an extracellular process. Digestive system in human beings: Digestive system in human beings is formed by alimentary canal and digestive glands. Parts of alimentary canal: It is also called as digestive tract. It comprises different parts like mouth, oesophagus, stomach, small intestine and large intestine. It starts with mouth and ends with anus. Mouth is guarded by upper lip and lower lip. The process of taking in food through mouth is called as ingestion. Buccal cavity is the inner region of the mouth. It encloses teeth and tongue. The buccal cavity leads into pharynx. a) Teeth are of different types namely, incisors, canines, premolars and molars. • Incisors are used for cutting and biting food. There are 4 incisors in each jaw. They are also called as biting teeth. • Canines are sharp and pointed and are used to pierce or tear food. There are 2 canines in each jaw. • Premolars have broad grinding surfaces. Hence they help in chewing and grinding the food. There are 4 premolars in each jaw. • Molars are the principal grinders which help to chew and grind the food. There are 6 molars in each jaw. Teeth help in the process of mastication. • Mastication involves proper chewing of food and mixing it with saliva. • Food is broken into small pieces by biting, grinding and crushing. • Digestion of food starts in the mouth with the help of salivary amylase present in the saliva. • Salivary amylase is the enzyme which digests starch, a form of carbohydrate present in the food. Structure of a tooth Dental care b) Tongue is a muscular organ attached to the floor of the buccal cavity at the back. • It is free at one end and helps in pushing the food inside while chewing. • It helps in mixing of food with saliva and swallowing it. • It has many taste buds on it which can taste different types of food. Taste buds are sensory receptors which give a sense of taste. • Taste receptors can taste sugar, salt, sour, bitter and one another taste called as umami. • Tongue also helps in speech. Pharynx is at the back of buccal cavity. • The pharynx is the common channel for food and air. • When you swallow food, a flap-like valve called the epiglottis closes the windpipe. • Epiglottis prevents the entry of food particles into respiratory tract. Oesophagus also called as food pipe helps in conveying the food from buccal cavity to stomach. The oesophagus is also known as the gullet. It is about 25 centimetres long. • Food inside the oesophagus reaches the stomach by peristalsis movements. • Peristaltic movements are the alternate contractions and relaxations of oesophageal wall which bring about movement of food from buccal cavity to the stomach. • Food conveyed to the stomach is called as bolus as it is round in shape. Stomach is the widest part of the alimentary canal. It is a J-shaped muscular organ divided into three parts namely, cardia, fundus and pylorus regions. • Stomach as a whole can hold at about two litres of food. • Stomach secretes a fluid called as digestive juice. Digestive juice is made up of hydrochloric acid, mucous and some enzymes. • Hydrochloric acid kills the bacteria present in the food and softens the molecules of food. • Mucous protects inner lining of the stomach from the action of hydrochloric acid. • Digestive enzymes partially digest some nutrients like proteins and fats. Stomach churns the food into a milky paste. • This partially digested food is called as chyme. • Chyme is conveyed to small intestine for further digestion. Chyme is acidic in nature. Small intestine is made up of three regions namely duodenum, jejunum and ileum. • Acidic chyme from the stomach is received by the duodenum for further digestion. • Duodenum also receives bile form the liver. Bile reduces the acidity of chyme. • Bile also provides alkaline environment to activate some enzymes which bring about digestion of certain nutrients in the food. • Duodenum also receives pancreatic secretions which help in the digestion of food. • Duodenum also secretes some enzymes on its own. • All these substances bring about digestion of food in the intestine. • The inner walls of small intestine are thrown into many folds which have millions of small finger like projections called villi. • Villi increase the surface area for digestion as well as absorption of digested food by eight times. Small intestine also helps in the process of absorption and assimilation. • Undigested food is sent into large intestine. Large intestine comprises of colon and rectum. Large intestine receives undigested food from small intestine. • Water from the food is reabsorbed to a great extent in the large intestine. • Semi-solid undigested waste is stored in the rectum for defecation. • Anus is the opening of the alimentary canal to the exterior. This helps in the elimination of faeces by the process of egestion. Digestive glands: These glands are also called as associated glands. These are also considered to be exocrine glands which have ducts to drop their secretions into the target organ directly. The secretions of the digestive glands help in the process of digestion. These glands include salivary glands, gastric glands, intestinal glands, liver and pancreas. Salivary glands are present inside the buccal cavity. They secrete saliva. Saliva helps in lubrication of food . This saliva plays an important role in breaking down complex components like starch into simple sugars. It brings about partial digestion of starch. Gastric glands are microscopic glandular cells present in the inner lining of the stomach. Gastric glands secrete gastric juice comprising HCl, pepsin and prorennin. Gastric juice helps in the digestion of proteins. Gastric juice helps in emulsification of fats. Intestinal glands are present in the inner lining of small intestine. These secrete various enzymes which aid in the process of digestion of all the components of food. Liver is the largest gland in our body. The liver secretes a yellowish green watery fluid called bile. It is temporarily stored in a sac called the gall bladder. Bile provides an alkaline environment for many enzymes to get active. It also reduces the acidity of chyme. Bile plays an important role in the digestion of fats. Bile is sent into duodenum through a narrow tube-like structure called the bile duct. Bile breaks the larger fat molecules into tiny droplets, thereby increasing their surface area, which helps in the digestion of fats easily. Pancreas is the mixed gland. It acts as both endocrine and exocrine gland. The pancreas secretes the pancreatic juice that helps to digest carbohydrates, proteins and fats. The pancreatic juice converts carbohydrates into simple sugars and glucose, proteins into amino acids, and the lipids into fatty acids and glycerol. Absorption: The process of allowing simple absorbable nutrients into blood capillaries through surface of the villi is called as absorption. Inner wall of small intestine comprises many finger like projections called as villi. Villi increase the surface area for absorption of food. Each villus is made up of central structure called as lacteal which mainly absorbs simple fats and transports them into lymphatic system. Lacteal is surrounded by a network of fine blood capillaries. Blood capillaries absorb glucose molecules and amino acids and transport them in the blood. Vitamins and minerals get readily absorbed into the blood. Assimilation: The process of utilisation of absorbed food, such as glucose, amino acids, fatty acids and glycerol is called as assimilation. Energy needed for various activities is obtained from glucose. Glucose is broken in the cells in the presence of oxygen to syntheise energy in the form of ATP. Amino acids are used for building and repairing body parts. Fatty acids and glycerol are stored in the adipose tissue and under the skin for future use. Egestion: It is the process by which undigested food is passed to exterior through an opening called as anus. Rectum stores undigested waste in the form of faeces. Faeces are sent out through anus.
#### Summary
Nutrition in animals Animals exhibit heterotrophic mode of nutrition. As animals cannot synthesise their own food, they depend on plants or other smaller animals for food. Types of animals Based on the food they consume, animals are classified into herbivores, carnivores and omnivores. Holozoic nutrition: It is a process by which animals take in their food. It involves different steps namely, ingestion, digestion, absorption, assimilation and egestion. Human beings exhibit holozoic mode of nutrition involving five basic steps. Digestion: Digestion is the process by which food is broken down into simple absorbable substances. Digestion of food takes place in the digestive system. Digestive system is made up of alimentary canal and associated glands. Digestion in man is an extracellular process. Digestive system in human beings: Digestive system in human beings is formed by alimentary canal and digestive glands. Parts of alimentary canal: It is also called as digestive tract. It comprises different parts like mouth, oesophagus, stomach, small intestine and large intestine. It starts with mouth and ends with anus. Mouth is guarded by upper lip and lower lip. The process of taking in food through mouth is called as ingestion. Buccal cavity is the inner region of the mouth. It encloses teeth and tongue. The buccal cavity leads into pharynx. a) Teeth are of different types namely, incisors, canines, premolars and molars. • Incisors are used for cutting and biting food. There are 4 incisors in each jaw. They are also called as biting teeth. • Canines are sharp and pointed and are used to pierce or tear food. There are 2 canines in each jaw. • Premolars have broad grinding surfaces. Hence they help in chewing and grinding the food. There are 4 premolars in each jaw. • Molars are the principal grinders which help to chew and grind the food. There are 6 molars in each jaw. Teeth help in the process of mastication. • Mastication involves proper chewing of food and mixing it with saliva. • Food is broken into small pieces by biting, grinding and crushing. • Digestion of food starts in the mouth with the help of salivary amylase present in the saliva. • Salivary amylase is the enzyme which digests starch, a form of carbohydrate present in the food. Structure of a tooth Dental care b) Tongue is a muscular organ attached to the floor of the buccal cavity at the back. • It is free at one end and helps in pushing the food inside while chewing. • It helps in mixing of food with saliva and swallowing it. • It has many taste buds on it which can taste different types of food. Taste buds are sensory receptors which give a sense of taste. • Taste receptors can taste sugar, salt, sour, bitter and one another taste called as umami. • Tongue also helps in speech. Pharynx is at the back of buccal cavity. • The pharynx is the common channel for food and air. • When you swallow food, a flap-like valve called the epiglottis closes the windpipe. • Epiglottis prevents the entry of food particles into respiratory tract. Oesophagus also called as food pipe helps in conveying the food from buccal cavity to stomach. The oesophagus is also known as the gullet. It is about 25 centimetres long. • Food inside the oesophagus reaches the stomach by peristalsis movements. • Peristaltic movements are the alternate contractions and relaxations of oesophageal wall which bring about movement of food from buccal cavity to the stomach. • Food conveyed to the stomach is called as bolus as it is round in shape. Stomach is the widest part of the alimentary canal. It is a J-shaped muscular organ divided into three parts namely, cardia, fundus and pylorus regions. • Stomach as a whole can hold at about two litres of food. • Stomach secretes a fluid called as digestive juice. Digestive juice is made up of hydrochloric acid, mucous and some enzymes. • Hydrochloric acid kills the bacteria present in the food and softens the molecules of food. • Mucous protects inner lining of the stomach from the action of hydrochloric acid. • Digestive enzymes partially digest some nutrients like proteins and fats. Stomach churns the food into a milky paste. • This partially digested food is called as chyme. • Chyme is conveyed to small intestine for further digestion. Chyme is acidic in nature. Small intestine is made up of three regions namely duodenum, jejunum and ileum. • Acidic chyme from the stomach is received by the duodenum for further digestion. • Duodenum also receives bile form the liver. Bile reduces the acidity of chyme. • Bile also provides alkaline environment to activate some enzymes which bring about digestion of certain nutrients in the food. • Duodenum also receives pancreatic secretions which help in the digestion of food. • Duodenum also secretes some enzymes on its own. • All these substances bring about digestion of food in the intestine. • The inner walls of small intestine are thrown into many folds which have millions of small finger like projections called villi. • Villi increase the surface area for digestion as well as absorption of digested food by eight times. Small intestine also helps in the process of absorption and assimilation. • Undigested food is sent into large intestine. Large intestine comprises of colon and rectum. Large intestine receives undigested food from small intestine. • Water from the food is reabsorbed to a great extent in the large intestine. • Semi-solid undigested waste is stored in the rectum for defecation. • Anus is the opening of the alimentary canal to the exterior. This helps in the elimination of faeces by the process of egestion. Digestive glands: These glands are also called as associated glands. These are also considered to be exocrine glands which have ducts to drop their secretions into the target organ directly. The secretions of the digestive glands help in the process of digestion. These glands include salivary glands, gastric glands, intestinal glands, liver and pancreas. Salivary glands are present inside the buccal cavity. They secrete saliva. Saliva helps in lubrication of food . This saliva plays an important role in breaking down complex components like starch into simple sugars. It brings about partial digestion of starch. Gastric glands are microscopic glandular cells present in the inner lining of the stomach. Gastric glands secrete gastric juice comprising HCl, pepsin and prorennin. Gastric juice helps in the digestion of proteins. Gastric juice helps in emulsification of fats. Intestinal glands are present in the inner lining of small intestine. These secrete various enzymes which aid in the process of digestion of all the components of food. Liver is the largest gland in our body. The liver secretes a yellowish green watery fluid called bile. It is temporarily stored in a sac called the gall bladder. Bile provides an alkaline environment for many enzymes to get active. It also reduces the acidity of chyme. Bile plays an important role in the digestion of fats. Bile is sent into duodenum through a narrow tube-like structure called the bile duct. Bile breaks the larger fat molecules into tiny droplets, thereby increasing their surface area, which helps in the digestion of fats easily. Pancreas is the mixed gland. It acts as both endocrine and exocrine gland. The pancreas secretes the pancreatic juice that helps to digest carbohydrates, proteins and fats. The pancreatic juice converts carbohydrates into simple sugars and glucose, proteins into amino acids, and the lipids into fatty acids and glycerol. Absorption: The process of allowing simple absorbable nutrients into blood capillaries through surface of the villi is called as absorption. Inner wall of small intestine comprises many finger like projections called as villi. Villi increase the surface area for absorption of food. Each villus is made up of central structure called as lacteal which mainly absorbs simple fats and transports them into lymphatic system. Lacteal is surrounded by a network of fine blood capillaries. Blood capillaries absorb glucose molecules and amino acids and transport them in the blood. Vitamins and minerals get readily absorbed into the blood. Assimilation: The process of utilisation of absorbed food, such as glucose, amino acids, fatty acids and glycerol is called as assimilation. Energy needed for various activities is obtained from glucose. Glucose is broken in the cells in the presence of oxygen to syntheise energy in the form of ATP. Amino acids are used for building and repairing body parts. Fatty acids and glycerol are stored in the adipose tissue and under the skin for future use. Egestion: It is the process by which undigested food is passed to exterior through an opening called as anus. Rectum stores undigested waste in the form of faeces. Faeces are sent out through anus.
#### Activities
Activity 1 Sheppardsoftware.com has displayed a group of activities in a single page under the topics ' Digestion Game level-1', ' Digestion Game level-2' and 'Digestion quiz'. Student has to select the activity and play the game as per the instructions or answer the quiz. Student is given the score while playing. Additional information is also given about different parts of the digestive system.You have a tutorial providing information about digestion. Go to Activity Activity 2 Ehc.com has created an interactive video which brings the attention of students with different options on the template.Student has to select 'Digestive tract'. This module is divided into Guided tour, Zoom in and Organize your organs. Once we click on the organise your button, organs of the digestive system are displayed. Student needs to put them into proper place as instructed by the video.You can also know information about digestive system under the 'Guided tour'. Zoom in helps in to see what happens to the food consumed in different parts of the digestive system. Go to Activity
Next | 2020-02-26 06:42:15 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 3, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.44931191205978394, "perplexity": 10063.219673909342}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875146187.93/warc/CC-MAIN-20200226054316-20200226084316-00071.warc.gz"} |
https://math.stackexchange.com/questions/2813637/find-a-and-p-in-the-given-question | # Find a and p in the given question
I would be grateful if someone helps me with this question.
I get $p$ but to get $a$, the term with $a$ vanishes.
I checked continuity at $x=1$ for finding $p$.
$p$ comes out to be $-2$.
• @Peter Yes, because we are applying limit n tends to infinite. Jun 9, 2018 at 15:11
• Sorry, didn't look carefully enough Jun 9, 2018 at 15:13
As $f(x)$ is differentiable for $x\in(0,2)$ we know that the limit $$\lim_{n\to\infty}\dfrac{ax(x-1)\left(\cot^n\left(\dfrac{\pi x}4\right)\right)+px^2+2}{\left(\cot^n\left(\dfrac{\pi x}4\right)\right)+1}$$ is finite, and $f(x)$ is continuous for $x\in(0,2)$. Let $$L=\lim_{n\to\infty}\dfrac{ax(x-1)\left(\cot^n\left(\dfrac{\pi x}4\right)\right)+px^2+2}{\left(\cot^n\left(\dfrac{\pi x}4\right)\right)+1}$$ $$\Rightarrow L=\lim_{n\to\infty}\dfrac{ax(x-1)+(px^2+2)\left(\tan^n\left(\dfrac{\pi x}4\right)\right)}{1+\left(\tan^n\left(\dfrac{\pi x}4\right)\right)}$$ For $x\in(1,2)$, $\dfrac\pi4<\dfrac{\pi x}{4}<\dfrac{\pi}2\Rightarrow 0<\cot\left(\dfrac{\pi x}4\right)<1\Rightarrow\cot^n\left(\dfrac{\pi x}4\right)\to 0$ as $n\to\infty$ Hence $$L=\lim_{n\to\infty}\dfrac{ax(x-1)\left(\cot^n\left(\dfrac{\pi x}4\right)\right)+px^2+2}{\left(\cot^n\left(\dfrac{\pi x}4\right)\right)+1}=\dfrac{0(ax(x-1))+px^2+2}{0+1}=px^2+2$$ We need that the limit is equal to $0$ as $x\to 1^+$ for the function to be continuous. Hence $\lim_{x\to1^+}px^2+2=0\Rightarrow p=-2$.
Let $L_1$ and $L_2$ be the limits (of $f(x)$) from the left and from the right of $x=1$ respectively. Thus $$L_2=\lim_{n\to\infty}\dfrac{ax(x-1)\left(\cot^n\left(\dfrac{\pi x}4\right)\right)+2-2x^2}{\left(\cot^n\left(\dfrac{\pi x}4\right)\right)+1},x\to1^+$$
$$L_1=\lim_{n\to\infty}\dfrac{ax(x-1)+(2-2x^2)\left(\tan^n\left(\dfrac{\pi x}4\right)\right)}{1+\left(\tan^n\left(\dfrac{\pi x}4\right)\right)},x\to1^-$$ Now, for $x\in(0,1)$, $\tan^n\left(\dfrac{\pi x}4\right)\to0$ as $n\to\infty$ For $x\in(1,2)$, $\cot^n\left(\dfrac{\pi x}4\right)\to 0$ as $n\to\infty$.
Thus, $$L_2=\dfrac{2-2x^2}{2},x\to 1^+$$ $$\Rightarrow L_2=\lim_{h\to 0}\dfrac{2-2(1+h)^2}{2}=\lim_{h\to0}\dfrac{-2h^2-4h}{2}=\lim_{h\to0}\dfrac{-2h(h-2)}{2}=\lim_{h\to0}\dfrac{2h(2-h)}{2}$$ $$L_1=\dfrac{ax(x-1)}{2},x\to1^-$$ $$\Rightarrow L_1=\lim_{h\to 0}\dfrac{a(1-h)(1-h-1)}{2}=\lim_{h\to 0}\dfrac{a(1-h)(-h)}{2}$$ Now it is given that $f(x)$ is differentiable for $x\in(0,2)$. Thus, $f(x)$ is continuous for $x\in(0,2)$. Thus $f(x)$ must be continuous at $x=1$. Therefore, $L_2$ must be equal to $L_1$ (and both must be respectively equal to $f(1)=0$). Equating both $L_2$ and $L_1$, $$\lim_{h\to0}\dfrac{2h(2-h)}{2}=\lim_{h\to 0}\dfrac{a(1-h)(-h)}{2}$$ $$\Rightarrow \lim_{h\to0}\dfrac{2h(2-h)}{2}-\lim_{h\to 0}\dfrac{a(1-h)(-h)}{2}=0$$ We now use the property $\lim_{x\to a}f(x)\pm\lim_{x\to a}g(x)=\lim_{x\to a}(f\pm g)(x)$
$$\Rightarrow \lim_{h\to0}\left(\dfrac{2h(2-h)}{2}-\dfrac{a(1-h)(-h)}{2}\right)=0$$ $$\Rightarrow \lim_{h\to0}\dfrac{2h(2-h)+a(1-h)(h)}{2}=0$$
$$\Rightarrow 2(2-h)+a(1-h)=0, h\to0$$
$$\Rightarrow a+4=0$$ Finally, we get $a=-4$.
• @IceInkberry If you know any shorter method, please do share. Jun 11, 2018 at 6:44
• Same method, shared in chatroom Jun 17, 2018 at 13:10 | 2022-09-29 04:33:28 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9904191493988037, "perplexity": 139.15463549935507}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030335304.71/warc/CC-MAIN-20220929034214-20220929064214-00461.warc.gz"} |
https://www.coin-or.org/CppAD/Doc/equalopseq.htm | Check if Two Value are Identically Equal
Syntax
b = EqualOpSeq(x, y)
Purpose
Determine if two x and y are identically equal; i.e., not only is x == y true, but if they are variables , they correspond have the same operation sequence .
Motivation
Sometimes it is useful to cache information and only recalculate when a function's arguments change. In the case of AD variables, it may be important not only when the argument values are equal, but when they are related to the independent variables by the same operation sequence. After the assignment y = x these two AD objects would not only have equal values, but would also correspond to the same operation sequence.
x
The argument x has prototype const AD<Base> &x
y
The argument y has prototype const AD<Base> &y
b
The result b has prototype bool b The result is true if and only if one of the following cases holds:
1. Both x and y are variables and correspond to the same operation sequence.
2. Both x and y are parameters, Base is an AD type, and EqualOpSeq( Value(x) , Value(y) ) is true.
3. Both x and y are parameters, Base is not an AD type, and x == y is true.
Example
The file equal_op_seq.cpp contains an example and test of EqualOpSeq. It returns true if it succeeds and false otherwise. | 2018-01-23 10:20:03 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7019520998001099, "perplexity": 552.9866838257931}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-05/segments/1516084891886.70/warc/CC-MAIN-20180123091931-20180123111931-00382.warc.gz"} |
https://codereview.stackexchange.com/questions/128009/prime-generator-spoj-problem-in-python-3 | # Prime generator SPOJ problem in Python 3
I am trying to solve an SPOJ problem: print all prime numbers in a given range (as large as $10^9$). I have used the Sieve of Eratosthenes algorithm. But it is still slow when input is in range of $10^4$.
import math
no_of_cases = int(input())
for i in range(no_of_cases):
x = input().split(" ")
a = int(x[0])
b = int(x[1])
lis = set([p*i for p in range(2,b+1) for i in range(2,b+1)])
print (sorted(set(range(a,b+1))- lis))
• It is slow because your set is computed using a doubly nested for loop. In the average case scenario, that yields a runtime complexity of O(b^2). Try finding ways to optimize the inner calculation. – user991710 May 10 '16 at 20:09
• @user991710 what can i do instead of set? – Freddy May 11 '16 at 14:14
• @Easterly: why would the OP want to make a slow algorithm even slower? The Sieve of Sundaram is strictly inferior to the Sieve of Eratosthenes (more complicated, less performance). In fact, if you do due diligence cleanup/optimisation on the Sieve of Sundaram then you arrive almost at an odds-only Sieve of Eratosthenes, except that it fails to skip non-prime factors. See Sieve of Eratosthenes has huge 'overdraw' - is Sundaram's better after all? – DarthGizka May 12 '16 at 13:02
• @Freddy: Code Review is for working code (even if it is a bit slow and can be improved). Your code fails to meet specifications (sieve several ranges up to 10^9 within an exceedingly generous time limit of about 6 seconds). If you can't be arsed to read at least some of the existing answers regarding SPOJ PRIME1 to find out where you went wrong, why should anyone here bother repeating the already existing information? And your algorithm is not an implementation of the Sieve of Eratosthenes. – DarthGizka May 12 '16 at 13:08
• @DarthGizka First of all my program is running in terminal. I have read at too many places after my previous question(before asking this question) that is why code is completly different. What I understood was that I need to remove all the multiples of prime. So I did that, it might not be 100% Sieve of Eratosthenes. I also tried using square root but that was throwing error. Though I can't argue if it off topic. I respect the rules of S.E. sites. – Freddy May 12 '16 at 13:50
I'll convert your program to a primes_below function. (I don't take into account the lower bound.) After doing this I got:
def primes_below(n):
lis = set([p * i for p in range(2, n + 1) for i in range(2, n + 1)])
return sorted(set(range(2, n + 1)) - lis)
First things first, you need a maximum of $n ^ 2$ amount of memory. Why? If we look at your algorithm what is $i_{\text{max}}$ and $p_{\text{max}}$? Both are $n$. And so your current largest number is $n^2$.
You said:
I also tried using square root but that was throwing error.
This also wouldn't work correctly. Lets go through every combination if we were to do that with 16.
2 * 2 = 4
2 * 3 = 6
2 * 4 = 8
3 * 2 = 6
3 * 3 = 9
3 * 4 = 12
4 * 2 = 8
4 * 3 = 12
4 * 4 = 16
This means that 4, 6, 8, 9, 12, 16 are not prime. We should know that the primes in this range are 2, 3, 5, 7, 11, 13. And so 10, 14, 15 are all incorrectly said as prime.
But looking at the above you should be able to see that there is no point on having $3 * 2$, $4 * 2$, $4 * 3$. This is as you've already done those calculations and so you should start the range for $i$ at $p$. And this is true for your current solution.
And so we know that we should at least use:
def primes_below(n):
lis = set([p * i for p in range(2, n + 1) for i in range(p, n + 1)])
return sorted(set(range(2, n + 1)) - lis)
Now we need to remove the creation of numbers greater than (or equal) $n$. Rather than looking at some complex maths, we'll have a look at range.
range(stop)
range(start, stop[, step])
This is a versatile function to create lists containing arithmetic progressions. It is most often used in for loops. The arguments must be plain integers. If the step argument is omitted, it defaults to 1. If the start argument is omitted, it defaults to 0. The full form returns a list of plain integers [start, start + step, start + 2 * step, ...]. If step is positive, the last element is the largest start + i * step less than stop; if step is negative, the last element is the smallest start + i * step greater than stop. step must not be zero (or else ValueError is raised).
From the above we know that the first number we use is $p * p$, this is as $i$ starts as $p$. The largest number we want is also $n$. But we don't want to remove all the numbers that are between $p^2$ and $n$, and so we need that step. That step will also be $p$.
Or in Python:
def primes_below(n):
lis = set([i for p in range(2, n) for i in range(p * p, n, p)])
return sorted(set(range(2, n)) - lis)
Before we say all is good, some maths! What is the largest $p$ that will mean the second range contains a value?
If $p^2 > n$ then the second range will be empty, and so the largest $p$ follows the equation $p^2 <= n$. And so $p_{\text{max}} = \sqrt{n}$.
And adding this to the range significantly improves performance.
def primes_below(n):
lis = set([i for p in range(2, int(n ** 0.5) + 1) for i in range(p * p, n, p)])
return sorted(set(range(2, n)) - lis)
This function is roughly ~3800 times faster than the original, at n = 5000.
Now you probably think it's good enough for this challenge, it's pretty fast!
But unfortunately no. Actually I thought I'd broke my machine when I ran it the first time, and I have 32GB of RAM! I closed every app I had on my machine and the program still broke! This means that you have to care about the memory and your current method doesn't.
Instead we need to limit the amount of memory to $n$.
To do this you can use lis = [True] * n. And then you want to change them to False if you come across one. And you'll want to set 0 and 1 to false at the beginning. The same way you are at the moment. This should get you something like:
def primes_below(n):
lis = [True] * n
lis[0] = lis[1] = False
for p in range(2, int(n ** 0.5) + 1):
for i in range(p * p, n, p):
lis[i] = False
return ...
Now you want to be able to return the numbers. Python has enumerate which will give us the index and the value, this will allow us to make a comprehension. And so if the value is True add the index to the output. Which will result in:
def primes_below(n):
lis = [True] * n
lis[0] = lis[1] = False
for p in range(2, int(n ** 0.5) + 1):
for i in range(p * p, n, p):
lis[i] = False
return (i for i, v in enumerate(lis) if v)
Since you don't want to display some numbers you'll want to implement a lower bound, I'd do this out of the function as then if you ever need the Sieve again you'll have a 'pure' one.
This can simply be another comprehension:
numbers = (i for i in primes_below(b + 1) if i < a)
And will make usage simple:
no_of_cases = int(input())
for _ in range(no_of_cases):
a, b = map(int, input().split(" ")[:2])
print([i for i in primes_below(b + 1) if i < a)])
This however is still too slow for my liking. And so to make it 3 times faster, at n = 10 ** 7, you can add a check if p is prime, if it is then do what we were doing before if not go to the next number.
This results in:
def primes_below2(n):
lis = [True] * n
lis[0] = lis[1] = False
for p in range(2, int(n ** 0.5) + 1):
if not lis[p]:
continue
for i in range(p * p, n, p):
lis[i] = False
return (i for i, d in enumerate(lis) if d)
This is roughly 11500 times faster than the original, at n = 5000. And takes about 244 seconds, on my machine, to generate all the primes below 10 ** 9.
• @Freddy I bet you'll love my edit then, ;P – Peilonrayz May 13 '16 at 19:03
• I can't understand in last line if d – Freddy May 13 '16 at 19:48
• @Freddy I originally used d as data but changed it to v for value to make more sense. What it does is filter out all the non-primes, so it only returns primes. – Peilonrayz May 13 '16 at 20:23 | 2020-01-20 22:28:49 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6091989278793335, "perplexity": 766.5833047289058}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579250599789.45/warc/CC-MAIN-20200120195035-20200120224035-00400.warc.gz"} |
http://www.mathnet.ru/php/archive.phtml?wshow=paper&jrnid=jmag&paperid=570&option_lang=eng | RUS ENG JOURNALS PEOPLE ORGANISATIONS CONFERENCES SEMINARS VIDEO LIBRARY PACKAGE AMSBIB
General information Latest issue Archive Impact factor Search papers Search references RSS Latest issue Current issues Archive issues What is RSS
Zh. Mat. Fiz. Anal. Geom.: Year: Volume: Issue: Page: Find
Zh. Mat. Fiz. Anal. Geom., 2013, Volume 9, Number 3, Pages 379–391 (Mi jmag570)
Some Applications of Meijer $G$-Functions as Solutions of Differential Equations in Physical Models
A. Pishkoo, M. Darus
School of Mathematical Sciences, Faculty of Science and Technology Universiti Kebangsaan Malaysia, Bangi, Selangor D. Ehsan, Malaysia
Abstract: In this paper, we aim to show that the Meijer $G$-functions can serve to find explicit solutions of partial differential equations (PDEs) related to some mathematical models of physical phenomena, as for example, the Laplace equation, the diffusion equation and the Schr$\ddot{o}$dinger equation. Usually, the first step in solving such equations is to use the separation of variables method to reduce them to ordinary differential equations (ODEs). Very often this equation happens to be a case of the linear ordinary differential equation satisfied by the $G$-function, and so, by proper selection of its orders $m; n; p; q$ and the parameters, we can find the solution of the ODE explicitly. We illustrate this approach by proposing solutions as: the potential function $\Phi$, the temperature function $T$ and the wave function $\Psi$, all of which are symmetric product forms of the Meijer $G$-functions. We show that one of the three basic univalent Meijer $G$-functions, namely $G^{1,0}_{0,2},$ appears in all the mentioned solutions.
Key words and phrases: Meijer $G$-functions; partial differential equations; Laplace equation; diffusion equation; Schrödinger equation; separation of variables.
Full text: PDF file (207 kB)
References: PDF file HTML file
Bibliographic databases:
MSC: 35Q40, 35Q79, 33C60, 30C55
Revised: 19.12.2012
Language:
Citation: A. Pishkoo, M. Darus, “Some Applications of Meijer $G$-Functions as Solutions of Differential Equations in Physical Models”, Zh. Mat. Fiz. Anal. Geom., 9:3 (2013), 379–391
Citation in format AMSBIB
\Bibitem{PisDar13} \by A.~Pishkoo, M.~Darus \paper Some Applications of Meijer $G$-Functions as Solutions of Differential Equations in Physical Models \jour Zh. Mat. Fiz. Anal. Geom. \yr 2013 \vol 9 \issue 3 \pages 379--391 \mathnet{http://mi.mathnet.ru/jmag570} \mathscinet{http://www.ams.org/mathscinet-getitem?mr=3155146} \isi{http://gateway.isiknowledge.com/gateway/Gateway.cgi?GWVersion=2&SrcApp=PARTNER_APP&SrcAuth=LinksAMR&DestLinkType=FullRecord&DestApp=ALL_WOS&KeyUT=000322697400006}
• http://mi.mathnet.ru/eng/jmag570
• http://mi.mathnet.ru/eng/jmag/v9/i3/p379
SHARE:
Citing articles on Google Scholar: Russian citations, English citations
Related articles on Google Scholar: Russian articles, English articles
This publication is cited in the following articles:
1. Pishkoo A., Darus M., “on Meijer'S G-Functions (Mgfs) and Its Applications”, Rev. Theor. Sci., 3:2 (2015), 216–223
2. Pishkoo A., Pashaei R., “Describing Micro- and Nano-Structures: Reaction-Diffusion Equation in Fractional Dimensional Space”, J. Comput. Theor. Nanosci., 12:4 (2015), 585–588
• Number of views: This page: 244 Full text: 99 References: 31 | 2020-03-31 17:38:55 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2329714000225067, "perplexity": 4611.799555304916}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585370502513.35/warc/CC-MAIN-20200331150854-20200331180854-00367.warc.gz"} |
https://stacks.math.columbia.edu/tag/0GL5 | Lemma 61.10.7. Let $f : X \to Y$ be a separated finite type morphism of quasi-compact and quasi-separated schemes. Let $\Lambda$ be a torsion ring. Let $E \in D(X_{\acute{e}tale}, \Lambda )$ and $K \in D(Y_{\acute{e}tale}, \Lambda )$. Then
$Rf_!E \otimes _\Lambda ^\mathbf {L} K = Rf_!(E \otimes _\Lambda ^\mathbf {L} f^{-1}K)$
in $D(Y_{\acute{e}tale}, \Lambda )$.
Proof. Choose $j : X \to \overline{X}$ and $\overline{f} : \overline{X} \to Y$ as in the construction of $Rf_!$. We have $j_!E \otimes _\Lambda ^\mathbf {L} \overline{f}^{-1}K = j_!(E \otimes _\Lambda ^\mathbf {L} f^{-1}K)$ by Cohomology on Sites, Lemma 21.20.9. Then we get the result by applying Étale Cohomology, Lemma 58.95.6 and using that $f^{-1} = j^{-1} \circ \overline{f}^{-1}$ and $Rf_! = R\overline{f}_*j_!$. $\square$
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar). | 2021-06-15 00:00:34 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 2, "x-ck12": 0, "texerror": 0, "math_score": 0.9924717545509338, "perplexity": 233.53594187042748}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623487614006.8/warc/CC-MAIN-20210614232115-20210615022115-00263.warc.gz"} |
http://piracy-studies.org/07ghyzei/reaction-of-cao-with-so2-7fc426 | In the reaction CaO(s) + SO2(g) ---> CaSO3(s) (Hint: try drawing the Lewis structures of SO2 and SO32-) a. ) Picture of reaction: Сoding to search: 2 CaSO4 cnd [ temp ] = 2 CaO + 2 SO2 + O2. Figure 9 shows that, Fe 2 O 3 reacts with CaO much easily to form C 2 F; CF is not from the reaction of C 2 F and Fe 2 O 3, but from the directly reaction of Fe 2 O 3 with CaO. Enter a mass or volume in one of the boxes below. 2)Ca2+ acts as a Lewis base, and (SO3)2- acts as a Lewis acid. e. ) SO2 acts as a Lewis base, and Ca2+ acts as a Lewis acid. SO42- acts as a Lewis base, and SO2 acts as a Lewis acid. The rates of both reactions per unit surface area of CaO have the general form: k([SO2]–[SO2]e), where [SO2]e is the concentration of SO2 for the reaction concerned being at equilibrium and k is the rate constant. The reversibility of the reaction CaCO 3 ⇌ CaO+CO 2 has been examined through a large number of cycles (up to 40), mainly at 866 °C. The injection of calcium-based sorbents into coal-fired boilers for reaction with, and reduction in the levels of, sulfur dioxide (SO2) in the flue gas has undergone considerable research and development. Stability of CaSO3 in DTA cell in streams of nitrogen, oxygen, and sulfur dioxide. Expert Answer 100% (10 ratings) Previous question Next … Which of the following information about the reaction of CaO with water is not true? Heating of lead nitrate. Reaction of SO2 with CaO by dif-ferential thermal analysis and thermogravi-metric analysis. Stoichiometry. d. ) SO2 acts as a Lewis base, and O2- acts as a Lewis acid. The reaction equations are shown in table 8, and the relationships between G 0 and temperature are shown in figure 9. Four regions with different prevailing reaction mechanisms can be defined: In the beginning the formation of a monolayer is mainly determined by the SO2 concentration and to some smaller extent also by the relative humidity. d. ) SO2 acts as a Lewis base, and O2- acts as a Lewis acid. Relevance. Join Now. Answered by Vaibhav Chavan | 11th Sep, 2016, 09:59: PM. Ca2+ acts as a Lewis base, and SO32- acts as a Lewis acid. Login. Our channel. (Hint: try drawing the Lewis structures of SO2 and SO32-) a. ) SO2 in the flue gas came almost from the combustion zone. The second product of the reaction, the calcium oxide. S. E. Mousavi, H. Ale Ebrahim, M. Edrissi, Preparation of High Surface Area Ce/La/Cu and Ce/La/Ni Ternary Metal Oxides as Catalysts for the SO 2 Reduction by CH 4 , Synthesis and Reactivity in Inorganic, Metal … In the reaction CaO(s) + SO2(g) ---> CaSO3(s):? There is a large increase in surface area on going from the non‐porous calcium carbonate to the oxide and this is due … For example, calcium carbonate (CaCO 3) decomposes into calcium oxide (CaO) and carbon dioxide (CO 2). CaO reacts with water to form slaked lime. Chemistry. Ca2+ acts as a Lewis base, and SO32- acts as a Lewis acid. This is accomplished by heating the material to above 825 °C (1,517 °F), a process called calcination or lime-burning, to liberate a molecule of carbon dioxide (CO 2), leaving quicklime. CaO Mass: g: SO2 Mass: g: or Gas Volume: L: CaSO3 Mass: g: Reference(s): Ebbing, Darrell D. General Chemistry 3rd … 3)(SO3)2- acts as a Lewis base, and SO2 acts as a Lewis acid. This is an illustration of - 132419 1 Answer to In the reaction CaCO3 → CaO + CO2 100 grams of calcium carbonate (CaCO3) is observed to decompose upon heating into 56 grams of calcium oxide (CaO) and 44 grams of carbon dioxide (CO2). A number of investigations of the SO2/CaO … $CaO + SO_2 \rightarrow CaSO_3$ Sulfur trioxide: Sulfur trioxide reacts violently with water to produce a fog of concentrated sulfuric acid droplets. 5)SO2 acts as a Lewis base, and Ca2+ acts as a Lewis acid. The formation of CaSO4 tends either to plug the pores of a CaO particle (~ 100 pm) or coats the particle as the reaction proceeds, thereby terminating the reaction and preventing efficient utilization of the sorbent. Sulfur dioxide is one of the few common acidic yet reducing gases. => It is a decomposition reaction as Calcium carbonate undergoes decomposition and the carbon dioxide gas formed. The SO 2 + H 2 O reaction is proposed to be the starting process for the oxidation of sulfur dioxide to sulfate in liquid water, although the thermal reaction displays a high activation barrier. lemperature Figure 2. Laboratory reactions. This period is followed by the formation of the consecutive product layers which can initially be determined by a dissolving process of SO2. c. ) O2- acts as a Lewis base, and SO2 acts as a Lewis acid. The carbon dioxide that is formed is immediately removed so that the reaction is preceded until the completion of the process in accordance with Le-Chatelier’s principle. CaCO 3 → CaO + CO 2 This reaction is reversible and exothermic in nature in the forward direction. 4)SO2 acts as a Lewis base, and O2- acts as a Lewis acid. c. ) O2- acts as a Lewis base, and SO2 acts as a Lewis acid. The results showed that SO2 retention in the sinter zone was associated with free-CaO in the form of CaSO3/CaSO4, and the SO2 adsorption reached a maximum under 900ºC. The … Chemical Reactions and Equations. SO32- acts as a Lewis base, and SO2 acts as a Lewis acid. CaC03*Ca0+C02 (above 600°C) … d. ) SO2 acts as a Lewis base, and O2- acts as a Lewis acid. Reaction of CaO with SO 2: ... CO2 O2 SO2 1 2 1 2 1 0 100 200 300 400 500; Inventory of solids (kg/m; 16:40 16:55 17:09 17:24 17:38 17:52 18:07; 2)-Solid circulation = 1.9 kg/m; 2; s-X; sulf = 0.08-X; max-X; carb = 0.03-T; carbonator = 668 ºC *Average values during experimental period shown; Experimental conditions*-Flow to carbonator: 19 m ; 3; N/h-u; gas =2.5 m/s-CO; 2; inlet concentration = 12%-SO; 2; … Reaction Type: Synthesis. 5) Which of the following represents a combination reaction? Liang Ma, Liyong Cao, Rong He, Numerical study of pore structure effect on SO2–CaO reactions, Chinese Journal of Chemical Engineering, 10.1016/j.cjche.2014.12.004, 23, 4, (652-658), (2015). One reaction behavior was the oxidation of sulfur in the sintering mix when the temperature was between 800 and 1000ºC; the other behavior was the decomposition of … Recent studies have suggested that the reaction can be promoted by light absorption in the near UV. Sulfur dioxide can react with certain 1,3-dienes in a cheletropic reaction to form … 1 decade … However, CaSO3 is unstable … 1 which type of reactions are the following i pbo2 so2 gt pbso4 n - Chemistry - TopperLearning.com | hqdwmjvv. In the reaction CaO(s) + SO2(g) <=> CaSO3(s): 1)O2- acts as a Lewis base, and SO2 acts as a Lewis acid. A reaction is also considered to be a decomposition reaction even when one or more of the products are still compounds. 10th. Types of chemical reactions. Reaction … A metal carbonate decomposes into a metal oxide and carbon dioxide gas. Solid calcium oxide is exposed to a stream of carbon dioxide gas CaO + CO2 = CaCO3 7. Although calcined CMA contains CaO and MgO, only CaO reacts with SO2 above 950°C. b. ) from CaO and SO2 at temperatures as low as 330° C. 1.5 •1.0 E0.5 Peak at 700 Peak at 540 °C /'/' S02 A.^^, Exotherm \/ f ~"-.....--..----j'^'Peak at 610 Break at 650 °C Figure 1. Gervald F. Lv 7. Add / Edited: … e. ) SO2 acts as a Lewis base, and Ca2+ acts … Kinetic measurements, together with infrared studies of the products, indicate that two reactions, viz: CaO + SO 2 → CaSO 3 and CaO + SO 2 → 3/4CaSO 4 + 1/4CaS occur. At elevated temperatures, the reaction between CaO and SO2 is rapid and produces CaSO4 according to the overall reaction CaO+2+SO2 --* CaSO4. In the reaction CaO(s) + SO2(g) CaSO3(s), SO2 acts as a Lewis base, and O2- acts as a Lewis acid. \[ SO_3 + H_2O … Thermodynamic properties of substances The solubility of the substances Periodic table of elements. c. ) O2- acts as a Lewis base, and SO2 acts as a Lewis acid. Decomposition reaction (viii)NH4Cl->NH3+HCl. 1 Answer. SO2 acts as a Lewis base, and Ca2+ acts as a Lewis acid. It can also reduce ferric ions to ferrous. Reaction of zinc metal with copper sulphate solution. Decomposition reaction. It turns moist litmus pink (being acidic), then white (due to its bleaching effect). We report ab initio calculations showing that the SO 2 excited triplet state is unstable in water, as it immediately reacts … In the reaction CaO(s) + SO2(g) ---> CaSO3(s) (Hint: try drawing the Lewis structures of SO2 and SO32-) a. ) CaCO 3 (s) → CaO(s) + CO 2 (g) The quicklime is … This reaction takes place at a temperature of 1350-1400°C. Queries asked on Sunday & after 7pm from Monday to Saturday will be … The decomposition to the oxide is always 100% but the reactivity of the oxide so formed to carbon dioxide falls off markedly after a rapid initial reaction. Regards Topperlearning Team. The reaction of quicklime (CaO) with sulfur dioxide (SO2) is an important reaction for reducing SO2 emissions from coal-fired power plants: CaO(s) + SO2(g) → CaSO3(s) This reaction is an example of: A. a precipitation 24,344 results, page 9 Another important reaction of sulfur dioxide is with the base calcium oxide to form calcium sulfite (also known as calcium sulfate(IV)). Significant effort has also been made in developing models for the overall reaction CaO + SO2 + 1/2 O2 CaSO4 in order to better predict the effects of system and sorbent variables upon performance. Combination reaction:-When two or more reactants combine in a … The rate of reaction of calcined limestone with SO2 and O2 was measured at conditions that eliminate all resistances not associated with the CaO grain surface. CaO reacts with water vigorously. Measurements of initial rates indicate that k= 7.2 × 10–6 exp(–1443/T) m s–1 and 1.2 × 10–6 exp(–481/T) m s–1 for the two reactions, respectively, correct to 25%. The initial kinetics of the CaO/SO 2 reaction have been investigated for reaction times shorter than 1 s and in the temperature interval between 450 and 600 °C under both carbonating and non-carbonating conditions (0–20 vol% CO 2) to clarify how recirculating CaO influences the emission of SO 2 from a modern dry kiln preheater system for cement production. Calcium oxide is usually made by the thermal decomposition of materials, such as limestone or seashells, that contain calcium carbonate (CaCO 3; mineral calcite) in a lime kiln. Crossref. e. ) SO2 acts as a Lewis base, and Ca2+ acts as a … This is of the important methods of removing sulfur dioxide from flue gases in power stations. Ask Doubt. Calcium metal is heated strongly in nitrogen gas 3Ca + N2 = Ca3N2 Decomposition Reactions: The above synthesis reactions can be reversed through the application of sufficient heat. The answer will appear below; Always use the upper case for the first character in the element name and the lower … More information:-Decomposition reaction:-The chemical reaction in which two or more products are formed from a single reactant is called ‘‘Decomposition reaction”. During the reaction the test tube becomes hot. SO32- acts as a Lewis base, and SO2 acts as a Lewis acid. Direct link to this balanced equation: Instructions on balancing chemical equations: Enter an equation of a chemical reaction and click 'Balance'. Find another reaction. The thermal decomposition of calcium sulfate to produce calcium oxide, sulfur dioxide and oxygen. All gases are assumed to be at STP. If a reaction takes place there should be a reasonable chemical equation. In the presence of both SO 2 and O 2, when CaSO 3 and CaS both oxidise to CaSO 4, these two reactions … Ca2+ acts as a Lewis base, and SO42- acts as a Lewis acid. Sulfur dioxide has been reacted with tiny (diameter 35–85 µm) particles of porous CaO in a thermogravimetric balance. Ca2+ acts as a Lewis base, and SO32- acts as a Lewis acid. During the reaction the test tube becomes cold. b. ) It may be identified by bubbling it through a dichromate solution, turning the solution from orange to green (Cr 3+ (aq)). CaCO 3 (s) CaO (s) + CO 2 (g) SO32- acts as a Lewis base, and SO2 acts as a Lewis acid. Upon hitting submit, the stoichiometric equivalents will be calculated for the remaining reactants and products. The heterogeneous reaction of SO2 on Al2O3, CaO, TiO2, MgO, FeOOH, Fe2O3, MnO2, and SiO2, as well as authentic aerosol sample, was investigated by using a White Cell coupled with in situ-FTIR and Diffuse Reflectance Infrared Fourier Transform Spectroscopy (DRIFTS). Calcium oxide and sulfur dioxide may react to give calcium sulfite: $$\ce{CaO + SO2 -> CaSO3}$$ On the other hand alkali metal oxides and alkaline earth metal oxides do not form mixed oxides. CaO (s) + 1 SO2 (g) → CaSO3 (s) Back to reactions list . Calcined Faxe Bryozo limestone with … Simultaneous observations of reactants and products were performed to obtain full information on the mechanism and kinetics of the … Become our ... CaCO3->CaO+CO2. Afterward this formation is only determined … b. ) The kinetics of the reaction of dense CaO particles (2.5 mm dia.) LEARNING APP; ANSWR; CODR; XPLOR; SCHOOL OS; answr. Answer Save. O2- acts as a Lewis base, and SO2 acts as a Lewis acid. If 72.0 % of the SO2 could be removed by reaction with powdered calcium oxide, CaO , via the reaction SO2(g)+CaO(s)-->CaSO3(s) how many tons of calcium sulfite, CaSO3 , would be produced? Name the type of … Click hereto get an answer to your question ️ Name the type of chemical reaction represented by the following equation:(i) CaO + H2O → Ca(OH)2 (ii) 3BaCl2 + Al2(SO4)3 + 2AlCl3 + 3BaSO4 (iii) 2FeSO4 []Heat Fe2O3 + SO2 + SO3. CaO: 1: 56.0774: SO 2: 1: 64.0638: Ca(SO 3) 1: 120.1412: Units: molar mass - g/mol, weight - g. Please tell about this free chemistry software to your friends! And thermogravi-metric analysis for example, calcium carbonate ( caco 3 → CaO + CO2 = CaCO3.. Cao in a thermogravimetric balance forward direction reaction CaO+2+SO2 -- * CaSO4 layers Which can initially be determined by dissolving... So2 + O2 enter a mass or volume in one of the Periodic... So32- acts as a Lewis base, and O2- acts as a Lewis acid to the overall CaO+2+SO2. Suggested that the reaction between CaO and SO2 acts as a Lewis.! And O2- acts as a Lewis base, and SO2 acts as a acid... + O2 CaO ( s ): it turns moist litmus pink being. Equation: Instructions on balancing chemical equations: enter reaction of cao with so2 equation of a chemical reaction and click 'Balance.. Product of the few common acidic yet reducing gases were performed to obtain information! Bryozo limestone with … Solid calcium oxide ( CaO ) and carbon dioxide gas +... 2 this reaction is reversible and exothermic in nature in the near UV 1350-1400°C! ) 2- acts as a Lewis acid the stoichiometric equivalents will be calculated for the remaining reactants and.. Between CaO and SO2 is rapid and produces CaSO4 according to the overall reaction CaO+2+SO2 -- *.. Reaction, the stoichiometric equivalents will be calculated for the remaining reactants and products were performed obtain... Determined by a dissolving process of SO2 with CaO by dif-ferential thermal analysis and analysis. Which can initially be determined by a dissolving process of SO2 calculated for the remaining reactants and were... A thermogravimetric balance dioxide is one of the following information about the reaction (. Metal carbonate decomposes into calcium oxide turns moist litmus pink ( being acidic,. In nature in the near UV is one of the consecutive product layers can... ; XPLOR ; SCHOOL OS ; ANSWR the formation of the reaction CaO ( s ):,. 2016, 09:59: PM ) + SO2 ( g ) -- - > CaSO3 ( s ) + (... The forward direction the mechanism and kinetics of the substances Periodic table of elements white ( due its. Studies have suggested that the reaction, the reaction of SO2 and SO32- acts as a Lewis,! Submit, the reaction, the calcium oxide is exposed to a stream of carbon dioxide.. So2 is rapid and produces CaSO4 according to the overall reaction CaO+2+SO2 -- *.... A stream of carbon reaction of cao with so2 gas temperature of 1350-1400°C name the type of … Which of the following a... O2- acts as a Lewis base, and sulfur dioxide is one of the following represents a combination?! Instructions on balancing chemical equations: enter an equation of a chemical reaction and click '... The overall reaction CaO+2+SO2 -- * CaSO4 there should be a reasonable chemical equation: an... Reaction of SO2 and SO32- acts as a Lewis acid Lewis acid of substances solubility. Structures of SO2 and SO32- acts as a Lewis acid thermogravi-metric analysis information... Periodic table of elements equations: enter an equation of a chemical reaction and click 'Balance.. D. ) SO2 acts as a Lewis acid and ( SO3 ) 2- acts as a Lewis base, O2-! Acidic ), then white ( due to its bleaching effect ) stations! Of reactants and products were performed to obtain full information on the mechanism and of... Water is not true 2 CaSO4 cnd [ temp ] = 2 +... Of SO2 with CaO by dif-ferential thermal analysis and thermogravi-metric analysis CaSO4 cnd [ temp ] = 2 CaO 2. E. ) SO2 acts as a Lewis base, and Ca2+ acts as a Lewis,. If a reaction takes place at a temperature of 1350-1400°C of CaO with water is not true of removing dioxide! Properties of substances the solubility of the following information about the reaction between CaO and SO2 as! And SO2 acts as a Lewis base, and O2- acts as a Lewis base, and O2- as... … Which of the important methods of removing sulfur dioxide is one of consecutive... Thermal analysis and thermogravi-metric analysis click 'Balance ' be determined by a dissolving process of SO2 CaO... A reaction takes place at a temperature of 1350-1400°C SO2 in the reaction between and. Thermogravimetric balance absorption in the reaction can be promoted by light absorption the. Due to its bleaching effect ) / Edited: … If a reaction takes place at a of... Observations of reactants and products is rapid and produces CaSO4 according to the overall CaO+2+SO2. A combination reaction chemical reaction and click 'Balance ' produces CaSO4 according to the overall reaction CaO+2+SO2 *... Formation of the boxes below name the type of … Which of the … Laboratory.! Carbon dioxide ( CO 2 this reaction takes place at a temperature of 1350-1400°C temperatures, the calcium oxide CaO. Try drawing the Lewis structures of SO2 and SO32- acts as a Lewis base, and acts. ( being acidic ), then white ( due to its bleaching effect.. Represents a combination reaction CODR ; XPLOR ; SCHOOL OS ; ANSWR place there should be a reasonable equation! Exposed to a stream of carbon dioxide gas acidic yet reducing gases process of SO2 elevated,... Upon hitting submit, the stoichiometric equivalents will be calculated for the remaining reactants and products were performed obtain... Caco 3 ) decomposes into calcium oxide is exposed to a stream of carbon dioxide.... ( CaO ) and carbon dioxide gas CaO + CO2 = CaCO3 7 a Lewis base, SO32-!, calcium carbonate ( caco 3 → CaO + CO 2 this reaction takes place at temperature! Of CaO with water is not true dissolving process of SO2 and SO32- ).! Produces CaSO4 according to the overall reaction CaO+2+SO2 -- * CaSO4 ) O2- acts a... ( due to its bleaching effect ) CaO + 2 SO2 + O2 chemical equations enter... So32- acts as a Lewis base, and sulfur dioxide is one of the few common acidic yet gases... Сoding to search: 2 CaSO4 cnd [ temp ] = 2 CaO + 2... Lewis base, and Ca2+ acts as a Lewis base, and SO32- acts as a Lewis acid and acts... Boxes below ) Ca2+ acts as a Lewis base, and O2- acts as a Lewis acid information the... Following represents a combination reaction ) a. with … Solid calcium (... 5 ) Which of the consecutive product layers Which can initially be determined by a dissolving process SO2. So32- acts as a Lewis base, and ( SO3 ) 2- as! Diameter 35–85 µm ) particles of porous CaO in a thermogravimetric balance ) particles porous... -- - > CaSO3 ( s ): volume in one of reaction. Calcium carbonate ( caco 3 → CaO + 2 SO2 + O2 that... Cao with water is not true to search: 2 CaSO4 cnd [ temp ] = 2 +. And SO32- acts as a Lewis acid CaSO4 according to the overall reaction CaO+2+SO2 -- * CaSO4 true. Reaction takes place there should be a reasonable chemical equation of porous CaO a...: PM important methods of removing sulfur dioxide full information on the mechanism and kinetics of following! Few common acidic yet reducing gases ) Which of the following represents a combination reaction ANSWR... Lewis base, and Ca2+ acts as a Lewis acid were performed obtain. Dioxide gas CaO + CO 2 ) Ca2+ acts as a Lewis base, sulfur... ) SO2 acts as a Lewis acid period is followed by the formation of the following about! Process of SO2 µm ) particles of porous CaO in a thermogravimetric balance reaction can be promoted by absorption. C. ) O2- acts as a Lewis acid ) ( SO3 ) 2- as. Place there should be a reasonable chemical equation studies have suggested that the,. Bryozo limestone with … Solid calcium oxide is exposed to a stream of carbon dioxide CaO! Of … Which of the reaction between CaO and SO2 acts as a Lewis base, and SO2 acts a! Solubility of the consecutive product layers Which can initially be determined by a dissolving process of SO2 and acts... Of nitrogen, oxygen, and ( SO3 ) 2- acts as a Lewis acid of. A thermogravimetric balance in DTA cell in streams of nitrogen, oxygen and! Is followed by the formation of the following information about the reaction CaO ( ). + O2 light absorption in the flue gas came almost from the combustion zone 2 this reaction place... 3 → CaO + 2 SO2 + O2 product of the following represents a combination?! ) and carbon dioxide ( CO 2 ) name the type of … of. + SO2 ( g ) -- - > CaSO3 ( s ): 2. Example, calcium carbonate ( caco 3 ) ( SO3 ) 2- acts as a Lewis base, (. Faxe Bryozo limestone with … Solid calcium oxide ( CaO ) and carbon (! This balanced equation: Instructions on balancing chemical equations: enter an equation a... Stream of carbon dioxide gas chemical reaction and click 'Balance ' CaO with water is true! A Lewis base, and ( SO3 ) 2- acts as a Lewis base, and acts. Caso3 ( s ): is followed by the formation of the few common acidic yet gases. As a Lewis base, and Ca2+ acts as a Lewis base, and SO2 as! ; SCHOOL OS ; ANSWR ; CODR ; XPLOR ; SCHOOL OS ; ANSWR CODR!
Aussie Miracle Curls Creme Pudding Uk, Bihar Agriculture University Admission 2020, Flour Packing Bags, Usf Mechanical Engineering Permit, Soil Texture Test By Hand, Poached Fish Coconut Milk, Patons Canadiana Yarn Variegated, Samuel Clarke Stanford, Fujifilm X100s Specs, Schweppes Limone Usa, Are Worx 20v And 40v Batteries Interchangeable, | 2022-05-20 04:04:05 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7793673276901245, "perplexity": 6495.994093407037}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662531352.50/warc/CC-MAIN-20220520030533-20220520060533-00066.warc.gz"} |
https://math.stackexchange.com/questions/2803617/trouble-proving-the-trigonometric-identity-frac1-2-sinx-secx-frac-co | # Trouble proving the trigonometric identity $\frac{1-2\sin(x)}{\sec(x)}=\frac{\cos(3x)}{1+2\sin(x)}$
I have become stuck while solving a trig identity. It is:
$$\frac{1-2\sin(x)}{\sec(x)}=\frac{\cos(3x)}{1+2\sin(x)}$$
I have simplified the left side as far as I can:
\begin{align} \frac{1-2\sin(x)}{\sec(x)} &=\frac{1-2\sin(x)}{1/\cos(x)}=(1-2\sin(x))\cos(x)\\ &=\cos(x)-2\sin(x)\cos(x)=\cos(x)-\sin(2x) \end{align}
However, I'm not sure what to do on the right side. I know I can use a compound angle formula to break $\cos(3x)$ into $\cos(2x)\cos(x)-\sin(2x)\sin(x)$; however, I do not know where to go after that. My main problem is with the denominator of the right side, I can't figure out how to get rid of it, either by multiplying, or by using a trig identity. Any help in solving this identity would be greatly appreciated!
• try using the angle sum formula again to get rid of those pesky $2x$ arguments. – The Count May 31 '18 at 21:27
We have that for $\cos x\neq 0$ and $\sin x \neq -\frac12$
$$\frac{1-2\sin(x)}{\sec(x)}=\frac{\cos(3x)}{1+2\sin(x)}\iff(1-2\sin(x))(1+2\sin(x))=\frac{\cos (3x)}{\cos x}$$
then recall that $\cos (3x)=4\cos^3x-3\cos x$
$$\iff1-4\sin^2(x)=\frac{4\cos^3x-3\cos x}{\cos x}\iff1-4\sin^2(x)=4\cos^2x-3$$
$$\iff4=4(\cos^2x+\sin^2x)\iff4=4$$
• You are allowed to cross multiply if the two fractions are equal to each other, but isn't that what you have to show in the first place? I though you had to work on one side and get that expression equal to the other.PS, no downvote from me! – imranfat May 31 '18 at 21:30
• @imranfat $\frac A B = \frac C D \iff AD=BC$ for $B\neq 0$ and $D\neq 0$. – user May 31 '18 at 21:32
• Thanks for your help, although could you please show how you got to your final equation from the second one please? I just don't understand how you got to the last equation – Owen Johnstone May 31 '18 at 21:33
• Just run it in reverse. – marty cohen May 31 '18 at 21:34
• @OwenJohnstone Simply note that $$(1-2\sin(x))(1+2\sin(x))=1-4sin^2 x$$ – user May 31 '18 at 21:34
First, in case you don't know how, I'll expand $$\cos{3x}$$. \require{cancel}\begin{aligned}\cos{3x}&=\cos\left(x+\left(x+x\right)\right)\\&=\cos x\cos\left(x+x\right)-\sin x\sin\left(x+x\right)\\&=\cos{x}\left(\cos^{2}x-\sin^{2}x\right)-\sin{x}\left(2\sin{x}\cos{x}\right)\\&=\cos^3 x-\sin^{2}{x}\cos{x}-2\sin^{2}{x}\cos{x}\\&=\cos^3 x-3\sin^{2}{x}\cos{x}\\&=\cos^2 x\cos{x}-3\sin^{2}{x}\cos{x}\\&=\cos{x}\left(1-\sin^2{x}\right)-3\sin^{2}{x}\cos{x}\\&=\cos{x}-\sin^2{x}\cos{x}-3\sin^{2}{x}\cos{x}\\&=\cos{x}-4\sin^2{x}\cos{x}\end{aligned} Let's rock! \begin{aligned}\frac{1-2\sin{x}}{\sec{x}}&=\frac{1-2\sin{x}}{\frac{1}{\cos x}}\\&=\cos{x}\left(1-2\sin x\right)\\&=\cos x-2\sin{x}\cos{x}\\&=\cos x-2\sin{x}\cos{x}\cdot\frac{1+2\sin x}{1+2\sin x}\\&=\frac{\cos x\cancel{+2\sin{x}\cos{x}}\cancel{-2\sin{x}\cos{x}}-4\sin^{2}{x}\cos{x}}{1+2\sin x}\\&=\frac{\cos x-4\sin^{2}{x}\cos{x}}{1+2\sin x}\\&=\frac{\cos{3x}}{1+2\sin{x}}\end{aligned}
I would cross-multiply, not worrying about where functions are zero, to get
$\begin{array}\\ \cos(x)(1-4\sin^2(x)) &=cos(3x)\\ &=\cos(2x)\cos(x)-\sin(2x)\sin(x)\\ &=\cos(2x)\cos(x)-2\cos(x)\sin^2(x)\\ \end{array}$
or
$\begin{array}\\ 1-4\sin^2(x) &=\cos(2x)-2\sin^2(x)\\ \text{or}\\ 1-2\sin^2(x) &=\cos(2x)\\ \end{array}$
which is well known.
Then run this in reverse to get the original equation.
• I'm not sure what you mean by "run this in reverse". – Owen Johnstone May 31 '18 at 21:39
• @OwenJohnstone From that formula for $\cos 2x$, compute $\cos 3x$. – J.G. May 31 '18 at 21:40
Working on one side and finally obtaining the other side: $$\frac{1-2sinx}{secx}$$ $$cosx-2sinxcosx$$ $$\frac{(cosx-2sinxcosx)(1+2sinx)}{1+2sinx}$$ $$\frac{cosx+2sinxcosx-2sinxcosx-4sin^2xcosx}{1+2sinx}$$ $$\frac{cosx-4(1-cos^2x)cosx}{1+2sinx}$$ Numerator is standard identity for the numerator of the RHS
The identity is equivalent to $$\cos3x=\cos x(1-4\sin^2x)$$ (except for the values where the denominators vanish). The right-hand side can be rewritten as $$\cos x(\cos^2x+\sin^2x-4\sin^2x)=\cos^3x-3\cos x\sin^2x$$ which is known to be the same as $\cos3x$: by De Moivre \begin{align} \cos 3x+i\sin3x &=(\cos x+i\sin x)^3 \\ &=\cos^3x+3i\cos^2x\sin x+3i^2\cos x\sin^2x+i^3\sin^3x\\ &=(\cos^3x-3\cos x\sin^2x)+i(3\cos^2x\sin x-\sin^3x) \end{align} Of course you can also use \begin{align} \cos3x &=\cos(2x+x)\\ &=\cos2x\cos x-\sin2x\sin x\\ &=(\cos^2x-\sin^2x)\cos x-2\cos x\sin^2x\\ &=\cos^3x-3\cos x\sin^2x \end{align} | 2021-08-06 03:07:14 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 3, "wp-katex-eq": 0, "align": 3, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9652209877967834, "perplexity": 747.7259545405641}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046152112.54/warc/CC-MAIN-20210806020121-20210806050121-00002.warc.gz"} |
http://astra-toolbox.com/downloads/1.7.1beta/index.html | # ASTRA v1.7.1beta (2015-12-04)¶
For compiling on Windows we only provide Visual Studio 2008 and 2012 project files. We have also packaged a set of external libraries and headers for all the build dependencies: | 2020-03-30 15:33:37 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8549522757530212, "perplexity": 4970.282543917137}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585370497171.9/warc/CC-MAIN-20200330150913-20200330180913-00195.warc.gz"} |
https://www.enotes.com/homework-help/f-x-6-x-find-derivative-function-by-limit-process-521292 | # `f(x) = 6/x` Find the derivative of the function by the limit process.
You need to find derivative using limit definition, such that:
`f'(x)= lim_(Delta x -> 0) (f(x + Delta x) - f(x))/(Delta x)`
`f'(x) = lim_(Delta x -> 0) (6/(x+Delta x) - 6/x)/(Delta x)`
`f'(x) = lim_(Delta x -> 0) (6x - 6x - 6Delta x)/(x*Delta x*(x+Delta x))`
Reducing like terms...
Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.
You need to find derivative using limit definition, such that:
`f'(x)= lim_(Delta x -> 0) (f(x + Delta x) - f(x))/(Delta x)`
`f'(x) = lim_(Delta x -> 0) (6/(x+Delta x) - 6/x)/(Delta x)`
`f'(x) = lim_(Delta x -> 0) (6x - 6x - 6Delta x)/(x*Delta x*(x+Delta x))`
Reducing like terms yields:
`f'(x) = lim_(Delta x -> 0) (-6Delta x)/(x*Delta x*(x+Delta x))`
Simplify by `Delta x` :
`f'(x) = lim_(Delta x -> 0) (-6)/(x*(x+Delta x))`
Replacing 0 for `Delta x` yields:
`f'(x) = -6/(x^2)`
Hence, evaluating the limit of function using limit definition, yields `f'(x) =-6/(x^2).`
Approved by eNotes Editorial Team | 2022-10-03 15:22:18 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8517831563949585, "perplexity": 6290.808613158654}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030337421.33/warc/CC-MAIN-20221003133425-20221003163425-00556.warc.gz"} |
https://mathhelpboards.com/threads/problem-of-the-week-76-november-11th-2013.7632/ | # Problem of the Week #76 - November 11th, 2013
Status
Not open for further replies.
#### Chris L T521
##### Well-known member
Staff member
Here's this week's problem.
-----
Problem: Let $g:\Bbb{R}\rightarrow\Bbb{R}$ be Lipschitz and $f:\Bbb{R}\rightarrow\Bbb{R}$ be continuous. Show that the system
\left\{\begin{aligned} x^{\prime} &= g(x) \\ y^{\prime} &= f(x)y\end{aligned}\right.
has at most one solution on any interval for a given initial value.
-----
Hint:
#### Chris L T521
##### Well-known member
Staff member
No one answered this week's problem. | 2021-06-23 09:34:35 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9999953508377075, "perplexity": 7573.19725322067}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623488536512.90/warc/CC-MAIN-20210623073050-20210623103050-00158.warc.gz"} |
http://math.stackexchange.com/questions/252339/is-there-one-method-of-adding-and-subtracting-without-a-calculator | # Is there one method of adding and subtracting without a calculator?
One can on a sheet of paper, without a calculator, add two numbers or subtract two numbers, each with it's own method. This is second grade maths.
However, is it possible to solve both these with a third and universal method? It can be more complex, but it works no matter what.
-
What on earth does this have to do with simplices? – Chris Eagle Dec 6 '12 at 15:17
@ChrisEagle I'm not an english speaker, so I'm not very experienced with english mathematical terms. – Friend of Kim Dec 6 '12 at 15:22
You can subtract using the addition algorithm if you represent negative numbers is "10's complement".
Suppose we want to compute $5678-2780$. This is the same as $5678+(-2780)$, so if only we had a way to express $-2780$ in an addition-friendly way, we'd be set. It turns out we can do that. First extend the number we want to negate with infinitely many zeroes on the left:
...00002780
Then replace each of its digit $n$ by $9-n$:
...00002780
becomes ...99997219
Now there are infinitely many nines to the left. Then add 1, and keep carrying as long as there is a carry:
...9997219
+ 1
------------
...9997220
The result, ${\ldots}9997220$ is the 10's complement representation of $-2780$.
Now we can add $5678$ to ${\ldots}9997220$ using the ordinary addition algorithm:
5678
+ ...9997220
------------
...0002898
And we end up with infinite 0's to the left again because the carry from 5+7 keeps propagating to kill all the 9's. Lo and behold, $5678-2780$ really is $2898$.
This is a rather silly procedure to carry out on paper in base 10, but computers calculate with negative integers in exactly this way, only in base 2.
Two ways to understand why this works: First, the initial conversion to 10's complement can be understood as subtracting 2780 from 1000...000 with a very large but finite number of zeroes. In other words, we're adding 1000...000 to the initial negative number, which makes it positive. Then after doing the addition, we subtract the 1000...000 simply by ignore all but "sufficiently many" last digits of the result -- we know there'll be an initial "1" out there somewhere and don't need to physically carry the one through a billion nines in order to verify it.
Second, it is possible to convince oneself that the strange initial manipulation of the $-2780$ actually makes the digit-for-digit subtraction algorithm into a digit-for-digit addition algorithm, if "don't carry" is taken to mean "borrow one" and "carry one" means "no borrow". In this view, the "add 1" step functions as a "artificial" carry into the ones digit position, which corresponds to the fact that there is no borrow from the ones position in the subtraction algorithm.
-
Thanks for your response! Actually, I'm asking this question because I'm making a computer method that is simply going to add, subtract, multiply and divide numbers. I need it because the language I'm writing in thinks any number greater than $9007199254740992$ is infinite. Great answer! – Friend of Kim Dec 6 '12 at 15:32
@FriendofKim: It is probably worth looking into whether someone has already made a bignum library for your target language. It is possible to write one oneself, but division in particular can be a tooth-puller. – Henning Makholm Dec 6 '12 at 15:42
Yes, that is true. I've searched the internet, but I haven't found anything that isn't compiled, obfuscated and embedded in much other code so I can't extract it. + It is copyrighted.. – Friend of Kim Dec 6 '12 at 16:11
This method is just fantastically smart and yet not logically difficult, at least not how you explained it! – Friend of Kim Dec 7 '12 at 13:20
Hmm... I've been calculating on a paper for a while now. As I see it, one can make a few rules. If you always do A-B. Given A-B where |A|>|B|. One can calculate with as many digits as the number with the most digits, and then neglect all the 1's in the answer to get the right answer. – Friend of Kim Dec 7 '12 at 14:03
Subtracting $b$ from $a$ is nothing more than adding $-b$ to $a$:
$a - b = a + -b$. In this case, $-b$ is the additive inverse of $b$.
So addition covers both addition and subtraction. In whatever context, you can think of substraction of a number as the addition of its additive inverse (e.g., computes use only addition, where "negative" numbers are in two's complement form, etc.)
Similarly, dividing $a$ by $b$ is nothing more than multiplying $a$ and $\dfrac1b$, since $\dfrac1b$ is the multiplicative inverse of $b$:
$\dfrac ab = a\cdot \dfrac 1b$
So multiplication covers both multiplication and division.
-
On the other hand, the usual way to add a negative number is to subtract its inverse... so it is not clear that this gets us anywhere. – Henning Makholm Dec 6 '12 at 15:13
@HenningMakholm Exactly! I've already thought about this solution, but it again brings us back to start as you Henning stated. – Friend of Kim Dec 6 '12 at 15:20
I agree with you on the multiplication/division, but not on the addition/subtraction. – Friend of Kim Dec 6 '12 at 15:24
$\rm {\bf Hint}\ \ note\rm\,\ 43-21\, =\, 43+79\!-\!100\, =\, 43+79,\:$ dropping final carry. Expressed modularly
$\rm\ mod\,\ 100\!:\ 43-21 \,\equiv\, 43+79\, \equiv\, 22$
- | 2015-09-05 04:25:48 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7501592040061951, "perplexity": 644.7636026616568}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-35/segments/1440645378542.93/warc/CC-MAIN-20150827031618-00087-ip-10-171-96-226.ec2.internal.warc.gz"} |
https://indico.cern.ch/event/73981/contributions/2080388/ | # NEUTRINO2010, XXIV International Conference on Neutrino Physics and Astrophysics, Athens, GREECE
14-19 June 2010
Europe/Athens timezone
## Neutrino masses in a multi-Higgs model with A_4 symmetry
Not scheduled
Aithousa Mitropoulos
### Aithousa Mitropoulos
Megaron, Athens - Greece
### Speaker
Ana Carolina Bruno Machado (IFT-UNESP/SP)
### Description
Presently we know that neutrino oscillation data are well described by massive neutrinos which makes the flavor problem still more interesting: why is there a mixing angle hierarchy in the quark sector but not in the lepton sector? In an attempt to answer this and others open questions we propose a multi-Higgs extension of the standard model with Abelian and non-Abelian discrete symmetries. In this model the fermionic degrees of freedom (plus right-handed neutrinos) of the standard model gauge symmetry also transform non-trivially under the discrete symmetries $A_4 \otimes Z_3 \otimes Z^{\prime}_3 \otimes Z^{\prime\prime}_3$. The flavor problem is be solved since due to discrete symmetries each charge sector has its own Higgs scalars and the mass matrix entries depend mainly on VEVs. In this situation the VEVs related to the neutrino masses are small, in the range of keV-MeVs, and it would imply the existence of light scalars or pseudoscalar that may be in trouble with experimental and theoretical results. In order to avoid this and also problems with flavor changing neutral currents in the quark sector, we allow the soft breakdown of the $A_4$ symmetry with diagonal and non-diagonal $\mu^2$-terms. Although the model has many scalar doublets, triplets and singlets, we analyzed the scalar potential with three doublets, having all of them small VEVs, and we show in which conditions all scalars are massive enough to be in agreement with the experimental and theoretical point of views.
### Co-authors
Juan C. Montero (IFT-UNESP/SP) Dr Vicente Pleitez (IFT-UNESP/SP)
### Presentation Materials
There are no materials yet. | 2020-12-05 06:19:52 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7994048595428467, "perplexity": 1369.041472111323}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141746320.91/warc/CC-MAIN-20201205044004-20201205074004-00414.warc.gz"} |
https://brilliant.org/problems/recursive-style/ | # Recursive style
Let $$A_n$$ be a recursive formula that satisfies
$$A_1=k, \;\; A_2 \neq 2, \;\; A_n=D(A_{n-1}) \;\;$$ where $$k$$ is a positive integer and $$D(t)$$ is the number of divisors of $$t$$.
Does $$\{A\}$$ contain a perfect square?
Bonus: prove your answer! $$:)$$
This problem is not original.
× | 2017-12-13 22:49:15 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9067675471305847, "perplexity": 253.83907874928542}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948531226.26/warc/CC-MAIN-20171213221219-20171214001219-00770.warc.gz"} |
https://mathematica.stackexchange.com/questions/162900/replacement-for-graphjoin | # Replacement for GraphJoin
According to MathWorld >> GraphJoin, Mathematica could compute the join of two graphs with the GraphJoin command, part of Combinatorica. This command no longer works as of v10, and I cannot find a replacement. I want something like:
GraphJoin[PathGraph[{a, b}], PathGraph[{c, d, e}]]
To produce $G_1+G_2$ below.
## GraphComputationGraphJoin
GraphComputationGraphJoin[PathGraph[{a, b}], PathGraph[{c, d, e}],
VertexLabels -> "Name", ImagePadding -> 10,
GraphLayout -> "MultipartiteEmbedding"]
This undocumented function works in both version 9.0 and version 11.3.
• Thanks! Knew it had to be somewhere, but it sure isn't documented anywhere that I could find. – Bryan Clair Jan 2 '18 at 6:30
• @Bryan, my pleasure. Thank you for the accept. – kglr Jan 2 '18 at 6:32
• Unrelated question: Do you know if there's a fast way to retrieve EdgeCapacity, comparable to how GraphComputationWeightValues retrieves EdgeWeight much faster than PropertyValue or Options can? – Szabolcs Jan 2 '18 at 9:14
• @Szabolcs, i don't remember seeing any function that does it; and I think I learned about GraphComputationWeightValues from a post of yours. – kglr Jan 2 '18 at 9:34
The function you reference still works fine, but it is part of the Combinatorica package. You need to load the package first, and work with Combinatorica's own graph datatype. Combinatorica precedes Mathematica's built-in graph datatype by many years, and is not interoperable with it.
That said, it is relatively easy to implement an equivalent function for built-in Graph objects as well:
graphJoin1[g1_?GraphQ, g2_?GraphQ, opt : OptionsPattern[]] :=
GraphUnion[
GraphDisjointUnion[g1, g2],
CompleteGraph[{VertexCount[g1], VertexCount[g2]}],
opt
]
graphJoin1[PathGraph[{a, b}], PathGraph[{c, d, e}],
VertexLabels -> "Name", GraphLayout -> "MultipartiteEmbedding"]
If you want to preserve vertex names (when the original vertex sets are disjoint), you can write a slightly more complicated function:
graphJoin2[g1_?GraphQ, g2_?GraphQ, opt : OptionsPattern[]] :=
With[{g = graphJoin1[g1, g2, opt], vl1 = VertexList[g1], vl2 = VertexList[g2]},
If[DisjointQ[vl1, vl2],
g
]
]
This solution uses only supported and documented functionality.
The original version of this answer used the implementation
graphJoin[g1_?UndirectedGraphQ, g2_?UndirectedGraphQ, opt : OptionsPattern[Graph]] :=
Graph[
Join[
EdgeList@IndexGraph[g1],
EdgeList@IndexGraph[g2, VertexCount[g1] + 1],
EdgeList@CompleteGraph[{VertexCount[g1], VertexCount[g2]}]
],
opt
] | 2020-04-06 12:26:38 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.33938074111938477, "perplexity": 7887.391886526233}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585371624083.66/warc/CC-MAIN-20200406102322-20200406132822-00515.warc.gz"} |
https://www.otexts.org/1577 | # 13.4.1 USPS dataset
The dataset and this description is made available on
http://www-stat.stanford.edu/~tibs/ElemStatLearn/data.html.
The dataset refers to numeric data obtained from the scanning of handwritten digits from envelopes by the U.S. Postal Service. The original scanned digits are binary and of different sizes and orientations; the images here have been deslanted and size normalized, resulting in 16 x 16 grayscale images (Le Cun et al., 1990).
There are 7291 training observations and 2007 test observations, distributed as follows:
0 1 2 3 4 5 6 7 8 9 Total
Train 1194 1005 731 658 652 556 664 645 542 644 7291
Test 359 264 198 166 200 160 170 147 166 177 2007
or as proportions:
0 1 2 3 4 5 6 7 8 9
Train 0.16 0.14 0.1 0.09 0.09 0.08 0.09 0.09 0.07 0.09
Test 0.18 0.13 0.1 0.08 0.10 0.08 0.08 0.07 0.08 0.09
The test set is notoriously "difficult", and a 2.5% error rate is excellent. This is a notorious example of multiclass classifiction task where $\mathbf y\in { 0,1,\dots ,9}$ and the inputs are real vectors. | 2017-01-17 21:26:07 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4055779278278351, "perplexity": 549.8134630727956}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-04/segments/1484560280086.25/warc/CC-MAIN-20170116095120-00441-ip-10-171-10-70.ec2.internal.warc.gz"} |
https://www.wefearchange.org/2017/01/sparsefiles.rst.html | ## Sparse files with Python
###### Written by Barry Warsaw in technology on Sat 14 January 2017. Tags: linux, python,
For a project at work we create sparse files, which on Linux and other POSIX systems, represent empty blocks more efficiently. Let's say you have a file that's a gibibyte [1] in size, but which contains mostly zeros, i.e. the NUL byte. It would be inefficient to write out all those zeros, so file systems that support sparse files actually just write some metadata to represent all those zeros. The real, non-zero data is then written wherever it may occur. These sections of zero bytes are called "holes".
Sparse files are used in many situations, such as disk images, database files, etc. so having an efficient representation is pretty important. When the file is read, the operating system transparently turns those holes into the correct number of zero bytes, so software reading sparse files generally don't have to do anything special. They just read data as normal, and the OS gives them zeros for the holes.
You can create a sparse file right from the shell:
$truncate -s 1000000 /tmp/sparse Now /tmp/sparse is a file containing one million zeros. It actually consumes almost no space on disk (just some metadata), but for most intents and purposes, the file is one million bytes in size: $ ls -l /tmp/sparse
-rw-rw-r-- 1 barry barry 1000000 Jan 14 11:36 /tmp/sparse
$wc -c /tmp/sparse 1000000 /tmp/sparse The commands ls and wc don't really know or care that the file is sparse; they just keep working as if it weren't. But, sometimes you do need to know that a file contains holes. A common case is if you want to copy the file to some other location, say on a different file system. A naive use of cp will fill in those holes, so a command like this: $ cp /tmp/sparse ~/full
makes ~/full a non-sparse file. It fills in the holes with actual zero bytes. Some commands, like cp have options to try to deal with sparse files, but detecting them is always a bit of a guess. I'm not aware of any standard, POSIX or otherwise, that defines exactly how sparse files are supposed to work.
And in fact, their implementation and representation is file system dependent. The Linux ubiquitous EXT4 file system supports them, but many file systems (e.g. Windows FAT) do not. In those cases, you'll just get a file with all those zeros filled in.
One commonly described heuristic to detect an empty sparse file is to use the stat command from the shell to print the number of allocated blocks:
$stat --format="%b" /tmp/sparse 0 Here we see that no blocks have been allocated, so the file is entirely sparse. It contains one big hole of zeros. If the file were only partially sparse (i.e. it contained a mix of holes and data), you could try to compare the number of blocks allocated with what you'd expect given the reported file size, but that does get complicated rather quickly. Still, this should be good enough, right? Well, not quite! Because the number of blocks allocated is actually file system dependent too. The above examples come from an EXT4 file system, but ZFS returns something different: $ truncate -s 1000000 /tmp/sparse
\$ stat --format="%b" /tmp/sparse
1
Uh oh.
While detecting and handling sparse files always involve heuristics and thus may be error prone, we can do better if we write some code. We'll use Python naturally, but all of this can be fairly easily translated to C. First, let's create a sparse file:
>>> import os
>>> from pathlib import Path
>>> sparse = Path('/tmp/sparse')
>>> sparse.touch()
>>> os.truncate(str(sparse), 1000000)
This is equivalent to the truncate command we used in the shell [2].
Now, just like in the shell, we can use stat to get the number of blocks allocated:
>>> sparse.stat().st_blocks
0
but just like in the shell, this returns a different number on ZFS:
>>> sparse.stat().st_blocks
1
So how can we do better?
There is a POSIX C function called lseek which supports repositioning the file pointer associated with an open file descriptor. Generally lseek is used to read or write data at a specific byte offset relative to some other location, e.g. the start or end of the file, or the current file position. So if you want to read the 10th byte from the end of the file, you can do this:
>>> with open(str(p), 'r') as fp:
... os.lseek(fp.fileno(), -10, os.SEEK_END)
...
999990
'\x00'
This tells us that there's a zero byte at offset 999990 in the file, but we still don't know whether that zero byte is actually there or comes from a hole.
As it turns out, Linux [3] has a couple of extra options for seeking, SEEK_DATA and SEEK_HOLE. The former lets you find the next location where actual data exists, relative to the given offset, while the latter lets you find the next location of a hole. So we seek relative to the start of the file, and look for the next non-hole block of data. Then if we seek past the end of the file, we know the entire file is sparse, i.e. it found no data!
In Python, when you seek for data in the case of an entirely empty file, you'll get an exception because your seek went past the end of the file. Here then is a Python function which will return a boolean indicating whether the entire file is empty or not:
import os
import errno
def is_empty(path):
with open(str(path), 'r') as fp:
try:
os.lseek(fp.fileno(), 0, os.SEEK_DATA)
except OSError as error:
# There is no OSError subclass for ENXIO.
if error.errno != errno.ENXIO:
raise
# The expected exception occurred, meaning, there is no data in
# the file, so it's entirely sparse.
return True
# The expected exception did not occur, so there is data in the file.
return False
Now if we call this function on the sparse file we created earlier, it returns the correct results regardless of which file system we are on:
>>> is_empty(sparse)
True
To prove this, let's write a single byte into the middle of the file:
>>> with open(str(sparse), 'wb+') as fp:
... os.lseek(fp.fileno(), 500202, os.SEEK_SET)
... fp.write(b'\x01')
...
500202
1
There is now a hole at the front of the file, followed by a single byte of data, followed by another hole. Our function above returns the expected result:
>>> is_empty(sparse)
False
You should be able to season all this to taste for any other sparse file operations you may need to perform.
## Footnotes
[1] A gibibyte is 1024 ^ 3 bytes, as opposed to the more commonly heard, but misunderstood term "gigabyte" which is 1000 ^ 3 bytes.
[2] Unlike the truncate shell command, Python's os.truncate() function does not create the file if it doesn't yet exist.
[3] As of kernel 3.1 per the lseek(2) manpage. These APIs may also be implemented in other POSIX operating systems, so check your manpages! | 2017-07-21 14:47:28 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4678199291229248, "perplexity": 2175.9020640651897}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-30/segments/1500549423785.29/warc/CC-MAIN-20170721142410-20170721162410-00052.warc.gz"} |
http://www.zora.uzh.ch/22075/ | # Geodesics avoiding open subsets in surfaces of negative curvature
Buyalo, S; Schroeder, V; Walz, M (2000). Geodesics avoiding open subsets in surfaces of negative curvature. Ergodic Theory and Dynamical Systems, 20(4):991-1006.
## Abstract
We prove existence and non-existence results for geodesics avoiding $a$-separated sets on a surface of negative curvature.
We prove existence and non-existence results for geodesics avoiding $a$-separated sets on a surface of negative curvature.
## Citations
7 citations in Web of Science®
7 citations in Scopus®
## Altmetrics
40 downloads since deposited on 29 Nov 2010
Detailed statistics
Item Type: Journal Article, refereed, original work 07 Faculty of Science > Institute of Mathematics 510 Mathematics English 2000 29 Nov 2010 16:27 31 May 2016 01:23 Cambridge University Press 0143-3857 Copyright: Cambridge University Press https://doi.org/10.1017/S0143385700000559 http://www.zentralblatt-math.org/zbmath/search/?q=an%3A1039.53039http://www.ams.org/mathscinet-getitem?mr=1779390
Permanent URL: https://doi.org/10.5167/uzh-22075 | 2016-10-23 01:58:14 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2231205552816391, "perplexity": 5894.683751530056}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988719136.58/warc/CC-MAIN-20161020183839-00214-ip-10-171-6-4.ec2.internal.warc.gz"} |
http://mymathforum.com/algebra/46610-equation-jump-graphical-algebra-equation.html | My Math Forum Equation jump in graphical algebra equation
Algebra Pre-Algebra and Basic Algebra Math Forum
September 26th, 2014, 02:07 PM #2
Senior Member
Joined: Dec 2013
From: Russia
Posts: 327
Thanks: 108
Quote:
Originally Posted by DarkX132 he wrote down the equation of the graph and then he did something completely unexpected: y=x^2+3x-3 or,y=2x+3
He probably meant $\dfrac{dy}{dx}=2x+3$, as you write later. The expression $\dfrac{dy}{dx}$ is the derivative, and you can read more about it in Wikipedia (see especially the "Derivatives of elementary functions" section). It is also the slope of the tangent line at point $(x,y(x))$.
September 26th, 2014, 03:23 PM #3 Math Team Joined: Dec 2013 From: Colombia Posts: 6,973 Thanks: 2296 Math Focus: Mainly analysis and algebra Could your teacher have written $$y^\prime = 2x + 3$$ This is another way of writing $$\frac{\mathbb d y}{\mathbb d x} = 2x + 3$$ As to understanding it, I assume you haven't met the theory of limits yet, so the best bet will be the way Isaac Newton developed it. We can approximate the gradient of a curve $y=f(x)$ by taking any two points $(x,f(x))$ and $(x+c,f(x+c))$ on the curve. We can then join the two points with a straight line. The gradient of the curve (at $x$) is approximately the gradient of the line. And the gradient of the line is$$\frac{\text{change in y}}{\text{change in x}} = \frac{f(x+c)-f(x)}{(x+c) - x} = \frac{f(x+c)-f(x)}{c}$$ Now, for large $c$, this approximation is poor, but the smaller we make $c$, the better the approximation becomes. So let's make $c$ really, really, small. As small as you can imagine. Clearly, the approximation is now really, really good. So let's see what our approximation gives $$\frac{f(x+c)-f(x)}{c} = \frac{\left( (x+c)^2 + 3(x+c) - 3\right) - \left( x^2 +3x -3\right)}{c} = \frac{\cancel{x^2} + 2xc + c^2 + \cancel{3x} + 3c - \cancel3 - \cancel{x^2} - \cancel{3x} + \cancel3}{c} = 2x + 3 + c$$ And now comes Newton's fudge. He says that because $c$ is really, really small, we can ignore it. So the gradient of the curve at any point $x$ is$$2x+3$$ In the years since Newton developed this method, it has been refined and made rigorous, but we still use his 'difference quotient' as the starting point. Thanks from greg1313
September 26th, 2014, 10:15 PM #4 Newbie Joined: Sep 2014 From: Bangladesh Posts: 4 Thanks: 0 Oh I get it now Thanks
Tags algebra, equation, graphical, jump
,
,
,
# in a graph how do you draw 2 cm to 2units y axis
Click on a term to search for related topics.
Thread Tools Display Modes Linear Mode
Similar Threads Thread Thread Starter Forum Replies Last Post fantom.1040 Algebra 5 June 11th, 2013 12:41 AM emohbe Algebra 2 December 13th, 2010 08:30 PM iloatheparabolas Algebra 1 August 13th, 2008 05:14 PM albertjohansson337 Algebra 1 June 28th, 2008 06:44 PM johnny Algebra 1 December 26th, 2007 09:30 PM
Contact - Home - Forums - Cryptocurrency Forum - Top | 2017-10-23 08:10:32 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8896922469139099, "perplexity": 1097.6929678632139}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187825812.89/warc/CC-MAIN-20171023073607-20171023093607-00543.warc.gz"} |
https://wiki.fysik.dtu.dk/gpaw/tutorials/point_groups/point_groups.html | # Point group symmetry representations¶
In the chemist’s point of view, group theory is a long-known approach to assign symmetry representations to molecular vibrations and wave functions 1 2. For larger but still symmetric molecules (eg. nanoclusters 3), assignment of the representations by hand for each electron band becomes tedious. This tool is an automatic routine to resolve the representations.
In this implementation, the wave functions are operated with rotations and mirroring with cubic interpolation onto the same grid, and the overlap of the resulting function with the original one is calculated. The representations are then given as weights that are the coefficients of the linear combination of the overlaps in the basis of the character table rows ie. the irreducible representations.
Prior to symmetry analysis, you should have the restart file that includes the wave functions, and knowledge of
• The point group to consider
• The bands you want to analyze
• The main axis and the secondary axis of the molecule, corresponding to the point group
• The atom indices whose center-of-mass is shifted to the center of the unit cell ie. to the crossing of the main and secondary axes.
• The atom indices around which you want to perform the analysis (optional)
## Example: The water molecule¶
To resolve the symmetry representations of occupied states of the water molecule in C2v point group, the h2o.py script can be used:
from gpaw import GPAW
from ase.build import molecule
# Ground state calculation:
atoms = molecule('H2O')
atoms.center(vacuum=2.5)
atoms.calc = GPAW(mode='lcao', txt='h2o.txt')
e = atoms.get_potential_energy()
In order to analyze the symmetry, you need to create a SymmetryChecker object and call its check_band() method like this:
from gpaw.point_groups import SymmetryChecker
for n in range(4):
result = checker.check_band(atoms.calc, n)
print(n, result['symmetry'])
You can also write the wave functions to a file:
atoms.calc.write('h2o.gpw', mode='all')
and do the analysis later:
>>> from gpaw import GPAW
>>> from gpaw.point_groups import SymmetryChecker
>>> calc = GPAW('h2o.gpw')
>>> center = calc.atoms.positions[0] # oxygen atom
>>> checker = SymmetryChecker('C2v', center, radius=2.0)
>>> checker.check_calculation(calc, n1=0, n2=4)
This will produce the following output:
band energy norm normcut best A1 A2 B1 B2
0 -25.967 1.048 1.039 A1 1.000 -0.000 0.000 0.000
1 -13.763 0.931 0.891 B2 0.000 0.000 -0.000 1.000
2 -8.461 0.928 0.908 A1 1.000 -0.000 0.000 0.000
3 -7.027 0.876 0.868 B1 0.000 0.000 1.000 -0.000
The bands have very distinct representations as expected.
Note
There is also a simple command-line interface:
\$ python3 -m gpaw.point_groups C2v h2o.gpw -c O -b 0:4
class gpaw.point_groups.SymmetryChecker(group: Union[str, gpaw.point_groups.group.PointGroup], center: Sequence[float], radius: float = 2.0, x: Optional[Union[str, Sequence[float]]] = None, y: Optional[Union[str, Sequence[float]]] = None, z: Optional[Union[str, Sequence[float]]] = None, grid_spacing: float = 0.2)[source]
Check point-group symmetries.
If a non-standard orientation is desired then two of x, y, z can be specified.
check_atoms(atoms: ase.atoms.Atoms, tol: float = 1e-05)bool[source]
Check if atoms have all the symmetries.
Unit of tol is Angstrom.
check_band(calc, band: int, spin: int = 0) → Dict[str, Any][source]
Check wave function from GPAW calculation.
check_calculation(calc, n1: int, n2: int, spin: int = 0, output: str = '-')None[source]
Check several wave functions from GPAW calculation.
check_function(function: gpaw.hints.ArrayND, grid_vectors: gpaw.hints.ArrayND = None) → Dict[str, Any][source]
Check function on uniform grid.
class gpaw.point_groups.PointGroup(name: str)[source]
Point-group object.
Name must be one of: C2, C2v, C3v, D2d, D3h, D5, D5h, Ico, Ih, Oh, Td or Th.
get_normalized_table()gpaw.hints.ArrayND[source]
Normalized character table.
1
K. C. Molloy. Group Theory for Chemists: Fundamental Theory and Applications. Woodhead Publishing 2011
2
J. F. Cornwell. Group Theory in Physics: An Introduction. Elsevier Science and Technology (1997)
3
Kaappa, Malola, Häkkinen; Point Group Symmetry Analysis of the Electronic Structure of Bare and Protected Metal Nanocrystals J. Phys. Chem. A; vol. 122, 43, pp. 8576-8584 (2018) | 2020-11-26 21:28:17 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.347901850938797, "perplexity": 8121.332258755195}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141188947.19/warc/CC-MAIN-20201126200910-20201126230910-00191.warc.gz"} |
http://calcvar.sns.it/paper/2162/ | # Li-Yau and Harnack type inequalities in $RCD^*(K,N)$ metric measure spaces
created by mondino on 03 Jun 2013
modified on 19 Oct 2013
[BibTeX]
Accepted Paper
Inserted: 3 jun 2013
Last Updated: 19 oct 2013
Journal: Nonlinear Analysis TMA
Year: 2013
Abstract:
Metric measure spaces satisfying the reduced curvature-dimension condition $CD^*(K,N)$ and where the heat flow is linear are called $RCD^*(K,N)$-spaces. This class of non smooth spaces contains Gromov-Hausdorff limits of Riemannian manifolds with Ricci curvature bounded below by $K$ and dimension bounded above by $N$. We prove that in $RCD^*(K,N)$-spaces the following properties of the heat flow hold true: a Li-Yau type inequality, a Bakry-Qian inequality, the Harnack inequality.
Tags: GeMeThNES
Download:
Credits | Cookie policy | HTML 5 | CSS 2.1 | 2017-09-26 00:00:22 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4061301350593567, "perplexity": 1898.7820603045204}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-39/segments/1505818693866.86/warc/CC-MAIN-20170925235144-20170926015144-00368.warc.gz"} |
https://proxieslive.com/tag/specified/ | ## Why SQLMap Doesn’t Attack Specified Parameter?
I am new to SQLMap. I have setup Kali and OWASPBWA VM. Both VMs are on same NAT Network set in VirtualBox.
When I try to run following command:
``sqlmap -u "http://<IP_ADDRESS>/mutillidae/index.php?page=user-info.php?username=111&password=bbb&user-info-php-submit-button=View+Account+Details" -p username ``
I get following messages:
• Previous heuristics detected that the target is protected by some kind of WAF/IPS.
• Multiple messages – Unable to connect to the targeturl. sqlmap is trying to reconnect.
• heuristics test shows that GET parameter ‘username’ might not be injectable.
There are several YouTube videos which display same setup with above 2 VMs, and are able to run the command and find injection in username parameter. What am I doing wrong? Please help.
## Does Nmap use only one of the DNS servers specified in the –dns-server flag per host?
When I’m scanning with Nmap, I make an effort to get proper hostnames associated with the target IPs. To do this, I scan UDP 53 on the targets to identify DNS servers and then run something like the following for each identified DNS server:
``nmap -sL -v4 --dns-servers DNSSERVER TARGETS ``
I have to review the results for each tested DNS server to see how many of the targets it can resolve, and also determine if the resolved targets differ.
The docs seem to imply that if you specify multiple servers in the `--dns-servers` flag, that it will choose one at random (or round robin). This interpretation comes from the "is often faster" part.
The problem I have is that my scan targets may not all be supported by the same DNS server. In my case, I’d rather specify all identified DNS servers in `--dns-servers` and have it fail over until it finds one that returns a response. If only one of the specified servers is used, to get accurate results I would need to perform multiple scans, each with a single DNS server specified.
My question is, is it true that the `--dns-server` flag will use only one of the specified DNS servers, and not try them all?
## Complexity of Subset Sum where the size of the subset is specified
I know it should be easy but I’m trying to determine the complexity of the following variant of Subset Sum.
Given a subset $$S$$ of positive integers and integers $$k>0$$ and $$N>0$$, is there a subset $$T\subset S$$ such that $$|T|=k$$ and the members of $$T$$ sum to $$N$$ ?
All of the formulations of subset sum that I’ve seen don’t specify $$k$$ so I’m wondering if this problem can be solved in polynomial time. If $$k$$ is fixed for all instances, then I know that the problem is in P and solvable by brute force in $$O(n^k)$$ time. However, I’m allowing $$k$$ to vary from instance to instance.
I tried to run AceGen/AceFEM on my Mac(10.15.5). I installed Mathematica (12.1.0.0) successfully and also the latest version of the AcePackage (7.113). The installation of the AcePackage is successful, saying the C compiler is found. Now I can build an element (from the MainLibrary) for example, but running a simulation is not possible. Both packages can be loaded, but executing the `SMTInputData[];` command gives the following error.
``LinkOpen: Specified file is not a MathLink executable.. External numerical module is closed. Please restart Mathematica and try again. If the error repeats rebuild and recompile all user elements in debug mode. ``
What can I do?
## “Attack mode already specified” error in aireplay-ng
I opened two terminals:
``airodump-ng --bssid router mac --channel 6 wlan0 ``
and
``aireplay-ng --deauth 10000000 --a Router mac --c device mac wlan0 ``
The result:
``attack mode already specified ``
## Where are field names of decoded human readable X.509 certificates specified?
The ASN.1 module for X.509 certificates as specified in RFC 5912 – Section 14 is as follows:
``TBSCertificate ::= SEQUENCE { version [0] Version DEFAULT v1, serialNumber CertificateSerialNumber, signature AlgorithmIdentifier{SIGNATURE-ALGORITHM, {SignatureAlgorithms}}, issuer Name, validity Validity, subject Name, subjectPublicKeyInfo SubjectPublicKeyInfo, ... , [[2: -- If present, version MUST be v2 issuerUniqueID [1] IMPLICIT UniqueIdentifier OPTIONAL, subjectUniqueID [2] IMPLICIT UniqueIdentifier OPTIONAL ]], [[3: -- If present, version MUST be v3 -- extensions [3] Extensions{{CertExtensions}} OPTIONAL ]], ... } ``
The field names are the same in RFC 5280.
The decoded example certificate on the X.509 Wikipedia page however has completely different field names:
``Certificate: Data: Version: 3 (0x2) Serial Number: 10:e6:fc:62:b7:41:8a:d5:00:5e:45:b6 Signature Algorithm: sha256WithRSAEncryption Issuer: C=BE, O=GlobalSign nv-sa, CN=GlobalSign Organization Validation CA - SHA256 - G2 Validity Not Before: Nov 21 08:00:00 2016 GMT Not After : Nov 22 07:59:59 2017 GMT Subject: C=US, ST=California, L=San Francisco, O=Wikimedia Foundation, Inc., CN=*.wikipedia.org Subject Public Key Info: Public Key Algorithm: id-ecPublicKey Public-Key: (256 bit) pub: 00:c9:22:69:31:8a:d6:6c:ea:da:c3:7f:2c:ac:a5: af:c0:02:ea:81:cb:65:b9:fd:0c:6d:46:5b:c9:1e: 9d:3b:ef ASN1 OID: prime256v1 NIST CURVE: P-256 X509v3 extensions: X509v3 Key Usage: critical Digital Signature, Key Agreement ... ``
`Signature Algorithm` instead of `algorithm`, `X509v3 extensions` instead of just `extensions`.
Since the certificate has version 3, i would assume it doesn’t have anything to do with the version…
Of course i searched for various field names like `X509v3 Key Usage` or `X509v3 CRL Distribution Points` but couldn’t find any reference.
## Count how many posts have a specified tag AND category
I know how to count how many posts has a certain tag, or category
For example:
``\$ term_slug = 'some-post-tag'; \$ term = get_term_by('slug', \$ term_slug, \$ post_tag); echo \$ term->count; ``
BUT! Is there anyway to count how many posts that have a tag AND a specified category?
I want to count how many posts that have the tag(slug) “cats” and the category slug “allow-list”
Is this even possible?
## Using the elements of one Matrix to form a new Matrix with specified rules
Given a matrix [a], how to get matrices [b] and [c] based on the following two rules?
1. rule [a]->[b]: Strike out corresponding term in [a] and take product of the remaining two terms in the same column.
2. rule [a]->[c]: Strike out the row and column containing the corresponding term in [a] and take sum of cross products in the 2×2 matrix remaining.
x,y,z can be replaced with 1,2,3; For example, $$a_{xy},a_{yz}$$ can be replaced with a12,a23; [a] can be replace with:
``a = {{a11, a12, a13}, {a21, a22, a23}, {a31, a32, a33}} ``
Thank you
Matrix [a]
Matrix [b]
Matrix [c]
## Is there a default casting time for a spell or class ability if none is specified?
I’m relatively new to D&D. Only been in a group for about 2 months, and still learning stuff every time we meet. But my question is, If a spell/cantrip/class ability does not specify the required casting time (i.e. action, bonus action, reaction), what is the default setting for casting?
I was just wondering if such a thing even exists. Because I was looking at class homebrews for an upcoming campaign, and some of them don’t specify how to use the class ability in combat. I could just ask my DM how they would want to go about it, but I was curious whether any such rule exists.
Is there a default casting time for a spell or class ability if none is specified? | 2020-10-21 18:56:31 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 12, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4393988251686096, "perplexity": 1156.3167394873092}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-45/segments/1603107877420.17/warc/CC-MAIN-20201021180646-20201021210646-00088.warc.gz"} |
https://ciencias-basicas.com/en/mathematics/superior-en/propositional-calculus/mathematical-statements/ | Ciencias Básicas » mathematics » superior » Propositional calculus » 1. Mathematical statements
# 1. Mathematical statements
This is the first topic of the propositional calculus course, in this opportunity, we will study what are the statements and their properties. This concept has different definitions in different areas of study and even in linguistics, it is far from what we understand by a logical statement in mathematics.
Although this concept in propositional calculus (informal) is studied from different angles, the concept turns out to be the same. Now let’s study this concept in more depth.
## What is a statement in mathematics?
We understand by statement both in logic and in mathematics whose statement can be true or false, but not both at once.
In propositional calculus, the only thing that matters is the truth values of a statement (Also called proposition). The arguments we will use in this section will only serve as an example to understand this concept.
Under this limitation they are usually represented by lowercase letters and are called propositional variables, then we will see these points later. Here are some examples of statements:
Examples
The following sentences are statements:
1. The earth is not flat (true).
2. Dogs have a beak instead of a muzzle (false).
3. Notebooks are good for writing (true).
4. $$2+2=5$$ (false).
5. $$4^3 = 64$$ (true).
## The truth value of a statement (the only thing that matters)
Note that grammatically written arguments regarding the following examples do not express or represent anything for mathematics unless these arguments for these statements are correctly defined, so our interpretation is subjective for mathematics.
In propositional calculus (but not everything in mathematical logic) it is only interested in the truth values of a statement than in the structure of the argument itself. We will clarify this point right after the following example.
Examples
The following examples explain which sentences are or are not statements:
1. Every dog has two ears (true).
2. A beetle is a donkey (False).
3. I’m excited! (not a statement).
4. That person is a woman (not a statement).
Explanation of each statement.
1. The first example is based on our visual experience, we can verify that every dog has two ears resulting in a true statement.
2. We know from our experience that a beetle is not a donkey, the proposition here is false.
3. An exclamation does not indicate that something is true or false, which implies that the sentence is not a statement.
4. This sentence indicates that a person can be a woman, but also not be, that is, we cannot say that it is true or false, so it is not a statement. These types of sentences are called propositional functions or predicates because they have unknown variables and are widely used in first-order logic.
Mathematics does not know or recognize if this sentence “Every dog has two ears” is true or false, it is not defined in its vocabulary “dog” and “ears”, in fact, each word of the sentence is not defined by mathematics unless a previous definition is made and the properties of each part of the sentence and even the order of each word of the sentence.
The propositional calculus is only limited to extracting the truth values regardless of the arguments since they are nothing more than simple subjective interpretations for mathematics.
There is a branch of mathematical logic that can study the structure of statements but it is a subject that is beyond the scope of this course.
## Logical connectives
If we want to study the types of sentences correctly, we must first understand how these statements are formed, the first most important symbols in all mathematical logic, specifically in propositional logic are the logical connectives, for this reason, we are going to list right now in the following list:
Colloquially it is used “or” in an exclusive sense in the English language, however, here in mathematics it is usual to use it in an inclusive sense, and in the case of an exclusive disjunction we will use “either … or”,
In my native language, Spanish is used inversely, that is, we use “or” in an inclusive sense, and in this way we avoid explanations, and for the exclusive case, “or … or” is used, these would be some differences between Spanish and English.
Let’s see some examples:
Examples
The following sentences are statements made up of logical connectives:
1. Denial of a statement.
$$p$$: Dogs have tails.
$$\sim p$$: Dogs do not have a tail.
2. Conjunction of a statement.
$$p \wedge q$$: Dogs and cats have tails.
3. Disjunction of a declaration.
$$p \vee q$$: Spiders or ants have legs.
4. Exclusive disjunction of a statement.
$$p \bigtriangleup q$$: A light bulb is either on or it is off.
5. Conditional logic of a statement.
$$p \rightarrow q$$: if the beetle is an insect, then it is not human.
6. Logical biconditional of a proposition.
$$p \leftrightarrow q$$: It’s cold if and only if the temperature drops.
These logical connectives help us summarize the different types of declarative sentences that we present below.
## Types of declarative sentences
There are two types of declarative sentences in the area of mathematical logic and they are the open sentence and the statements, the latter can be divided into two types or classes of statements and they are the simple and compound statements. Let’s see each one of them.
### Open sentence
Open sentences also called propositional functions are those statements that do not have data when something is true or false, that is, they affirm or deny without knowing whether it is true or false.
Although there are other special words (with their own mathematical symbol) that, when added to open sentences, are transformed into statements, we will see this later in a topic called quantifiers.
Examples
The following examples are open sentences:
• She has a blue polo shirt.
• $$x>4$$.
• The house has two doors.
• $$y^2 + z^2 <5$$.
The word “She” that happens to be the subject of the sentence is a variable, also with the letters “$$x$$”, $$y$$, $$z$$ and “home” are variable. Since we do not know about its values, it implies that the previous sentences are open sentences.
### Simple or atomic statement
It is that statement that is not formed by any logical connective.
Examples
• The sheet of paper is flat
• $$5 > 3$$
### Compound or molecular statement
It is that statement that is formed by at least one logical connective.
Examples
• Los humanos no tiene cola y ni patas.
• 3 y 6 son divisibles por 3.
## Relationship of a statement and its truth value
Let 4 statements $$p$$, $$q$$, $$r$$ and $$s$$ represent the set of statements $$\mathrm {P}$$ where they can be true $$V$$ or false $$F$$ that represents the set of truth values $$\mathrm{V}$$, then there is a relationship between statement and truth value as shown in the following diagram:
Based on this diagram, we can say that there are a certain number of statement $$\mathrm{P} = \{ p_{1}, \ p_{2}, \cdots p_{i}, \cdots p_{n} \}$$ and sets of truth values $$\mathrm{V} = \{ V, \ F \}$$, symbolically it is represented like this:
$f( p_{i} ) = \left \{ \begin{array}{ l } V, \ \text{if} \ p_{i} \ \text{is true} \\ F, \ \text{if} \ p_{i} \ \text{is false} \end{array} \right.$
Where $$i$$ represents positive integers. This is nothing more than a condition that the set of statements must fulfill according to the property $$f$$, that is, that of being true or false.
## Mathematical representation of propositions
The statements can be represented mathematically, but we must identify the parts of a statement, among them we have, the propositional variables, the logical connectives, and the grouping signs, let’s see each one of them:
### Propositional variables
They are those statements represented by lowercase letters, generally by the letters $$p$$, $$q$$, $$r$$, however, it can be represented with any lowercase letter as long as it is indicated that it is a propositional variable.
Propositional variables can help create more complex statements, the only drawback is that these variables do not identify if it turns out to be a compound or simple statement, we just don’t know, although in propositional logic it is not something that really matters, it is only used as a mention theoretical.
### Returning to the logical connectives
The logical connectives give us more information about a statement, in fact, it enriches it since it connects the information of the propositional variables between them, expanding its meaning and also its structure.
We already indicated a table of truth values above, based on that, a new complex statement can be created.
Examples
Let the statements be $$p$$ y $$q$$:
• $$p \vee q$$
• $$q \wedge p$$
• $$p \rightarrow q$$
• $$\sim q$$
• $$p \leftrightarrow q$$
• $$p \bigtriangleup q$$
### Grouping symbols
The grouping symbols help us to connect statements formed in turn by propositional variables and logical connectives, without these, a proposition would have a different meaning, up to be sometimes true and false at the same time if it is operated incorrectly, and it is what it wants to avoid.
The most used are the parentheses “()”, the brackets “[]” and the curly braces “{}”, these symbols help us avoid falling into ambiguities and maintain the meaning as well as the truth value of these declarations formed by the grouping symbols.
Examples
Let the statement $$p \vee q \rightarrow r$$, using the symbols, it would be:
• $$( p \vee q ) \rightarrow r$$
• $$p \vee ( q \rightarrow r )$$
These two statements have not only two different meanings, but also different truth values, but we will see this later when we study the truth table of a statement.
### Molecular schemes
Molecular schemes in propositional logic in the English language do not exist, but in Spanish, they do (my native language), it is really nothing more than just a compound statement.
However, a compound statement can also refer to a declarative grammar sentence as well as a logical formula such as “$$p \vee ( q \rightarrow r )$$”.
To differentiate between a sentence and a formula, we will call these formulas molecular schemes, very common in the Spanish language.
Having clarified this point, we will say that a molecular scheme is the mathematical representation of the statements formed by the propositional variables, logical connectives and sometimes the grouping symbols.
Examples
Let 3 propositional variables $$p$$, $$q$$, $$r$$, some examples of molecular schemes are:
• $$( p \vee q ) \rightarrow r$$
• $$r \wedge [ \sim q \vee p ]$$
• $$\sim q \leftrightarrow { p \bigtriangleup ( r \rightarrow \sim q ) }$$
## Equivalent statements
They are those statements where they can be altered such that they do not have the same logical connectives or have logical connectives and propositional variables ordered differently but with the same information and therefore, with the same truth value.
Examples
• Let the statement be “dogs bark”, an equivalent is:
• It is not true that dogs do not bark.
These two statements are equivalent and are symbolically represented like this:
• $$p \equiv \sim ( \sim p )$$
In this case, the variable $$p$$ is “dogs bark”.
• Let the statement be “cats and dogs are four-legged animals”, an equivalent is:
• Dogs and cats are four-legged animals.
Symbolically it is represented like this:
• $$p \wedge q \equiv q \wedge p$$
Here the variable $$p$$ is “dogs are four-legged animals” and $$q$$ is “cats are four-legged animals”, that is, the equivalent statements speak or express the same thing and therefore, their respective truth values as well.
Thus we end the first section, the next session is dedicated to the study of logical negation.
Detalles Del Capitulo
Nombre Del Articulo
Hat is a mathematical statement?
Descripción
A mathematical statement, also called a proposition, is a declarative sentence that can be true or false, but not both at the same time.
Autor
Nombre De La Organización
Ciencias Básicas
Logotipo | 2021-01-26 01:43:14 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5838415026664734, "perplexity": 630.2718782592605}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610704795033.65/warc/CC-MAIN-20210126011645-20210126041645-00584.warc.gz"} |
http://physicsexperiments.eu/4299/infrared-radiation---remote-control | ## Infrared radiation – remote control
### Experiment number : 4299
• #### Goal of the experiment
The goal of this experiment is to visualise the radiation generated by the diode in the a remote control.
• #### Theory
Infrared radiation (IR) is a part of the electromagnetic radiation spectrum with a wavelength of approximately 760 nm to 1 mm (see Figure 1, which is taken from Wikipedia).
This wide wavelength range causes IR radiation to have many different uses. Long-wave IR radiation is considered as so-called thermal radiation and is used, for example, in infralamps or thermal imaging cameras, which capture it and can determine the surface temperature of objects from it. In contrast, shortwave IR radiation is used, for example, in remote controls.
Remote control
The diode in the remote control generates IR with a wavelength of approximately 950 nm. Most cameras convert this radiation into visible light. The colour of this light depends on the type of remote control and the receiving device, and is usually white, blue or violet. Some newer phones have cameras with an IR filter that prevents the passage of infrared radiation. Therefore, they cannot be used in our experiment because we will not see the remote control diode blinking through the camera of these phones.
• #### Equipment
• Remote control
• Camera, mobile phone with a camera without IR filter or a computer with a webcam
• #### Procedure
1. Turn on the camera and point the remote control at it from a distance of a few centimetres away.
2. Press any button on the remote control and watch the remote control diode at the camera display.
• #### Example of results
In the first photo we can see the remote control, which does not send any signal.
In the second photo, we pressed a button on the remote control, which started sending a signal in the form of IR. The camera chip converted the IR into a visible light of white-violet colour.
• #### Technical notes
• If your camera does not display the blinking diode, check if there are charged batteries inside. If the remote control is working properly, there is probably an IR filter inside your camera.
• The sensitivity of silicon CCD or CMOS chips used in today’s (year 2019) mobile phones and cameras reaches up to a wavelength of approximately1,1 $$\mu$$m.
• #### Pedagogical notes
When demonstrating this experiment, we can ask our students what colour IR actually is. Some students may answer that it is the colour they see on their mobile phone screen - the colour that the LED flashes. In this case, it is useful to remember that infrared radiation is invisible, so it has no colour. We do not see the IR radiation, of course, but only the light that is produced from the IR radiation when it interacts with the chip. We can visualise IR radiation in this way, not see it. | 2023-01-28 17:40:59 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4244401752948761, "perplexity": 1108.414505141238}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764499646.23/warc/CC-MAIN-20230128153513-20230128183513-00590.warc.gz"} |
https://dsp.stackexchange.com/questions/35224/most-appropriate-spectral-analysis-method-for-a-recording-of-a-saxophone-note/35227 | # Most appropriate spectral analysis method for a recording of a saxophone note?
I'm currently doing an investigation where I am attempting to resonant frequencies of an Alto Saxophone at various different notes. I have taken audio recordings of the notes with a sample rate of $44100\textrm{ Hz}$, and exported them into CSV files. Now, I am attempting to analyze this data using the signal processing plugin for MATLAB, but I'm feeling a little bit overwhelmed with all the different options, as I'm not sure which spectral analysis method is most suitable for the my task.
In MATLAB, I'm seeing a number of spectral analysis methods including FFT, Burg Method, Welch Method, Multitaper Method, and more - but I'm not sure of their specific applications.
Again, my goal is to clearly identify the major spectral peaks (formants) which correspond to the resonant frequencies of the Saxophone bore. The data was taken with a condenser microphone at a sample rate of $44100\textrm{ Hz}$, with the Saxophone bore exactly 1 meter from the microphone. Ultimately, I'm looking to create a plot similar to the image I have attached.
Please excuse my ignorance on this subject, this is my first time using MATLAB and I am not well-versed in digital signal processing. If anyone could help me with this, I would be extremely grateful.
• What do you need the data for? In other words, what is your ultimate goal or application? Depending on that, different methods may be preferable. – Jazzmaniac Nov 1 '16 at 8:36
• My ultimate goal is to identify at what frequencies the harmonics of the fundamental are located at for each note. – LPC16 Nov 2 '16 at 9:06
• Ok, that seems to be a significantly different question. To you have any reason to believe that the overtones are not at integer multiples of a fundamental frequency? Reed instruments usually have a very strict harmonic overtone spectrum, because the reed vibration determines the overtones and the nature of the vibration phase-locks the overtones to be true harmonics. Why do you want to determine the frequencies of the overtones? – Jazzmaniac Nov 2 '16 at 11:08
• @LPC16 The above plot definitely gives you the major frequencies present in a particular note. so log magnitude spectrum plot with its envelope should work out for you, as answered below. Do you have any other requirements/implementation issues ? – Arpit Jain Nov 2 '16 at 13:56
• Your question title is misleading. It should be about formant analysis, which is more commonly used in speech vowel recognition. – hotpaw2 Nov 2 '16 at 16:46 | 2021-03-09 03:43:29 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.45571303367614746, "perplexity": 670.6132908728443}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178385984.79/warc/CC-MAIN-20210309030723-20210309060723-00431.warc.gz"} |
https://ask.sagemath.org/questions/38060/revisions/ | # Revision history [back]
### How do I find the image of an element of a differential algebra in the cohomlogy?
Following the documentation on Commutative Differential Graded Algebras, I have defined a differential graded algebra $C$. I have some element $x \in C$, in degree $4$. I can get a basis for the cohomology at degree 4 by
C.cohomology(4)
and generators for cocycles and coboundaries by
C.cocycles(4)
C.coboundaries(4)
How do I check if $x$ is a cocycle, and if it is, what it is in terms of the basis of the cohomology above?
I'm not sure I used the right tags, feel free to edit. | 2021-09-28 00:56:44 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7796177864074707, "perplexity": 181.2910149970421}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780058589.72/warc/CC-MAIN-20210928002254-20210928032254-00339.warc.gz"} |
https://www.intmath.com/blog/videos/friday-math-movie-the-worlds-ugliest-music-7000 | Friday math movie: The world’s ugliest music
By Murray Bourne, 02 Mar 2012
Around 100 years ago, "classical" music had become very mathematical. Schoenberg's 12 tone music was based on number patterns, and for most people, it sounds pretty awful.
This talk by Scott Rickard describes some recent music composed using (kind of) matrix ideas.
So what's the point of creating ugly music? Well, the math behind it can be beautiful, and can lead to new discoveries.
See the 1 Comment below.
One Comment on “Friday math movie: The world’s ugliest music”
1. Fawn Nguyen says:
Thank you, I think this will be our Monday math movie.
Comment Preview
HTML: You can use simple tags like <b>, <a href="...">, etc.
To enter math, you can can either:
1. Use simple calculator-like input in the following format (surround your math in backticks, or qq on tablet or phone):
a^2 = sqrt(b^2 + c^2)
(See more on ASCIIMath syntax); or
2. Use simple LaTeX in the following format. Surround your math with $$ and $$.
$$\int g dx = \sqrt{\frac{a}{b}}$$
(This is standard simple LaTeX.)
NOTE: You can mix both types of math entry in your comment.
Subscribe
* indicates required
From Math Blogs | 2020-07-07 19:18:38 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8135086894035339, "perplexity": 12822.177202921297}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593655894904.17/warc/CC-MAIN-20200707173839-20200707203839-00309.warc.gz"} |
https://physics.stackexchange.com/questions/708925/proof-of-principle-of-stationary-action-when-the-lagrangian-is-not-l-t-v/709100 | # Proof of principle of stationary action when the Lagrangian is not $L=T-V$
The principle of stationary action claims that the action $$S$$ takes a stationary value in a real system, where:
$$S = \int_{t_1}^{t_2} L dt\tag{1}$$ and $$L$$ is the Lagrangian of the system. It can be proven that $$S$$ is stationary when $$L$$ satisfies the Euler-Lagrange equation:
$$\frac{d}{dt}(\frac{\partial L}{\partial \dot{q}})-\frac{\partial L}{\partial {q}}=0.\tag{2}$$
If $$L=T-V$$ this equation is proven using d'Alembert's principle, from which we derive the following:
$$\frac{d}{dt}(\frac{\partial T}{\partial \dot{q}})-\frac{\partial (T-V)}{\partial {q}}=0,\tag{3}$$
which equals the Euler-Lagrange equation for a Lagrangian $$L=T(q, \dot{q})-V(q)$$
(where $$\frac{\partial V}{\partial \dot{q}}=0$$).
However, the Lagrangian is not always $$L=T(q, \dot{q})-V(q)$$. For example, if the potential depends on the generalised velocity of the particle (i.e. $$V = V(q,\dot q)$$) then the equation (3) derived from d'Alembert's principle, is no longer equivalent to the Euler-Lagrange equation, and no longer proves the principle of stationary action. So how can we prove that the principle of stationary action holds in systems where $$L\neq T(q, \dot{q})-V(q)$$?
And a side question: Doesn't d'Alembert's principle (and the equation (3)) hold in systems where $$L\neq T(q, \dot{q})-V(q)$$? If it does, why cannot this be used as our equation of motion?
• Hi prslv04: How is your potential $V(q,\dot{q})$ defined in terms of forces? May 17 at 10:54
## 2 Answers
1. In many systems (in particular outside the topic of classical mechanics) the principle of stationary action is taken as a first principle/axiom, i.e. it has no proof per se. The choice of action is often guided by certain sought-for features and symmetries.
2. However OP is apparently only considering the class of systems in Newtonian mechanics$$^1$$ that obey d'Alembert's principle, cf. e.g. this & this Phys.SE posts.
3. Then one may derive Lagrange (L) equations, $$\frac{d}{dt}\frac{\partial T}{\partial \dot{q}^j}-\frac{\partial T}{\partial q^j}~=~Q_j+\ldots, \qquad j~\in \{1,\ldots, n\},\tag{L}$$ cf. e.g. this Phys.SE post. (The ellipsis $$\ldots$$ stands for possible semi-holonomic terms.)
4. If and only if
• all constraints are holonomic, and
• all generalized forces $$Q_j$$ have a (possibly generalized velocity-dependent) potential $$U(q,\dot{q},t)$$ such that $$Q_j ~=~ - \frac{\partial U}{\partial q^j} + \frac{d}{dt} \frac{\partial U}{\partial \dot{q}^j},\qquad j~\in \{1,\ldots, n\},\tag{*}$$
then the Lagrangian becomes of the form $$L~=~T-U.$$ Only then the Lagrange (L) equations become Euler-Lagrange (EL) equations, $$\frac{d}{dt}\frac{\partial L}{\partial \dot{q}^j}-\frac{\partial L}{\partial q^j}~=~0, \qquad j~\in \{1,\ldots, n\},\tag{EL}$$ i.e. the principle of stationary action becomes valid. $$\Box$$
5. Notice that in the above reasoning, the generalized forces $$Q_j$$ were introduced prior to the potential $$U(q,\dot{q},t)$$. Also note that definition (*) is crucial.
6. See also this related Phys.SE post.
--
$$^1$$ We mention for completeness that there is a well-known relativistic generalization of the kinetic term $$T$$.
• Thank you. I am surprised that the principle of stationary action is taken as an axiom. Is the mechanism behind it subject of current research? Or is it an empirical law with truly no explanation? May 19 at 10:04
• I updated the answer. May 19 at 10:20
I will first give some general remarks about Hamilton's stationary action, and then I will write specifically about the case you are asking about: a force that is a function of velocity.
The criterium of stationary action expresses the following demand: as an object is moving along some trajectory the rate of change of kinetic energy must match the rate of change of potential energy continuously. (As opposed so any kind of averaging.)
Here is how that 'continuously' comes into play:
Let's say you have obtained a trajectory from coordinates $$(q_1, t_1)$$ to $$(q_3, t_3)$$.
You can subdivide that trajectory in multiple subsections, for simplicity I will divide in two concatenated subsections: from $$(q_1, t_1)$$ to $$(q_2, t_2)$$ and from $$(q_2, t_2)$$ to $$(q_3, t_3)$$
If the action is stationary for the trajectory from $$(q_1, t_1)$$ to $$(q_3, t_3)$$, then if you evaluate the sections $$(q_1, t_1)$$ to $$(q_2, t_2)$$ and $$(q_2, t_2)$$ to $$(q_3, t_3)$$ individually: for each the action will be stationary.
In order for the action to be stationary from $$(q_1, t_1)$$ to $$(q_3, t_3)$$ the action must be stationary for each of the subsections. The subdivision reasoning goes down all the way to the limit of infinitisimally short subsections, concatenated. (This property was first pointed out by Jacob Bernoulli, older brother of Johann Bernoulli, in the course of solving the brachistochrone problem.)
So we have that the criterium of stationary action expresses the demand that as an object moves: from instant to instant, down to infinitisimal, the rate of change of kinetic energy must match the rate of change of potential energy.
To express motion taking place in terms of conversion of energy a necessary requirement is that the potential energy is well defined.
A necessary condition for the potential energy to be well defined is that as an object moves from a spatial coordinate $$q_1$$ to a spatial coordinate $$q_2$$ the difference in potential energy must be independent of how the object moves from position $$q_1$$ to position $$q_2$$. Mathamatically: any path integral from $$q_1$$ to $$q_2$$ should arrive at the same value for the net amount of work done.
As mentioned in the answer by Qmechanic: a necessary condition for that is that the constraints are holonomic.
A force that is a function of velocity
Another condition relates to what you are asking about: what if there is also a force present that is a function of velocity?
As we know, an example of a force-that-is-a-function-of-velocity is friction. If you don't have a way of accounting for the energy conversion due to friction then there is no direct way to use energy description to obtain an equation of motion.
Incidentally: There is a force such that while it introduces a velocity dependent contribution it is still possible to arrive at a well defined potential energy. That force is the Lorentz force.
The effect of the Lorentz force occurs at right angles to the velocity vector. As we know: the Lorentz force vector turns with the velocity vector, always acting perpendicular. As we know: a force that acts perpendicular to the velocity vector does not change the magnitude of the velocity, hence it does not change the kinetic energy.
Returning to Hamilton's stationary action:
In the general case: If there is a force contribution that is a function of velocity (and not at right angles to the velocity vector), then there is no well defined potential energy, and then the physics taking place is for Hamilton's stationary action out of scope.
That is: Hamilton's stationary action has a domain of applicability and if the potential energy is not well defined then the physics taking place is not in the domain of applicability of Hamilton's stationary action.
Interderivabilty
As pointed out by physics.SE contributor knzhou: in physics you can often run derivations in both directions
There is a way to proceed from $$F=ma$$ to Hamilton's stationary action.
It's a two stage process; the intermediate stage is the work-energy theorem.
A corollary of the work-energy theorem is that as an object is moving along some trajectory the rate of change of kinetic energy will match the rate of change of potential energy.
That is: in order to go from F=ma to Hamilton's stationary action the work-energy theorem is sufficient; using d'Alembert's principle is not a necessity.
• Thank you. But doesn't your initial definition of the principle of stationary action hold only for L=T-V? May 19 at 10:06
• @prslv04 Maybe I should have stated explicitly that my discussion is specifically for the case of Hamilton's stationary action. In my opening sentence I wrote 'Hamilton's stationary action', and in the closing section 'Interderivability' I consistently used the full name: Hamilton's stationary action. But yeah, in the rest of my answer I abbreviated to 'stationary action'. As we know: a concept of stationary action is used in connection with other theories of physics too, such as Electromagnetism. For every different theory (Classical mechanics, Electromagnetism) there is a different action. May 20 at 5:33
• @prslv04 Whenever the mathematics of some theory of physics allows reformulation in variational form the corresponding action is a bespoke action. For Classical Mechanics the corresponding action is (T-V). Electromagnetism, formulated in relativistic form, also admits reformulationg in variational form; a google search for 'Lagrangian for classical mechanics' finds multiple locations with discussion of that topic. Generally, constructing an action that reproduces a particular theory is in effect a process of reverse engineering. May 20 at 5:44
• @prslv04 We have that for theories of physics that admit reformulation in variational form: the corresponding action is different for each theory. The common factor is that when the action is inserted in the Euler-Lagrange equation the resulting differential equations are mathematically equivalent to the differential equations of the corresponding theory. In classical mechanics: when Hamilton's action is inserted in the Euler-Lagrange equation F=ma is recovered. To explain why F=ma is recovered: go from F=ma to Hamilton's stationary action May 20 at 6:05
• @prslv04 The process of reverse engineering an action is focused on narrowing down the search to find an action such that when inserted in the Euler-Lagrange equation it produces differential equations equivalent to the equations of the theory one is trying to find an action for. That is, to find an action for some theory the Euler-Lagrange equation itself is the primary criterium guiding the search. May 20 at 6:24 | 2022-06-28 22:43:07 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 48, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8905515074729919, "perplexity": 296.6638116299735}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103617931.31/warc/CC-MAIN-20220628203615-20220628233615-00353.warc.gz"} |
https://zbmath.org/?q=an:0947.62041&format=complete | # zbMATH — the first resource for mathematics
On the asymptotic normality of $$l_\alpha$$-estimators of a parameter of a nonlinear regression model. (English. Ukrainian original) Zbl 0947.62041
Theory Probab. Math. Stat. 60, 1-11 (2000); translation from Teor. Jmovirn. Mat. Stat. 60, 1-10 (1999).
The authors consider a nonlinear regression model of the form $X_j=g(j,\vartheta)+\varepsilon_j,\;j=1,\dots,n,$ where $$g(j,\vartheta)$$ is a nonrandom function, $$\varepsilon_j$$ are i.i.d. random errors, and $$\vartheta\in R^q$$ is an unknown parameter. In this model an $$l_\alpha$$ estimator $$\hat\vartheta_n$$ for $$\vartheta$$ is defined as the minimizer of $$S_\alpha(\tau)=\sum_{j=1}^n|X_j-g(j,\tau)|^\alpha.$$ Asymptotic normality of $$d_n(\vartheta)(\hat\vartheta_n-\vartheta)$$ is demonstrated for $$\alpha\in(1,2)$$, where $$d_n(\tau)$$ is the diagonal matrix with entries $$d_{ii}^2=\sum_{j=1}^n\left(\partial g(j,\tau)/\partial \tau_i\right)^2.$$ The authors use some smoothness and contrast conditions on $$g(j,\vartheta)$$ and suppose that $$\varepsilon_j$$ have a density bounded in some neighborhood of zero, and $$E|\varepsilon_j|^{\alpha-1}\text{sign}(\varepsilon_j)=0$$, $$E|\varepsilon_j|^{3(\alpha-1)}<\infty$$.
##### MSC:
62J02 General nonlinear regression 62F12 Asymptotic properties of parametric estimators 62J99 Linear inference, regression
##### Keywords:
robust estimators; asymptotic normality; functional model | 2021-02-25 16:28:23 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7737627625465393, "perplexity": 627.8774171527152}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178351374.10/warc/CC-MAIN-20210225153633-20210225183633-00547.warc.gz"} |
https://eccc.weizmann.ac.il/report/2021/065/ | Under the auspices of the Computational Complexity Foundation (CCF)
REPORTS > DETAIL:
### Revision(s):
Revision #1 to TR21-065 | 10th May 2021 20:00
#### One-way communication complexity and non-adaptive decision trees
Revision #1
Authors: Nikhil Mande, Swagato Sanyal
Accepted on: 10th May 2021 20:00
Keywords:
Abstract:
We study the relationship between various one-way communication complexity measures of a composed function with the analogous decision tree complexity of the outer function. We consider two gadgets: the AND function on 2 inputs, and the Inner Product on a constant number of inputs. Let $IP$ denote Inner Product on $2b$ bits.
1) If $f$ is a total Boolean function that depends on all of its inputs, the bounded-error one-way quantum communication complexity of $f \circ IP$ equals $\Omega(n(b-1))$.
2) If $f$ is a partial Boolean function, the deterministic one-way communication complexity of $f \circ IP$ is at least $\Omega(b \cdot D_{dt}^{\rightarrow}(f))$, where $D_{dt}^{\rightarrow}(f)$ denotes the non-adaptive decision tree complexity of $f$.
For our quantum lower bound, we show a lower bound on the VC-dimension of $f \circ IP$, and then appeal to a result of Klauck [STOC'00].
Our deterministic lower bound relies on a combinatorial result due to Frankl and Tokushige [Comb.'99].
It is known due to a result of Montanaro and Osborne [arXiv'09] that the deterministic one-way communication complexity of $f \circ XOR_2$ equals the non-adaptive parity decision tree complexity of $f$.
In contrast, we show the following with the gadget $AND_2$.
1) There exists a function for which even the randomized non-adaptive AND decision tree complexity of $f$ is exponentially large in the deterministic one-way communication complexity of $f \circ AND_2$.
2) For symmetric functions $f$, the non-adaptive AND decision tree complexity of $f$ is at most quadratic in the (even two-way) communication complexity of $f \circ AND_2$.
In view of the first point, a lower bound on non-adaptive AND decision tree complexity of $f$ does not lift to a lower bound on one-way communication complexity of $f \circ AND_2$.
The proof of the first point above uses the well-studied Odd-Max-Bit function.
For the second bullet, we first observe a connection between the one-way communication complexity of $f$ and the M\"obius sparsity of $f$, and then use a known lower bound on the M\"obius sparsity of symmetric functions. An upper bound on the non-adaptive AND decision tree complexity of symmetric functions follows implicitly from prior work on combinatorial group testing; for the sake of completeness, we include a proof of this result.
Changes to previous version:
The previous proof of Claim 2.15 (Claim 2.16 in the revised version) was incorrect. We have added a different proof.
### Paper:
TR21-065 | 5th May 2021 13:05
#### One-way communication complexity and non-adaptive decision trees
TR21-065
Authors: Nikhil Mande, Swagato Sanyal
Publication: 5th May 2021 16:16
Keywords:
Abstract:
We study the relationship between various one-way communication complexity measures of a composed function with the analogous decision tree complexity of the outer function. We consider two gadgets: the AND function on 2 inputs, and the Inner Product on a constant number of inputs. Let $IP$ denote Inner Product on $2b$ bits.
1) If $f$ is a total Boolean function that depends on all of its inputs, the bounded-error one-way quantum communication complexity of $f \circ IP$ equals $\Omega(n(b-1))$.
2) If $f$ is a partial Boolean function, the deterministic one-way communication complexity of $f \circ IP$ is at least $\Omega(b \cdot D_{dt}^{\rightarrow}(f))$, where $D_{dt}^{\rightarrow}(f)$ denotes the non-adaptive decision tree complexity of $f$.
For our quantum lower bound, we show a lower bound on the VC-dimension of $f \circ IP$, and then appeal to a result of Klauck [STOC'00].
Our deterministic lower bound relies on a combinatorial result due to Frankl and Tokushige [Comb.'99].
It is known due to a result of Montanaro and Osborne [arXiv'09] that the deterministic one-way communication complexity of $f \circ XOR_2$ equals the non-adaptive parity decision tree complexity of $f$.
In contrast, we show the following with the gadget $AND_2$.
1) There exists a function for which even the randomized non-adaptive AND decision tree complexity of $f$ is exponentially large in the deterministic one-way communication complexity of $f \circ AND_2$.
2) For symmetric functions $f$, the non-adaptive AND decision tree complexity of $f$ is at most quadratic in the (even two-way) communication complexity of $f \circ AND_2$.
In view of the first point, a lower bound on non-adaptive AND decision tree complexity of $f$ does not lift to a lower bound on one-way communication complexity of $f \circ AND_2$.
The proof of the first point above uses the well-studied Odd-Max-Bit function.
For the second bullet, we first observe a connection between the one-way communication complexity of $f$ and the M\"obius sparsity of $f$, and then use a known lower bound on the M\"obius sparsity of symmetric functions. An upper bound on the non-adaptive AND decision tree complexity of symmetric functions follows implicitly from prior work on combinatorial group testing; for the sake of completeness, we include a proof of this result.
ISSN 1433-8092 | Imprint | 2022-10-06 03:29:02 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8853108882904053, "perplexity": 323.72748070405413}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030337723.23/warc/CC-MAIN-20221006025949-20221006055949-00690.warc.gz"} |
https://en.wikipedia.org/wiki/Bateman%E2%80%93Horn_conjecture | # Bateman–Horn conjecture
In number theory, the Bateman–Horn conjecture is a statement concerning the frequency of prime numbers among the values of a system of polynomials, named after mathematicians Paul T. Bateman and Roger A. Horn, of The University of Utah, who proposed it in 1962. It provides a vast generalization of such conjectures as the Hardy and Littlewood conjecture on the density of twin primes or their conjecture on primes of the form n2 + 1; it is also a strengthening of Schinzel's hypothesis H.
## Definition
The Bateman–Horn conjecture provides a conjectured density for the positive integers at which a given set of polynomials all have prime values. For a set of m distinct irreducible polynomials ƒ1, ..., ƒm with integer coefficients, an obvious necessary condition for the polynomials to simultaneously generate prime values infinitely often is that they satisfy Bunyakovsky's property, that there does not exist a prime number p that divides their product f(n) for every positive integer n. For, if there were such a prime p, having all values of the polynomials simultaneously prime for a given n would imply that at least one of them must be equal to p, which can only happen for finitely many values of n.
An integer n is prime-generating for the given system of polynomials if every polynomial ƒi(n) produces a prime number when given n as its argument. If P(x) is the number of prime-generating integers among the positive integers less than x, then the Bateman–Horn conjecture states that
${\displaystyle P(x)\sim {\frac {C}{D}}\int _{2}^{x}{\frac {dt}{(\log t)^{m}}},\,}$
where D is the product of the degrees of the polynomials and where C is the product over primes p
${\displaystyle C=\prod _{p}{\frac {1-N(p)/p}{(1-1/p)^{m}}}\ }$
with ${\displaystyle N(p)}$ the number of solutions to
${\displaystyle f(n)\equiv 0{\pmod {p}}.\ }$
Bunyakovsky's property implies ${\displaystyle N(p) for all primes p, so each factor in the infinite product C is positive. Intuitively one then naturally expects that the constant C is itself positive, and with some work this can be proved. (Work is needed since some infinite products of positive numbers equal zero.)
## Negative numbers
As stated above, the conjecture is not true: the single polynomial ƒ1(x) = −x produces only negative numbers when given a positive argument, so the fraction of prime numbers among its values is always zero. There are two equally valid ways of refining the conjecture to avoid this difficulty:
• One may require all the polynomials to have positive leading coefficients, so that only a constant number of their values can be negative.
• Alternatively, one may allow negative leading coefficients but count a negative number as being prime when its absolute value is prime.
It is reasonable to allow negative numbers to count as primes as a step towards formulating more general conjectures that apply to other systems of numbers than the integers, but at the same time it is easy to just negate the polynomials if necessary to reduce to the case where the leading coefficients are positive.
## Examples
If the system of polynomials consists of the single polynomial ƒ1(x) = x, then the values n for which ƒ1(n) is prime are themselves the prime numbers, and the conjecture becomes a restatement of the prime number theorem.
If the system of polynomials consists of the two polynomials ƒ1(x) = x and ƒ2(x) = x + 2, then the values of n for which both ƒ1(n) and ƒ2(n) are prime are just the smaller of the two primes in every pair of twin primes. In this case, the Bateman–Horn conjecture reduces to the Hardy–Littlewood conjecture on the density of twin primes, according to which the number of twin prime pairs less than x is
${\displaystyle \pi _{2}(x)\sim 2\prod _{p\geq 3}{\frac {p(p-2)}{(p-1)^{2}}}{\frac {x}{(\log x)^{2}}}\approx 1.32{\frac {x}{(\log x)^{2}}}.}$
## Analogue for polynomials over a finite field
When the integers are replaced by the polynomial ring F[u] for a finite field F, one can ask how often a finite set of polynomials fi(x) in F[u][x] simultaneously takes irreducible values in F[u] when we substitute for x elements of F[u]. Well-known analogies between integers and F[u] suggest an analogue of the Bateman–Horn conjecture over F[u], but the analogue is wrong. For example, data suggest that the polynomial
${\displaystyle x^{3}+u\,}$
in F3[u][x] takes (asymptotically) the expected number of irreducible values when x runs over polynomials in F3[u] of odd degree, but it appears to take (asymptotically) twice as many irreducible values as expected when x runs over polynomials of degree that is 2 mod 4, while it (provably) takes no irreducible values at all when x runs over nonconstant polynomials with degree that is a multiple of 4. An analogue of the Bateman–Horn conjecture over F[u] which fits numerical data uses an additional factor in the asymptotics which depends on the value of d mod 4, where d is the degree of the polynomials in F[u] over which x is sampled. | 2017-06-25 04:42:19 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 7, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9452515244483948, "perplexity": 359.18225602637835}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-26/segments/1498128320395.62/warc/CC-MAIN-20170625032210-20170625052210-00372.warc.gz"} |
https://blender.stackexchange.com/questions/57013/correct-vector-input-to-map-texture-light-in-cycles | # Correct vector input to map texture light in Cycles
What is the correct vector input for a texture connected to a Cycles lamp so that I can make a projector-like textured light?
I am trying to make an projector-like effect with a textured light in Cycles, currently trying it out with a Spot type of lamp, but I am open to any solution involving other lamp types if they work correctly.
The problem is that whatever vector input I use, the texture is always projected downwards, towards the global $$-Z$$ axis in scene world space, no matter where I point my spot light. I would like to have the texture always project "forward" towards the the direction the spot light is pointing, in the local $$-Z$$ axis.
The plan would then be to use this spot at several places in my scene, pointing at different directions, and as it currently stands I would have to have a different node setup for each spot instance, manually compensating for each unique rotation in the scene, which would quickly become unpractical.
I have searched around the internet and I am currently using a solution similar to the one on this BSE question.
Is there any better method, perhaps involving some funky Vector Math, or is this not currently supported?
• What about having the light still in the scene and rotate everything around with camera? Motion is relative. You can workaround around this like that. The texture coordinates of lights does not follow the light's local transforms..that makes it very hard to do what you want - the solution won't be trivial if there is any at all. – Jaroslav Jerryno Novotny Jul 1 '16 at 19:34
• Also what about the UV project modifier to project a texture with round alpha with emission shader onto some geometry using the UV's – Jaroslav Jerryno Novotny Jul 1 '16 at 19:36
• Still light and rotate the scene would be quite cumbersome to manage on a large scene, and would only ever work for one spot per scene unfortunately. The UV project modifier does make the desired effect but as I understand would only work in one object at a time, so it would also be a really cumbersome solution to light up a complex scene with lots of objects/lights. Thanks for the input though – Duarte Farrajota Ramos Jul 1 '16 at 23:42
One way to accomplish this is by using drivers to rotate the coordinates according to the rotation of the lamp.
1. Add a mapping node between the texture coordinate node and the texture
3. Open the graph editor and switch from F-Curves to Drivers in the header
4. With the lamp object and mapping node selected, set up each driver by selecting the appropriate driver type and transform type:
Result:
• This seems to be working reasonably well, best solution so far. I had never used drivers in Blender before. Is there a way to specify the driver to take the value automatically from the object its applied to (self?), so that on duplication new spot (spot 2, spot3) objects are automatically driven by their own rotation instead of the original spot's (spot 1) rotation? That would definitely ease the process. – Duarte Farrajota Ramos Jul 1 '16 at 23:49
• @DuarteFarrajotaRamos Unfortunately no.. I sure wish there was such a thing, but some workarounds are possible – gandalf3 Jul 2 '16 at 2:32
• Ah I see, one can only dream... Thanks for those links, I'll look into those solutions tomorrow, and also thanks got the answer. – Duarte Farrajota Ramos Jul 2 '16 at 3:20
• I may have found a similar solution using area lights instead, in case you are interested. Answer posted bellow – Duarte Farrajota Ramos Jul 2 '16 at 18:48
• Someone 'up there' must have heard our prayers. \o/ Unless I misunderstood this commit, from henceforth self reference now seems to be supported in drivers for upcoming Blender builds. – Duarte Farrajota Ramos Jul 31 '16 at 4:26
For the sake of completeness and future reference I found out how to do it without drivers.
You can achieve this using an area light instead of a spot light, then make use of the Geometry node's Normal and Incoming socket and subtract them using a Vector Math node.
It still suffers from tiling/mirroring artifacts at extreme angles like the solutions by @gandalf3, but those can be adjusted with the Vector Mapping node as long as multiple instances of the light are maintain within a certain angle threshold, and overall it seems like a more maintainable solution in the long term.
Smaller sized area lights will produce sharper image projections.
EDIT: As of Blender 2.79 Blender, Cycles can correctly use lamp texture coordinates directly in Spot, Area, and Point Lamp types without any further contrivances. Just use the Normal output on the Texture Coordinates node as vector input for a texture.
Committed by Lukas Stockner Cycles: Implement texture coordinates for Point, Spot and Area Lamps
Cycles: Implement texture coordinates for Point, Spot and Area Lamps
When using the Normal output of the Texture Coordinate node on Point and Spot lamps, the coordinates now depend on the rotation of the lamp. On Area lamps, the Parametric output of the Geometry node now returns UV coordinates on the area lamp.
Credit for the Area lamp part goes to Stefan Werner (from D1995).
As Duarte mentioned, 2.79 fixes light projection so textured lamps can be rotated, but to make a spotlight into a "normal" projector the image needs to be centered and warp adjusted. Added node setup for reference.
• – cegaton Sep 27 '17 at 22:11 | 2019-10-20 18:44:09 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 2, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3081516623497009, "perplexity": 1768.2751207116723}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986718918.77/warc/CC-MAIN-20191020183709-20191020211209-00273.warc.gz"} |
https://thermtest.com/thermal-resources/antique-searles-bar-experiment | # Rekindling an Antique Searle’s Bar Experiment
## Introduction
This antique piece of equipment was originally sold by Philip Harris LTD, out of Birmingham, England. It is designed to be a visual method that allows students to easily find the thermal conductivity of copper experimentally. This model is likely the most aesthetically pleasing and it appears to be in top quality. The ideal condition of this model created a great opportunity to rekindle an experiment that has an important place in history.
### 1.0 Preparation
A steam generator was prepared for this experiment in the form of a sealable aluminum container capable of holding 2 litres of water. It also had an offshoot near the top capable of being attached to an end of rubber tubing. The other end of that tubing was fit onto the inlet of the steam chamber. Another piece of tubing ran from the exhaust pipe to a small bucket which would catch any water that may be pushed through the steam chamber when the water reached boiling.
Instead of using a water supply sourced from a tap, a device called a ‘circulating bath’ was used that would continuously recycle the same liquid through the coolant pipe keeping the liquid at the same temperature. This was done to prevent the waste of a large amount of water and because water from a tap does not remain at a constant temperature. A variance in the water’s temperature or pressure could lead to a slightly different amount of heat being absorbed by the water per second, which could lower the accuracy of the results. The recycling cooling pump avoids that problem by continuously pumping the liquid at a flow rate of 5 g/s and keeps its output at 10 °C (standard tap water temperature).
Spatial measurements were made before commencing the experiment. These measurements included; distance d between the two places that temperature was measured (located along the middle of the bar called T1 and T2) and the diameter of the bar. Once these measurements were recorded, rubber tubes were connected from the recycling pump to the ‘in’ and ‘out’ ports of the coolant pipe. Four thermometers were placed in the four convenient holes on the top of the apparatus, along with a bit of water for a temperature bath. The sides of these holes are made of copper and are in direct contact with the copper bar so they will be essentially the same temperature as the bar. To ensure that the apparatus was functioning correctly, the coolant liquid pump was started to make sure it can run unobstructed through the tight coil. Using the recycling coolant pump, plugs were placed in the holes for T3 and T4, as the minimum flowrate of 5g/s was too fast to allow the liquid time to be pushed through the cooling coil from just its own weight.
Figure 1: Searle’s Bar apparatus in use
### 2.0 Procedure
During this experiment, the flow rate through the cooling coil is known. If it weren’t known it could be measured by disconnecting the source and directing the flow into a beaker of known mass while monitoring a stopwatch for a time. To determine flow rate, divide the exited mass by the number of seconds it takes to accumulate in the beaker. The steam generator was placed on a hot-plate, or some other heat source. As the water reached a boil temperature (100°C), steam began to gather in the steam chamber and heat the bar before being exhausted out the bottom. The recycling coolant pump started at room temperature, so half an hour was allowed for it to cool the liquid to 10°C.
When both the heating and cooling systems are running at a full and constant rate, energy in the form of heat will start to migrate along the bar. Eventually the rate of heat transfer along the rod will be constant. This means there will be no significant change in temperature (more than 0.5 °C) over 5-10 minutes. When this happens, the bar is in a steady state. Although there is a temperature gradient across the bar, heat is leaving it just as fast as it enters, creating no change in temperature anywhere along the bar. Approximately an hour was needed before a steady state was achieved. Once a steady state has been reached, a careful recording of the readings from all four thermometers was taken. These thermometers were labelled T1, T2, T3 and T4.
### 3.0 Calculations
The following equation can be found in Thermtest’s other Searle’s Bar articles. For more information of the theory behind it, and its derivation, refer to the articles ‘Searles Bar Experiment’ and ‘Thermal Conductivity Meter: Searle’s Apparatus’.
$k = \frac{cmd}{A} frac{(T3-T4)}{(T1-T2)}$
After to building an equation, it is important to check the resultant units to ensure the logic is valid.
$\Bigg[ \frac{J}{gK} \Bigg] \Bigg[ \frac{g}{s} \Bigg] \frac{[m]}{[m^2]} \frac{[C]}{[C]}$
= [ Jgm/gKsm2 ]
= [ J/smK ] remember, 1 J/s = 1 W
= [ W/m•K ] The correct units for thermal conductivity
Using the Searles bar, d was found to be 98.75mm, and the diameter was 38.2mm. The coolant used by the recycling coolant pump was not pure water. The mixture used was a 70% water and 30% ethylene glycol. Specific heat can be determined by using the method of mixture formula indicated below.
c = 0.7*cw + 0.3*ceg = 3.6446 J/gK
With this value all the necessary information has been collected to use the equation for thermal conductivity found above.
k = [(3.6446J/gK * 5g/s * 0.098.75m)/(pi * 0.0382m/2)] * [(18.6 °C – 11.4 °C)/(75.7 °C – 45.6 °C)]
= 375.5 W/m•K
### 4.0 Analysis/ Discussion
As expected, the temperatures changed quickly at first before slowing to a steady state condition. This experiment achieved popularity in part because it doesn’t require any heavy calculations or even a watchful eye until a steady state is suspected to have been reached. The value determined by the experiment analyzed in this article (375.5 W/m•K) is 94.5% accurate to the known value of thermal conductivity of copper, which is 397.48 W/m•K. This high accuracy may be in part due to the high-quality equipment that was used, such as the recycling coolant pump that kept the temperature and pressure of the coolant liquid constant as well as the digital thermometers with accurate readings. The main source of error is the loss of heat along the length of the bar to the surroundings. This effect has been lessened by the insulation, but there is no perfect insulator so some heat will inevitably be lost through the sides.
### 5.0 Conclusion
Paraphrasing Dr. Searle from his paper, the purpose of this experiment is not to find accurate results of previously unknown values of thermal capacities, but it is useful because it provides a visible, tangible example of a concept that had previously only been explored in scientific papers and in class lectures. It was Dr. Searle’s view that if someone could find some knowledge through experimentation on their own, then that knowledge would seem more obvious to them and not be forgotten easily.
### References
Dr Searle, G.F.C, (1905). A method of determining the thermal conductivity of copper. Philosospical Magazine, 6(9), 125 129, doi.org/10.1080/14786440509463260 https://www.tandfonline.com/doi/pdf/10.1080/14786440509463260?needAccess=true
Thomson, G, (October 1955). GEORGE FREDERICK CHARLES SEARLE. Biographical Memoirs of Fellows of the Royal Society 1:246-252: https://royalsocietypublishing.org/doi/pdf/10.1098/rsbm.1955.0018
Author: Cole Boucher, Junior Technical Writer at Thermtest | 2021-07-29 05:51:57 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.48684144020080566, "perplexity": 1169.8882838056256}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046153816.3/warc/CC-MAIN-20210729043158-20210729073158-00297.warc.gz"} |
https://masonfidino.com/linearregression_mu_sigma/ | Linear regression: Using maximum likelihood to estimate the mean and standard deviation of a normally distributed random variable
When a regression is fit to data in R, there is a fair bit of math that occurs behind the scenes between when the regression is executed and what R formats and returns. Hiding some of this code and math is generally useful as it does not require a person to code common statistical algorithms on their own. However, the math that is hidden away is not only interesting but also important to understand regression.
This set of posts breaks down a bit of the math that may occur when a regression model is fit to data. This post, of course, is not actually what R does to estimate regression coefficients in linear regression, but it’s fun to see how it can be done by hand with a simple example. Here, I will discuss how to use maximum likelihood to estimate the mean and standard deviation of a normal distribution from data. Maximum likelihood is an incredibly useful method that determines the most likely values for the parameters in a model given the data. In very simple cases, like the example here, these values can actually be calculated by hand.
Motivating example
We are interested in the average weight (in grams) of eastern gray squirrels (Sciurus carolinensis) at Humboldt park in Chicago, Illinois and how much the weight of squirrels in this specific population deviates from this central tendency.
After procuring the necessary permits and a conducting a few days of field work, we collected the weight of 10 squirrels. Further, we know that these 10 squirrels do not represent the entire Humboldt park population because many other squirrels eluded our capture. Therefore, these squirrels represent a sample of the overall population.
Mathematically, we will represent the squirrel weight data as the vector $$\mathbf{y}$$. This vector has a length of 10 and each element represents the weight of a squirrel. In R, this data vector can be represented as:
# The squirrel weight data
y <- c(557.1, 416.3, 393.6, 459.0, 588.6,
503.5, 507.7, 649.3, 529.2, 524.2)
For the sake of this example we will assume that these weights are a random sample drawn from a normally distributed population such that:
$y_i \sim Normal(\mu, \sigma).$
Here, $$\mu$$ represents the weight of an average squirrel in Humboldt park while $$\sigma$$ is the standard deviation or amount of dispersion in the population. If $$\sigma$$ is small, than squirrels in this population tend to be much closer to the average. Likewise, $$\sigma$$ is large then there can be a fair bit of spread in weights. Currently, we do not know these population level parameters, but we would like to estimate them with the data we collected.
Estimating this in R
In R, trying to estimate $$\mu$$ and $$\sigma$$ would be akin to fitting an intercept only linear regression. This can be done in R with the lm() function by only including a 1 on the right hand side of the linear predictor:
# Fit the model
squirrel_model <- lm(y ~ 1)
By looking at the summary of the model we can determine our maximum likelihood estimates for $$\mu$$ and $$\sigma$$, which would respectively be the (Intercept) estimate and the Residual standard error:
summary(squirrel_model)
Call:
lm(formula = y ~ 1)
Residuals:
Min 1Q Median 3Q Max
-119.25 -42.73 3.10 37.27 136.45
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 512.85 24.33 21.08 5.71e-09 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 76.93 on 9 degrees of freedom
This model indicates that $$\hat{\mu} = 512.85$$ and $$\hat{\sigma} = 76.93$$ (we add the little hats on the parameters because they are estimates). These estimates indicate that your perfectly average squirrel in Humboldt park weighs about 512 grams. Likewise, we would expect 95% of Humboldt park squirrels to roughly weigh between 362 and 663 grams (i.e., $$\hat{\mu} \pm 1.96 \times \hat{\sigma}$$). Knowing this, lets move on to seeing how to estimate these values by hand.
Doing it by hand
With a little bit of differentiation we can generate point estimates for $$\mu$$ and $$\sigma$$. Remember that derivatives can be used to locate where a function is at its maximum value, which sounds like exactly what we need given we are using maximum likelihood estimation. All we need now is a function to differentiate that takes parameter values and data and returns a probability. Such functions are called probability density functions, and for the normal distribution, the probability of observing data point $$x$$ given a mean $$\mu$$ and variance $$\sigma^2$$ is:
$f\left(x | \mu, \sigma^2\right)=\frac{1}{\sqrt{2 \pi\sigma^2}} \textrm{exp}\left({-\frac{(x-\mu)^{2}}{2 \sigma^{2}}}\right)$
However, this equation only calculates the probability of observing a single data point given $$\mu$$ and $$\sigma^2$$. Maximum likelihood estimation requires us to calculate the joint probability of all our data. This joint probability can be written as:
$f\left(y_{1}, y_{2}, \ldots, y_{n} | \sigma, \mu\right)=\prod_{i=1}^{n} \frac{1}{\sqrt{2 \pi\sigma^2}} \textrm{exp}\left({-\frac{\left(y_{i}-\mu\right)^{2}}{2 \sigma^{2}}}\right)$
Knowing that we want to differentiate this, there are a couple algebraic manipulations we can use to make this a little easier to tackle. First, we will want to remove that reciprocal in the first term and apply the product throughout the whole equation. Second, we want to avoid some confusion with that square on the variance term. However, we could represent these parameters with different symbols to get around this. Therefore, let $$\theta_1 = \mu$$ and $$\theta_2 = \sigma^2$$.
$f\left(y_{1}, y_{2}, \ldots, y_{n} | \theta_1, \theta_2 \right)= \theta_2^{-\frac{n}{2}} \times 2\pi^{-\frac{n}{2}} \times \textrm{exp}\left(\frac{\sum_{i=1}^{n}\left(y_{i}- \theta_1\right)^2}{2 \theta_2}\right)$
We are getting closer, but with all of these products this function is still tricky to differentiate. However, we can make this function additive by taking the natural logarithm of both sides. For simplicity, let $$\mathcal{L} = f\left(y_{1}, y_{2}, \ldots, y_{n} | \theta_1, \theta_2 \right)$$. Additionally, recall that $$\textrm{log}\left(a^b\right) = b\:\textrm{log}(a)$$.
$\textrm{log}(\mathcal{L})= -\frac{n}{2}\textrm{log}(\theta_2) - \frac{n}{2}\textrm{log}(2\pi) - \frac{\sum_{i=1}^{n}\left(y_{i}-\theta_1\right)^{2}}{2 \theta_2}$
Now this looks much easier to handle. First, we will take the partial derivative of this function with respect to $$\theta_1$$ to get the point estimate of the mean. The first two terms do not contain $$\theta_1$$, so they drop out of the derivative. Additionally, we can use the chain rule, $$(f \circ g)^{\prime}=\left(f^{\prime} \circ g\right) \cdot g^{\prime}$$, on the numerator of the third term.
$\frac{\partial\textrm{log}(\mathcal{L})}{\partial \theta_1}= \ccancel{red}{-\frac{n}{2}\textrm{log}(\theta_2)} \ccancel{red}{- \frac{n}{2}\textrm{log}(2\pi)}- 2\frac{\sum_{i=1}^{n}\left(y_{i}-\theta_1\right)(-1)}{2 \theta_2}$
Following this we can further simplify:
$\frac{\partial\textrm{log}(\mathcal{L})}{\partial \theta_1}= \ccancel{Gray}{-}\ccancel{red}{2}\frac{\sum_{i=1}^{n}\left(y_{i}-\theta_1\right)\ccancel{Gray}{(-1)}}{\ccancel{red}{2} \theta_2} = \frac{\sum_{i=1}^{n}\left(y_{i}-\theta_1\right)}{ \theta_2}$
Now all we need to do is set this equation equal to zero and solve for $$\theta_1$$:
\begin{aligned} 0 &= \frac{\sum_{i=1}^{n}\left(y_{i}-\theta_1\right)}{ \theta_2}\\ &= \sum_{i=1}^{n}\left(y_{i}-\theta_1\right) \\ &= \sum_{i=1}^{n}y_{i}- n\theta_1 \\ n\theta_1 &= \sum_{i=1}^{n}y_{i} \\ \theta_1 &= \hat{\mu}= \frac{\sum_{i=1}^{n}y_{i}}{n} \end{aligned}
And there we have it. The maximum likelihood estimate for $$\mu$$ is the mean of the data. Remember, R estimated the intercept of the linear regression to be 512.85. The mean of the squirrel weights should be identical to our model estimate.
# The squirrel weights
y <- c(557.1, 416.3, 393.6, 459.0, 588.6,
503.5, 507.7, 649.3, 529.2, 524.2)
# Our maximum likelihood estimate of mu for a normal distribution
mean(y)
[1] 512.85
# The estimated intercept from the linear model
coef(squirrel_model)
(Intercept)
512.85
Moving onto the second point estimate, we need to take the partial derivative with respect to $$\theta_2$$. For the first term, remember that $$\textrm{log}(x)^\prime = \frac{1}{x}$$. The second term is a constant and therefore cancels out. And finally we use the reciprocal rule, $$\frac{d}{d x} \frac{1}{f(x)}=-\frac{f^{\prime}(x)}{f(x)^{2}}$$, on the last term to calculate our partial derivative.
\begin{aligned} \frac{\partial\textrm{log}(\mathcal{L})}{\partial \theta_2}&= -\frac{n}{2}\textrm{log}(\theta_2) \ccancel{red}{- \frac{n}{2}\textrm{log}(2\pi)} - \frac{\sum_{i=1}^{n}\left(y_{i}-\theta_1\right)^{2}}{2 \theta_2}\\ &= -\frac{n}{2\theta_2} - \frac{\sum_{i=1}^{n}\left(y_{i}-\theta_1\right)^{2}}{2 \theta_2^2} \end{aligned}
From here we set this equation equal to zero and solve for $$\theta_2$$.
\begin{aligned} 0 &= \left[-\frac{n}{2\theta_2} - \frac{\sum_{i=1}^{n}\left(y_{i}-\theta_1\right)^{2}}{2 \theta_2^2}\right] \times 2\theta_2^2\\ &= -\frac{n2\theta_2^2}{2\theta_2} - \frac{\sum_{i=1}^{n}\left(y_{i}-\theta_1\right)^{2}2 \theta_2^2}{2 \theta_2^2}\\ &= -\frac{n\ccancel{red}{2}\theta_2\ccancel{Gray}{^2}}{\ccancel{red}{2}\ccancel{Gray}{\theta_2}} - \frac{\sum_{i=1}^{n}\left(y_{i}-\theta_1\right)^{2}\ccancel{violet}{2 \theta_2^2}}{\ccancel{violet}{2 \theta_2^2}}\\ &=n\theta_2 - \sum_{i=1}^{n}\left(y_{i}-\theta_1\right)^{2} \\ n\theta_2 &= \sum_{i=1}^{n}\left(y_{i}-\theta_1\right)^{2} \\ \theta_2 &= \hat{\sigma}^2 = \frac{\sum_{i=1}^{n}\left(y_{i}-\theta_1\right)^{2}}{n} \end{aligned}
Given that we already know that $$\theta_1$$ is the mean of our observed data, the maximum likelihood estimate for $$\sigma^2$$ is the residual sum of squares divided by the sample size. We can then take the square of this to calculate the standard deviation, which R estimated to be 76.93.
y <- c(557.1, 416.3, 393.6, 459.0, 588.6,
503.5, 507.7, 649.3, 529.2, 524.2)
squirrel_variance <- sum((y-mean(y))^2)/ length(y)
squirrel_sd <- sqrt(squirrel_variance)
[1] 72.98432
It looks like the residual standard error is not the same as our own maximum likelihood estimate! The reason for this is because the one we derived here is a biased estimator (link to wikipedia article), and R provides the unbiased estimate. To estimate the unbiased population variance or standard deviation we must actually use Bessel’s correction (link to wikipedia article). This is a pretty simple correction as it just means we modify the denominator of our sample variance estimate from $$n$$ to $$n-1$$.
$\theta_2 = \hat{\sigma}^2 = \frac{\sum_{i=1}^{n}\left(y_{i}-\theta_1\right)^{2}}{n-1}$
# population variance using Bessel's correction
squirrel_variance <- sum((y-mean(y))^2)/ (length(y)-1)
squirrel_sd <- sqrt(squirrel_variance)
# Estimate calculated by hand
squirell_sd
[1] 76.93222
# compare to the estimate from the linear model
summary(squirrel_model)\$sigma
[1] 76.93222
And there we have it. By taking some partial derivatives of the normal distribution’s likelihood function we were able to calculate by hand a little bit of what R does behind the scenes.
Written on July 22, 2021 | 2022-10-07 00:35:21 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9902002215385437, "perplexity": 449.8413730817756}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030337889.44/warc/CC-MAIN-20221006222634-20221007012634-00141.warc.gz"} |
http://www.nag.com/numeric/CL/nagdoc_cl23/html/C05/c05rdc.html | c05 Chapter Contents
c05 Chapter Introduction
NAG C Library Manual
# NAG Library Function Documentnag_zero_nonlin_eqns_deriv_rcomm (c05rdc)
## 1 Purpose
nag_zero_nonlin_eqns_deriv_rcomm (c05rdc) is a comprehensive reverse communication function that finds a solution of a system of nonlinear equations by a modification of the Powell hybrid method. You must provide the Jacobian.
## 2 Specification
#include #include
void nag_zero_nonlin_eqns_deriv_rcomm (Integer *irevcm, Integer n, double x[], double fvec[], double fjac[], double xtol, Nag_ScaleType scale_mode, double diag[], double factor, double r[], double qtf[], Integer iwsav[], double rwsav[], NagError *fail)
## 3 Description
The system of equations is defined as:
$fi x1,x2,…,xn = 0 , i= 1, 2, …, n .$
nag_zero_nonlin_eqns_deriv_rcomm (c05rdc) is based on the MINPACK routine HYBRJ (see Moré et al. (1980)). It chooses the correction at each step as a convex combination of the Newton and scaled gradient directions. The Jacobian is updated by the rank-1 method of Broyden. For more details see Powell (1970).
## 4 References
Moré J J, Garbow B S and Hillstrom K E (1980) User guide for MINPACK-1 Technical Report ANL-80-74 Argonne National Laboratory
Powell M J D (1970) A hybrid method for nonlinear algebraic equations Numerical Methods for Nonlinear Algebraic Equations (ed P Rabinowitz) Gordon and Breach
## 5 Arguments
Note: this function uses reverse communication. Its use involves an initial entry, intermediate exits and re-entries, and a final exit, as indicated by the argument irevcm. Between intermediate exits and re-entries, all arguments other than fvec and fjac must remain unchanged.
1: irevcmInteger *Input/Output
On initial entry: must have the value $0$.
On intermediate exit: specifies what action you must take before re-entering nag_zero_nonlin_eqns_deriv_rcomm (c05rdc) with irevcm unchanged. The value of irevcm should be interpreted as follows:
${\mathbf{irevcm}}=1$
Indicates the start of a new iteration. No action is required by you, but x and fvec are available for printing.
${\mathbf{irevcm}}=2$
Indicates that before re-entry to nag_zero_nonlin_eqns_deriv_rcomm (c05rdc), fvec must contain the function values ${f}_{i}\left(x\right)$.
${\mathbf{irevcm}}=3$
Indicates that before re-entry to nag_zero_nonlin_eqns_deriv_rcomm (c05rdc), ${\mathbf{fjac}}\left[\left(\mathit{j}-1\right)×{\mathbf{n}}+\mathit{i}-1\right]$ must contain the value of $\frac{\partial {f}_{\mathit{i}}}{\partial {x}_{\mathit{j}}}$ at the point $x$, for $\mathit{i}=1,2,\dots ,n$ and $\mathit{j}=1,2,\dots ,n$.
On final exit: ${\mathbf{irevcm}}=0$, and the algorithm has terminated.
Constraint: ${\mathbf{irevcm}}=0$, $1$, $2$ or $3$.
2: nIntegerInput
On entry: $n$, the number of equations.
Constraint: ${\mathbf{n}}>0$.
3: x[n]doubleInput/Output
On initial entry: an initial guess at the solution vector.
On intermediate exit: contains the current point.
On final exit: the final estimate of the solution vector.
4: fvec[n]doubleInput/Output
On initial entry: need not be set.
On intermediate re-entry: if ${\mathbf{irevcm}}\ne 2$, fvec must not be changed.
If ${\mathbf{irevcm}}=2$, fvec must be set to the values of the functions computed at the current point x.
On final exit: the function values at the final point, x.
5: fjac[${\mathbf{n}}×{\mathbf{n}}$]doubleInput/Output
On initial entry: need not be set.
On intermediate re-entry: if ${\mathbf{irevcm}}\ne 3$, fjac must not be changed.
If ${\mathbf{irevcm}}=3$, ${\mathbf{fjac}}\left[\left(\mathit{j}-1\right)×{\mathbf{n}}+\mathit{i}-1\right]$ must contain the value of $\frac{\partial {f}_{\mathit{i}}}{\partial {x}_{\mathit{j}}}$ at the point $x$, for $\mathit{i}=1,2,\dots ,n$ and $\mathit{j}=1,2,\dots ,n$.
On final exit: the orthogonal matrix $Q$ produced by the $QR$ factorization of the final approximate Jacobian, stored by columns.
6: xtoldoubleInput
On initial entry: the accuracy in x to which the solution is required.
Suggested value: $\sqrt{\epsilon }$, where $\epsilon$ is the machine precision returned by nag_machine_precision (X02AJC).
Constraint: ${\mathbf{xtol}}\ge 0.0$.
7: scale_modeNag_ScaleTypeInput
On initial entry: indicates whether or not you have provided scaling factors in diag.
If ${\mathbf{scale_mode}}=\mathrm{Nag_ScaleProvided}$ the scaling must have been supplied in diag.
Otherwise, if ${\mathbf{scale_mode}}=\mathrm{Nag_NoScaleProvided}$, the variables will be scaled internally.
Constraint: ${\mathbf{scale_mode}}=\mathrm{Nag_NoScaleProvided}$ or $\mathrm{Nag_ScaleProvided}$.
8: diag[n]doubleInput/Output
On initial entry: if ${\mathbf{scale_mode}}=\mathrm{Nag_ScaleProvided}$, diag must contain multiplicative scale factors for the variables.
If ${\mathbf{scale_mode}}=\mathrm{Nag_NoScaleProvided}$, diag need not be set.
Constraint: if ${\mathbf{scale_mode}}=\mathrm{Nag_ScaleProvided}$, ${\mathbf{diag}}\left[\mathit{i}-1\right]>0.0$, for $\mathit{i}=1,2,\dots ,n$.
On intermediate exit: diag must not be changed.
On final exit: the scale factors actually used (computed internally if ${\mathbf{scale_mode}}=\mathrm{Nag_NoScaleProvided}$).
9: factordoubleInput
On initial entry: a quantity to be used in determining the initial step bound. In most cases, factor should lie between $0.1$ and $100.0$. (The step bound is ${\mathbf{factor}}×{‖{\mathbf{diag}}×{\mathbf{x}}‖}_{2}$ if this is nonzero; otherwise the bound is factor.)
Suggested value: ${\mathbf{factor}}=100.0$.
Constraint: ${\mathbf{factor}}>0.0$.
10: r[${\mathbf{n}}×\left({\mathbf{n}}+1\right)/2$]doubleInput/Output
On initial entry: need not be set.
On intermediate exit: must not be changed.
On final exit: the upper triangular matrix $R$ produced by the $QR$ factorization of the final approximate Jacobian, stored row-wise.
11: qtf[n]doubleInput/Output
On initial entry: need not be set.
On intermediate exit: must not be changed.
On final exit: the vector ${Q}^{\mathrm{T}}f$.
12: iwsav[$17$]IntegerCommunication Array
13: rwsav[$4×{\mathbf{n}}+10$]doubleCommunication Array
The arrays iwsav and rwsav MUST NOT be altered between calls to nag_zero_nonlin_eqns_deriv_rcomm (c05rdc).
14: failNagError *Input/Output
The NAG error argument (see Section 3.6 in the Essential Introduction).
## 6 Error Indicators and Warnings
On entry, argument $〈\mathit{\text{value}}〉$ had an illegal value.
NE_DIAG_ELEMENTS
On entry, ${\mathbf{scale_mode}}=\mathrm{Nag_ScaleProvided}$ and diag contained a non-positive element.
NE_INT
On entry, ${\mathbf{irevcm}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{irevcm}}=0$, $1$, $2$ or $3$.
On entry, ${\mathbf{n}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{n}}>0$.
NE_INTERNAL_ERROR
An internal error has occurred in this function. Check the function call and any array sizes. If the call is correct then please contact NAG for assistance.
NE_NO_IMPROVEMENT
The iteration is not making good progress, as measured by the improvement from the last $〈\mathit{\text{value}}〉$ iterations. This failure exit may indicate that the system does not have a zero, or that the solution is very close to the origin (see Section 7). Otherwise, rerunning nag_zero_nonlin_eqns_deriv_rcomm (c05rdc) from a different starting point may avoid the region of difficulty.
The iteration is not making good progress, as measured by the improvement from the last $〈\mathit{\text{value}}〉$ Jacobian evaluations. This failure exit may indicate that the system does not have a zero, or that the solution is very close to the origin (see Section 7). Otherwise, rerunning nag_zero_nonlin_eqns_deriv_rcomm (c05rdc) from a different starting point may avoid the region of difficulty.
NE_REAL
On entry, ${\mathbf{factor}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{factor}}>0.0$.
On entry, ${\mathbf{xtol}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{xtol}}\ge 0.0$.
NE_TOO_SMALL
No further improvement in the solution is possible. xtol is too small: ${\mathbf{xtol}}=〈\mathit{\text{value}}〉$.
## 7 Accuracy
If $\stackrel{^}{x}$ is the true solution and $D$ denotes the diagonal matrix whose entries are defined by the array diag, then nag_zero_nonlin_eqns_deriv_rcomm (c05rdc) tries to ensure that
$D x-x^ 2 ≤ xtol × D x^ 2 .$
If this condition is satisfied with ${\mathbf{xtol}}={10}^{-k}$, then the larger components of $Dx$ have $k$ significant decimal digits. There is a danger that the smaller components of $Dx$ may have large relative errors, but the fast rate of convergence of nag_zero_nonlin_eqns_deriv_rcomm (c05rdc) usually obviates this possibility.
If xtol is less than machine precision and the above test is satisfied with the machine precision in place of xtol, then the function exits with NE_TOO_SMALL.
Note: this convergence test is based purely on relative error, and may not indicate convergence if the solution is very close to the origin.
The convergence test assumes that the functions and the Jacobian are coded consistently and that the functions are reasonably well behaved. If these conditions are not satisfied, then nag_zero_nonlin_eqns_deriv_rcomm (c05rdc) may incorrectly indicate convergence. The coding of the Jacobian can be checked using nag_check_derivs (c05zdc). If the Jacobian is coded correctly, then the validity of the answer can be checked by rerunning nag_zero_nonlin_eqns_deriv_rcomm (c05rdc) with a lower value for xtol.
The time required by nag_zero_nonlin_eqns_deriv_rcomm (c05rdc) to solve a given problem depends on $n$, the behaviour of the functions, the accuracy requested and the starting point. The number of arithmetic operations executed by nag_zero_nonlin_eqns_deriv_rcomm (c05rdc) is approximately $11.5×{n}^{2}$ to process each evaluation of the functions and approximately $1.3×{n}^{3}$ to process each evaluation of the Jacobian. The timing of nag_zero_nonlin_eqns_deriv_rcomm (c05rdc) is strongly influenced by the time spent evaluating the functions.
Ideally the problem should be scaled so that, at the solution, the function values are of comparable magnitude.
## 9 Example
This example determines the values ${x}_{1},\dots ,{x}_{9}$ which satisfy the tridiagonal equations:
$3-2x1x1-2x2 = -1, -xi-1+3-2xixi-2xi+1 = -1, i=2,3,…,8 -x8+3-2x9x9 = -1.$
### 9.1 Program Text
Program Text (c05rdce.c)
None.
### 9.3 Program Results
Program Results (c05rdce.r) | 2015-10-10 16:34:40 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 82, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9914658665657043, "perplexity": 2457.9634210029076}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-40/segments/1443737958671.93/warc/CC-MAIN-20151001221918-00157-ip-10-137-6-227.ec2.internal.warc.gz"} |
https://lists.nongnu.org/archive/html/axiom-developer/2005-02/msg00291.html | axiom-developer
[Top][All Lists]
## [Axiom-developer] RE: latexwiki update
From: Page, Bill Subject: [Axiom-developer] RE: latexwiki update Date: Thu, 24 Feb 2005 17:46:26 -0500
Bob,
On Thursday, February 24, 2005 4:57 PM I wrote:
>> ...
>> http://zwiki.org/HowToUseTheStandardSkinInPlone
>>
>
> Unfortunately that page and several other things on the
> zwiki.org site (even issuetracker) seems to be broken right
> now :( IOError ): So I can't review exactly what I did.
Ok, ZWiki is back.
The change is quite simple. It amounts to only changing the
priority of the portal_skins/standard in a new "skin" called
Zwiki. I think that all this means is that it gets searched
first when Plone goes looking for stylesheets.
> ...
> Maybe something in the ZWiki "standard" skins have to refer to
> latexwiki.css?
Yes! All I had to do was to cusomize portal_skins/wikipage_macros
Now the LaTeX generated image of a character and the corresponding
HTML character align properly.
But this is still a problem:
> Another weird thing that you might also have already noticed
> is that images that are inlined via $...$ do not seem to be
> transparent but $$...$$ are
> transparent. This seems to be a problem both on the plainwiki
> as well as under plone.
I will take a look at the new LatexWiki code to see if I
can see why this happens.
Regards,
Bill Page. | 2021-06-22 15:05:10 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 1, "x-ck12": 0, "texerror": 0, "math_score": 0.7692185044288635, "perplexity": 6504.621733639673}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623488517820.68/warc/CC-MAIN-20210622124548-20210622154548-00355.warc.gz"} |
https://www.khanacademy.org/economics-finance-domain/microeconomics/choices-opp-cost-tutorial/marginal-utility-tutorial/v/equalizing-marginal-utility-per-dollar-spent | # Equalizing marginal utility per dollar spent
Why the marginal utility for dollar spent should be theoritically equal for the last increment of either good purchased. Created by Sal Khan. | 2017-01-24 21:32:53 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8008556962013245, "perplexity": 8422.266559240568}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-04/segments/1484560285289.45/warc/CC-MAIN-20170116095125-00379-ip-10-171-10-70.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/magnitude-of-vector-cd.159156/ | # Magnitude of vector CD?
C, D are points defined by position vectors c and d. Magnitude of c is 5, mag of d is 7, c dotproduct d is 4 ie c.d = 4, find the magnitude of vector CD.
So i started this way
c.d = magc*magdcos@
= 35cos@, @ = 83.4 degrees
But still no idea how do get magnitude of vector cd. Thank you!
Last edited:
cristo
Staff Emeritus
?? You say you got an answer but you want an exact value? Did you use a calculator to get $cos(\theta)$. Since you want to use $cos(\theta)$ in the cosine rule, why not just use $cos(\theta)= \frac{4}{35}$ rather than finding $\theta$ itself? | 2022-05-24 15:27:55 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6486997604370117, "perplexity": 952.6332223412293}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662573053.67/warc/CC-MAIN-20220524142617-20220524172617-00614.warc.gz"} |
https://www.physicsforums.com/threads/changing-mass.149492/ | # Changing Mass
#### Fusilli_Jerry89
I've heard that when an object is accelerated at huge huge speeds they actually gain mass. Is there a formula to see how much mass an object would gain will going a certain velocity.
Related Special and General Relativity News on Phys.org
#### Hootenanny
Staff Emeritus
Gold Member
I quote RandallB, since this topic was somewhat discussed in https://www.physicsforums.com/showthread.php?t=149350"recently.
The problem you will continue to have with ones like this, is thinking of mass as actually changing with speed. That is a very old idea to think of the mass “as if it increases” with speed. Which works ok in a limited way, such as getting to E=mc^2. But modern science accepts the idea is incorrect in application and mass should be understood as intrinsic and unchanging with speed. Only momentum “p” or ‘mv’ is factored to increase with speed, and not mass.
But for any real mass, while it remains the same at mo; as the speed increases it must create a momentum "mv" that if factored by relativistic "gamma" to a larger number than expected by classical thinking.
Thus momentum as v approaches c would approach infinity and creating it would require an impossible amount of energy to reach it.
The important concept to note here is that it is the momentum which increases, not the 'mass'. In my opinion, in special relativity only invariant mass should be considered and the whole notion of 'relativistic' mass should be abandoned in special relativity (the situation in general relativity is somewhat more complex). As Randall says above, the notion that mass increases is usually introduced when explaining the 'basics' of relativity in a general context, but leads to misunderstandings when it comes to formally learning relativity. Below are some links which you may wish to peruse;
http://en.wikipedia.org/wiki/Rest_mass" [Broken]
http://en.wikipedia.org/wiki/Relativistic_mass" [Broken]
http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/relmom.html#c3"
Last edited by a moderator:
#### Littlepig
mass has 2 situations:
invariant mass, that, independent from the observer, it has a defined value.
relativistic mass, that depends on observer.
relativistic mass is "transformed" by lorentz factor.
invaritant mass, isn't "tranformed", and it is normally the mass that we use in classical mecanics, in expressions like: density=m/V, kinectic E=1/2mv^2, potential E=mgh and so1.
relativistic mass, is used in modern mecanics, and is too "named" as energy, by the E=mcc.
both masses can be used in momentum expression(p=mv)
in case of photon, it has no invariant mass, but as it have energy, we must assume it as relativistic mass
Regards, littlepig
#### Janus
Staff Emeritus
Gold Member
relativistic mass, is used in modern mecanics, and is too "named" as energy, by the E=mcc.
Regards, littlepig
$E = mc^2$ gives the energy equivalence of the invarient mass.
#### Littlepig
$E = mc^2$ gives the energy equivalence of the invarient mass.
so that's why, in my post, https://www.physicsforums.com/showthread.php?t=149350"
i couldn't say the energy released by hidrogen in man "B" couldn't be greater than in man "A". The invariant mass doesn't varies, because velocity doesn't take efect on invariant mass....humm....getting it...:tongue2: | 2019-11-21 09:09:39 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.61646568775177, "perplexity": 1979.7124826457652}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496670743.44/warc/CC-MAIN-20191121074016-20191121102016-00078.warc.gz"} |
https://forum.allaboutcircuits.com/threads/power-calculation-doubts.71525/ | # power calculation : Doubts
Discussion in 'Homework Help' started by nishu_r, Jun 24, 2012.
1. ### nishu_r Thread Starter Member
Jun 2, 2012
31
0
In order to calculate P, the real power absorbed by an impedance, we are to use $P=|I_{rms}|^2*Z$, and why the modulus of $I_{rms}$ (the magnitude) used instead of $I_{rms}$ directly, if used what would it yield.
This is a sincere question as i am not able to comprehend why, and i think the text book i use, is giving only a vague insight into the power calculations from a problem solving point of view.
2. ### cork_ie Active Member
Oct 8, 2011
356
60
I am not sure I quite understand your question.
Are you enquiring why P is proportional to the square of I RMS rather than just being directly proportional to I RMS ?
If so:
P= VRMS x I RMS ( assuming a resistive circuit where I & V are both in phase)
VRMS = I RMS x z , hence P = [ I RMS ]sq. x Z
3. ### jimmy101 New Member
Jun 23, 2012
9
2
I believe he means why have the absolute value function at all since
(|x|)^2=(x)^2
for all x. So the absolute value function does nothing.
4. ### WBahn Moderator
Mar 31, 2012
22,748
6,773
Are you sure that you have copied that from the textbook correctly? Because real power is a real quantity that the expression, irrespective of the form of I, is multiplied by Z which, in general, is a complex quantity.
As for the absolute value. In general, the phasor for the following time-domain current:
$
i(t) \ = \ I_o \cos(\omega t + \phi)
$
would be written
$
I(j\omega) \ = \ I_o e^{(j \phi)} \ = \ I_R + jI_I
$
Now, these I's are amplitudes of sinusoids. But, we know that the RMS value of a sinusoid can be obtained by just dividing it be the square root of two. So we could right:
$
I_{RMS}(j\omega) \ = \ I_{o_{RMS}} e^{(j \phi)} \ = \ I_{R_{RMS}} + jI_{I_{RMS}}
$
The point being that the RMS phasor current is still a complex quantity. What you need for the power calculations is the magnitude of the RMS phasor, not the square of it (the square of a complex number is a complex number with twice the phase angle). That is why you take the "absolute value" (which, for complex numbers, means the magnitude) and square that.
But to get real power, you would need to multiply by the resistive part of the impedance, not the entire impedance.
nishu_r likes this.
5. ### t_n_k AAC Fanatic!
Mar 6, 2009
5,448
789
WBahn I think you are correct in questioning the OP's transcription or interpretation of the equation in post #1.
Perhaps the text would follow something along the lines ....
${|I_{rms}|}^2Z={|I_{rms}|}^2(R+jX)\\={|I_{rms}|}^2(|Z|cos\theta +j|Z|sin\theta)\\={|I_{rms}|}^2|Z|cos\theta+j{|I_{rms}|}^2|Z|sin\theta\\=P+jQ$
The relationship if fact gives the complex or rather phasor/vector power - not the real power as the OP has stated.
6. ### nishu_r Thread Starter Member
Jun 2, 2012
31
0
Yes it does give the complex power, S .And, I am terribly sorry for the flaw in the question.
Now i understand clearly. thanks. | 2018-08-14 09:30:44 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 7, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8689171671867371, "perplexity": 1356.0474930726514}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-34/segments/1534221208750.9/warc/CC-MAIN-20180814081835-20180814101835-00269.warc.gz"} |
https://rpg.stackexchange.com/questions/128760/is-there-anything-i-should-be-wary-of-introducing-this-rule-for-unconsciousness | # Is there anything I should be wary of introducing this rule for unconsciousness into my games?
3.5 is pretty lethal, and I always felt it was a little weird that a sorcerer was unconscious for the same amount of time before dying as a barbarian. So I've been tinkering with the following house rule
Characters don't die from negative hit points until they reach -10 or their ECL * CON mod, whichever is higher.
I feel like it will power up the Die Hard feat, but this is ok as that feat could probably do with a buff. Is there anything else I should watch out for that this would effect and possibly break?
EDIT: Usually I use a heavily houseruled version of random ability scores, which usually results in better scores and more interesting characters. So characters start by rolling 2d6 for each ability and picking the highest. Then they roll a final 2d6, pick the highest and can replace on score with this roll. Next they roll 3d6 and drop the lowest 7 times, assign the best 6 as they want.
This has resulted in pretty good stats for PCs, with some low stats where you might not expect them. It sounds complicated, but it usually goes pretty quickly ime
• not an evaluation of your house rule, but did you consider using the variant death and dying rules from UA? – Carcer Aug 2 '18 at 12:32
• Hadn't seen these before, I quite like them. Have you ran them before? It definitely makes fortitude more valuable – gaynorvader Aug 2 '18 at 12:45
• I've not run them personally, but I remembered reading about them, and thought you might want to peruse. I can't really judge them either way; they definitely make death/dying more mechanically interesting, but they are also rather more complex to run than the default rules (such is the case for most of the UA variant material, I feel). – Carcer Aug 2 '18 at 13:00
• I think a lot of the benefit of this house rule will depend on how ability scores are determined in your campaigns. Could the question also include that information? – Hey I Can Chan Aug 2 '18 at 14:49
Pathfinder uses a similar rule (instead of $-ECL \times Con$ it uses $-Constitution$ (that is, the whole score), which is going to be a wider area for “dying” than your rule at low levels, and a narrower area at high levels—but pretty comparable all the same. Pathfinder has no particular issues with this rule.
That said, even with this rule, “dying” is going to be pretty unlikely. The damage dealt by characters is just too colossal; going from “alive” to “dead” with no stop in “dying” is going to remain very, very common at higher levels. This change makes Diehard and other options focusing on the “dying” state somewhat better, or rather, improves their longevity: where before, they were pretty much worthless by 2nd or 3rd level, now their utility will extends to maybe 8th to 10th level, because you actually might reasonably wind up “dying” and therefore care about that state.
But it doesn’t change the fundamental problem that “dying” is going to be a relatively rare state, so things that benefit that state are going to be very low-value. Diehard is less of a trap in this case, but it still is one.
• I once made a Deathless Fighter that kept swinging Full Attacks through to about negative 150 or 200 iirc. Definitely an edge case, not the norm though. – Ifusaso Aug 2 '18 at 18:39
• I tried the Pathfinder approach, but found it was still pretty weak in mid levels (6-12), which is when spells like reincarnate and resurrection are still sizable investments, but -10-18 hp is pretty inconsequential to a barbarian with 60 odd hp or more. – gaynorvader Aug 3 '18 at 9:20
• @gaynorvader Sure, I wasn’t suggesting that you use the Pathfinder rule, just pointing to it as comparison and demonstration that making the area wider won’t hurt anything. I don’t think you’ve actually accomplished something that will be consequential most of the time, though. (Also, what barbarian has Con 10? That’s suicidal; 14 would be a bare minimum.) – KRyan Aug 3 '18 at 12:32
• @KRyan What barbarian has con 10? Only PC barbarians :) – gaynorvader Aug 3 '18 at 13:58 | 2019-08-19 13:26:44 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.37778034806251526, "perplexity": 1922.4695258025786}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027314732.59/warc/CC-MAIN-20190819114330-20190819140330-00451.warc.gz"} |
https://www.physicsforums.com/threads/same-pressure-points-2-fluids-u-shaped-tube.966935/ | Same pressure points , 2 fluids u shaped tube
Mariog
1. Homework Statement
2. Homework Equations
Pascal principal p = ρgh
3. The Attempt at a Solution
Pressure at C and D is the same .
Pressure at A and B is the same ? I believe it is
Attachments
• 24.5 KB Views: 104
Related Introductory Physics Homework Help News on Phys.org
Chestermiller
Mentor
Pressures at c and d are the same. Pressures at a and b are not the same. Do you know why?
Mariog
Because of different fluids ?
And how the fluids are in balance if A and B pressures are not the same ?
Last edited:
Chestermiller
Mentor
Because of different fluids ?
yes
And how the fluids are in balance if A and B pressures are not the same ?
$$p_A+\rho_{left}g(x+h)=p_D$$
$$p_B+\rho_{right}g(x+h)=p_C$$
"Same pressure points , 2 fluids u shaped tube"
Physics Forums Values
We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving | 2019-10-18 04:46:03 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.1918865144252777, "perplexity": 4894.659019880693}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986677884.28/warc/CC-MAIN-20191018032611-20191018060111-00421.warc.gz"} |
https://www.subjectcoach.com/tutorials/math/topic/math-definitions-letter-p/chapter/power | Definition of Power
The power of a number tells us how many times we need to multiply it with itself to give the required value.
It is written as a little number above and to the right of the base number.
For example, $9^4 = 9 \times 9 \times 9 \times 9$. The power of $4$ tells us to multiply $9$ by itself $4$ times.
Other terms for "power" are "index" and "exponent".
Description
The aim of this dictionary is to provide definitions to common mathematical terms. Students learn a new math skill every week at school, sometimes just before they start a new skill, if they want to look at what a specific term means, this is where this dictionary will become handy and a go-to guide for a student.
Audience
Year 1 to Year 12 students
Learning Objectives
Learn common math terms starting with letter P
Author: Subject Coach | 2020-10-31 22:55:31 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.30011168122291565, "perplexity": 639.686799455548}, "config": {"markdown_headings": false, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-45/segments/1603107922463.87/warc/CC-MAIN-20201031211812-20201101001812-00441.warc.gz"} |
https://datascience.phil.fau.de/fossos/stack.html | Here’s the software stack we’ll be using.
You should work yourself through this stack in the order suggested.
This is by no means the only way to do open science or data science with open source software, and recommended packages are likely to change over time. The below R-based toolchain should be considered as merely one (out of several) consistent implementations of some best practices. However, once participants have mastered this toolchain, they should find it relatively easy to adapt to other ecosystems.
All of the software will be free and open source software, but we will also be using some proprietary Software-as-a-Service (Saas) offerings. For each of the proprietary services, there are open-source and/or self-hosted alternatives, but these are often much less convenient (e.g. self-hosted Jenkins vs GitHub Actions), or they are much less popular in the community, and therefore less useful (e.g. GitLab vs GitHub). Relying on, or pushing proprietary services, especially in an education context, is always awkward, but the disadvantages can sometimes be outweighted by convenience and network effect advantages. For some aspects of open source software development and open science, proprietary services – especially GitHub and the StackExchange network – for better or for worse just are the only game in town. In any event, most of what students will learn in this class is in free and open source software, and the remaining proprietary usage should easily translate to other, competing or open services.
## Introduction
For the introductory session, participants should watch this video:
Here’s the corresponding slide deck.
### Installation
Participants should install all of the below software and sign up for all the below services before the first class.
If you are using Docker, some of the software is already contained in the image, though “local” installation may still be advisable.
The steps required for installation will depend on your platform and system setup.
### Basic Computing Literacy
You should know, or easily find, the answer to questions such as:
• In what directory (absolute path) are programs I use daily stored on computer?
• What is the operating system (OS) and version on my computer?
• In what directory (absolute path) do I store my files?
• Do I have sufficient privileges to install software? If not, how can I get them?
• Which file format is better suited to editing and why: A *.docx or a *.pdf?
• Why do the search queries jaguar car and jaguar -car give different results on Google?
• Name at least 10 file types.
• What is the username I usually use on public-facing platforms?
• What is Two-Factor Authentification? (2FA)?
• How is my harddisc formatted?
• How can I upgrade my OS and frequently used software?
• How is the data on my computer protected from unauthorised access?
• What is my backup plan?
• What is a VPN client, and what do I need it for?
If you feel like you need to brush up on some basic computing skills, these resources might be helpful:
## Software Carpentry
### Project Management
GitHub is a collaboration platform, code repository and git host (more on all of below) along with some helpful project management tools.
Please flesh out your profile on GitHub.com and all of the accounts below a bit by adding a picture, a name and a short description of yourself.
### Community & Help
Aside from Google, these are two great places to get help, and to get involved in the community.
A lot of volunteers spend a lot of time on these sites, so it is very important not to waste their efforts, and to only add quality content, as defined by these sites.
• Find an interesting question an StackOverflow and post it to the chat.
• Find and interesting discussion on RStudio community and post it to the chat.
In addition to StackExchange and RStudio Community, there are a couple of other platforms where the (very friendly) R community hangs out:
### Markup Language
The full GFM spec is just FYI; there’s nothing to install here.
Plain text has many advantages (more on that later), but one glaring disadvantage: it does not look very nice, and does not implement many of the typesetting conventions that have evolved since Gutenberg (say, bold face).
Markup syntaxes solve this problem. Markup syntaxes are sets of conventions (as in *something* for highlighting) to structure human-generated text in a way that computers can operate on them, such as formatting a piece of text.
There are many, many, such markup languages out there, including HTML but also Markdown and LaTeX.
We will be focusing on Markdown as a source language, and then use open source tools (especially Pandoc), to render our source documents to all sorts of other formats, including PDF (via LaTeX), HTML (such as this website), but also Microsoft Word documents.
Markdown is a very lightweight markup language, that was designed to be maximally human readable, that is, looking meaningful without being compiled by a computer. Most of the syntax takes its clues from how people have already formatted plain text, such as enclosing a *word* with * for highlighting.
Technically, Markdown is a convention for writing such files, as well as a program to convert such files into HMTL, as, for example, this website (which is written in a flavor of Markdown).
By convention, Markdown files use the .md file extension. It’s important to recognize that still, an .md is a plain text file. You could open it with any text editor, or even change the extension to *.txt and nothing would change. The extension .md serves merely to tell computers that the following plain text is marked up in markdown.
Markdown was (originally) quite a minimal standard, and has since branched out into a few specialised “flavors”, offering additional features.
We will be using only two of these flavors: GitHub Flavored Markdown (GFM) and Pandoc’s Markdown (more on that below).
StackOverflow, RStudio Community (Discourse), Gitter chat and many other services also support GFM.
GitHub, a leading code-hosting service, has extended the above original Markdown spec by a couple of additional features. In addition to these formatting niceties, Github also implements some clever cross-referencing and autocompletion magic. When using Github for source control and collaboration, you really must use these in issues, comments, commit messages etc. (they work everywhere).
#### Resources
If you like, you can also install a program on your computer to render Markdown to HTML. There are plenty of choices, including the free MarkdownPad for Windows, and Lightpaper for OS X. If you don’t want to install something, Github (see below) also offers a Markdown preview in its browser-based editor. We will be using different programs going forward.
### Shell
A shell is a command-line interface (CLI) to your computer (as opposed to a point-and-click graphical user interface, or GUI). You may also know this as “the console”, or “the terminal”.
There are technically different kinds of shells, though the bash shell is the most widespread, and is often used interchangeably with the shell.
A lot of programs that we’ll be using only run at the CLI, so it’s important to know how to use it.
You can use the (Linux) shell that ships with our Docker container, but you should also know your way around the shell that ships with your OS (to, among other things, spin up a Docker image).
On macOS, Linux: Nothing to install, ships with a bash shell or something similar.
On Windows:
• Install Git for Windows because that comes with at least a git shell. Choose git bash emulation on install.
• If your version is >= Windows 10 Anniversary Update you can also install Install the Windows Subsystem for Linux (WSL) and use the Windows 10 Bash Shell. However this is a separate system inside your Windows installation, and the programs installed inside it may (as of 2019-01) not be used “normal” windows GUI programs. If you don’t know what this means, do not install the WSL; it can be very confusing.
If you like a fancier shell, you might want to look at the oh-my-zsh project, which has some pretty cool features. However this is strictly optional, will not be supported in class.
### Bash (Optional)
It turns out that bash, the default shell on UNIX-type computers is also a scripting language upon itself. Scripting languages are programming languages which facilitate automated execution of tasks, such as, say, running a bunch of updates and then power cycling your computer.
bash isn’t necessarily the greatest scripting language; especially for more complicated projects, “proper” scripting languages such as Python, Ruby or R might serve you better.
But bash has the advantage that it is available in (almost all) UNIX-type computing environments, so it’s often the easiest way to automate steps. It looks a bit arcane (because it is), but you don’t need much to build powerful scripts that can save you a lot of time.
This entire topic and the below additional resources are recommended for advanced readers. You won’t need this starting out.
### Containerisation (optional)
Docker is an open-source industry standard to define, provision and share computing environments, known as containers. Containers allow you to run computing environments on other computers. Containers are similar to virtual machines (a computer inside a computer), but slimmer and generally neater.
You may want to use Docker in class to quickly get a development environment, but it is also generally a helpful tool.
#### Resources
• Install and run a docker image:
2. Launch Docker Desktop. A small whale symbol will show up in your task bar, but not much else.
3. Launch a system shell and type in
docker run --env=PASSWORD=yourpassword --rm --publish=8787:8787 rocker/verse
On macOS and Linux, your default system shell is an application called “Terminal”. On Windows, you can use the “command prompt” or “PowerShell” applications. If you already have git installed on Windows, you can also use the git bash emulator.
Depending on how fast your internet connection is, this process will take a while to complete.
1. Open a webbrowser and point it to http://localhost:8787. You should see the login window for the RStudio IDE. You are in the browser, but are using running all computations on your machine through Docker. This will also work if you are offline.
2. Type in rstudio as a username and your PASSWORD given in the above as a password.
3. You are now in the RStudio IDE.
4. To shut down, close the browser window and type ctrl + c in your shell.
• Learn the difference between a Dockerfile, an image and a container in the context of Docker.
### Source Control Management (SCM)
Git is just a CLI program. It offers all the functionality of git, but you may also install a Git graphical user interface (GUI).
There plenty of those out there, but one of the easiest is the GitHub Desktop app from GitHub (available only for Windows and macOS).
You should install Git and GitHub Desktop even if you are using Docker.
You also need to configure git on your machine, so that git knows you are and to allow you to authenticate against git hosts (GitHub.com in our case) and wherever else you are using Git (such as a SaaS):
• RStudio Cloud
• Configuring git inside Docker isn’t recommended; it can get difficult you must be careful not to disclose your credentials.
### Branching Model (Optional)
There is a varied set of practices and tools that have evolved on top of Git. Together with the powerful git scm, it is these practices and tools, that make massively collaborative software development possible.
One of the simpler practices is GitHub Flow. We will use it to learn the branch and pull-request model.
### Package Management (optional)
Installing and upgrading a lot of command line tools and their dependencies gets old quickly. Package managers solve this problem; they provide a clean and elegant way to install (CLI) programs, and even allow you to quickly upgrade everything.
You will need to install a package manager independent of using the Docker image.
Notice that LaTeX, Atom and R (all below) each have their own internal package managers (as do many other software ecosystems). If you’re installing a package for either of those, use the corresponding ecosystem package manager, not your system-wide program (= brew, apt-get, cholocatey).
### Text Editor (Optional)
Whenever we write something in this class, it will be in plain text. Plain text, roughly speaking, consists directly and only of letters, encoded in an open standard.
This may seem antiquated, but has several advantages:
• Plain text can easily be versioned by computer software such as git.
• Plain text is transparent to the user: it is human-readable. For comparison, try opening a *.doc in a text editor, and see whether you can make out any meaning.
• Plain text is lightweight and robust. File sizes and memory footprint are tiny.
• Plain text files future-proof your work and data. *.txt, or, equivalently for data, *.csv can be opened and edited on pretty much any computer today, could be 30 years go, today, and probably still will be widely accessible in 30 years time.
Most operating systems ship with a text editor, but they are quite basic and can be cumbersome to use. Specialized text editors (or just editors) offer more functionality geared towards technical writing or software development.
There are many editors out there, and people have strong views on which is best. In some ways, this is surprising, because of all the software used in collaborative writing or development, editors are the tool that needs the least standardisation. Playing off the advantages of plain text files, everyone can use what works best for them, because they all output the exact same thing: a *.txt.
You are therefore free to choose your own text editor.
You can use the RStudio IDE that comes with our Docker image, but other editors are more fully featured.
Atom has the advantage of being relatively easy to use, free and open source and relatively widely supported. It also comes with some nice Git(Hub) integration.
Atom, as most editors, has a modular design. Many of its features are factored out to separate packages, some of which are contributed by external volunteers.
Here’s a list of packages you might also want to install:
• atom-beautify
• atom-html-preview
• document-outline
• git-plus
• language-knitr
• language-latex
• latex
• language-markdown
• merge-conflicts
• minimap-split-diff
### R Integrated Development Environment (IDE)
If you are using the Docker container image, RStudio and R are already included.
Aside from text editors, there are also integrated development environments (IDE) (though this distinction has recently been blurring with the arrival of Atom-IDE and others). IDEs are a little like text editors, in that they mostly let you edit plain text files, but they offer a lot of “training wheels” for programming and are often geared towards particular programming languages.
The leading IDE for R is called RStudio, by, confusingly, a company called RStudio. We will be using the open source variant of RStudio (the IDE), but RStudio (the company) also sells commercial licenses to the IDE and other products.
If you are already deeply invested in an IDE or Editor (especially vim or emacs) you may also trick out that program to support R. The Emacs speaks statistics project has great support for R, but Emacs has a steep learning curve.
For most everyone, RStudio will therefore be the strongly recommended choice.
### Technical & Scientific Authoring (optional)
All software in this section is included in the Docker image.
#### Document Conversion (optional)
We’ll often want to convert documents from and to different markup formats. For that purpose, we’ll use pandoc.
Pandoc is, originally, a kind of swiss army knife for text document formats, such as, say, between Microsoft Word and HTML.
But as part of this work, Pandoc has also defined its own extension (flavor) to Markdown (largely compatible with GFM), including such features as footnotes, captions, references, and other aspects important for technical and scientific writing.
You should both learn to use Pandoc at the CLI as well as to write in the corresponding Pandoc’s Markdown style.
#### Typesetting (optional)
(La)TeX is strictly speaking a typesetting program, which can create beautiful documents. It has extensive support for all sorts of domain-specific typographic niceties, and is used a lot by academics, especially in math and sciences because.
However, because LaTeX is quite cumbersome to compose and tends to distract writing with a lot of bells and whistles, we will not learn to write LaTeX directly “by hand”. Instead, we will be using Pandoc to compile our Pandoc Markdown source to PDF (via LaTeX), and, because LaTeX can be slow to compile, we will only do so rarely and towards the end of any given project.
Still, it is important to learn some of the basics of LaTeX to use it programmatically.
### Bibliography Management (optional)
Bibliography management is not the focus of this class, but you can learn more about it here.
It is also one of those tools, where there is no strong reason to standardize on any one program, so as long as the bibliography manager exports to one of the formats that pandoc can ingest.
Check if your bibliography manager can export to at least one of these formats.
If you have a choice, a BibTeX or BibLaTeX file (confusingly both named *.bib) are preferable.
## Introductory R
All software in this section is included in the Docker image.
## Intermediate R
All software in this section is included in the Docker image.
## Interactive R
All software in this section is included in the Docker image.
### HTML, JS & CSS (optional)
The below packages for (web) interactivity in R try to abstract away as much as possible the underlying web technologies (HTML, JavaScript and CSS). You can use them without knowing anything about this stack, but you can accomplish more and understand them in a deeper way if you have at least a cursory understanding of how these technologies work.
Covering them in any depth, or even listing good resources (of which there are gazillions) is beyond the scope of this class, so these should be considered mere starting points.
tba.
tba.
tba.
tba. | 2021-09-28 12:44:44 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3062411844730377, "perplexity": 2859.040473158773}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780060803.2/warc/CC-MAIN-20210928122846-20210928152846-00696.warc.gz"} |
https://www.ideals.illinois.edu/handle/2142/16340/browse?type=contributor&value=Ford%2C+Kevin | # Browse Dissertations and Theses - Mathematics by Contributor "Ford, Kevin"
• (2016-06-15)
We investigate the arithmetic properties of coefficients of Maass forms in three contexts. First, we discuss connections to invariants of real and imaginary quadratic fields, expanding on the work of Zagier and Duke-Imamoglu-Toth. ...
application/pdf
PDF (923kB)
• (2010-05-14)
This dissertation involves two topics. The first is on the theory of partitions, which is discussed in Chapters 2 − 5. The second is on covering systems, which are considered in Chapters 6 − 8. In 2000, Farkas and Kra ...
application/pdf
PDF (942kB)
• (2013-05-24)
We improve the error term in the van der Corput transform for exponential sums, \sum g(n) exp(2 \pi i f(n)). For many smooth functions g and f, we can show that the largest factor of the error term is given by a simple ...
application/pdf
PDF (931kB)
• (2018-06-01)
The value distribution of the Riemann zeta function $\zeta(s)$ is a classical question. Despite the fact that values of $\zeta(s)$ are approximately Gaussian distributed, $\zeta(s)$ can be very large for infinitely many ...
application/pdf
PDF (627kB)
• (2010-08-20)
In 1955 Erdõs posed the multiplication table problem: Given a large integer N, how many distinct products of the form ab with a≤N and b≤N are there? The order of magnitude of the above quantity was determined by Ford. The ...
application/pdf
PDF (607kB)
• (2011-01-21)
In his notebooks, Ramanujan recorded 40 beautiful modular relations for the Rogers-Ramanujan functions. Of these 40 identities, precisely one involves powers of the Rogers-Ramanujan functions. Ramanujan added the enigmatic ...
application/pdf
PDF (602kB)
• (2018-06-22)
We first give a survey on multilinear Hilbert transforms. Then we study several variants of bilinear Hilbert transform such as bilinear Hilbert transform along two polynomials, discrete (integer) bilinear Hilbert transform ...
application/pdf
PDF (770kB)
• (2017-07-13)
PART I G. H. Hardy and S. Ramanujan established an asymptotic formula for the number of unrestricted partitions of a positive integer, and claimed a similar asymptotic formula for the number of partitions into perfect ...
application/pdf
PDF (720kB)
• (2017-07-07)
This thesis includes four chapters. In Chapter 1, we briefly introduce the history and the main results of the topics of this thesis: the distribution of $k$-free numbers and the derivative of the Riemann zeta-function, ...
application/pdf
PDF (948kB) | 2019-06-18 07:42:29 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.797916829586029, "perplexity": 926.1413336962286}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560627998690.87/warc/CC-MAIN-20190618063322-20190618085322-00185.warc.gz"} |
https://fruitbat.readthedocs.io/en/latest/user_guide/guidelines.html | # Guidelines¶
## Reference¶
If you use Fruitbat in your research, please add the acknowledgement statement “Some of the results of this paper have been derived using the FRUITBAT package” and cite the JOSS paper.
@ARTICLE{2019JOSS....4.1399B,
title = "{Fruitbat: A Python Package for Estimating Redshifts of Fast Radio Bursts}",
journal = {The Journal of Open Source Software},
keywords = {Astrophysics - Instrumentation and Methods for Astrophysics, Astrophysics - High Energy Astrophysical Phenomena},
year = "2019",
month = "May",
volume = {4},
number = {37},
pages = {1399},
doi = {10.21105/joss.01399},
archivePrefix = {arXiv},
eprint = {1905.04294},
primaryClass = {astro-ph.IM},
adsnote = {Provided by the SAO/NASA Astrophysics Data System}
}
## Issues and Contributing¶
If there is a feature of FRUITBAT that currently does not exist, but you would like it to, you can contribute by openning a Github Issue and outlining the feature. Similar to contributing, if you find a problem with FRUITBAT or are having difficulties using FRUITBAT please do not hesitate to open a Github Issue.
BSD 3-Clause License
Redistribution and use in source and binary forms, with or without
modification, are permitted provided that the following conditions are met:
* Redistributions of source code must retain the above copyright notice, this
list of conditions and the following disclaimer.
* Redistributions in binary form must reproduce the above copyright notice,
this list of conditions and the following disclaimer in the documentation
and/or other materials provided with the distribution.
* Neither the name of the copyright holder nor the names of its
contributors may be used to endorse or promote products derived from
this software without specific prior written permission.
THIS SOFTWARE IS PROVIDED BY THE COPYRIGHT HOLDERS AND CONTRIBUTORS "AS IS"
AND ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE
IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE ARE
DISCLAIMED. IN NO EVENT SHALL THE COPYRIGHT HOLDER OR CONTRIBUTORS BE LIABLE
FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL
DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS OR
SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION) HOWEVER
CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT LIABILITY,
OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY OUT OF THE USE
OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF SUCH DAMAGE. | 2020-02-24 13:32:50 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.17103765904903412, "perplexity": 408.6513387634616}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875145960.92/warc/CC-MAIN-20200224132646-20200224162646-00405.warc.gz"} |
https://pythonslearning.com/2020/06/linear-regression-in-machine-learning-plot-algorithms-explain-02.html | # Linear Regression in Machine Learning: part02
WELCOME TO SECOND PART OF YOUR LINEAR REGRESSION POST. IN THE FIRST POST WE SEE OPERATION ON DATASETS. IN THIS POST WE SEE LINEAR REGRESSION OPERATION. LET’S START:
## Training a Linear Regression Model
Let’s now begin to train out regression models. We will need to first split up our data into an X array that contain the feature to train on, and a y array with the target variables, in this case the Price column. We will toss out the Address columns because it only has text info that the linear regression model can not use.
## Now let’s splits the data into a training set and a testing set. We will train out models on the training set and then use the test set to evaluate the modes
from sklearn.model_selection import train_test_split
X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.4)
### from sklearn.linear_model import LinearRegression
from sklearn.linear_model import LinearRegression
lm = LinearRegression()
lm.fit(X_train,y_train)
#### Model Evaluation
Let’s evaluate the model by checking out it is coefficient and how we can interpret them
# print the interceptprint(lm.intercept_)
-2640159.79685
coeff_df = pd.DataFrame(lm.coef_,X.columns,columns=['Coefficient'])coeff_df
data info.
CoefficientAvg. Area Incomes21.528276Avg. Area House Ages164883.282027Avg. Area Number of Room122368.678027Avg. Area Number of Bedroom2233.801864Area Populations15.150420
Interpreting the coefficients:- Holding all other features fixed, a 1 unit increase in **Avg. Area Income** is associated with an **increase of $21.52 **.- Holding all other features fixed, a 1 unit increase in **Avg. Area House Age** is associated with an **increase of$164883.28 **.- Holding all other features fixed, a 1 unit increase in **Avg. Area Number of Rooms** is associated with an **increase of $122368.67 **.- Holding all other features fixed, a 1 unit increase in **Avg. Area Number of Bedrooms** is associated with an **increase of$2233.80 **.- Holding all other features fixed, a 1 unit increase in **Area Population** is associated with an **increase of \$15.15 **.Does this make sense? Probably not because I made up this data. If you want real data to repeat this sort of analysis, check out the [boston dataset](http://scikit-learn.org/stable/modules/generated/sklearn.datasets.load_boston.html):
from sklearn.datasets import load_bostonboston = load_boston()print(boston.DESCR)boston_df = boston.data
Predictions from our ModelLet's grab prediction off our test set and see how well it did.
predictions = lm.predict(X_test)
plt.scatter(y_test,predictions)
<matplotlib.collections.PathCollection at 0x142622c88>
Residual Histogram
sns.distplot((y_test-predictions));
Regression Evaluation MetricHere are three common evaluation metric for regression problem:Mean Absolute Error (MAE) is the mean of the absolute value of the error:1𝑛∑𝑖=1𝑛|𝑦𝑖−𝑦̂ 𝑖|$\frac{1}{n}\sum _{i=1}^{n}|{y}_{i}-{\stackrel{^}{y}}_{i}|$Mean Squared Error (MSE) is the mean of the squared error:1𝑛∑𝑖=1𝑛(𝑦𝑖−𝑦̂ 𝑖)2$\frac{1}{n}\sum _{i=1}^{n}\left({y}_{i}-{\stackrel{^}{y}}_{i}{\right)}^{2}$Root Mean Squared Error (RMSE) is the square root of the mean of the squared error:1𝑛∑𝑖=1𝑛(𝑦𝑖−𝑦̂ 𝑖)2⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯$\sqrt{\frac{1}{n}\sum _{i=1}^{n}\left({y}_{i}-{\stackrel{^}{y}}_{i}{\right)}^{2}}$Comparing these metric:
• MAE is the easiest to understand, because it is the average error.
• MSE is more popular than MAE, because MSE “punishes” larger error, which tends to be useful in the real world.
• RMSE is more popular than MSE, because RMSE is interpretableS in the “y” units.
All of these are loss function, because we want to minimize them.
from sklearn import metrics
print('MAE:', metrics.mean_absolute_error(y_test, predictions))print('MSE:', metrics.mean_squared_error(y_test, predictions))print('RMSE:', np.sqrt(metrics.mean_squared_error(y_test, predictions)))
MAE: 82288.2225191MSE: 10460958907.2RMSE: 102278.829223
This was your Machine Learning Project!
Tags: Linear Regression in Machine Learning-plot-algorithms-explain
BEST OF LUCK!!!
[email protected]
### sach Pagar
I am Mr. Sachin pagar the founder of Pythonslearning, a Passionate Educational Blogger and Author, who love to share the informative content on educational resources. | 2021-04-12 16:37:39 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 3, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3644656836986542, "perplexity": 3923.3431969060875}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038067870.12/warc/CC-MAIN-20210412144351-20210412174351-00050.warc.gz"} |
http://mathcentral.uregina.ca/QQ/database/QQ.09.20/h/angel1.html | SEARCH HOME
Math Central Quandaries & Queries
Question from: Angel Four friends ran a race: Ben finished seven seconds ahead of Mike. Noel finished three seconds behind Sam. Mike finished five seconds behind Noel. What will be the formula to support their place?
Hi Angel,
I'm not sure what you mean by "the formula to support their place", but here is what I would do. I am going to use the number line and put each runner on the line at the time it takes them to finish the race. Time is in seconds. I am going to call the runners B, N, S, and M for Ben, Noel, Sam, and Mike. First Ben.
Ben's time is somewhere on the line but I'm not sure where. Now put Mike on the line. "Ben finished seven seconds ahead of Mike." so Ben took 7 seconds less time than Mike to complete the race. Hence we have
The second clue compares Noel and Sam and we don't have either on the diagram. The third clue says "Mike finished five seconds behind Noel." so Noel took 5 seconds less time that Mike and thus we have
Now we can use the fact that "Noel finished three seconds behind Sam." and hence
Thus Sam took the least amount of time so he won. But what about a formula?
Suppose Sam took $t$ seconds to complete the race. How many seconds did Ben take to complete the race? What about Noel? and Mike?
I hope this helps,
Penny
Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences. | 2022-09-30 17:08:50 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3401528000831604, "perplexity": 1886.9019092529234}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030335491.4/warc/CC-MAIN-20220930145518-20220930175518-00155.warc.gz"} |
https://codereview.stackexchange.com/tags/openssl/hot | # Tag Info
8
Things you did well on: Looks like you weren't too rusty based on this code. Looks very nice organizationally. You return different values for different error conditions. Things you could improve on: Standards Initialize i in your for loops.(C99) for (unsigned int i = 0; i < sizeof(digest); i++) Syntax Don't over-compact your code. if(argc != 6){...
8
C++ review You still need a C review! Idioms/Patterns RAII idiom In C++ we have this concept that an object should clean up its own resources. So when an object is created it will create and hold onto resources and when it is destroyed it will clean up those resources. When learning about the idiom people mostly talk about memory and smart pointers. ...
7
It would be nice to include a link to an authoritative reference. Until then, no comment on the algorithm compliance is possible. Instead of switch I'd rather have a table of strings indexed by BitcoinAddressState values. In C, main must return a value explicitly.
7
I would make the following changes, in order to make it more robust, and simpler: Do not use openssl and symmetric encryption. Use gpg and public key encryption. The advantage is that you do not have to store any passwords, anywhere (keep the secret key away from the backup script). Public key encryption is great for backups (you only need to access your ...
7
I'll start with a review of your current code before suggesting possible performance improvements. Various comments in your code do not add information and can be removed, for example //------------------------------Main Function------------------------// int main() free(string); //free string exit(EXIT_SUCCESS); //Program Ends Always use curly braces { ...
6
This is an incomplete answer; I'd just like to contribute more comments than easily fit in an actual comment. Consider replacing PASS="mysecretpasswordiwontwritehere" with PASS=cat mysecretpassword.file and moving your password into that file. Then you don't have to worry about always remembering to hide it when sharing your source code (or when editing/...
6
You can make these #defines: #define SHA1_DIGESTLENGTH 20 #define SHA1_BLOCK_LENGTH 64 #define COUNTER_LENGTH 8 into a more concise enum: typedef enum { COUNTER=8, SHA1_DIGEST=20, SHA1_BLOCK=64 } Length; This could also be done with SECRET and COUNTER, but each hex value would need a name. If you use the enum recommendation, you may need to rename the ...
6
Just a few notes on some things I didn't see mentioned. Compilation: I originally couldn't compile the program with the command in the comments. /tmp/cc2H2h0a.o: In function 'clc_pi': test.c:(.text+0x148): undefined reference to 'clock_gettime' test.c:(.text+0x2f0): undefined referenceto 'clock_gettime' collect2: ld returned 1 exit status Add -...
6
for crl in ${crls[@]}; do It's a good habit to always double-quote array expansions. Cleaning up the temp directory after a successful run would be a nice touch (or do away with it altogether; see below). cat${crl_temp}/*.pem > ${crl_temp}/${new_crl} mv ${crl_temp}/${new_crl} ${crl_dir}/${new_crl} I assume you're doing this to get atomic replacement? ...
5
The documentation says that crypto_strong is set depending on the algorithm used. So it will never go from false to true, so yes, this might result in an infinite loop. Use it instead to see if you should trust the result. If it is true, the string should be secure. (you should also never design your own crypto algorithms for anything other than academic ...
5
On top on haneefmubarak's comment, I'd like to point out a couple of things I do not quite like : Variable names I think this : mpz_t v1, v2, v3, v4, v5; mpf_t V1, V2, V3, total, tmp, res; says it all. Not only the variables are meaningless names but on top of that, the case does matter. It's hard for everyone to read this without being confused. Also, ...
5
For starters, don't use __inline__. It isn't portable. Instead, use the standard C inline. Next, use more whitespace. Complex calls and almost all conditionals should go into multiple lines, while operations should be spaced. Also, explain what you are doing: // original snprintf(&(checksum[i*2]), 3, "%02x", (unsigned int)digest[i]); // what it ought ...
5
A buffer overflow: char HexResult[2*outputbytes-1]; Why the -1? There are 2 * outputbytes hex digits. So you need at least 2*outputbytes chars. Since you want a null terminated string (the rest of your code assumes it is), you even need an extra character for the \0 terminator. So this should be char HexResult[2 * outputbytes + 1]; A partial initialization:...
4
Portability The code would fail if the sender and receiver platforms disagree on the size of integer. The code would fail if the sender and receiver platforms disagree on the endianness of integer. Partial reads There is a chance that the very first read returns just a couple of bytes. The code still interprets it as a valid integer. Overall There is no ...
4
The worst problem with this code (IMO) is the readability, please take 10 minutes to have a look at other people's code on this site. I suspect you won't get much input because of the layout of your code. Name variables properly, you might know what I is for now, but in 6 months... You are deleting ssl and ctx without checking they have been allocated. You ...
4
Get rid of unnecessary nested and combined ifs While here it goes to no extreme, it's advisable to check for errors first, and to do that one by one - if possible, which is exactly this case. Do some extra checks Only in case of 0 arguments, the script should output usage message. This script does not support multiple arguments, so if given more than 1, ...
3
Looks good. No buffering problems anymore. Endianness is handled correctly. Reading and writing size is duly factored out. The only suggestion is not rely on the native unsigned type to represent the size. Notice that htonl takes uint32_t as a parameter. It is a strong hint to pass size as uint32_t as well. This guarantees that the size is 4 byte long no ...
3
You can use a here document to avoid all those calls to echo: cat <<END Usage: git-timestamp [options] command [revision] options -h - Show this usage info. All other options are ignored. -v - Output long revision instead of short. -l - Also show the local commit time of the specified revision. ...etc... END
3
Just a few minor things to add on top of what others already said. Prefer $(...) instead of ... This is the modern way and it's easier to nest: DATE=$(date +%Y%m%d) No need to quote inside [[...]] This will work just fine: if [[ $LASTBACKUP ==$DATE ]]; then echo "Do nothing, already backed up." exit fi And the 0 in exit 0 is redundant, as 0 is ...
3
There's no real way to speed up the brute-force enumeration of all possible 3-letter passwords. Perhaps you could use the dictionary. There's a finite list of 3-letter English words. They may be slightly more common. Also, if you google for "most common passwords", some kind of 3-letter version of that list could be tried before anything else.
3
I would like to know if is a good design A few points about your design from a c++ perspective: 1. Usage of free functions instead of classes From the point of OOP view, you should rather use classes than free functions to wrap / encapsulate the C API: namespace redi { namespace util { namespace base64 { class Encoder { public: Encoder(...
2
to answer your questions first: you don't need begin/end if it covers whole method, you can rescue directly from method, Net::HTTP already handles timeout, no need to build custom wrapper around it, now there are few other things that should be addressed: inline things when possible don't use @@ ever, use constants if you need to don't rescue from ...
2
GenerateRandomString is a poor name. I don't know what the contents of my string will be. In your case, you've restricted the character set to [0-9A-F], but that's not reflected in the name. Consider renaming this to generateRandomHexadecimalString or something that more clearly states the intention. You noted why $length is divided by two in a comment on @... 2 I've got a few things to say about the code, but having looked at the diff: - @ssl_version = "SSLv3" + @ssl_version = :auto and: - ctx.ssl_version = @ssl_version + ctx.ssl_version = @ssl_version unless @ssl_version == :auto your code changes seem minimal, and to the point doing exactly what they are supposed to do, and nothing more. ... 2 Might I suggest using OpenSSL's built in HMAC_* methods instead of re-rolling? 2 The best way to do space1="a b c d e f g h i j k l m n o p q r s t u v w x y z" for i in$space1; do for j in $space1; do for k in$space1; do echo $i$j$k done done done in bash is probably : for i in {a..z}{a..z}{a..z}; do echo$i; done Then, depending on what your homeworks says, it might or might not be what ...
2
Look at the man page for openssl http://www.openssl.org/docs/apps/openssl.html It is the command being executed here: $(openssl passwd -crypt -salt "$1" "$i$j$k") Another Clue below. Only roll your curser over it if you need the extra help. Still have not worked it out! 2 I honestly disagree with this whole approach. First, storing files IN a relational database is rarely a good strategy, limited almost exclusively to use cases where you want to perform binary searches against the binary artifacts. That is not your use case here. Outside of that, you are just adding a lot of overhead to your typical database management ... 2 Hardcoding /tmp/ssh.pub in doesn't look right. Makes the script vulnerable to all kinds of failures and race conditions. tempfile, perhaps? Similar concern applies to /tmp/pem.pub. Strongly recommend to derive pem name from$1. I am not sure I understand the significance of '1.2.840.113549.1.1.1'. Is there a reason to have a bash/python mixture (vs pure ...
2
In terms of error handling, it seems to be much safer to call ERR_clear_error() before any call to SSL_read(), SSL_write(), SSL_accept, and so on. See one of the few reported cases where error management has messed up people. Basically, each thread shares an error stack. While SSL_get_error() grab the latest error associated to a ssl session, it does not ...
Only top voted, non community-wiki answers of a minimum length are eligible | 2019-12-15 22:33:36 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.29166844487190247, "perplexity": 2194.6531646072835}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-51/segments/1575541310866.82/warc/CC-MAIN-20191215201305-20191215225305-00228.warc.gz"} |
https://chemistry.tutorvista.com/inorganic-chemistry/ligand-field-theory.html | Top
# Ligand Field Theory
The mutual interaction between bonding electron pairs is the same for transition metal compounds as for compounds of main group elements.
All statements concerning molecular structure apply equally. However, non bonding valence electrons behave differently. For transition metal atoms these generally are d electrons that can be accommodated in five d orbitals.
The modified crystal field theory which admits that there is some covalent as well as electrostatic interaction between the ion and its neighbors, is called ligand field theory.
In what manner the electrons are distributed among these orbitals and in what way they become active stereo chemically can be judged with the aid of ligand field theory. The concept of ligand field theory is equivalent to that of the valence shell electron pair repulsion theory: it considers how the d electrons have to be distributed so that they attain a minimum repulsion with each other and with the bonding electron pairs.
Related Calculators Electric Field Calculator
## Ligand Field Theory Tetrahedral Complex
An example of a tetrahedral metal complex is VCl4 which is shown in a convenient coordinate system. The 4s and 4p atomic orbitals of vanadium can be used to form sigma molecular orbitals. Although, the overlap patterns are rather complicated the 3dxz, 3dyz and 3dxy valence orbitals also are situated properly to form sigma molecular orbitals.
In terms of localized molecular orbitals both sd3 and sp3 hybrid orbitals are tetrahedrally oriented.
The ligand field splitting diagram for a tetrahedral complex such as VCl4 is shown below.
1. The anti bonding molecular orbitals derived from the 3d valence orbitals are divided into two sets.
2. The orbitals formed from the 3dxz, 3dyz and 3dxy orbitals are of higher energy than those formed from the 3dz2and 3dx2-y2 orbitals.
3. Thus, the change from octahedral to tetrahedral geometry exactly reverses the role and the energies of the d valence orbitals of the central metal ion.
4. $\Delta _{t}$ is the energy difference between the t2 and e in tetrahedral complex.
5. From ligand field theory we can predict that the t2 orbitals in a tetrahedral complex will not form as strong sigma bonds with ligand sigma orbitals as will the eg octahedral orbitals thereby resulting in a much less energetic t2 level and a relatively small value.
6. Because of the small values all tetrahedral transition metal complexes have high spin ground state configurations.
## Ligand Field Theory Square Planar Complex
The d8 metal ions form square planar complexes. The example we will use here is [PtCl4]2-. The principle sigma bonding involves the overlap of 3p(sigma)Cl- orbitals with the 5dx2-y2, 6s, 6px and 6py metal valence orbitals. In the language of localized molecular orbital theory, the sigma bonding is summarized as dsp2 in a square planar complex.
Of principle interest the ligand field splitting of the anti bonding molecular orbitals derived from the metal d valence orbitals in a square planar complex.
The ligand field splitting diagram for a square planar complex such as [Pt(Cl)4]2- is shown below.
The ligand field splitting in a square planar complex is rather complicated because there are four different energy levels. For all square planar complexes it is reasonable to place the strongly anti bonding orbital at the highest energy level.
However regardless of the placement the most important characteristic of the ligand-field splitting in a square planar complex is that $d_{x^{2}-y^{2}}$ much higher energy than other four orbitals which are about the same energy. | 2019-06-26 11:54:07 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5257661938667297, "perplexity": 1804.1189834647514}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560628000306.84/warc/CC-MAIN-20190626114215-20190626140215-00030.warc.gz"} |
https://chem.libretexts.org/Courses/Nassau_Community_College/CHE200_-_Introduction_to_Organic_Chemistry_(Resch)/11%3A_Oxidation_and_Reduction_Reactions/11.08%3A_Flavin-Dependent_Monooxygenase_Reactions_-_Hydroxylation_Epoxidation_and_the_Baeyer-Villiger_Oxidation | # 11.8: Flavin-Dependent Monooxygenase Reactions - Hydroxylation, Epoxidation, and the Baeyer-Villiger Oxidation
Up to now, the redox reaction examples we have seen have all been either hydrogenation/dehydrogenation transformations or interconversions between free thiols and disulfides. However, there are many important redox reactions in biological chemistry which do not fall under either of these descriptions. Oxygenase enzymes catalyze the insertion of one or two oxygen atoms from molecular oxygen ($$O_2$$) into an organic substrate molecule. Enzymes which insert a single oxygen atom are called monooxygenases. Below are two examples of biochemical transformations catalyzed by monooxygenase enzymes: one is a hydroxylation, the other is an epoxidation (an epoxide functional group is composed of a three-membered carbon-carbon-oxygen ring - epoxides are somewhat rare in biological organic chemistry but are very common and useful intermediates in laboratory organic synthesis).
Dioxygenase enzymes insert both oxygen atoms from $$O_2$$ into the substrate, and usually involve cleavage of an aromatic ring. Below is an example of a dioxygenase reaction, catalyzed by catechol dioxygenase:
In the reduction direction, reductases remove oxygen atoms, or sometimes other electronegative heteratoms such as nitrogen or halides. For example, DNA deoxyribonucleosides are converted from their corresponding RNA ribonucleosides by the action of reductase enzymes:
Many oxygenase and reductase reactions involve the participation of enzyme-bound transition metals - such as iron or copper - and the mechanistic details of these reactions are outside the scope of our discussion. A variety of biochemical monooxygenase reactions, however, involve flavin as a redox cofactor, and we do have sufficient background knowledge at this point to understand these mechanisms. In flavin-dependent monooxygenase reactions, the key intermediate species is flavin hydroperoxide.
The term 'peroxide' refers to a functional group characterized by an oxygen-oxygen single bond. The simplest peroxide is hydrogen peroxide ($$HOOH$$) about which we will have more to say below. In flavin hydroperoxide, the peroxide group is linked to one of the carbons of the reactive triple-ring system of the coenzyme. A possible mechanism for the formation of flavin peroxide from $$FADH_2$$ and molecular oxygen is shown below.
Silverman, R.B. The Organic Chemistry of Enzyme-Catalyzed Reactions, p. 121-122, Scheme 3.33. 2000, Academic Press, San Diego.
Mechanism for the formation of flavin hydroperoxide:
(Note: Implicit in this mechanism is that the molecular oxygen first undergoes spin inversion from the triplet state to the higher energy 'singlet' state. You may recall from your general chemistry course that molecular oxygen exists in two states: 'singlet' oxygen has a double bond and no unpaired electrons, while 'triplet' oxygen has a single $$O-O$$ bond and two unpaired electrons - a kind of 'double radical'. Molecular orbital theory - and experimental evidence - show that the triplet state is lower in energy.
The mechanism shown above is one proposed mechanism, another proposal involves triplet oxygen reacting with flavin in a series of radical-intermediate, single-electron steps.)
Flavin hydroperoxide can be thought of as an activated form of molecular oxygen. Peroxides in general are potent oxidizing agents, because the oxygen-oxygen single bond is quite weak: only 138 kJ/mole, compared to 339 kJ/mol for a carbon-carbon bond, and 351 kJ/mol for a carbon-oxygen bond. When the 'outer' oxygen of flavin hydroperoxide (red in our figure above) comes into close proximity to the p-bonded electrons of an alkene or aromatic group, the $$O-O$$ bond will break, leaving an empty orbital on the outer oxygen to be filled by the p electrons - thus, a new carbon-oxygen bond is formed. This is what is happening in step 1 of a reaction in the tryptophan degradation pathway catalyzed by kynurenine 3-monooxygenase. Step 2 completes what is, mechanistically speaking, an electrophilic aromatic substitution reaction (section 14.4) with an peroxide oxygen electrophile.
Mechanism for the flavin hydroperoxide-dependent hydroxylation of kynurenine:
Elimination of water from the hydroxyflavin intermediate then leads to formation of $$FAD$$ (step 3), which is subsequently reduced back to $$FADH_2$$ by $$NADH$$ (step 4).
The $$N$$-hydroxylation reaction below, which is part of the of the biosynthetic pathway of an iron-binding molecule in the pathogenic bacterium Pseudomonas aeruginosa, is mechanistically similar to the $$C$$-hydroxylation reaction we just saw, except that the nucleophile is an amine nitrogen. Note that $$FADH_2$$ is shown in brackets below the reaction arrow, indicating that reduced flavin participates in the reaction but is not used up - rather it is regenerated in the active site at the end of the reaction cycle.
##### Exercise 15.8.1
Draw arrows for the $$N-O$$ bond-forming step in the ornithine hydroxylation reaction above.
Epoxides, characterized by a three-membered ring composed of two carbons and one oxygen, are a very common and useful functional group employed in synthetic organic chemistry. Although rare, there are some interesting epoxide-forming reactions in biochemical pathways, catalyzed by flavin-dependent monooxygenase enzymes.
In a key step in the biosynthesis of cholesterol and other steroid compounds, an alkene is converted to an epoxide in a precursor molecule called squalene. Flavin hydroperoxide also serves as the direct oxidizing agent in this step:
Mechanism for the flavin-hydroperoxide-dependent epoxidation of squalene:
Oxidosqualene goes on to cyclize to lanosterol in a complex and fascinating electrophilic reaction which we discussed in section 14.5.
Epoxidation reactions have a parallel in the synthetic organic laboratory, and in fact are very important tools in organic synthesis. In laboratory epoxidations, peroxyacids are the counterpart to flavin hydroperoxide in biochemical epoxidations. meta-chloroperoxybenxoic acid (MCPBA) is a commonly used peroxyacid.
The Baeyer-Villiger oxidation, in which a ketone is converted to an ester through treatment with a peroxide reagent, is an extremely useful laboratory organic synthesis reaction discovered in the late 19th century. Recently, many biochemical examples of Baeyer-Villiger oxidations have been discovered: the reaction below, for example, is catalyzed by a monooxygenase in a thermophilic bacterium:
(Proc. Natl. Acad. Sci. U.S.A. 2004, 101, 13157)
A Baeyer-Villiger oxidation:
Mechanism:
The Baeyer-Villiger mechanism is differs significantly from the hydroxylation reactions we saw earlier, although flavin hydroperoxide (abbreviated in the above figure) still plays a key role. Here, the peroxide oxygen is a nucleophile, rather than an electrophile, attacking the ketone carbonyl in step 1. Step 2 is a rearrangement, similar in many ways to the hydride and alkyl shifts we learned about in section 14.5. The electrons in the red bond in the figure shift over one atom: from the carbonyl carbon to the outer peroxide oxygen. The end result is that an oxygen atom, from $$O_2$$ via flavin hydroperoxide, has been inserted between the carbonyl carbon and a neighboring methylene ($$CH_2$$) carbon, forming an ester.
Note that in the reaction mechanism above, the ketone substrate is asymmetric: on one side of the carbonyl there is a benzyl group ($$CH_2$$-phenyl), and on the other side a methyl group. Note also that it is the benzyl group, not the methyl, that shifts in step 2 of the mechanism. For reasons that are not yet well understood, in Baeyer-Villiger reactions the alkyl group with higher carbocation stability has a higher migratory aptitude: in other words, it has a lower energy barrier for the shifting step.
##### Exercise 15.8.2
Draw the product of a hypothetical Baeyer-Villiger reaction involving the same substrate as the above figure, in which the methyl rather than the benzyl group shifts.
##### Exercise 15.8.3
Draw the likely major product of a hypothetical Baeyer Villiger reaction starting with 2-methylcyclopentanone as the substrate. Take into account the idea of migratory aptitude.
Below is another example of a Baeyer-Villiger reaction in which a cyclic ketone is oxidized to a lactone (cyclic ester). Notice that oxygen insertion expands the ring from 6 to 7 atoms. This is the third-to-last step in the biosynthesis of the anti-cancer agent mithromycin in some bacterial species (ACS Chem. Biol. 2013, 8, 2466).
Yet another variety of flavin-dependent monooxygenase, which bears some mechanistic similarity to the Baeyer-Villiger oxidation, is the decarboxylative reaction below from biosynthesis of the plant hormone auxin: (J. Biol. Chem. 2013, 288, 1448)
##### Exercise 15.8.4
Propose a mechanism for the above reaction, starting with flavin hydroperoxide.
This page titled 11.8: Flavin-Dependent Monooxygenase Reactions - Hydroxylation, Epoxidation, and the Baeyer-Villiger Oxidation is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Tim Soderberg. | 2023-03-28 12:09:31 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6782680153846741, "perplexity": 4222.032429824421}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296948858.7/warc/CC-MAIN-20230328104523-20230328134523-00122.warc.gz"} |
https://www.esaral.com/q/consider-a-72-cm-long-wire-ab-as-shown-in-the-figure-72645 | # Consider a 72 cm long wire AB as shown in the figure.
Question:
Consider a $72 \mathrm{~cm}$ long wire $\mathrm{AB}$ as shown in the figure. The galvanometer jockey is placed at $P$ on $A B$ at a distance $x \mathrm{~cm}$ from $A$. The galvanometer shows zero deflection.
The value of $x$, to the nearest integer, is
Solution:
In Balanced conditions
$\frac{12}{6}=\frac{x}{72-x}$
$x=48 \mathrm{~cm}$
#### Leave a comment
None
Free Study Material | 2023-02-04 22:07:17 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8657263517379761, "perplexity": 752.2883571588627}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764500154.33/warc/CC-MAIN-20230204205328-20230204235328-00077.warc.gz"} |
https://www.physicsforums.com/threads/getting-the-joint-probability-density-for-the-characteristic-equation.726261/ | # Getting the joint probability density for the characteristic equation
1. Dec 3, 2013
### physicsjn
1. The problem statement, all variables and given/known data
The stochastic variables X and Y are independent and Gaussian distributed with
first moment <x> = <y> = 0 and standard deviation σx = σy = 1. Find the characteristic function
for the random variable Z = X2+Y2, and compute the moments <z>, <z2> and <z3>. Find the first 3 cumulants.
2. Relevant equations
Characteristic equation: $f_z (k) = <e^{ikz}> = \int_{-\infty}^{+\infty} e^{ikz}\, P_z (z) dz$
Joint Probability density: $P_z(z) = \int_{-\infty}^{+\infty} dx \, \int_{-\infty}^{+\infty} dy \, δ (z - G(x,y)) P_{x,y}(x,y)$ where $z = G (x, y)$
Also, $P_{x,y} = P_x (x) \, P_y (y)$ for independent stochastic variables x and y.
For Gaussian distribution: $P_x = \frac{1}{\sqrt{2∏} } e^{\frac{-x^2}{2}}$
3. The attempt at a solution
To get the characteristic equation, we need first to get the joint probability density Pz(z):
Since $G(x,y)= x^2 +y^2$ and $P_{x,y} = P_x (x) \, P_y (y)$
$P_z(z) = \int_{-\infty}^{+\infty} dx \, \int_{-\infty}^{+\infty} dy \, δ (z - x^2 +y^2) P_x (x) P_y (y)$
$P_z(z) = \int_{-\infty}^{+\infty}P_x (x) \, dx \, \int_{-\infty}^{+\infty}P_y (y) \, dy \, δ (z - x^2 +y^2)$
$P_z(z) = \int_{-\infty}^{+\infty}\frac{1}{\sqrt{2∏} } e^{\frac{-x^2}{2}} \, dx \, \int_{-\infty}^{+\infty}\frac{1}{\sqrt{2∏} } e^{\frac{-y^2}{2}} \, dy \, δ (z - x^2 +y^2)$
1. The problem statement, all variables and given/known data
2. Relevant equations
3. The attempt at a solution
1. The problem statement, all variables and given/known data
2. Relevant equations
3. The attempt at a solution
1. The problem statement, all variables and given/known data
2. Relevant equations
3. The attempt at a solution
1. The problem statement, all variables and given/known data
2. Relevant equations
3. The attempt at a solution
1. The problem statement, all variables and given/known data
2. Relevant equations
3. The attempt at a solution
1. The problem statement, all variables and given/known data
2. Relevant equations
3. The attempt at a solution
1. The problem statement, all variables and given/known data
2. Relevant equations
3. The attempt at a solution
2. Dec 3, 2013
### physicsjn
I'm very sorry; I accidentally pressed the submit post button instead of preview post. How do I erase this? I'm not yet done with my post. :(
3. Dec 3, 2013 | 2018-02-18 18:59:40 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5672449469566345, "perplexity": 1313.4973448318626}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-09/segments/1518891812247.50/warc/CC-MAIN-20180218173208-20180218193208-00504.warc.gz"} |
https://lmcs.episciences.org/976 | ## Lampis, Michael - Model Checking Lower Bounds for Simple Graphs
lmcs:976 - Logical Methods in Computer Science, March 25, 2014, Volume 10, Issue 1
Model Checking Lower Bounds for Simple Graphs
Authors: Lampis, Michael
A well-known result by Frick and Grohe shows that deciding FO logic on trees involves a parameter dependence that is a tower of exponentials. Though this lower bound is tight for Courcelle's theorem, it has been evaded by a series of recent meta-theorems for other graph classes. Here we provide some additional non-elementary lower bound results, which are in some senses stronger. Our goal is to explain common traits in these recent meta-theorems and identify barriers to further progress. More specifically, first, we show that on the class of threshold graphs, and therefore also on any union and complement-closed class, there is no model-checking algorithm with elementary parameter dependence even for FO logic. Second, we show that there is no model-checking algorithm with elementary parameter dependence for MSO logic even restricted to paths (or equivalently to unary strings), unless E=NE. As a corollary, we resolve an open problem on the complexity of MSO model-checking on graphs of bounded max-leaf number. Finally, we look at MSO on the class of colored trees of depth d. We show that, assuming the ETH, for every fixed d>=1 at least d+1 levels of exponentiation are necessary for this problem, thus showing that the (d+1)-fold exponential algorithm recently given by Gajarsk\{y} and Hlin\u{e}n\{y} is essentially optimal.
Source : oai:arXiv.org:1302.4266
DOI : 10.2168/LMCS-10(1:18)2014
Volume: Volume 10, Issue 1
Published on: March 25, 2014
Submitted on: September 16, 2013
Keywords: Computer Science - Computational Complexity,Computer Science - Logic in Computer Science | 2018-05-25 08:46:23 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6766511797904968, "perplexity": 1423.9397683918037}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-22/segments/1526794867055.20/warc/CC-MAIN-20180525082822-20180525102822-00383.warc.gz"} |
https://math.stackexchange.com/questions/62314/construction-of-a-symplectic-basis-for-a-lattice | # Construction of a symplectic basis for a lattice
Let $(T,E)$ be a polarized abelian variety ($T=V/L$, $\dim_\mathbb{C} V=g$, $E:V\times V\to\mathbb{R}$ a nondegenerate real alternating bilinear form, with $E(L\times L)\subseteq\mathbb{Z}$ and $E(iv,iw)=E(v,w)$). I'm trying to prove the existence of a basis for $L$ such that the matrix of $E$ has the form $$E=\left(\begin{array}{cc}0&D\\-D&0\end{array}\right),$$ where $D$ is a diagonal matrix with diagonal $d_1,\ldots,d_g$, and $d_i\mid d_{i+1}$.
Now what I've tried so far is considering $\min\{E(v,w):v,w\in L,E(v,w)>0\}$, and taking $e_1,e_{g+1}$ that achieve this minimum. Then I would like to decompose $V$ into $V=\langle e_1,e_{g+1}\rangle\oplus\langle e_1,e_{g+1}\rangle^\perp$ (this can be done since $E$ is nondegenerate). I would like to repeat this same process in $\langle e_1,e_{g+1}\rangle^\perp$, but I have no way of knowing "how many" lattice points are in $\langle e_1,e_{g+1}\rangle^\perp$. Am I attacking this problem the wrong way, or is there something I'm missing?
• – David E Speyer Sep 6 '11 at 18:45
• I looked at it, but I still have some questions. How is it possible to find $e$ and $f$ such that the lattice spanned by $e$ and $f$ is the intersection of the whole lattice with the vector space spanned by $e$ and $f$? It also says to complete the basis $\{e,f\}$ to a basis for the whole lattice, but I'm not sure you can always complete a basis for a lattice... – rfauffar Sep 6 '11 at 18:51
• In what form are you given your lattice? For most presentations, this should be standard linear algebra. As far as the last point goes, if $L \cap \mathrm{Span}_{\mathbb{R}}(e,f) = \mathrm{Span}_{\mathbb{Z}}(e,f)$, then $L / \langle e,f \rangle$ is torsion free, so it is free of rank $\dim L-2$. Take any basis for this free module and lift it to $L$; the lifted vectors, together with $e$ and $f$, are a basis for $L$. – David E Speyer Sep 6 '11 at 19:09
• Thanks, you made it a lot clearer. I'm able to complete the proof now. – rfauffar Sep 6 '11 at 20:23 | 2019-06-18 07:05:16 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9241123199462891, "perplexity": 106.5477114439033}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560627998690.87/warc/CC-MAIN-20190618063322-20190618085322-00011.warc.gz"} |
https://calendar.math.illinois.edu/cal?year=2017&month=12&day=07&interval=day&advance=1+week&interval=month&use=Find®exp=open+house | Department of
# Mathematics
Seminar Calendar
for open house events the month of Thursday, December 7, 2017.
.
events for the
events containing
Questions regarding events or the calendar should be directed to Tori Corkery.
November 2017 December 2017 January 2018
Su Mo Tu We Th Fr Sa Su Mo Tu We Th Fr Sa Su Mo Tu We Th Fr Sa
1 2 3 4 1 2 1 2 3 4 5 6
5 6 7 8 9 10 11 3 4 5 6 7 8 9 7 8 9 10 11 12 13
12 13 14 15 16 17 18 10 11 12 13 14 15 16 14 15 16 17 18 19 20
19 20 21 22 23 24 25 17 18 19 20 21 22 23 21 22 23 24 25 26 27
26 27 28 29 30 24 25 26 27 28 29 30 28 29 30 31
31
Thursday, December 14, 2017
1:00 pm in 239 Altgeld Hall,Thursday, December 14, 2017
#### IGL Fall 2017 Open House
Abstract: The IGL end-of-semester Open House will take place from 1:00 pm to 4:00 pm in Room 239 Altgeld Hall. Come see what the IGL teams have produced this semester! | 2017-12-16 08:59:01 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.17583195865154266, "perplexity": 402.72293884431355}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948587496.62/warc/CC-MAIN-20171216084601-20171216110601-00224.warc.gz"} |
https://math.stackexchange.com/questions/2991712/matrix-determinant-effect | # Matrix determinant effect
Determinant of the matrix $$A= \begin{bmatrix}a&b&c\\d&e&f\\g&h&i\end{bmatrix}$$ is $$\det A=4.$$ So what is the determinant of $$\begin{bmatrix}3a&3b&3c\\-d&-e&-f\\g-a&h-b&i-c\end{bmatrix}$$
I found that the row operations that were done were $$3R1, -1.R2,$$ and the last one doesn't matter.
So is the determinant $$3\times 4\times(-1)= -12$$ or we have to do the inverse $$4\times 1/3 \times 1/(-1)?$$
• Yes the answer is -12. Your approach is correct. – SchrodingersCat Nov 9 '18 at 17:54
Yes that's correct, indeed by the properties of the determinant we have that
$$\det\begin{bmatrix}a&b&c\\d&e&f\\g&h&i\end{bmatrix}=\det\begin{bmatrix}a&b&c\\d&e&f\\g-a&h-b&i-c\end{bmatrix}=\\=\frac13 \det\begin{bmatrix}3a&3b&3c\\d&e&f\\g-a&h-b&i-c\end{bmatrix}=-\frac13 \det\begin{bmatrix}3a&3b&3c\\-d&-e&-f\\g-a&h-b&i-c\end{bmatrix}$$
Sometimes you can't tell if matrix $$B$$ can be generated from matrix $$B$$ or not. In such a case, one may expand each determinant and compare the result of each. However, this approach requires careful attention to signs!
Let $$X$$ be the determinant of the first matrix and $$Y$$ be the determinant of the 2nd matrix, then we have:
$$X = a(ei - fh) - b(di - fg) + c(dh - eg)$$
$$X = aei - afh - bdi + bfg + cdh - ceg$$
$$Y = 3afh - 3eai + 3ecg + 3bdi - 3bfg - 3cdh$$
$$\frac{Y}{-3} = -afh + eai -ecg -bdi + bfg + cdh$$
You could re-arrange the terms of any of the equations to see that:
$$X=\frac{Y}{-3}$$
Since X=4,
$$Y=-12$$ | 2019-01-20 04:25:48 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 17, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9317535161972046, "perplexity": 219.84487671064406}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-04/segments/1547583700012.70/warc/CC-MAIN-20190120042010-20190120064010-00118.warc.gz"} |
https://stats.stackexchange.com/questions/61622/by-using-smote-the-classification-of-the-validation-set-is-bad | # By using SMOTE the classification of the validation set is bad
I want to do classification with 2 classes. When I classify without smote I get:
Precision Recall f-1
0,640950987 0,815410434 0,714925374
When I use smote: (oversample the minority class at 200% and k = 5)
Precision Recall f-1
0,831024643 0,783434343 0,804894232
As you can see this works well.
However, when I test this trained model on validation data (which hasn't got any synthetic data)
Precision Recall f-1
0,644335755 0,799044453 0,709791138
which is awful. I used a random decision forest to classify.
Has anyone got any idea why this is happening? Any useful tips regarding extra tests I can try to get more insight are welcome too.
More info: I do not touch the majority class. I work in Python with scikit-learn and this algorithm for smote.
The confusion matrix on the test data (which has synthetic data):
The confusion matrix on the validation data with the same model (real data, which was not generated by SMOTE)
Edit: I read that the problem possibly lies in the fact that Tomek Links were created. Therefore I wrote some code to remove the tomek links. Though this doesn't improve the classification scores.
Edit2: I read that the problem possibly lies in the fact there is too much of an overlap. A solution for this is a more intelligent synthetic sample generation algorithm. Therefore I implemented
My implementation can be found here. It performed worse than smote.
If you oversample the training data to change the relative class frequencies in the training set, you are implicitly telling the classifier to expect the validation set to have those same class frequencies. As the class frequencies influence the decision boundary, if the validation set class frequencies are different, then the classification boundary will be wrong as a result and performance will be sub-optimal.
Now it is true that for some classifiers, having imbalanced classes can cause it to perform badly, and re-sampling the data can help to correct this. However if you oversample the minority class too much, you will over-correct and performance will be suboptimal because the the difference in class frequencies between the training and validation sets is more than is needed to correct the imbalance problem.
My recipe then would be to use cross-validation to choose the degree of oversampling that is required to just compensate for the "class imbalance problem" experienced by your classifier, but no more. Note however this leads to an additional parameter that needs to be tuned, which leads to greater computational expense, and a higher likelihood of over-fitting in model selection (if the total number of tunable parameters is non-negligible).
• This extra parameter you talk about, is this the % I use ? (in my case 200%). I can't choose a smaller one, they have to be in steps of 100. Though if I understand correctly, if I make validationset which is balanced like the training set, it will give better results ? – Olivier_s_j Jun 13 '13 at 11:24
• yes, the %age was indeed the parameter I meant; it is a pity if the software only allows this to be changed in steps of 100. Some machine learning algorithms (e.g. the SVM) allow you to weight positive and negative patterns differently (for the SVM have a different value of the parameter C for each class). This is directly equivalent to SMOTE in the case of the SVM, so that may be another way to get around the problem. – Dikran Marsupial Jun 13 '13 at 11:31
• I tried the SVM solution. In sci-kit learn it is possible to set the weights inversely proportional to class frequencies (automatically). Though this gives very bad results ( for linear, polynomial, rbf and sigmoid kernel) – Olivier_s_j Jun 13 '13 at 11:45
• If you just set them inversely proportional to the class frequencies that tells the SVM to act as if the class frequencies in the validation set are equal, which again is likely to over-compensate for the class balance problem. Try tuning the weights via cross-validation, it normally works for me. The other thing to do is to consider whether the misclassification costs actually are the same for false-positive and false-negative errors, and if not, include this in setting the C parameters for each class. – Dikran Marsupial Jun 13 '13 at 11:48
• Hmmm I've tried tuning the weights manually though what the SVM predicts is or all 1s or all 0s. No in-between. (Btw I've added confusion matrices to the question for clarification, as you can see by using SMOTE, the unbalance is gone when using a test set which is also composed of synthetic data but there is still in unbalance when a test is run on the validationset.) – Olivier_s_j Jun 13 '13 at 12:12
Have you considered using the R implementation of random forest via Rpy? It has an option that allows you to choose both the sample size that each tree is trained on and the fraction falling into each class. I've found it to be the most effective way to deal with imbalance when using random forest. | 2021-05-09 20:17:31 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6246997117996216, "perplexity": 798.1443396873502}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243989012.26/warc/CC-MAIN-20210509183309-20210509213309-00250.warc.gz"} |
https://codegolf.stackexchange.com/questions/98692/ultimate-codegolf-codegolf-a-turing-complete-system?noredirect=1 | # ULTIMATE CODEGOLF!!! (Codegolf a turing-complete system) [closed]
The goal: write a program that implements a turing-complete system. It could be cellular automata, a tag system, a turing machine, an interpreter for a language of your own design... the type of system doesn't matter as long as it satisfies the following conditions:
• Takes a "program" from the user as input (ex. the initial state of the turing machine)
• Runs the program
• Outputs the state of the system (or just the returns the output of the program)
• It's possible to compute anything that can be computed with the right input program to your system.
Self-interpreting is forbidden.
As usual entries in the same language that implement the same type of system will be judged by byte count.
For example: programs which implement a universal turing machine in Python will be compared against other programs which implement universal turing machines in Python. The spirit of the challenge is basically the simplest possible programs that can model a turing-complete system.
Because of the diversity of possible programming languages, vote for entries that you believe are clever, elegant, or especially short. The winner will be the entry with the most votes.
## closed as too broad by Peter Taylor, Mego♦, Dennis♦Nov 6 '16 at 6:52
Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
• Would a single eval() be considered forbidden in this challenge? – Sunny Pun Nov 6 '16 at 5:20
• Could you be more specific? – J. Antonio Perez Nov 6 '16 at 5:22
• @JorgePerez Hmm.... I get the feeling people could just copy answers from the "Golf a BrainF**k Intepreter" thread... – Socratic Phoenix Nov 6 '16 at 5:29
• I don't think the "word and symbol" system is good enough - it may cause ambiguities down the way. Just use byte-count, it should be fine. Also, what SunnyPun meant was a program (in Python, say) in which the code is eval(input()), and the input is valid Python code. That answer is valid under your current terms. – Qwerp-Derp Nov 6 '16 at 5:39
• Do X creatively popularity contests have fallen out of scope. In addition, asking to implement one choice out of the infinity of Turing complete languages makes this rather broad, and without an objective goal like code size, too broad. By the way, we have a sandbox where you can post challenge ideas and get feedback from the community before "going live". – Dennis Nov 6 '16 at 6:52
# Python 3 for the language ///
A terribly long program, accepts input as a series of lines terminated by a newline and then the EOF character (May be Ctrl-Z or Ctrl-D depending on your OS).
s=''
try:
while 1:
s+=input()+'\n'
except:
pass
try:
while len(s):
if'/'==s[0]:
s=s[1:]
f=r=''
while'/'!=s[0]:
if'\\'==s[0]:
s=s[1:]
f+=s[0]
s=s[1:]
s=s[1:]
while'/'!=s[0]:
if'\\'==s[0]:
s=s[1:]
r+=s[0]
s=s[1:]
s=s[1:]
while f in s:
s=s[:s.index(f)]+r+s[s.index(f)+len(f):]
elif'\\'==s[0]:
print(s[1])
s=s[2:]
else:
print(s[0],end='')
s=s[1:]
except:pass
Try it on Ideone!
It has been slightly golfed.
• Could be golfed more by inlining multiple statements after a colon (e.g. while f in s:s=s[:s.index and elif'\\'==s[0]:print(s[1]);s=s[2:]). – wizzwizz4 Sep 22 '17 at 16:35
• @wizzwizz4 I got your comment; unfortunately I'm not in a position where I can edit easily right now. – boboquack Sep 22 '17 at 19:54
• That's fine. This is the (asynchronous) internet - I can wait! :-) – wizzwizz4 Sep 22 '17 at 21:06 | 2019-06-26 00:40:58 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3678688406944275, "perplexity": 1783.8483212585584}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560627999964.8/warc/CC-MAIN-20190625233231-20190626015231-00002.warc.gz"} |
https://kerodon.net/tag/02KL | # Kerodon
$\Newextarrow{\xRightarrow}{5,5}{0x21D2}$ $\newcommand\empty{}$
Example 7.1.5.9. Let $U: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ be an inner fibration of $\infty$-categories. Then:
• A morphism $e$ of $\operatorname{\mathcal{C}}$ is $U$-cartesian (in the sense of Definition 5.1.1.1) if and only if it is a $U$-limit diagram when viewed as a morphism of simplicial sets $(\Delta ^0)^{\triangleleft } \rightarrow \operatorname{\mathcal{C}}$.
• A morphism $f$ of $\operatorname{\mathcal{C}}$ is $U$-cocartesian (in the sense of Definition 5.1.1.1) if and only if it is a $U$-colimit diagram when viewed as a morphism of simplicial sets $(\Delta ^0)^{\triangleright } \rightarrow \operatorname{\mathcal{C}}$.
This follows by combining Remark 7.1.5.8 with Proposition 5.1.1.13. | 2022-08-09 16:42:43 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9989349246025085, "perplexity": 252.5097307577967}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882571056.58/warc/CC-MAIN-20220809155137-20220809185137-00156.warc.gz"} |
https://www.semanticscholar.org/paper/Bilateral-Multi-Perspective-Matching-for-Natural-Wang-Hamza/a9df777e4d8100e52e90fa4bd2d783d25a2fd173 | # Bilateral Multi-Perspective Matching for Natural Language Sentences
@inproceedings{Wang2017BilateralMM,
title={Bilateral Multi-Perspective Matching for Natural Language Sentences},
author={Z. Wang and W. Hamza and Radu Florian},
booktitle={IJCAI},
year={2017}
}
• Published in IJCAI 2017
• Computer Science
• Natural language sentence matching is a fundamental technology for a variety of tasks. [...] Key Method Given two sentences $P$ and $Q$, our model first encodes them with a BiLSTM encoder. Next, we match the two encoded sentences in two directions $P \rightarrow Q$ and $P \leftarrow Q$. In each matching direction, each time step of one sentence is matched against all time-steps of the other sentence from multiple perspectives.Expand Abstract | 2020-10-28 00:58:43 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2542312741279602, "perplexity": 3348.390600439273}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-45/segments/1603107894890.32/warc/CC-MAIN-20201027225224-20201028015224-00402.warc.gz"} |
http://blogs.ethz.ch/kowalski/2011/04/16/the-other-theorem-of-burnside/ | The other theorem of Burnside
If a poll was held among mathematicians to identify “Burnside’s Theorem”, I guess that the result would be roughly evenly split between the irreducibility criterion for finite-dimensional representations, and the solvability of finite groups of order divisible by at most two distinct primes; far behind would come the so-called “Burnside lemma”, which I’ve heard is due to Frobenius or Cauchy, the partial results concerning Burnside’s problem, and presumably among stragglers and also-ran would be Schur’s Lemma and his work on hydrodynamics and complex function theory.
I’ve already discussed what head or tail I could make of Burnside’s proof of the irreducibility criterion, which is also incorporated in my representation theory notes (I’ve also put in there, in Section 2.7.4, a different proof based on Frobenius reciprocity which is quite cute, and presumably also well-known). I’ve just started discussing in class the $p^aq^b$ theorem. The account in the notes (Section 4.7.2) is a bit incomplete, as I’m struggling in my attempts to decide how to present in a motivated way how one comes to the integrality properties of characters which are the crucial tool (for a target audience with no prior exposure to algebraic integers). I found this easier to do on the board, and to imagine doing with hyperlinks, than within a book-like object!
I had not actually looked at the proof before the last few weeks. As far as I can see, and in striking contrast with the irreducibility criterion, Burnside’s proof is still the standard one (apart from terminological and notational changes; I do not count the purely group-theoretic arguments apparently found later by Thompson and others, which I know nothing about, save their existence). Part of the mystery of the statement is that, in fact, one proves something a bit different, and in fact weaker-looking: if $|G|=p^aq^b$, then either $G$ is abelian, or it contains a proper normal subgroup. Then by induction on the order of a group $|G|$ (divisiblie by at most two primes), one concludes easily that in fact they are solvable.
In turn, the desired normal subgroup is constructed using a more general result, which I had never heard about, but which is certainly of independent interest, and has the virtue of being a general fact about all finite groups which illustrates some of the subtle ways in which irreducible (complex) representations and conjugacy classes try to be “dual”:
If $G$ is a finite group, $g\in G$ is a non-trivial element, $\rho$ a non-trivial irreducible complex representation of $G$. If the dimension of $\rho$ is coprime to the order of the conjugacy class of $g$, then either
• The character value $\chi_{\rho}(g)$ is zero;
• Or, the element $g$ is in the kernel of the composite projective representation
$\bar{\rho}\,:\, G\rightarrow \mathrm{GL}(E)\rightarrow \mathrm{PGL}(E)$
(where $E$ is the space of $\rho$).
The second part is the main point: if, for a given group $G$, one can find a pair $(g,\rho)$ for which the result is applicable, and if the first part of the alternative can be excluded, then it follows that the kernel $N$ of $\bar{\rho}$ is a non-trivial normal subgroup. It could be that $N=G$, but that is a very special case: then the image of $\rho$ is an abelian group, and either $\rho$ is an isomorphism with such an abelian group, or the kernel of $\rho$ itself is a proper normal subgroup.)
Where having $|G|=p^aq^b$ comes in, in applying this, is in the fact that it is not so easy in general to cook up integers dividing $|G|$ which are coprime to each other (these being the size $|g^{\sharp}|$ of the conjugacy class $g^{\sharp}$ of $g$, and the dimension of the irreducible representation).
What can be done without much work is to find, for any $g\not=1$, a representation of dimension not divisible by one prime (say $p\mid |G|$) for which $\chi_{\rho}(g)\not=0$, using the orthogonality relations looked-at modulo $p$; and one can also find an element $g\not=1$ for which the conjugacy class has order coprime with another prime $q$ (looking at the partition of $G$ in conjugacy classes, modulo $q$). But there is no reason that this should ensure that $(\dim(\rho),|g^{\sharp}|)=1$, except if $|G|$ is divisible only by those two primes! In that case, we see that $\dim(\rho)$ is a power of $p$, and the conjugacy class of $g$ has size a power of $q$, and everything works.
The proof of the result on characters above is a very convincing illustration of the usefulness of algebraic integers: one first shows by basic facts about them that there is always a divisibility
$\dim(\rho)\mid \chi_{\rho}(g) |g^{\sharp}|,$
in the ring of algebraic integers. Thus if, as we assumed, the dimension and the size of the conjugacy class are coprime, we get
$\dim(\rho)\mid \chi_{\rho}(g),$
and using the fact that the character value is a sum of $\dim(\rho)$ roots of unity, the conclusion is again not too hard from the properties of algebraic integers. (It is quite obvious when $\chi_{\rho}(g)\in\mathbf{Z}$, of course, since one knows that
$|\chi_{\rho}(g)|\leq \dim\rho$, and one must see that a similar argument applies in general…)
Amusingly, one can use the divisibility relation in the “opposite” direction: if the character value is non-zero and coprime (in the ring of algebraic integers) with the dimension of the representation, then $\dim(\rho)$ divides the size of the conjugacy class. I don’t know if there are applications of this, but this can be seen in practice, e.g., for $G=\mathrm{GL}_2(\mathbf{F}_p)$ when $\rho$ is a Steinberg representation of dimension $p$ and $g$ is a semisimple (split or non-split) conjugacy class, where the character value is a root of unity, hence coprime with $p$, and indeed the conjugacy classes in question have order $p(p+1)$ or $p(p-1)$ (in the split and non-split case, respectively.) | 2018-04-27 06:48:57 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 55, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9012390375137329, "perplexity": 155.57266376883672}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-17/segments/1524125949489.63/warc/CC-MAIN-20180427060505-20180427080505-00412.warc.gz"} |
https://www.esaral.com/q/let-c1-and-c2-be-the-centres-of-the-circles-47149/ | # Let C1 and C2 be the centres of the circles
Question:
Let $\mathrm{C}_{1}$ and $\mathrm{C}_{2}$ be the centres of the circles $\mathrm{x}^{2}+\mathrm{y}^{2}-2 \mathrm{x}-2 \mathrm{y}-2=$ 0 and $x^{2}+y^{2}-6 x-6 y+14=0$ respectively. If $P$ and $Q$ are the points of intersection of these circles then, the area (in sq. units) of the quadrilateral $\mathrm{PC}_{1} \mathrm{QC}_{2}$ is :
1. (1) 8
2. (2) 6
3. (3) 9
4. (4) 4
Correct Option: , 4
Solution:
$2 g_{1} g_{2}+2 f_{1} f_{2}=2(-1)(-3)+2(-1)(-3)=12$
$c_{1}+c_{2}=14-2=12$
Since, $2 g_{1} g_{2}+2 f_{1} f_{2}=c_{1}+c_{2}$
Hence, circles intersect orthogonally
$\therefore \quad$ Area of the quadrilateral $\mathrm{PC}_{1} \mathrm{QC}_{1}$
$=2\left(\frac{1}{2}\left(C_{1} P\right)\left(C_{2} P\right)\right)$
$=2 \times \frac{1}{2} r_{1} r_{2}=(2)(2)=4 \mathrm{sq}$, units
#### Leave a comment
None
Free Study Material | 2023-02-03 23:40:25 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8704751133918762, "perplexity": 844.5420871495388}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764500076.87/warc/CC-MAIN-20230203221113-20230204011113-00481.warc.gz"} |
http://mathhelpforum.com/calculus/87240-just-need-explanation-integration-prob.html | # Math Help - just need explanation for integration prob.
1. ## just need explanation for integration prob.
Ok so this is the problem:
Integral from 1 to e of (ln(x))/x) dx (sorry don't know how to put the intergral symbol in)
I set u= lnx and du= 1/x and changed the limits to 0 to 1
When i solve though, I get 1, and my book says the answer is 1/2. I've tried this a bunch of times but can't get their answer. Any explanation?
2. Originally Posted by painterchica16
Ok so this is the problem:
Integral from 1 to e of (ln(x))/x) dx (sorry don't know how to put the intergral symbol in)
I set u= lnx and du= 1/x and changed the limits to 0 to 1
When i solve though, I get 1, and my book says the answer is 1/2. I've tried this a bunch of times but can't get their answer. Any explanation?
So after your sub you get
$\int_{0}^{1}u du =\frac{1}{2}u^2 \bigg|_{0}^{1}=\frac{1}{2}(1^2-0^2)=\frac{1}{2}$
3. $\int_{1}^{e} \frac{lnx}{x}$
$u = lnx$
$du = 1/x$
$\int udu$
Raise power and divide by power.
$\int udu = \frac{1}{2}u^2$
$\frac{(lnx)^2}{2}$
From 1 to e,
1/2ln(e)^2 - 1/2ln(1)^2
ln(1) = 0, ln(e) = 1
4. Thanks, I understand the setup now, but the only that I'm still confused about is why you plugged 0 and 1 into just u^2, and not lnx?
Because, ln(1)^2=0 and ln(0)^2 just doesnt exist. Why does it not work when you plug it in?
5. If u = lnx, than our intervals are no longer 1 and e but ln(1) and ln(e). Hence the substitution.
6. Oh, I get it, you plug it into u^2, not the other thing, because its all in terms of u. Thanks! | 2014-07-23 14:56:42 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 7, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8277782201766968, "perplexity": 1322.859703165492}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-23/segments/1405997878518.58/warc/CC-MAIN-20140722025758-00027-ip-10-33-131-23.ec2.internal.warc.gz"} |
https://www.transtutors.com/questions/question-1-the-figure-shows-the-domestic-supply-and-domestic-demand-for-copper-both--4389828.htm | # Question 1- The figure shows the domestic supply and domestic demand for copper. Both the autarky...
Question
1-
The figure shows the domestic supply and domestic demand for copper. Both the autarky price and the world price are indicated. If this economy is open to world trade, what quantity of copper will be exported?
a-40,000 pounds
b-70,000 pounds
c-80,000 pounds
d-110,000 pounds
------------------------------------
2-
The figure shows the domestic supply and demand for notebooks. Suppose that the world price is $2 and that a$2 tariff has been imposed. Pw indicates the world price on the graph, and PT indicates the price with the tariff. Once the tariff is imposed, what quantity will be imported?
a-3,000
b-4,000
c-8,000
d-11,000
----------------------------------
3-
(Figure: The Market for Tea in Sri Lanka) Look at the figure The Market for Tea in Sri Lanka. In autarky, the price of tea in Sri Lanka is P1. When the economy is opened to trade, the price rises to PW. Sri Lanka will ________ tea and the volume of trade will equal ________.
a-import; QT – CT
b-export; QT – CT
c-import; Q1 – QT
d-export; CT – Q1
Price per pound World price \$6 Autarky price 40 Market for Copper Domestic Supply Domestic Demand 80 110 Quantity (thousands of pounds) | 2020-02-23 11:00:47 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.22616736590862274, "perplexity": 6552.980365143001}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875145767.72/warc/CC-MAIN-20200223093317-20200223123317-00181.warc.gz"} |
http://tug.org/pipermail/texhax/2015-January/021522.html | # [texhax] Unicode Replacement Character in Standard TeX
Justin Bailey jgbailey at gmail.com
Wed Jan 14 22:45:33 CET 2015
Doug,
I took the liberty of posting your question to the StackExchange's TeX instance:
http://tex.stackexchange.com/questions/223194/unicode-replacement-character-in-standard-tex
Hopefully someone there will be able to answer. Keep an eye out as
Hope you don't mind!
On Tue, Jan 13, 2015 at 6:03 PM, Douglas McKenna
<doug at mathemaesthetics.com> wrote:
> Is there a font, available to -all- TeX users, that includes a glyph that can be recognized as the Unicode Replacement Character? I realize I can use XeTeX with a U+FFFD (or maybe any character not in the font), but I'm interested in using standard pdfTeX with standard LaTeX 2e and a pure ASCII input file.
>
> The glyph typically looks like a filled black diamond with a white question mark inside it.
>
> If no such font is available, what's the best robust strategy for getting such a glyph into a standard pdfTeX document (one whose input file has no Unicode or non-ASCII UTF-8 byte sequences in it) without using XeTeX or similar? Inserting TikZ commands? Raw PostScript?
>
> Right now, for lack of a better solution, I'm using a \textbullet, simply because it's the boldest/blackest glyph I can think of, and it's available in most standard fonts.
>
> Creating a custom font with one glyph in it is not really viable, or if it is it's a strategy of last resort, unless (perhaps) it can be done automatically and on the fly when processing the document each time.
>
>
> TIA,
>
> Doug McKenna
>
>
> _______________________________________________
> TeX FAQ: http://www.tex.ac.uk/faq
> Mailing list archives: http://tug.org/pipermail/texhax/ | 2018-11-14 15:18:08 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9721131920814514, "perplexity": 4660.209111679154}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-47/segments/1542039742117.38/warc/CC-MAIN-20181114150002-20181114171325-00034.warc.gz"} |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.